Difference between revisions of "2022 AMC 12A Problems/Problem 22"
MRENTHUSIASM (talk | contribs) m |
MRENTHUSIASM (talk | contribs) m (→Solution 1) |
||
Line 26: | Line 26: | ||
c & = z_1 + z_2 \\ | c & = z_1 + z_2 \\ | ||
& = z_1 + \bar z_1 \\ | & = z_1 + \bar z_1 \\ | ||
− | & = 2 {\rm Re} | + | & = 2 {\rm Re}(z_1), |
\end{align*} | \end{align*} | ||
</cmath> | </cmath> | ||
where the first equality follows from Vieta's formula. | where the first equality follows from Vieta's formula. | ||
− | Thus, <math>{\rm Re} | + | Thus, <math>{\rm Re}(z_1) = \frac{c}{2}</math>. |
We have | We have | ||
<cmath> | <cmath> | ||
\begin{align*} | \begin{align*} | ||
− | \frac{1}{z_1} & = \frac{1}{10} \frac{10}{z_1} \\ | + | \frac{1}{z_1} & = \frac{1}{10}\cdot\frac{10}{z_1} \\ |
− | & = \frac{1}{10} \frac{z_1 z_2}{z_1} \\ | + | & = \frac{1}{10}\cdot\frac{z_1 z_2}{z_1} \\ |
& = \frac{z_2}{10} \\ | & = \frac{z_2}{10} \\ | ||
− | & = \frac{\bar z_1}{10} . | + | & = \frac{\bar z_1}{10}. |
\end{align*} | \end{align*} | ||
</cmath> | </cmath> | ||
Line 47: | Line 47: | ||
<cmath> | <cmath> | ||
\begin{align*} | \begin{align*} | ||
− | \frac{1}{z_2} & = \frac{1}{10} \frac{10}{z_2} \\ | + | \frac{1}{z_2} & = \frac{1}{10}\cdot\frac{10}{z_2} \\ |
− | & = \frac{1}{10} \frac{z_1 z_2}{z_2} \\ | + | & = \frac{1}{10}\cdot\frac{z_1 z_2}{z_2} \\ |
− | & = \frac{z_1}{10} | + | & = \frac{z_1}{10}. |
\end{align*} | \end{align*} | ||
</cmath> | </cmath> | ||
Line 58: | Line 58: | ||
\begin{align*} | \begin{align*} | ||
{\rm Area} \ Q | {\rm Area} \ Q | ||
− | & = \frac{1}{2} \left| {\rm Re} | + | & = \frac{1}{2} \left| {\rm Re}(z_1) \right| |
− | \cdot 2 \left| {\rm Im} | + | \cdot 2 \left| {\rm Im}(z_1) \right| |
\cdot \left( 1 - \frac{1}{10^2} \right) \\ | \cdot \left( 1 - \frac{1}{10^2} \right) \\ | ||
& = \frac{1}{2} |c| \sqrt{10 - \frac{c^2}{4}} \left( 1 - \frac{1}{10^2} \right) \\ | & = \frac{1}{2} |c| \sqrt{10 - \frac{c^2}{4}} \left( 1 - \frac{1}{10^2} \right) \\ |
Revision as of 19:24, 11 January 2023
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, 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
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.