2022 AMC 12A Problem 22

Problem 22 of 25HarderAlgebra Geometry

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?

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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}.

The two vertical sides of this trapezoid have lengths

2\sqrt{10}\sin\theta

and

\frac{2\sin\theta}{\sqrt{10}},

and their horizontal separation is

(\sqrt{10}-\frac{1}{\sqrt{10}})\cos\theta.

Hence its area is

\frac{99}{10}\sin\theta\cos\theta =\frac{99}{20}\sin2\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.