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2022 AMC 12A Problems/Problem 22

Revision as of 11:44, 12 November 2022 by Sigmapie (talk | contribs)

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 $Q$ in the complex plane. When the area of $Q$ obtains its maximum possible value, $c$ is closest to which of the following?

Solution

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} \frac{10}{z_1} \\ & = \frac{1}{10} \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} \frac{10}{z_2} \\ & = \frac{1}{10} \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

Because $z^2 - cz + 10 = 0$, notice that $|z_1||z_2|=|10|=10$. Furthermore, note that because $c$ is real, $z_2=\bar z_1$. Thus, $\frac{1}{z_1}=\frac{\bar z_1}{z_1\cdot{\bar z_1}}=\frac{z_2}{|z_1|^2}=\frac{z_2}{100}$. Similarly, $\frac{1}{z_2}=\frac{z_1}{100}$. On the complex coordinate plane, let $z_1=A_2$, $z_2=B_2$,$\frac{1}{z_2}=A_1$, $\frac{1}{z_1}=B_1$. Notice how $OA_1B_1$ is similar to $OA_2B_2$. Thus, the area of $A_1B_1B_2B_1$ is $(k)(OA_2B_2)$ for some constant $k$, and $OA_2B_2 =$ (In progress)

Video Solution

https://youtu.be/pcB2sg7Ag58

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

See Also

2022 AMC 12A (Problems • Answer Key • Resources)
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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