Problem:
Let c be a real number, and let z1β and z2β be the two complex numbers satisfying the equation z2βcz+10=0. Points z1β,z2β,z1β1β, and z2β1β are the vertices of (convex) quadrilateral Q in the complex plane. When the area of Q obtains its maximum possible value, c is closest to which of the following?
The quadrilateral in the complex plane will be degenerate unless c2β40β is imaginary, in which case the quadrilateral will be a trapezoid with vertical parallel sides whose real parts are 2cβ and 20cβ and whose lengths are 40βc2β and 101β40βc2β.
which is a constant times c40βc2β. This value is maximized when its square, c2(40βc2), is maximized, and this occurs when c2=20, which means c=20ββ20.25β=(A)4.5β.
OR
Set z1β=rcisΞΈ. By Vieta's Formulas, z1βz2β=10 and z1β+z2β=c. Hence z2β=r10βcis(βΞΈ). For the sum z1β+z2β to equal a real number, it must be that rsinΞΈ+r10βsin(βΞΈ)=0, which is equivalent to r=10β. Therefore the vertices of the quadrilateral are
z1β=10βcisΞΈ,z2β=10βcis(βΞΈ),z3β=z1β1β=10βcis(βΞΈ)β, and z4β=z2β1β=10βcisΞΈβ
Thinking geometrically, let O be the origin in the complex plane and let Z1β through Z4β be the points represented by the complex numbers z1β through z4β, respectively. Then Z4β lies on OZ1ββ with OZ1β=10β OZ4β and Z3β lies on OZ2ββ with OZ2β=10β OZ3β (where O is the origin). It follows that
