Question

Complex numbers $a,$ $b,$ and $c$ are the zeros of a polynomial $P(z) = z^3 + qz + r,$ 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 $h^2.$

Answer

Solution:

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

By the parallelogram law, $|a|^2 + |c|^2 = \frac{|a - c|^2 + |a + c|^2}{2},$ and $|b|^2 = |a + c|^2,$ so $250 = \frac{9\,|a + c|^2 + |a + c|^2}{2} + |a + c|^2 = 6\,|a + c|^2.$

Therefore $h^2 = |a - c|^2 = 9\,|a + c|^2 = \frac{9 \cdot 250}{6} = 375.$

2012 AIME I Problem 14

Solution:

Problem 14 in Other Years