Difference between revisions of "2022 AMC 12A Problems/Problem 22"
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With the restriction that <math>a^2+b^2=(a-b)^2+2ab=10</math>, <math>ab</math> is maximized when <math>a=b=\sqrt{5}</math>. | With the restriction that <math>a^2+b^2=(a-b)^2+2ab=10</math>, <math>ab</math> is maximized when <math>a=b=\sqrt{5}</math>. | ||
− | Remember, <math>c</math> | + | Remember, <math>c</math> he sum of the roots, so <math>c=z_1+z_2=2a=2\sqrt5=\sqrt{20}\approx\boxed{4.5}</math> ~quacker88 |
+ | |||
+ | |||
+ | ==Solution 4 (Fast)== | ||
+ | Like the solutions above we can know that <math>|z_1| = |z_2| = \sqrt{10}</math> and <math>z_1=\overline{z_2}</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>. | ||
+ | |||
+ | According to 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>. | ||
+ | |||
+ | Thus, <math>c= 2 {\rm Re} \ z_1 =2 \sqrt{10} cos\frac{\pi}{4}=\sqrt{20} \approx \boxed{\textbf{(A) 4.5}} </math> ~PluginL | ||
==Video Solution by Punxsutawney Phil== | ==Video Solution by Punxsutawney Phil== |
Revision as of 23:00, 13 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, he sum of the roots, so
~quacker88
Solution 4 (Fast)
Like the solutions above we can know that and
.
Let where
, then
,
,
.
According to 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.