Difference between revisions of "2022 AMC 12A Problems/Problem 22"

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Problem

Let $c$ be a real number, and let $z_1$ and $z_2$ be the two complex numbers satisfying the equation $z^2 - cz + 10 = 0$ . Points $z_1$ , $z_2$ , $\frac{1}{z_1}$ , and $\frac{1}{z_2}$ are the vertices of (convex) quadrilateral $\mathcal{Q}$ in the complex plane. When the area of $\mathcal{Q}$ obtains its maximum possible value, $c$ is closest to which of the following?

$\textbf{(A) }4.5 \qquad\textbf{(B) }5 \qquad\textbf{(C) }5.5 \qquad\textbf{(D) }6\qquad\textbf{(E) }6.5$

Solution 1

Because $c$ is real, $z_2 = \bar z_1$ . We have $\begin{align*} 10 & = z_1 z_2 \\ & = z_1 \bar z_1 \\ & = |z_1|^2 , \end{align*}$ where the first equality follows from Vieta's formula.

Thus, $|z_1| = \sqrt{10}$ .

We have $\begin{align*} c & = z_1 + z_2 \\ & = z_1 + \bar z_1 \\ & = 2 {\rm Re}(z_1), \end{align*}$ where the first equality follows from Vieta's formula.

Thus, ${\rm Re}(z_1) = \frac{c}{2}$ .

We have $\begin{align*} \frac{1}{z_1} & = \frac{1}{10}\cdot\frac{10}{z_1} \\ & = \frac{1}{10}\cdot\frac{z_1 z_2}{z_1} \\ & = \frac{z_2}{10} \\ & = \frac{\bar z_1}{10}. \end{align*}$ where the second equality follows from Vieta's formula.

We have $\begin{align*} \frac{1}{z_2} & = \frac{1}{10}\cdot\frac{10}{z_2} \\ & = \frac{1}{10}\cdot\frac{z_1 z_2}{z_2} \\ & = \frac{z_1}{10}. \end{align*}$ where the second equality follows from Vieta's formula.

Therefore, $\begin{align*} {\rm Area} \ Q & = \frac{1}{2} \left| {\rm Re}(z_1) \right| \cdot 2 \left| {\rm Im}(z_1) \right| \cdot \left( 1 - \frac{1}{10^2} \right) \\ & = \frac{1}{2} |c| \sqrt{10 - \frac{c^2}{4}} \left( 1 - \frac{1}{10^2} \right) \\ & = \frac{1 - \frac{1}{10^2}}{4} \sqrt{c^2 \left( 40 - c^2 \right)} \\ & \leq \frac{1 - \frac{1}{10^2}}{4} \cdot \frac{c^2 + \left( 40 - c^2 \right)}{2} \\ & = \frac{1 - \frac{1}{10^2}}{4} \cdot 20 , \end{align*}$ where the inequality follows from the AM-GM inequality, and it is augmented to an equality if and only if $c^2 = 40 - c^2$ . Thus, $|c| = 2 \sqrt{5} \approx \boxed{\textbf{(A) 4.5}}$ .

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Solution 2 (Trapezoid)

Since $c$ , which is the sum of roots $z_1$ and $z_2$ , is real, $z_1=\overline{z_2}$ .

Let $z_1=a+bi$ . Then $z_2=a-bi$ . Note that the product of the roots is $10$ by Vieta's, so $z_1z_2=(a+bi)(a-bi)=a^2+b^2=10$ .

Thus, $\frac{1}{z_1}=\frac{1}{a-bi}\cdot\frac{a+bi}{a+bi}=\frac{a+bi}{a^2+b^2}=\frac{a+bi}{10}$ . With the same process, $\frac{1}{z_2}=\frac{a-bi}{10}$ .

So, our four points are $a+bi,\frac{a+bi}{10},a-bi,$ and $\frac{a-bi}{10}$ . WLOG let $a+bi$ be in the first quadrant and graph these four points on the complex plane. Notice how quadrilateral Q is a trapezoid with the real axis as its axis of symmetry. It has a short base with endpoints $\frac{a+bi}{10}$ and $\frac{a-bi}{10}$ , so its length is $\left|\frac{a+bi}{10} - \frac{a-bi}{10}\right| =\frac{b}{5}$ . Likewise, its long base has endpoints $a+bi$ and $a-bi$ , so its length is $|(a+bi)-(a-bi)|=2b$ .

The height, which is the distance between the two lines, is the difference between the real values of the two bases $\implies h= a-\frac{a}{10}=\frac{9a}{10}$ .

Plugging these into the area formula for a trapezoid, we are trying to maximize $\frac12\cdot\left(2b+\frac{b}{5}\right)\cdot\frac{9a}{10}=\frac{99ab}{100}$ . Thus, the only thing we need to maximize is $ab$ .

With the restriction that $a^2+b^2=(a-b)^2+2ab=10$ , $ab$ is maximized when $a=b=\sqrt{5}$ .

Remember, $c$ is the sum of the roots, so $c=z_1+z_2=2a=2\sqrt5=\sqrt{20}\approx\boxed{4.5}$

~quacker88

Solution 3 (Fast)

Like the solutions above we can know that $|z_1| = |z_2| = \sqrt{10}$ and $z_1=\overline{z_2}$ .

Let $z_1=\sqrt{10}e^{i\theta}$ where $0<\theta<\pi$ , then $z_2=\sqrt{10}e^{-i\theta}$ , $\frac{1}{z_1}=\frac{1}{\sqrt{10}}e^{-i\theta}$ , $\frac{1}{z_2}= \frac{1}{\sqrt{10}}e^{i\theta}$ .

On the basis of symmetry, the area $A$ of $\mathcal{Q}$ is the difference between two isoceles triangles,so

$2A=10\sin2\theta-\frac{1}{10}\sin2\theta\leq10-\frac{1}{10}$ . The inequality holds when $2\theta=\frac{\pi}{2}$ , or $\theta=\frac{\pi}{4}$ .

Thus, $c= 2 {\rm Re} \ z_1 =2 \sqrt{10} \cos\frac{\pi}{4}=\sqrt{20} \approx \boxed{\textbf{(A) 4.5}}$ .

~PluginL

Solution 5 (Calculus Finish)

Like in Solution 3, we find that $Q = \frac12\cdot\left(2b+\frac{b}{5}\right)\cdot\frac{9a}{10}= \frac{99}{100}ab$ , thus, $Q$ is maximized when $ab$ is maximized. $ab = a \cdot \sqrt{10 - a^2} = \sqrt{10a^2 - a^4}$ , let $f(a) = \sqrt{10a^2 - a^4}$ .

By the Chain Rule and the Power Rule, $f'(a) = \frac12 \cdot (10a^2 - a^4)^{-\frac12} \cdot (10(2a)-4a^3)) = \frac{20a-4a^3}{ 2\sqrt{10a^2 - a^4} } = \frac{10a-2a^3}{ \sqrt{10a^2 - a^4} }$

$f'(a) = 0$ , $10a-2a^3 = 0$ , $a \neq 0$ , $a^2 = 5$ , $a = \sqrt{5}$

$\because f'(a) = 0$ when $a = \sqrt{5}$ , $f'(a)$ is positive when $a < \sqrt{5}$ , and $f'(a)$ is negative when $a > \sqrt{5}$

$\therefore f(a)$ has a local maximum when $a = \sqrt{5}$ .

Notice that $ab = \frac{ (a+b)^2 - (a^2+b^2) }{2} = \frac{c^2 - 10}{2}$ , $\frac{c^2 - 10}{2} = 5$ , $c = \sqrt{2 \cdot 5 + 10} = \sqrt{20} \approx \boxed{\text{(A) 4.5}}$

~isabelchen

Video Solution by Math-X (Smart and Simple)

https://youtu.be/7yAh4MtJ8a8?si=1NLsu57rYrEPP1A2&t=6893 ~Math-X

Video Solution by Punxsutawney Phil

https://youtube.com/watch?v=bbMcdvlPcyA

Video Solution by Steven Chen

https://youtu.be/pcB2sg7Ag58

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

@@ Line 92: / Line 92: @@
 ~quacker88
-==Solution 4 (Fast)==
+==Solution 3 (Fast)==
 Like the solutions above we can know that <math>|z_1| = |z_2| = \sqrt{10}</math> and <math>z_1=\overline{z_2}</math>.

2022 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
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