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Difference between revisions of "1984 AHSME Problems/Problem 30"
m (Fixed formatting)
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==Problem==
 
==Problem==
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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\sin40^\circ </math>, then
+
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\sin40^\circ </math>, then <math> |w+2w^2+3w^3+...+9w^9|^{-1} </math> equals
  
−
<math> |w+2w^2+3w^3+...+9w^9|^{-1} </math>
+
<math> \textbf{(A) \ }\frac{1}{9}\sin40^\circ \qquad \textbf{(B) \ }\frac{2}{9}\sin20^\circ \qquad \textbf{(C) \ } \frac{1}{9}\cos40^\circ \qquad \textbf{(D) \ }\frac{1}{18}\cos20^\circ \qquad \textbf{(E) \ } \text{None of these} </math>
−
 
 
−
equals
 
−
 
 
−
<math> \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} </math>
 
  
 
==Solution==
 
==Solution==
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However, <math>\sum_{j=i}^{9} w^j(1-w)</math> is simply <math>w^i-w^{10}</math>. Therefore
 
However, <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>
  
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<cmath>(w+w^2+\cdots +w^9)(1-w)=w-w^{10}=w(1-w^{9})=0</cmath>
 
<cmath>(w+w^2+\cdots +w^9)(1-w)=w-w^{10}=w(1-w^{9})=0</cmath>
  
−
This shows that <math>S=\frac{-9w^{10}}{1-w}</math>. Note that <math>w^{10}=w\cdot w^9=w</math>, so <math>S=\frac{-9w}{1-w}</math>. It's not hard to show that <math>|S|^{-1}=|S^{-1}|=|-S^{-1}|</math>, so the number we seek is equal to <math>|\frac{1-w}{9w}|</math>.
+
This shows that <math>S=\frac{-9w^{10}}{1-w}</math>. Note that <math>w^{10}=w\cdot w^9=w</math>, so <math>S=\frac{-9w}{1-w}</math>.  
 +
 
 +
It's not hard to show that <math>\left|S\right|^{-1}=\left|S^{-1}\right|=\left|-S^{-1}\right|</math>, so the number we seek is equal to <math>\left|\frac{1-w}{9w}\right|</math>.
  
 
Now we plug <math>w</math> into the fraction:
 
Now we plug <math>w</math> into the fraction:
  
−
<cmath>\frac{1-w}{9w}=\frac{(1-\cos{40})-i\sin{40}}{9\cos{40}+9i\sin{40}}</cmath>
+
<cmath>\frac{1-w}{9w}=\frac{(1-\cos{40^{\circ}})-i\sin{40^{\circ}}}{9\cos{40}+9i\sin{40^{\circ}}}</cmath>
  
−
We multiply the numerator and denominator by <math>9\cos{40}-9i\sin{40}</math> and simplify to get
+
We multiply the numerator and denominator by <math>9\cos{40^{\circ}}-9i\sin{40^{\circ}}</math> and simplify to get
  
−
<cmath>\frac{1-w}{9w}=\frac{(\cos{40}-i\sin{40})((1-\cos{40})-i\sin{40})}{9}</cmath>
+
<cmath>\frac{1-w}{9w}=\frac{(\cos{40^{\circ}}-i\sin{40^{\circ}})((1-\cos{40^{\circ}})-i\sin{40^{\circ}})}{9}</cmath>
  
−
<cmath>=\frac{\cos{40}-\cos^2{40}-\sin^2{40}+i(-\sin{40}+\sin{40}\cos{40}-\sin{40}\cos{40})}{9}</cmath>
+
<cmath>=\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}</cmath>
  
−
<cmath>=\frac{(\cos{40}-1)-i\sin{40}}{9}</cmath>
+
<cmath>=\frac{(\cos{40^{\circ}}-1)-i\sin{40^{\circ}}}{9}</cmath>
  
 
The absolute value of this is
 
The absolute value of this is
  
−
<cmath>|\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}}</cmath>
+
<cmath>\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}}}</cmath>
  
−
Note that, from double angle formulas, <math>\cos{40}=\cos^2{20}-\sin^2{20}</math>, so <math>1-\cos{40}=\cos^2{20}+\sin^2{20}-(\cos^2{20}-\sin^2{20})=2\sin^2{20}</math>. Therefore
+
Note that, from double angle formulas, <math>\cos{40^{\circ}}=\cos^2{20^{\circ}}-\sin^2{20^{\circ}}</math>, so <math>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}}</math>. Therefore
  
−
<cmath>|\frac{1-w}{9w}|=\frac{\sqrt{2}}{9} \sqrt{2\sin^2{20}}</cmath>
+
<cmath>\left|\frac{1-w}{9w}\right|=\frac{\sqrt{2}}{9} \sqrt{2\sin^2{20^{\circ}}}</cmath>
  
−
<cmath>=\frac{2}{9}\sin{20}</cmath>
+
<cmath>=\frac{2}{9}\sin{20^{\circ}}</cmath>
  
−
Therefore the correct answer is <math>\mathrm{(B) \ }</math>.
+
Therefore the correct answer is <math>\boxed{\textbf{(B)}}</math>.
  
 
==See Also==
 
==See Also==
 
{{AHSME box|year=1984|num-b=29|after=Last Problem}}
 
{{AHSME box|year=1984|num-b=29|after=Last Problem}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 00:20, 20 February 2019

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

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

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 $\left|S\right|^{-1}=\left|S^{-1}\right|=\left|-S^{-1}\right|$, 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
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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