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

By Vieta's formula, the sum of the roots is equal to 0, or $a+b+c=0$ . Therefore, $\frac{(a+b+c)}{3}=0$ . Because the centroid of any triangle is the average of its vertices, the centroid of this triangle is the origin. Let one leg of the right triangle be $x$ and the other leg be $y$ . Without the loss of generality, let $\overline{ac}$ be the hypotenuse. The magnitudes of $a$ , $b$ , and $c$ are just $\frac{2}{3}$ of the medians because the origin, or the centroid in this case, cuts the median in a ratio of $2:1$ . So, $|a|^2=\frac{4}{9}\cdot((\frac{x}{2})^2+y^2)=\frac{x^2}{9}+\frac{4y^2}{9}$ because $|a|$ is two thirds of the median from $a$ . Similarly, $|c|^2=\frac{4}{9}\cdot(x^2+(\frac{y}{2})^2)=\frac{4x^2}{9}+\frac{y^2}{9}$ . The median from $b$ is just half the hypotenuse because the hypotenuse of any right triangle is just half the hypotenuse. So, $|b|^2=\frac{4}{9}\cdot\frac{x^2+y^2}{4}=\frac{x^2}{9}+\frac{y^2}{9}$ . Hence, $|a|^2+|b|^2+|c|^2=\frac{6x^2+6y^2}{9}=\frac{2x^2+2y^2}{3}=250$ . Therefore, $h^2=x^2+y^2=\frac{3}{2}\cdot250=\boxed{375}$ .

2012 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

Page

Toolbox

Search

2012 AIME I Problems/Problem 14

Problem 14

Solution

See also