2012 AIME I Problem 14

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14.

Complex numbers a,a, b,b, and cc are the zeros of a polynomial P(z)=z3+qz+r,P(z) = z^3 + qz + r, and a2+b2+c2=250.|a|^2 + |b|^2 + |c|^2 = 250. The points corresponding to a,a, b,b, and cc in the complex plane are the vertices of a right triangle with hypotenuse h.h. Find h2.h^2.

Answer: 375
Concepts:complex numberVieta’s Formulasright triangle
Difficulty rating: 3060
Solution:

Since P(z)P(z) has no z2z^2 term, a+b+c=0.a + b + c = 0. Say the right angle is at b;b; then the hypotenuse joins aa and c,c, so h=ac,h = |a - c|, and b=(a+c).b = -(a + c). The midpoint d=a+c2d = \frac{a + c}{2} of the hypotenuse is the circumcenter of the right triangle, so bd=h2.|b - d| = \frac{h}{2}. Since bd=32(a+c),b - d = -\frac{3}{2}(a + c), this gives ac=3a+c.|a - c| = 3\,|a + c|.

By the parallelogram law, a2+c2=ac2+a+c22,|a|^2 + |c|^2 = \frac{|a - c|^2 + |a + c|^2}{2}, and b2=a+c2,|b|^2 = |a + c|^2, so 250=9a+c2+a+c22+a+c2=6a+c2. \begin{aligned} 250 &= \frac{9\,|a + c|^2 + |a + c|^2}{2} \\ &\quad {}+ |a + c|^2 \\ &= 6\,|a + c|^2. \end{aligned}

Therefore h2=ac2=9a+c2h^2 = |a - c|^2 = 9\,|a + c|^2 =92506= \frac{9 \cdot 250}{6} =375.= 375.

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