Difference between revisions of "1984 AHSME Problems/Problem 30"
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==Problem== | ==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\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> \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> | |
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− | <math> \ | ||
==Solution== | ==Solution== | ||
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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> | ||
− | + | By the geometric series formula, <math>\sum_{j=i}^{9} w^j(1-w)</math> is simply <math>w^i-w^{10}</math>. Therefore | |
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<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>. |
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+ | 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>\ | + | 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}} |
Latest revision as of 21:21, 17 July 2020
Problem
For any complex number ,
is defined to be the real number
. If
, then
equals
Solution
Let . Note that
Now we multiply by
:
By the geometric series formula, is simply
. Therefore
A simple application of De Moivre's Theorem shows that is a ninth root of unity (
), so
This shows that . Note that
, so
.
It's not hard to show that , so the number we seek is equal to
.
Now we plug into the fraction:
We multiply the numerator and denominator by and simplify to get
The absolute value of this is
Note that, from double angle formulas, , so
. Therefore
Therefore the correct answer is .
See Also
1984 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 29 |
Followed by Last Problem | |
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All AHSME Problems and Solutions |
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