Difference between revisions of "2012 AIME I Problems/Problem 14"
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== Solution == | == Solution == | ||
+ | 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>: | ||
+ | |||
+ | <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 &= \sqrt{\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 <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 01:48, 17 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
Since and are real, at least one of and must be real, with the remaining roots being complex conjugates. Without loss of generality, we assume is real and and are and respectively, for real and 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
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 |