During AMC testing, the AoPS Wiki is in read-only mode. No edits can be made.

2012 AIME I Problems/Problem 14

Revision as of 11:57, 25 March 2012 by Numbertheorist17 (talk | contribs) (Solution)

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 2

This is a more geometric 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, $h^2=x^2+y^2=\frac{3}{2}\cdot250=\boxed{375}$.

See also

2012 AIME I (ProblemsAnswer KeyResources)
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
Invalid username
Login to AoPS