Difference between revisions of "2022 AMC 12A Problems/Problem 22"
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Let <math>c</math> be a real number, and let <math>z_1</math> and <math>z_2</math> be the two complex numbers satisfying the equation | Let <math>c</math> be a real number, and let <math>z_1</math> and <math>z_2</math> be the two complex numbers satisfying the equation | ||
− | <math>z^2 - cz + 10 = 0</math>. Points <math>z_1</math>, <math>z_2</math>, <math>\frac{1}{z_1}</math>, and <math>\frac{1}{z_2}</math> are the vertices of (convex) quadrilateral <math>Q</math> in the complex plane. When the area of <math>Q</math> obtains its maximum possible value, <math>c</math> is closest to which of the following? | + | <math>z^2 - cz + 10 = 0</math>. Points <math>z_1</math>, <math>z_2</math>, <math>\frac{1}{z_1}</math>, and <math>\frac{1}{z_2}</math> are the vertices of (convex) quadrilateral <math>\mathcal{Q}</math> in the complex plane. When the area of <math>\mathcal{Q}</math> obtains its maximum possible value, <math>c</math> is closest to which of the following? |
− | ==Solution== | + | <math>\textbf{(A) }4.5 \qquad\textbf{(B) }5 \qquad\textbf{(C) }5.5 \qquad\textbf{(D) }6\qquad\textbf{(E) }6.5</math> |
+ | |||
+ | ==Solution 1== | ||
Because <math>c</math> is real, <math>z_2 = \bar z_1</math>. | Because <math>c</math> is real, <math>z_2 = \bar z_1</math>. | ||
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c & = z_1 + z_2 \\ | c & = z_1 + z_2 \\ | ||
& = z_1 + \bar z_1 \\ | & = z_1 + \bar z_1 \\ | ||
− | & = 2 {\rm Re} | + | & = 2 {\rm Re}(z_1), |
\end{align*} | \end{align*} | ||
</cmath> | </cmath> | ||
where the first equality follows from Vieta's formula. | where the first equality follows from Vieta's formula. | ||
− | Thus, <math>{\rm Re} | + | Thus, <math>{\rm Re}(z_1) = \frac{c}{2}</math>. |
We have | We have | ||
<cmath> | <cmath> | ||
\begin{align*} | \begin{align*} | ||
− | \frac{1}{z_1} & = \frac{1}{10} \frac{10}{z_1} \\ | + | \frac{1}{z_1} & = \frac{1}{10}\cdot\frac{10}{z_1} \\ |
− | & = \frac{1}{10} \frac{z_1 z_2}{z_1} \\ | + | & = \frac{1}{10}\cdot\frac{z_1 z_2}{z_1} \\ |
& = \frac{z_2}{10} \\ | & = \frac{z_2}{10} \\ | ||
− | & = \frac{\bar z_1}{10} . | + | & = \frac{\bar z_1}{10}. |
\end{align*} | \end{align*} | ||
</cmath> | </cmath> | ||
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<cmath> | <cmath> | ||
\begin{align*} | \begin{align*} | ||
− | \frac{1}{z_2} & = \frac{1}{10} \frac{10}{z_2} \\ | + | \frac{1}{z_2} & = \frac{1}{10}\cdot\frac{10}{z_2} \\ |
− | & = \frac{1}{10} \frac{z_1 z_2}{z_2} \\ | + | & = \frac{1}{10}\cdot\frac{z_1 z_2}{z_2} \\ |
− | & = \frac{z_1}{10} | + | & = \frac{z_1}{10}. |
\end{align*} | \end{align*} | ||
</cmath> | </cmath> | ||
Line 56: | Line 58: | ||
\begin{align*} | \begin{align*} | ||
{\rm Area} \ Q | {\rm Area} \ Q | ||
− | & = \frac{1}{2} \left| {\rm Re} | + | & = \frac{1}{2} \left| {\rm Re}(z_1) \right| |
− | \cdot 2 \left| {\rm Im} | + | \cdot 2 \left| {\rm Im}(z_1) \right| |
\cdot \left( 1 - \frac{1}{10^2} \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}{2} |c| \sqrt{10 - \frac{c^2}{4}} \left( 1 - \frac{1}{10^2} \right) \\ | ||
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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
− | ==Video Solution== | + | ==Solution 2 (Trapezoid)== |
+ | Since <math>c</math>, which is the sum of roots <math>z_1</math> and <math>z_2</math>, is real, <math>z_1=\overline{z_2}</math>. | ||
+ | |||
+ | Let <math>z_1=a+bi</math>. Then <math>z_2=a-bi</math>. Note that the product of the roots is <math>10</math> by Vieta's, so <math>z_1z_2=(a+bi)(a-bi)=a^2+b^2=10</math>. | ||
+ | |||
+ | Thus, <math>\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}</math>. With the same process, <math>\frac{1}{z_2}=\frac{a-bi}{10}</math>. | ||
+ | |||
+ | So, our four points are <math>a+bi,\frac{a+bi}{10},a-bi,</math> and <math>\frac{a-bi}{10}</math>. WLOG let <math>a+bi</math> 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 <math>\frac{a+bi}{10}</math> and <math>\frac{a-bi}{10}</math>, so its length is <math>\left|\frac{a+bi}{10} - \frac{a-bi}{10}\right| =\frac{b}{5}</math>. Likewise, its long base has endpoints <math>a+bi</math> and <math>a-bi</math>, so its length is <math>|(a+bi)-(a-bi)|=2b</math>. | ||
+ | |||
+ | The height, which is the distance between the two lines, is the difference between the real values of the two bases <math>\implies h= a-\frac{a}{10}=\frac{9a}{10}</math>. | ||
+ | |||
+ | Plugging these into the area formula for a trapezoid, we are trying to maximize <math>\frac12\cdot\left(2b+\frac{b}{5}\right)\cdot\frac{9a}{10}=\frac{99ab}{100}</math>. Thus, the only thing we need to maximize is <math>ab</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>. | ||
+ | |||
+ | 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 | ||
+ | |||
+ | ==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>. | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | On the basis of symmetry, the area <math>A</math> of <math>\mathcal{Q}</math> is the difference between two isoceles triangles,so | ||
+ | |||
+ | <math>2A=10\sin2\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>. | ||
+ | |||
+ | 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 | ||
+ | |||
+ | ==Solution 4 (Calculus Finish)== | ||
+ | |||
+ | Like in Solution 3, we find that <math>Q = \frac12\cdot\left(2b+\frac{b}{5}\right)\cdot\frac{9a}{10}= \frac{99}{100}ab</math>, thus, <math>Q</math> is maximized when <math>ab</math> is maximized. <math>ab = a \cdot \sqrt{10 - a^2} = \sqrt{10a^2 - a^4}</math>, let <math>f(a) = \sqrt{10a^2 - a^4}</math>. | ||
+ | |||
+ | By the Chain Rule and the Power Rule, <math>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} }</math> | ||
+ | |||
+ | <math>f'(a) = 0</math>, <math>10a-2a^3 = 0</math>, <math>a \neq 0</math>, <math>a^2 = 5</math>, <math>a = \sqrt{5}</math> | ||
+ | |||
+ | <math>\because f'(a) = 0</math> when <math>a = \sqrt{5}</math>, <math>f'(a)</math> is positive when <math>a < \sqrt{5}</math>, and <math>f'(a)</math> is negative when <math>a > \sqrt{5}</math> | ||
+ | |||
+ | <math>\therefore f(a)</math> has a local maximum when <math>a = \sqrt{5}</math>. | ||
+ | |||
+ | Notice that <math>ab = \frac{ (a+b)^2 - (a^2+b^2) }{2} = \frac{c^2 - 10}{2}</math>, <math>\frac{c^2 - 10}{2} = 5</math>, <math>c = \sqrt{2 \cdot 5 + 10} = \sqrt{20} \approx \boxed{\text{(A) 4.5}}</math> | ||
+ | |||
+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen 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 | https://youtu.be/pcB2sg7Ag58 | ||
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
+ | |||
+ | == See Also == | ||
+ | |||
+ | {{AMC12 box|year=2022|ab=A|num-b=21|num-a=23}} | ||
+ | |||
+ | [[Category:Intermediate Algebra Problems]] | ||
+ | [[Category:Intermediate Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 16:19, 16 August 2024
Contents
Problem
Let be a real number, and let
and
be the two complex numbers satisfying the equation
. Points
,
,
, and
are the vertices of (convex) quadrilateral
in the complex plane. When the area of
obtains its maximum possible value,
is closest to which of the following?
Solution 1
Because is real,
.
We have
where the first equality follows from Vieta's formula.
Thus, .
We have
where the first equality follows from Vieta's formula.
Thus, .
We have
where the second equality follows from Vieta's formula.
We have
where the second equality follows from Vieta's formula.
Therefore,
where the inequality follows from the AM-GM inequality, and it is augmented to an equality if and only if
.
Thus,
.
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Solution 2 (Trapezoid)
Since , which is the sum of roots
and
, is real,
.
Let . Then
. Note that the product of the roots is
by Vieta's, so
.
Thus, . With the same process,
.
So, our four points are and
. WLOG let
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
and
, so its length is
. Likewise, its long base has endpoints
and
, so its length is
.
The height, which is the distance between the two lines, is the difference between the real values of the two bases .
Plugging these into the area formula for a trapezoid, we are trying to maximize . Thus, the only thing we need to maximize is
.
With the restriction that ,
is maximized when
.
Remember, is the sum of the roots, so
~quacker88
Solution 3 (Fast)
Like the solutions above we can know that and
.
Let where
, then
,
,
.
On the basis of symmetry, the area of
is the difference between two isoceles triangles,so
. The inequality holds when
, or
.
Thus, .
~PluginL
Solution 4 (Calculus Finish)
Like in Solution 3, we find that , thus,
is maximized when
is maximized.
, let
.
By the Chain Rule and the Power Rule,
,
,
,
,
when
,
is positive when
, and
is negative when
has a local maximum when
.
Notice that ,
,
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
~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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.