Difference between revisions of "2022 AMC 12A Problems/Problem 22"
m |
|||
Line 70: | Line 70: | ||
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
+ | |||
+ | ==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 12:44, 12 November 2022
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.