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 3 (Trapezoid)== | ==Solution 3 (Trapezoid)== |
Revision as of 16:18, 16 August 2024
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 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.