Question

Let $c$ be a real number, and let $z_1,z_2$ be the two complex numbers satisfying the quadratic $z^2-cz+10=0.$ Points $z_1,$ $z_2,$ $\dfrac{1}{z_1},$ and $\dfrac{1}{z_2}$ are the vertices of a (convex) quadrilateral $Q$ in the complex plane. When the area of $Q$ obtains its maximum value, $c$ is the closest to which of the following?

$4.5$

$5$

$5.5$

$6$

$6.5$

Solution:

If the roots are non-real, then $z_1=\sqrt{10}\,e^{i\theta}$ and $z_2=\overline{z_1},$ since $z_1z_2=10.$ Then $\dfrac{1}{z_1}=\dfrac{1}{\sqrt{10}}e^{-i\theta}$ and $\dfrac{1}{z_2}=\dfrac{1}{\sqrt{10}}e^{i\theta}.$

These four points form a trapezoid with two vertical sides symmetric about the real axis. Its area works out to $\frac{99}{20}\sin 2\theta,$ which is maximized at $\theta=45^\circ.$

Then $c=z_1+z_2=2\sqrt{10}\cos45^\circ=2\sqrt5\approx4.47,$ closest to $4.5.$

Thus, the correct answer is A.

2022 AMC 12A Problem 22

Solution:

Problem 22 in Other Years