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> | ||
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==Solution 1== | ==Solution 1== | ||
Revision as of 18:35, 3 August 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
Solutions
Solution 1
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, .
Solution 2
Assume and are real, so 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
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 |