2022 AMC 12A Problems/Problem 22
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 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 (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 3 (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 4 (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 ,
,
Solution 5 (calculus but it's bash)
Note that , so let
and
. Taking a look at the answer choices, they range between
to
, and in that range,
is always less than
. Thus,
for our possible answer choices; we can then rewrite
and
as
and
, respectively, with real coefficients.
Let us compute :
Then, while
.
In the complex plane, we can draw a rough sketch of :
Note that, here, our quadrilateral is an isoceles trapezoid. The shorter base length is .
The longer base length is .
The average of the two bases is .
The height of our trapezoid (which is horizontal parallel to the -axis in our diagram above) is simply
.
Since the area of a trapezoid is the product of the average of its bases and its height, we conclude that our trapezoid's area is , which is a function of
. Thus, let
.
Taking the derivative (FINALLY we get to the Calculus part haha, this definitely wasn't clickbait :3 trust me) with respect to , we find that
.
To find an extremum, we set the derivative equal to zero:
Clearly, this is very close to , so we are done. QED.
~Technodoggo (Minor Edits by dolphindesigner)
Solution 6 (AM-GM Inequality)
First and foremost, because is a complex number, and the value of
could be found in terms of
, the given quadratic equation could be solved. WLOG,
and
. If
, no quadrilateral will form. Thereby, it could be inferred that
.
Each complex number could be represented as a point.
Notice that vertices form a trapezoid. Therefore, the area of the trapezoid
could be represented as
Since we are finding the maximum value, any desired constant could be multiplied. Multiplying 80 and other constants to the equation above may provide a simplified form.
We can use AM-GM Inequality! According to the AM-GM Inequality,
reaches maximum when
. In another words,
. Using
for the approximation of
,
. Thus, the closest value in the answer choice is
.
~MaPhyCom
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.