Difference between revisions of "2012 AIME I Problems/Problem 14"

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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

Solution 1

Solution 2

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, $h^2=x^2+y^2=\frac{3}{2}\cdot250=\boxed{375}$ .

@@ Line 4: / Line 4: @@
 == Solution ==
 ===Solution 1===
-Since <math>q</math> and <math>r</math> are real, at least one of <math>a,</math> <math>b,</math> and <math>c</math> must be real, with the remaining roots being pairs of complex conjugates. Without loss of generality, we assume <math>a</math> is real and <math>b</math> and <math>c</math> are <math>x + yi</math> and <math>x - yi</math> respectively. By symmetry, the triangle described by <math>a,</math> <math>b,</math> and <math>c</math> must be isosceles and is thus an isosceles right triangle with hypotenuse <math>\overline{ab}.</math> Now since <math>P(z)</math> has no <math>z^2</math> term, we must have <math>a+b+c = a + (x + yi) + (x - yi) = 0</math> and thus <math>a = -2x.</math> Also, since the length of the altitude from the right angle of an isosceles triangle is half the length of the hypotenuse, <math>a-x=y</math> and thus <math>y=-3x.</math> We can then solve for <math>x</math>:
-<cmath>
-\begin{align*}
-|a|^2 + |b|^2 + |c|^2 &= 250\\
-|-2x|^2 + |x-3xi|^2 + |x+3xi|^2 &= 250\\
-x^2 + (x^2 + 9x^2) + (x^2 + 9x^2) &= 250\\
-x^2 &= \frac{250}{24}
-\end{align*}
-</cmath>
-Now <math>h</math> is the distance between <math>b</math> and <math>c,</math> so <math>h = 2y = -6x</math> and thus <math>h^2 = 36x^2 = 36 \cdot \frac{250}{24} = \boxed{375.}</math>
 ===Solution 2===

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