Difference between revisions of "2022 AMC 12A Problems/Problem 22"
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c^2&=20 \\ | c^2&=20 \\ | ||
c&=\sqrt{20} \\ | c&=\sqrt{20} \\ | ||
− | &\approx4. | + | &\approx4.47 \\ |
\end{align*} | \end{align*} | ||
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~Technodoggo | ~Technodoggo | ||
+ | (Minor Edits by dolphindesigner) | ||
==Video Solution by Math-X (Smart and Simple)== | ==Video Solution by Math-X (Smart and Simple)== |
Latest revision as of 22:54, 1 October 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 , ,
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:
\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} \\ &\approx4.47 \\ \end{align*}
Clearly, this is very close to , so we are done. QED.
~Technodoggo (Minor Edits by dolphindesigner)
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.