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Difference between revisions of "Mock AIME I 2012 Problems/Problem 8"

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-Solution by '''thecmd999'''
 
-Solution by '''thecmd999'''
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==Solution 3 (Very nice)==
 
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Note that the "<math>+1</math>" will make <math>z</math> maximized when it makes the RHS "minimized", or when it is pointing in the opposite direction of <math>z</math>. Therefore, <math>z</math> is pointing to the left, so <math>z=z^2-1</math>, <math>z^2-z-1=0</math>, and <math>z=\frac{\sqrt5+1}{2}</math>, <math>r + s + t = \boxed{054}</math>.
 

Latest revision as of 17:51, 7 May 2020

Problem

Suppose that the complex number $z$ satisfies $\left|z\right| = \left|z^2+1\right|$. If $K$ is the maximum possible value of $\left|z\right|$, $K^4$ can be expressed in the form $\dfrac{r+\sqrt{s}}{t}$. Find $r+s+t$.

Solution 1

We begin by dividing both sides by $|z|\ne 0$ to obtain $1 = \frac{\left|z^2+1\right|}{\left|z\right|} = \left|\frac{z^2+1}{z}\right| = \left|z+\frac{1}{z}\right|$. Now, consider that we may write $z = re^{i\theta}=r\left(\cos(\theta)+i\sin(\theta)\right)$ with $r$ a positive real number so that $|z| = r$ and $\theta$ some real number. Then, \[\left|z+\frac{1}{z}\right| = \left|re^{i\theta} + \frac{1}{r}\cdot e^{-i\theta}\right|= \left|r\left(\cos(\theta)+i\sin(\theta)\right)+\frac{1}{r}\left(\cos(-\theta)+i\sin(-\theta)\right)\right|\] \[=\left|r\left(\cos(\theta)+i\sin(\theta)\right)+\frac{1}{r}\left(\cos(\theta)-i\sin(\theta)\right)\right|=\left|\left(r+\frac{1}{r}\right) \cos(\theta) +\left(r-\frac{1}{r}\right)i\sin(\theta)\right|\] \[= \sqrt{\left(r+\frac{1}{r}\right)^2\cos^2(\theta)+\left(r-\frac{1}{r}\right)^2\sin^2(\theta)}=\sqrt{\left(r^2+\frac{1}{r^2}\right)\left(\cos^2(\theta)+\sin^2(\theta)\right)+2\left(\cos^2(\theta)-\sin^2(\theta)\right)}\] \[= \sqrt{r^2+\frac{1}{r^2} + 2\cos(2\theta)}.\] Since we need $\left|z+\frac{1}{z}\right| = 1$, we must have $r^2+\frac{1}{r^2} + 2\cos(2\theta) = 1$, or equivalently, $r^4 + \left(2\cos(2\theta) - 1\right)r^2 + 1 =0$. By the quadratic equation, this has roots \[r^2 = \frac{1-2\cos(2\theta)\pm \sqrt{ \left(2\cos(2\theta) - 1\right)^2-4}}{2},\] and to maximize $r$, we take the larger root \[r^2 = \frac{1-2\cos(2\theta)+\sqrt{ \left(2\cos(2\theta) - 1\right)^2-4}}{2}\] which is clearly maximized when $2\cos(2\theta)$ is minimized. Since $-1\le \cos(2\theta)\le 1$, the maximum value of $r$ will occur where $2\cos(2\theta) = -2$, so the maximum value of $r$ occurs where \[r^2 = \frac{3 + \sqrt{5}}{2}\] and finally we find that the maximum value of $|z|=r>0$ is \[r= \sqrt{\frac{3+\sqrt{5}}{2}} = \frac{\sqrt{6+2\sqrt{5}}}{2} = \frac{1+\sqrt{5}}{2},\text{ the golden ratio.}\] Taking the fourth power, the desired answer is $\dfrac{7+\sqrt{45}}{2} \implies r+s+t=\boxed{054}$.

Solution 2

Let

$z=re^{i\theta}$.

Note that

$|z|^2 = r^2$.

To compute $|z^2+1|^2$, connect $z^2$ to the origin and construct an altitude to the x-axis. Extend this line one unit above $z^2$ and connect that point to the origin to create $z^2+1$.

Now, we use the Law of Cosines on the triangle with vertices at the origin, $z^2$, and $z^2+1$, giving

$|z^2+1|^2 = r^4 + 1 - 2r^2 \cos(2\theta+90) = r^4 + 2r^2 \sin(2\theta)+1$

(or you could use the Pythagorean Theorem). Either way, we proceed from here as we did with Solution 1 to get $r + s + t = \boxed{054}$.

-Solution by thecmd999

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