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

(Solution 1)
(Solution 2)
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===Solution 2===
 
===Solution 2===
By Vieta's formula, the sum of the roots is equal to 0, or <math>a+b+c=0</math>. Therefore, <math>\frac{(a+b+c)}{3}=0</math>. 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 <math>x</math> and the other leg be <math>y</math>. Without the loss of generality, let <math>\overline{ac}</math> be the hypotenuse. The magnitudes of <math>a</math>, <math>b</math>, and <math>c</math> are just <math>\frac{2}{3}</math> of the medians because the origin, or the centroid in this case, cuts the median in a ratio of <math>2:1</math>. So, <math>|a|^2=\frac{4}{9}\cdot((\frac{x}{2})^2+y^2)=\frac{x^2}{9}+\frac{4y^2}{9}</math> because <math>|a|</math> is two thirds of the median from <math>a</math>. Similarly, <math>|c|^2=\frac{4}{9}\cdot(x^2+(\frac{y}{2})^2)=\frac{4x^2}{9}+\frac{y^2}{9}</math>. The median from <math>b</math> is just half the hypotenuse because the hypotenuse of any right triangle is just half the hypotenuse. So, <math>|b|^2=\frac{4}{9}\cdot\frac{x^2+y^2}{4}=\frac{x^2}{9}+\frac{y^2}{9}</math>. Hence, <math>|a|^2+|b|^2+|c|^2=\frac{6x^2+6y^2}{9}=\frac{2x^2+2y^2}{3}=250</math>. Therefore, <math>h^2=x^2+y^2=\frac{3}{2}\cdot250=\boxed{375}</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 = 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>:
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<cmath>
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\begin{align*}
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|a|^2 + |b|^2 + |c|^2 &= 250\\
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|-2x|^2 + |x-3xi|^2 + |x+3xi|^2 &= 250\\
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4x^2 + (x^2 + 9x^2) + (x^2 + 9x^2) &= 250\\
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x^2 &= \frac{250}{24}
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\end{align*}
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</cmath>
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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>
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2012|n=I|num-b=13|num-a=15}}
 
{{AIME box|year=2012|n=I|num-b=13|num-a=15}}

Revision as of 08:57, 1 April 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

Solution 1

Solution 2

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 $a+b+c = a + (x + yi) + (x - yi) = 0$ 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^2 &= \frac{250}{24} \end{align*}

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

See also

2012 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
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All AIME Problems and Solutions
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