2012 AIME I Problems/Problem 14

Problem 14

Complex numbers $a,$ $b,$ and $c$ are 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 $h^2.$

Solution

Since $q$ and $r$ are real, at least one of $a,$ $b,$ and $c$ must be real, with the remaining roots being complex conjugates. Without loss of generality, we assume $a$ is real and $b$ and $c$ are $x + yi$ and $x - yi$ respectively, for real $x$ and $y.$ By symmetry, the triangle described by $a,$ $b,$ and $c$ must be isosceles and is thus an isosceles right triangle with hypotenuse $\overline{ab}.$ Now since $P(z)$ has no $z^2$ term, we must have $a+b+c = 0$ and thus $a = -2x.$ Also, since the length of the altitude from the right angle of an isosceles triangle is half the length of the hypotenuse, $a-x=y$ and thus $y=-3x.$ We can then solve for $x$ :

$\begin{align*} |a|^2 + |b|^2 + |c|^2 &= 250\\ |-2x|^2 + |x-3xi|^2 + |x+3xi|^2 &= 250\\ 4x^2 + (x^2 + 9x^2) + (x^2 + 9x^2) &= 250\\ x &= \sqrt{\frac{250}{24}} \end{align*}$

Now $h$ is the distance between $b$ and $c,$ so $h = 2y = -6x$ and $h^2 = 36x^2 = 36 \cdot \frac{250}{24} = \boxed{375.}$

2012 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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2012 AIME I Problems/Problem 14

Problem 14

Solution

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