Difference between revisions of "2022 AMC 12A Problems/Problem 22"
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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>. | 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>Q</math> is the difference between two isoceles triangles,so | |
<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>. | <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>. |
Revision as of 00:37, 14 November 2022
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
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
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.