Difference between revisions of "2012 AIME II Problems/Problem 6"
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== Solution == | == Solution == | ||
− | {{ | + | Let's consider the maximization constraint first: we want to maximize the value of <math>|z^5 - (1+2i)z^3|</math> |
+ | Simplifying, we have | ||
+ | |||
+ | <math>|z^3| * |z^2 - (1+2i)|</math> | ||
+ | |||
+ | <math>=|z|^3 * |z^2 - (1+2i)|</math> | ||
+ | |||
+ | <math>=125|z^2 - (1+2i)|</math> | ||
+ | |||
+ | Thus we only need to maximize the value of <math>|z^2 - (1+2i)|</math>. | ||
+ | |||
+ | To maximize this value, we must have that <math>z^2</math> is in the opposite direction of <math>1+2i</math>. The unit vector in the complex plane in the desired direction is <math>\frac{-1}{\sqrt{5}} + \frac{-2}{\sqrt{5}} i</math>. Furthermore, we know that the magnitude of <math>z^2</math> is <math>25</math>, because the magnitude of <math>z</math> is <math>5</math>. From this information, we can find that <math>z^2 = \sqrt{5} (-5 - 10i)</math> | ||
+ | |||
+ | Squaring, we get <math>z^4 = 5 (25 - 100 + 100i) = -375 + 500i</math>. Finally, <math>c+d = -375 + 500 = \boxed{125}</math> | ||
+ | |||
+ | == Solution 2 == | ||
+ | |||
+ | WLOG, let <math>z_{1}=5(\cos{\theta_{1}}+i\sin{\theta_{1}})</math> and | ||
+ | |||
+ | <math>z_{2}=1+2i=\sqrt{5}(\cos{\theta_{2}+i\sin{\theta_{2}}})</math> | ||
+ | |||
+ | This means that | ||
+ | |||
+ | <math>z_{1}^3=125(\cos{3\theta_{1}}+i\sin{3\theta_{1}})</math> | ||
+ | |||
+ | <math>z_{1}^4=625(\cos{4\theta_{1}}+i\sin{4\theta_{1}})</math> | ||
+ | |||
+ | Hence, this means that | ||
+ | |||
+ | <math>z_{2}z_{1}^3=125\sqrt{5}(\cos({\theta_{2}+3\theta_{1}})+i\sin({\theta_{2}+3\theta_{1}}))</math> | ||
+ | |||
+ | And | ||
+ | |||
+ | <math>z_{1}^5=3125(\cos{5\theta_{1}}+i\sin{5\theta_{1}})</math> | ||
+ | |||
+ | Now, common sense tells us that the distance between these two complex numbers is maxed when they both are points satisfying the equation of the line <math>yi=mx</math>, or when they are each a <math>180^{\circ}</math> rotation away from each other. | ||
+ | |||
+ | Hence, we must have that <math>5\theta_{1}=3\theta_{1}+\theta_{2}+180^{\circ}\implies\theta_{1}=\frac{\theta_{2}+180^{\circ}}{2}</math> | ||
+ | |||
+ | Now, plug this back into <math>z_{1}^4</math>(if you want to know why, reread what we want in the problem!) | ||
+ | |||
+ | So now, we have that | ||
+ | <math>z_{1}^4=625(\cos{2\theta_{2}}+i\sin{2\theta_{2}})</math> | ||
+ | |||
+ | Notice that <math>\cos\theta_{2}=\frac{1}{\sqrt{5}}</math> and <math>\sin\theta_{2}=\frac{2}{\sqrt{5}}</math> | ||
+ | |||
+ | Then, we have that <math>\cos{2\theta_{2}}=\cos^2{\theta_{2}}-\sin^2{\theta_{2}}=-\frac{3}{5}</math> and <math>\sin{2\theta_{2}}=2\sin{\theta_{2}}\cos{\theta_{2}}=\frac{4}{5}</math> | ||
+ | |||
+ | Finally, plugging back in, we find that <math>z_{1}^4=625(-\frac{3}{5}+\frac{4i}{5})=-375+500i</math> | ||
+ | |||
+ | <math>-375+500=\boxed{125}</math> | ||
+ | |||
== See Also == | == See Also == | ||
{{AIME box|year=2012|n=II|num-b=5|num-a=7}} | {{AIME box|year=2012|n=II|num-b=5|num-a=7}} | ||
+ | {{MAA Notice}} |
Latest revision as of 01:34, 31 December 2020
Contents
Problem 6
Let be the complex number with and such that the distance between and is maximized, and let . Find .
Solution
Let's consider the maximization constraint first: we want to maximize the value of Simplifying, we have
Thus we only need to maximize the value of .
To maximize this value, we must have that is in the opposite direction of . The unit vector in the complex plane in the desired direction is . Furthermore, we know that the magnitude of is , because the magnitude of is . From this information, we can find that
Squaring, we get . Finally,
Solution 2
WLOG, let and
This means that
Hence, this means that
And
Now, common sense tells us that the distance between these two complex numbers is maxed when they both are points satisfying the equation of the line , or when they are each a rotation away from each other.
Hence, we must have that
Now, plug this back into (if you want to know why, reread what we want in the problem!)
So now, we have that
Notice that and
Then, we have that and
Finally, plugging back in, we find that
See Also
2012 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.