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

(Solution 2)
(Solution)
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== Solution ==
 
== Solution ==
This is a more geometric solution. 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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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>.
  
 
== 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 11:58, 25 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

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
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All AIME Problems and Solutions
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