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) | ||
− | ==Solution 2 | + | ==Solution 2 (Trapezoid)== |
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Since <math>c</math>, which is the sum of roots <math>z_1</math> and <math>z_2</math>, is real, <math>z_1=\overline{z_2}</math>. | Since <math>c</math>, which is the sum of roots <math>z_1</math> and <math>z_2</math>, is real, <math>z_1=\overline{z_2}</math>. | ||
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~quacker88 | ~quacker88 | ||
− | ==Solution | + | ==Solution 3 (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>. | Like the solutions above we can know that <math>|z_1| = |z_2| = \sqrt{10}</math> and <math>z_1=\overline{z_2}</math>. | ||
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~PluginL | ~PluginL | ||
− | Video Solution by Math-X (Smart and Simple) | + | ==Solution 4 (Calculus Finish)== |
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+ | Like in Solution 3, we find that <math>Q = \frac12\cdot\left(2b+\frac{b}{5}\right)\cdot\frac{9a}{10}= \frac{99}{100}ab</math>, thus, <math>Q</math> is maximized when <math>ab</math> is maximized. <math>ab = a \cdot \sqrt{10 - a^2} = \sqrt{10a^2 - a^4}</math>, let <math>f(a) = \sqrt{10a^2 - a^4}</math>. | ||
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+ | By the Chain Rule and the Power Rule, <math>f'(a) = \frac12 \cdot (10a^2 - a^4)^{-\frac12} \cdot (10(2a)-4a^3)) = \frac{20a-4a^3}{ 2\sqrt{10a^2 - a^4} } = \frac{10a-2a^3}{ \sqrt{10a^2 - a^4} }</math> | ||
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+ | <math>f'(a) = 0</math>, <math>10a-2a^3 = 0</math>, <math>a \neq 0</math>, <math>a^2 = 5</math>, <math>a = \sqrt{5}</math> | ||
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+ | <math>\because f'(a) = 0</math> when <math>a = \sqrt{5}</math>, <math>f'(a)</math> is positive when <math>a < \sqrt{5}</math>, and <math>f'(a)</math> is negative when <math>a > \sqrt{5}</math> | ||
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+ | <math>\therefore f(a)</math> has a local maximum when <math>a = \sqrt{5}</math>. | ||
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+ | Notice that <math>ab = \frac{ (a+b)^2 - (a^2+b^2) }{2} = \frac{c^2 - 10}{2}</math>, <math>\frac{c^2 - 10}{2} = 5</math>, <math>c = \sqrt{2 \cdot 5 + 10} = \sqrt{20} \approx \boxed{\text{(A) 4.5}}</math> | ||
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+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ||
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+ | ==Video Solution by Math-X (Smart and Simple)== | ||
https://youtu.be/7yAh4MtJ8a8?si=1NLsu57rYrEPP1A2&t=6893 ~Math-X | https://youtu.be/7yAh4MtJ8a8?si=1NLsu57rYrEPP1A2&t=6893 ~Math-X | ||
Latest revision as of 16:19, 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 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 ,
,
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.