Difference between revisions of "2022 AMC 12A Problems/Problem 22"
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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
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+ | ==Solution 2== | ||
+ | Because <math>z^2 - cz + 10 = 0</math>, notice that <math>|z_1||z_2|=|10|=10</math>. Furthermore, note that because <math>c</math> is real, <math>z_2=\bar z_1</math>. Thus, <math>\frac{1}{z_1}=\frac{\bar z_1}{z_1\cdot{\bar z_1}}=\frac{z_2}{|z_1|^2}=\frac{z_2}{100}</math>. Similarly, <math>\frac{1}{z_2}=\frac{z_1}{100}</math>. On the complex coordinate plane, let <math>z_1=A_2</math>, <math>z_2=B_2</math>,<math>\frac{1}{z_2}=A_1</math>, <math>\frac{1}{z_1}=B_1</math>. Notice how <math>OA_1B_1</math> is similar to <math>OA_2B_2</math>. Thus, the area of <math>A_1B_1B_2B_1</math> is <math>(k)(OA_2B_2)</math> for some constant <math>k</math>, and <math>OA_2B_2 = </math> | ||
+ | (In progress) | ||
==Video Solution== | ==Video Solution== |
Revision as of 11:44, 12 November 2022
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
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)
Video Solution
~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.