Difference between revisions of "2022 AMC 12A Problems/Problem 22"
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~PluginL | ~PluginL | ||
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+ | ==Solution 5 (Calculus Finish)== | ||
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+ | 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>. | ||
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+ | 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> | ||
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+ | <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> | ||
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+ | <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> | ||
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+ | <math>\therefore f(a)</math> has a local maximum when <math>a = \sqrt{5}</math>. | ||
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+ | 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> | ||
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+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ||
==Video Solution by Math-X (Smart and Simple)== | ==Video Solution by Math-X (Smart and Simple)== |
Revision as of 10:20, 1 January 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 (Incomplete)
Because , notice that . Furthermore, note that because is real, . Thus, . Similarly, . On the complex coordinate plane, let , ,, . Notice how is similar to . Thus, the area of is for some constant , and (In progress)
Solution 3 (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 4 (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 5 (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.