Problem:
Complex numbers a,b, and c are the zeros of a polynomial P(z)=z3+ qz+r, and ∣a∣2+∣b∣2+∣c∣2=250. The points corresponding to a,b, and c in the complex plane are the vertices of a right triangle with hypotenuse h. Find h2.
Solution:
Let a,b, and c correspond to points A,B, and C in the complex plane, respectively, and assume that ∠ABC is right. Let the midpoint D of AC correspond to the number d=2a+c. Because the coefficient of z2 in P(z) is 0,a+b+c=0 and b=−(a+c). Because D is the circumcenter of △ABC, the distances DA,DB, and DC are equal, so ∣b−d∣=2∣a−c∣. Thus
∣∣∣∣∣−(a+c)−2a+c∣∣∣∣∣=2∣a−c∣
implying that ∣a−c∣=3∣a+c∣. Note that ∣a∣2+∣c∣2=2∣a−c∣2+2∣a+c∣2 for any two complex numbers a and c. It follows that
250=2∣a−c∣2+2∣a+c∣2+∣a+c∣2=6∣a+c∣2
and h2=∣a−c∣2=9∣a+c∣2=69⋅250=375.
The problems on this page are the property of the MAA's American Mathematics Competitions