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Difference between revisions of "2012 AIME I Problems/Problem 14"
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== Solution ==
 
== Solution ==
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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 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, for real <math>x</math> and <math>y.</math> 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 = 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>:
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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 = 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>
 
<cmath>

Revision as of 03:44, 17 March 2012

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 pairs of complex conjugates. Without loss of generality, we assume $a$ is real and $b$ and $c$ are $x + yi$ and $x - yi$ respectively. 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 $h^2 = 36x^2 = 36 \cdot \frac{250}{24} = \boxed{375.}$

See also

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
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