2022 AMC 12A Problems/Problem 22

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 4 (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

Solution 5 (calculus but it's bash)

Note that $z=\dfrac c2\pm\dfrac{\sqrt{c^2-40}}2$, so let $z_1=\dfrac c2+\dfrac{\sqrt{c^2-40}}2$ and $z_2=\dfrac c2-\dfrac{\sqrt{c^2-40}}2$. Taking a look at the answer choices, they range between $c=4.5$ to $c=6.5$, and in that range, $c^2$ is always less than $40$. Thus, $c^2-40<0$ for our possible answer choices; we can then rewrite $z_1$ and $z_2$ as $\dfrac c2+\dfrac{\sqrt{40-c^2}}2i$ and $\dfrac c2-\dfrac{\sqrt{40-c^2}}2i$, respectively, with real coefficients.

Let us compute $\dfrac1z$:

\[\dfrac1z=\dfrac1{\frac c2\pm\frac{\sqrt{c^2-40}}2}=\dfrac{\frac c2\mp\frac{\sqrt{c^2-40}}2}{\left(\frac c2\right)^2-\left(\frac{\sqrt{c^2-40}}2\right)^2}=\dfrac{\frac c2\mp\frac{\sqrt{c^2-40}}2}{\frac{c^2}4-\frac{c^2-40}4}=\dfrac{2c\mp2\sqrt{c^2-40}}{40}=\dfrac{c\mp\sqrt{c^2-40}}{20}.\]

Then, $\dfrac1{z_1}=\dfrac{c-\sqrt{c^2-40}}{20}=\dfrac c{20}-\dfrac{\sqrt{40-c^2}}{20}i$ while $\dfrac1{z_2}=\dfrac{c+\sqrt{c^2-40}}{20}=\dfrac c{20}+\dfrac{\sqrt{40-c^2}}{20}i$.

In the complex plane, we can draw a rough sketch of $z_1,z_2,\dfrac1{z_1},\dfrac1{z_2}$:

[asy] import graph; unitsize(0.5cm);  /*xaxis(Ticks, xmin=-1,xmax=8); yaxis(Ticks, ymin=-11,ymax=11);*/ draw((-1,0)--(8,0)); draw((0,-11)--(0,11));  dot((7,10));dot((7,-10));dot((0.7,1));dot((0.7,-1));  draw((7,10)--(7,-10)--(0.7,-1)--(0.7,1)--cycle);  label("$z_1$", (7,10), E); label("$z_2$", (7,-10), E); label("$\frac1{z_2}$", (0.7,1), N); label("$\frac1{z_1}$", (0.7,-1), S); [/asy]

Note that, here, our quadrilateral is an isoceles trapezoid. The shorter base length is $\left(\dfrac{\sqrt{40-c^2}}{20}-\left(-\dfrac{\sqrt{40-c^2}}{20}\right)\right)=\dfrac1{10}\sqrt{40-c^2}$.

The longer base length is $\left(\dfrac{\sqrt{40-c^2}}2-\left(-\dfrac{\sqrt{40-c^2}}2\right)\right)=\sqrt{40-c^2}$.

The average of the two bases is $\dfrac{11}{20}\sqrt{40-c^2}$.

The height of our trapezoid (which is horizontal parallel to the $x$-axis in our diagram above) is simply $\dfrac c2-\dfrac c{20}=\dfrac9{20}c$.

Since the area of a trapezoid is the product of the average of its bases and its height, we conclude that our trapezoid's area is $\dfrac{99}{400}c\sqrt{40-c^2}$, which is a function of $c$. Thus, let $A(c)=\dfrac{99}{400}c\sqrt{40-c^2}$.

Taking the derivative (FINALLY we get to the Calculus part haha, this definitely wasn't clickbait :3 trust me) with respect to $c$, we find that $\dfrac{dA}{dc}=\dfrac{99}{400}\left(\sqrt{40-c^2}+c\left(\dfrac{-2c}{2\sqrt{40-c^2}}\right)\right)=\dfrac{99}{400}\left(\sqrt{40-c^2}-\dfrac{c^2}{\sqrt{40-c^2}}\right)$.

To find an extremum, we set the derivative equal to zero:

dAdc=99400(40c2c240c2)0=99400(40c2c240c2)40c2c240c2=040c2=c240c2(40c2)2=c2c2=40c22c2=40c2=20c=204.47

Clearly, this is very close to $\boxed{\textbf{(A)}~4.5}$, so we are done. QED.

~Technodoggo (Minor Edits by dolphindesigner)

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)

See Also

2022 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 21
Followed by
Problem 23
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

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