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

Revision as of 16:36, 12 November 2022

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)

Solution 3

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 74: / Line 74: @@
 Because <math>z^2 - cz + 10 = 0</math>, notice that <math>|z_1||z_2|=|10|=10</math>. Furthermore, note that because <math>c</math> is real, <math>z_2=\bar z_1</math>. Thus, <math>\frac{1}{z_1}=\frac{\bar z_1}{z_1\cdot{\bar z_1}}=\frac{z_2}{|z_1|^2}=\frac{z_2}{100}</math>. Similarly, <math>\frac{1}{z_2}=\frac{z_1}{100}</math>. On the complex coordinate plane, let <math>z_1=A_2</math>, <math>z_2=B_2</math>,<math>\frac{1}{z_2}=A_1</math>, <math>\frac{1}{z_1}=B_1</math>. Notice how <math>OA_1B_1</math> is similar to <math>OA_2B_2</math>. Thus, the area of <math>A_1B_1B_2B_1</math> is  <math>(k)(OA_2B_2)</math> for some constant <math>k</math>, and <math>OA_2B_2 = </math>
 (In progress)
+==Solution 3==
 ==Video Solution by Punxsutawney Phil==

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