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Difference between revisions of "1984 AHSME Problems/Problem 30"

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<cmath>S(1-w)=\sum_{i=1}^{9}\sum_{j=i}^{9} w^j(1-w)</cmath>
 
<cmath>S(1-w)=\sum_{i=1}^{9}\sum_{j=i}^{9} w^j(1-w)</cmath>
  
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However, <math>\sum_{j=i}^{9} w^j(1-w)</math> is simply <math>w^i-w^{10}</math>. Therefore
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By the geometric series formula, <math>\sum_{j=i}^{9} w^j(1-w)</math> is simply <math>w^i-w^{10}</math>. Therefore
 
<cmath>S(1-w)=\sum_{i=1}^{9} w^i-w^{10}=(w+w^2+\cdots +w^9)-9w^{10}</cmath>
 
<cmath>S(1-w)=\sum_{i=1}^{9} w^i-w^{10}=(w+w^2+\cdots +w^9)-9w^{10}</cmath>
  

Revision as of 20:21, 17 July 2020

Problem

For any complex number $w=a+bi$, $|w|$ is defined to be the real number $\sqrt{a^2+b^2}$. If $w=\cos40^\circ+i\sin40^\circ$, then $|w+2w^2+3w^3+...+9w^9|^{-1}$ equals

$\text{(A) }\frac{1}{9}\sin40^\circ \qquad \text{(B) }\frac{2}{9}\sin20^\circ \qquad \text{(C) } \frac{1}{9}\cos40^\circ \qquad \text{(D) }\frac{1}{18}\cos20^\circ \qquad \text{(E) } \text{None of these}$

Solution

Let $S=w+2w^2+3w^3+...+9w^9$. Note that

\[S=\sum_{i=1}^{9}\sum_{j=i}^{9} w^j\]

Now we multiply $S$ by $1-w$:

\[S(1-w)=\sum_{i=1}^{9}\sum_{j=i}^{9} w^j(1-w)\]

By the geometric series formula, $\sum_{j=i}^{9} w^j(1-w)$ is simply $w^i-w^{10}$. Therefore \[S(1-w)=\sum_{i=1}^{9} w^i-w^{10}=(w+w^2+\cdots +w^9)-9w^{10}\]

A simple application of De Moivre's Theorem shows that $w$ is a ninth root of unity ($w^9=1$), so

\[(w+w^2+\cdots +w^9)(1-w)=w-w^{10}=w(1-w^{9})=0\]

This shows that $S=\frac{-9w^{10}}{1-w}$. Note that $w^{10}=w\cdot w^9=w$, so $S=\frac{-9w}{1-w}$.

It's not hard to show that $\left|S\right|^{-1}=\left|S^{-1}\right|=\left|-S^{-1}\right|$, so the number we seek is equal to $\left|\frac{1-w}{9w}\right|$.

Now we plug $w$ into the fraction:

\[\frac{1-w}{9w}=\frac{(1-\cos{40^{\circ}})-i\sin{40^{\circ}}}{9\cos{40}+9i\sin{40^{\circ}}}\]

We multiply the numerator and denominator by $9\cos{40^{\circ}}-9i\sin{40^{\circ}}$ and simplify to get

\[\frac{1-w}{9w}=\frac{(\cos{40^{\circ}}-i\sin{40^{\circ}})((1-\cos{40^{\circ}})-i\sin{40^{\circ}})}{9}\]

\[=\frac{\cos{40^{\circ}}-\cos^2{40^{\circ}}-\sin^2{40^{\circ}}+i(-\sin{40^{\circ}}+\sin{40^{\circ}}\cos{40^{\circ}}-\sin{40^{\circ}}\cos{40^{\circ}})}{9}\]

\[=\frac{(\cos{40^{\circ}}-1)-i\sin{40^{\circ}}}{9}\]

The absolute value of this is

\[\left|\frac{1-w}{9w}\right|=\frac{1}{9}\sqrt{(\cos{40^{\circ}}-1)^2+\sin^2{40^{\circ}}}=\frac{1}{9}\sqrt{1-2\cos{40^{\circ}}+\cos^2{40^{\circ}}+\sin^2{40^{\circ}}}=\frac{\sqrt{2}}{9}\sqrt{1-\cos{40^{\circ}}}\]

Note that, from double angle formulas, $\cos{40^{\circ}}=\cos^2{20^{\circ}}-\sin^2{20^{\circ}}$, so $1-\cos{40^{\circ}}=\cos^2{20^{\circ}}+\sin^2{20^{\circ}}-(\cos^2{20^{\circ}}-\sin^2{20^{\circ}})=2\sin^2{20^{\circ}}$. Therefore

\[\left|\frac{1-w}{9w}\right|=\frac{\sqrt{2}}{9} \sqrt{2\sin^2{20^{\circ}}}\]

\[=\frac{2}{9}\sin{20^{\circ}}\]

Therefore the correct answer is $\boxed{\textbf{(B)}}$.

See Also

1984 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 29 		Followed by
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All AHSME Problems and Solutions

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