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

Revision as of 00:22, 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

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

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 $1-w$ :

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

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

@@ Line 2: / Line 2: @@
 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> \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>
+<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>
 ==Solution==

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

Revision as of 00:22, 20 February 2019

Problem

Solution

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