Difference between revisions of "2012 AIME I Problems/Problem 14"
(→Solution) |
(→Solution) |
||
Line 1: | Line 1: | ||
==Problem 14== | ==Problem 14== | ||
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 2 == | == Solution 2 == |
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 2
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 |