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

(Created page with "==Problem== For any complex number <math> w=a+bi </math>, <math> |w| </math> is defined to be the real number <math> \sqrt{a^2+b^2} </math>. If <math> w=\cos40^\circ+i\si...")
 
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==See Also==
 
{{AHSME box|year=1984|num-b=29|after=Last Problem}}
 
{{AHSME box|year=1984|num-b=29|after=Last Problem}}
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Revision as of 11:52, 5 July 2013

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

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

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 $1-w$:

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

However, $\sum_{j=i}^{9} w^j(1-w)$ 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 $|S|^{-1}=|S^{-1}|=|-S^{-1}|$, so the number we seek is equal to $|\frac{1-w}{9w}|$.

Now we plug $w$ into the fraction:

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

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

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

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

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

The absolute value of this is

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

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

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

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

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

See Also

1984 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 29 		Followed by
Last Problem
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 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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