Difference between revisions of "2019 AMC 12B Problems/Problem 17"
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==Solution== | ==Solution== | ||
− | Convert z and z^3 into <cmath>r\text{cis}\theta</cmath> | + | Convert z and z^3 into <cmath>r\text{cis}\theta</cmath> form, giving <cmath>z=r\text{cis}\theta</cmath> and <cmath>z^3=r^3\text{cis}(3\theta)</cmath>. Since the distance from 0 to z is r, the distance from 0 to z^3 must also be r, so r=1. Now we must find <cmath>\text{cis}(2\theta)=60</cmath>. From 0 < theta < pi/2, we have <cmath>\theta=\frac{\pi}{2}</cmath> and from pi/2 < theta < pi, we see a monotonic decrease of <cmath>\text{cis}(2\theta)</cmath>, from 180 to 0. Hence, there are 2 values that work for 0 < theta < pi. But since the interval pi < theta < 2pi is identical, because 3theta=theta, we have 4 solutions. There are not infinitely many solutions since the same four solutions are duplicated. (D) |
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+ | -FlatSquare | ||
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+ | Someone pls help with LaTeX formatting, thanks | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2019|ab=B|num-b=16|num-a=18}} | {{AMC12 box|year=2019|ab=B|num-b=16|num-a=18}} |
Revision as of 14:37, 14 February 2019
Problem
Solution
Convert z and z^3 into form, giving and . Since the distance from 0 to z is r, the distance from 0 to z^3 must also be r, so r=1. Now we must find . From 0 < theta < pi/2, we have and from pi/2 < theta < pi, we see a monotonic decrease of , from 180 to 0. Hence, there are 2 values that work for 0 < theta < pi. But since the interval pi < theta < 2pi is identical, because 3theta=theta, we have 4 solutions. There are not infinitely many solutions since the same four solutions are duplicated. (D)
-FlatSquare
Someone pls help with LaTeX formatting, thanks
See Also
2019 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 16 |
Followed by Problem 18 |
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 | |
All AMC 12 Problems and Solutions |