Difference between revisions of "2012 AIME I Problems/Problem 14"
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− | ==Problem | + | ==Problem== |
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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<math>\Rightarrow h^2 = \boxed{375}</math> | <math>\Rightarrow h^2 = \boxed{375}</math> | ||
− | + | == Video Solution by Richard Rusczyk == | |
https://artofproblemsolving.com/videos/amc/2012aimei/354 | https://artofproblemsolving.com/videos/amc/2012aimei/354 |
Latest revision as of 19:41, 24 January 2021
Contents
Problem
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 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 median 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
Solution 3 (Messy)
Let the roots , , and each be represented by complex numbers , , and . By Vieta's formulas, their sum is 0. Breaking into real and imaginary components, we get:
And, we know that the sum of the squares of the magnitudes of each is 250, so
Given the complex plane, we set each of these complex numbers to points: , , . WLOG let be the vertex opposite the hypotenuse.
If the three points form a right triangle, the vectors from to and 's dot product is 0.
Substituting and likewise, simplifying:
Rearranging we get:
The answer is the distance from to = . Substituting the equation equal to 250,
Taking our original equations summing to 0, and squaring each we get:
Adding, we get:
Substituting again we obtain:
Substituting the equivalence of :
Solving for , we find it equal to .
Substituting this value into our answer expression, we get:
, Answer = .
Solution 4 (clean)
As noted in the previous solutions, . Let , , and we have . Then the given translates to Note that in a right triangle, the sum of the squares of the three sides is equal to two times the square of the hypotenuse, by the pythagorean theorem. Thus, we have so and we may conclude. ~ rzlng
Solution 5 (vectors)
As shown in the other solutions, .
Without loss of generality, let be the complex number opposite the hypotenuse.
Note that there is an isomorphism between under and under .
Let , , and be the corresponding vectors to , , and .
Thus
Now implies that
Also note that because there is a right angle at , and are perpendicular.
Note that
.
Video Solution by Richard Rusczyk
https://artofproblemsolving.com/videos/amc/2012aimei/354
~ dolphin7
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.