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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...")
 
(→‎Added Solution 2)
 
(5 intermediate revisions by 3 users not shown)
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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> \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} </math>
  
−
equals
+
==Solution 1==
−
 
 
−
<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==
 
 
Let <math>S=w+2w^2+3w^3+...+9w^9</math>. Note that
 
Let <math>S=w+2w^2+3w^3+...+9w^9</math>. Note that
  
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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>
  
−
However, <math>\sum_{j=i}^{9} w^j(1-w)</math> is simply <math>w^i-w^{10}</math>. Therefore
+
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>
  
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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^{\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>\left|\frac{1-w}{9w}\right|=\frac{\sqrt{2}}{9} \sqrt{2\sin^2{20^{\circ}}}</cmath>
 +
 
 +
<cmath>=\frac{2}{9}\sin{20^{\circ}}</cmath>
 +
 
 +
Therefore the correct answer is <math>\boxed{\textbf{(B)}}</math>.
 +
 
 +
==Solution 2==
 +
Notice that <math>w = \cos40^\circ+i\sin40^\circ = \cos\left(\frac{2\pi}{9}\right)+i\sin\left(\frac{2\pi}{9}\right)</math> is the primitive root of the equation <math>x^9 = 1</math>.
 +
 
 +
Consider the simple geometric progression,
 +
 
 +
<cmath>f(x) = 1 + x + x^2 +\cdots + x^9 = \frac{x^{10} - 1}{x - 1}</cmath>
 +
 
 +
Differentiating this equation (using the [https://artofproblemsolving.com/wiki/index.php/Quotient_Rule Quotient Rule]), we obtain the following equation,
 +
 
 +
<cmath>f'(x) = 1 + 2x + 3x^2 + \cdots + 9x^8 = \frac{9x^{10} - 10x^9 + 1}{(x - 1)^2}</cmath>
 +
 
 +
Now, we can evaluate the given expression,
  
−
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
+
<cmath>w + 2w^2 + 3w^3 + \cdots + 9w^9 = w \cdot (1+2w+3w^2+...+9w^8)</cmath>
 +
<cmath> = w\cdot f'(w)</cmath>
 +
<cmath>= w\cdot\frac{9w^{10} - 10w^9 + 1}{(w - 1)^2}</cmath>
 +
<cmath> = \frac{9w}{w-1} \; (\because w^9 = 1)</cmath>
  
−
<cmath>|\frac{1-w}{9w}|=\frac{\sqrt{2}}{9} \sqrt{2\sin^2{20}}</cmath>
+
Finally, we can calculate the given modulus,
  
−
<cmath>=\frac{2}{9}\sin{20}</cmath>
+
<cmath>\lvert w+2w^2+3w^3+...+9w^9 \rvert^{-1} = \lvert \frac{1}{9}\cdot\left(1 - w^8) \right) \rvert</cmath>
 +
<cmath>= \lvert \frac{1}{9}\cdot\left(1 - \cos\left(320^\circ\right)+i\sin\left(320^\circ\right)) \right) \rvert</cmath>
 +
<cmath>= \frac{1}{9}\sqrt{\left(1 - \cos\left(320^\circ\right)\right)^2+\left(\sin\left(320^\circ\right)\right)^2}</cmath>
 +
<cmath>= \frac{1}{9}\sqrt{2 - 2\cos\left(320^\circ\right)}</cmath>
 +
<cmath>= \frac{2}{9}\sin\left(160^\circ\right)</cmath>
 +
<cmath>= \boxed{\frac{2}{9}\sin\left(20^\circ\right)}</cmath>
  
−
Therefore the correct answer is <math>\mathrm{(B) \ }</math>.
+
~ [https://artofproblemsolving.com/wiki/index.php/User:Plusone plusone]
  
 
==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}}

Latest revision as of 09:04, 8 September 2024

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 1

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 $\boxed{\textbf{(B)}}$.

Solution 2

Notice that $w = \cos40^\circ+i\sin40^\circ = \cos\left(\frac{2\pi}{9}\right)+i\sin\left(\frac{2\pi}{9}\right)$ is the primitive root of the equation $x^9 = 1$.

Consider the simple geometric progression,

\[f(x) = 1 + x + x^2 +\cdots + x^9 = \frac{x^{10} - 1}{x - 1}\]

Differentiating this equation (using the Quotient Rule), we obtain the following equation,

\[f'(x) = 1 + 2x + 3x^2 + \cdots + 9x^8 = \frac{9x^{10} - 10x^9 + 1}{(x - 1)^2}\]

Now, we can evaluate the given expression,

\[w + 2w^2 + 3w^3 + \cdots + 9w^9 = w \cdot (1+2w+3w^2+...+9w^8)\] \[= w\cdot f'(w)\] \[= w\cdot\frac{9w^{10} - 10w^9 + 1}{(w - 1)^2}\] \[= \frac{9w}{w-1} \; (\because w^9 = 1)\]

Finally, we can calculate the given modulus,

\[\lvert w+2w^2+3w^3+...+9w^9 \rvert^{-1} = \lvert \frac{1}{9}\cdot\left(1 - w^8) \right) \rvert\] \[= \lvert \frac{1}{9}\cdot\left(1 - \cos\left(320^\circ\right)+i\sin\left(320^\circ\right)) \right) \rvert\] \[= \frac{1}{9}\sqrt{\left(1 - \cos\left(320^\circ\right)\right)^2+\left(\sin\left(320^\circ\right)\right)^2}\] \[= \frac{1}{9}\sqrt{2 - 2\cos\left(320^\circ\right)}\] \[= \frac{2}{9}\sin\left(160^\circ\right)\] \[= \boxed{\frac{2}{9}\sin\left(20^\circ\right)}\]

~ plusone

See Also

1984 AHSME (Problems • Answer Key • Resources)
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Problem 29 		Followed by
Last Problem
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