Difference between revisions of "2012 AIME I Problems/Problem 14"
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== Solution == | == 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\\ | ||
| + | 4x^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=== | ||
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>. | 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 18:43, 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
Solution 1
Since 
and 
are real, at least one of and must be real, with the remaining roots being pairs of complex conjugates. Without loss of generality, we assume 
is real and 
and are 
and 
respectively. By symmetry, the triangle described by and must be isosceles and is thus an isosceles right triangle with hypotenuse 
Now since 
has no 
term, we must have 
and thus 
Also, since the length of the altitude from the right angle of an isosceles triangle is half the length of the hypotenuse, 
and thus 
We can then solve for 
:
Now 
is the distance between and 
so 
and thus 
Solution 2
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 | ||
