Difference between revisions of "2012 AIME I Problems/Problem 14"
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Complex numbers <math>a,</math> <math>b,</math> and <math>c</math> are zeros of a polynomial <math>P(z) = z^3 + qz + r,</math> and <math>|a|^2 + |b|^2 + |c|^2 = 250.</math> The points corresponding to <math>a,</math> <math>b,</math> and <math>c</math> in the complex plane are the vertices of a right triangle with hypotenuse <math>h.</math> Find <math>h^2.</math> | Complex numbers <math>a,</math> <math>b,</math> and <math>c</math> are zeros of a polynomial <math>P(z) = z^3 + qz + r,</math> and <math>|a|^2 + |b|^2 + |c|^2 = 250.</math> The points corresponding to <math>a,</math> <math>b,</math> and <math>c</math> in the complex plane are the vertices of a right triangle with hypotenuse <math>h.</math> Find <math>h^2.</math> | ||
| − | == 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>. | 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>. | ||
== 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 12:57, 25 March 2012
Problem 14
Complex numbers 
and 
are zeros of a polynomial 
and 
The points corresponding to and in the complex plane are the vertices of a right triangle with hypotenuse 
Find 
Solution
This is a more geometric solution. By Vieta's formula, the sum of the roots is equal to 0, or 
. Therefore, 
. 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 
and the other leg be 
. Without the loss of generality, let 
be the hypotenuse. The magnitudes of 
, 
, and are just 
of the medians because the origin, or the centroid in this case, cuts the median in a ratio of 
. So, 
because 
is two thirds of the median from . Similarly, 
. The median from is just half the hypotenuse because the hypotenuse of any right triangle is just half the hypotenuse. So, 
. Hence, 
. Therefore, 
.
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 | ||
