Difference between revisions of "2022 AMC 12A Problems/Problem 22"
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Technodoggo (talk | contribs) (crazy ahh bash solution :clown:) |
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~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ||
+ | |||
+ | ==Solution 5 (calculus but it's bash)== | ||
+ | |||
+ | Note that <math>z=\dfrac c2\pm\dfrac{\sqrt{c^2-40}}2</math>, so let <math>z_1=\dfrac c2+\dfrac{\sqrt{c^2-40}}2</math> and <math>z_2=\dfrac c2-\dfrac{\sqrt{c^2-40}}2</math>. Taking a look at the answer choices, they range between <math>c=4.5</math> to <math>c=6.5</math>, and in that range, <math>c^2</math> is always less than <math>40</math>. Thus, <math>c^2-40<0</math> for our possible answer choices; we can then rewrite <math>z_1</math> and <math>z_2</math> as <math>\dfrac c2+\dfrac{\sqrt{40-c^2}}2i</math> and <math>\dfrac c2-\dfrac{\sqrt{40-c^2}}2i</math>, respectively, with real coefficients. | ||
+ | |||
+ | Let us compute <math>\dfrac1z</math>: | ||
+ | |||
+ | <cmath>\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}.</cmath> | ||
+ | |||
+ | Then, <math>\dfrac1{z_1}=\dfrac{c-\sqrt{c^2-40}}{20}=\dfrac c{20}-\dfrac{\sqrt{40-c^2}}{20}i</math> while <math>\dfrac1{z_2}=\dfrac{c+\sqrt{c^2-40}}{20}=\dfrac c{20}+\dfrac{\sqrt{40-c^2}}{20}i</math>. | ||
+ | |||
+ | In the complex plane, we can draw a rough sketch of <math>z_1,z_2,\dfrac1{z_1},\dfrac1{z_2}</math>: | ||
+ | |||
+ | <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("$\frac1{z_1}$", (0.7,1), N); | ||
+ | label("$\frac1{z_2}$", (0.7,-1), S); | ||
+ | </asy> | ||
+ | |||
+ | Note that, here, our quadrilateral is an isoceles trapezoid. The shorter base length is <math>\left(\dfrac{\sqrt{40-c^2}}{20}-\left(-\dfrac{\sqrt{40-c^2}}{20}\right)\right)=\dfrac1{10}\sqrt{40-c^2}</math>. | ||
+ | |||
+ | The longer base length is <math>\left(\dfrac{\sqrt{40-c^2}}2-\left(-\dfrac{\sqrt{40-c^2}}2\right)\right)=\sqrt{40-c^2}</math>. | ||
+ | |||
+ | The average of the two bases is <math>\dfrac{11}{20}\sqrt{40-c^2}</math>. | ||
+ | |||
+ | The height of our trapezoid (which is horizontal parallel to the <math>x</math>-axis in our diagram above) is simply <math>\dfrac c2-\dfrac c{20}=\dfrac9{20}c</math>. | ||
+ | |||
+ | 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 <math>\dfrac{99}{400}c\sqrt{40-c^2}</math>, which is a function of <math>c</math>. Thus, let <math>A(c)=\dfrac{99}{400}c\sqrt{40-c^2}</math>. | ||
+ | |||
+ | Taking the derivative (FINALLY we get to the Calculus part haha, this definitely wasn't clickbait :3 trust me) with respect to <math>c</math>, we find that <math>\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)</math>. | ||
+ | |||
+ | To find an extremum, we set the derivative equal to zero: | ||
+ | |||
+ | \begin{align*} | ||
+ | \dfrac{dA}{dc}&=\dfrac{99}{400}\left(\sqrt{40-c^2}-\dfrac{c^2}{\sqrt{40-c^2}}\right) \ | ||
+ | 0&=\dfrac{99}{400}\left(\sqrt{40-c^2}-\dfrac{c^2}{\sqrt{40-c^2}}\right) \ | ||
+ | \sqrt{40-c^2}-\dfrac{c^2}{\sqrt{40-c^2}}&=0 \ | ||
+ | \sqrt{40-c^2}&=\dfrac{c^2}{\sqrt{40-c^2}} \ | ||
+ | \left(\sqrt{40-c^2}\right)^2&=c^2 \ | ||
+ | c^2&=40-c^2 \ | ||
+ | 2c^2&=40 \ | ||
+ | c^2&=20 \ | ||
+ | c&=\sqrt{20} \ | ||
+ | &\approx4.45. \ | ||
+ | \end{align*} | ||
+ | |||
+ | Clearly, this is very close to <math>\boxed{\textbf{(A)}~4.5}</math>, so we are done. QED. | ||
+ | |||
+ | ~Technodoggo | ||
==Video Solution by Math-X (Smart and Simple)== | ==Video Solution by Math-X (Smart and Simple)== |
Revision as of 02:17, 25 September 2024
Contents
[hide]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 , ,
Solution 5 (calculus but it's bash)
Note that , so let and . Taking a look at the answer choices, they range between to , and in that range, is always less than . Thus, for our possible answer choices; we can then rewrite and as and , respectively, with real coefficients.
Let us compute :
Then, while .
In the complex plane, we can draw a rough sketch of :
Note that, here, our quadrilateral is an isoceles trapezoid. The shorter base length is .
The longer base length is .
The average of the two bases is .
The height of our trapezoid (which is horizontal parallel to the -axis in our diagram above) is simply .
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 , which is a function of . Thus, let .
Taking the derivative (FINALLY we get to the Calculus part haha, this definitely wasn't clickbait :3 trust me) with respect to , we find that .
To find an extremum, we set the derivative equal to zero:
Clearly, this is very close to , so we are done. QED.
~Technodoggo
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.