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

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With the restriction that <math>a^2+b^2=(a-b)^2+2ab=10</math>, <math>ab</math> is maximized when <math>a=b=\sqrt{5}</math>.  
 
With the restriction that <math>a^2+b^2=(a-b)^2+2ab=10</math>, <math>ab</math> is maximized when <math>a=b=\sqrt{5}</math>.  
  
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Remember, <math>c</math> is the sum of the roots, so <math>c=z_1+z_2=2a=2\sqrt5=\sqrt{20}\approx\boxed{4.5}</math> ~quacker88
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Remember, <math>c</math> he sum of the roots, so <math>c=z_1+z_2=2a=2\sqrt5=\sqrt{20}\approx\boxed{4.5}</math> ~quacker88
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==Solution 4 (Fast)==
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Like the solutions above we can know that <math>|z_1| = |z_2| = \sqrt{10}</math> and <math>z_1=\overline{z_2}</math>.
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Let <math>z_1=\sqrt{10}e^{i\theta}</math> where <math>0<\theta<\pi</math>, then <math>z_2=\sqrt{10}e^{-i\theta}</math>, <math>\frac{1}{z_1}=\frac{1}{\sqrt{10}}e^{-i\theta} </math>, <math>\frac{1}{z_2}= \frac{1}{\sqrt{10}}e^{i\theta}</math>.
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According to symmetry, the area <math>A</math> of <math>Q</math> is the difference between two isoceles triangles,so
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<math>2A=10sin2\theta-\frac{1}{10}sin2\theta\leq10-\frac{1}{10}</math>. The inequality holds when <math>2\theta=\frac{\pi}{2}</math>, or <math>\theta=\frac{\pi}{4}</math>.
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Thus, <math>c= 2 {\rm Re} \ z_1 =2 \sqrt{10} cos\frac{\pi}{4}=\sqrt{20} \approx \boxed{\textbf{(A) 4.5}} </math> ~PluginL
  
 
==Video Solution by Punxsutawney Phil==
 
==Video Solution by Punxsutawney Phil==

Revision as of 22:00, 13 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 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} \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 (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, $c$ he sum of the roots, so $c=z_1+z_2=2a=2\sqrt5=\sqrt{20}\approx\boxed{4.5}$ ~quacker88


Solution 4 (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 $z_2=\sqrt{10}e^{-i\theta}$, $\frac{1}{z_1}=\frac{1}{\sqrt{10}}e^{-i\theta}$, $\frac{1}{z_2}= \frac{1}{\sqrt{10}}e^{i\theta}$.

According to symmetry, the area $A$ of $Q$ is the difference between two isoceles triangles,so

$2A=10sin2\theta-\frac{1}{10}sin2\theta\leq10-\frac{1}{10}$. 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

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 (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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