Difference between revisions of "Mock AIME I 2015 Problems/Problem 11"
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+ | ==Problem== | ||
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+ | Suppose <math>\alpha</math>, <math>\beta</math>, and <math>\gamma</math> are complex numbers that satisfy the system of equations <cmath> | ||
+ | |||
==Solution 1== | ==Solution 1== | ||
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==Solution 2== | ==Solution 2== | ||
− | Let <math>\alpha = a</math>, <math>\beta = b</math>, and <math>\ | + | Let <math>\alpha = a</math>, <math>\beta = b</math>, and <math>\gamma = c</math>. Then our system becomes |
− | <cmath>a + b + c</cmath> | + | <cmath>a + b + c = 6</cmath> |
<cmath>a^3 + b^3 + c^3 = 87</cmath> | <cmath>a^3 + b^3 + c^3 = 87</cmath> | ||
<cmath>(a + 1)(b + 1)(c + 1) = 33</cmath>. | <cmath>(a + 1)(b + 1)(c + 1) = 33</cmath>. | ||
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<cmath>6(6^2 - 3y) = 87 - 3x</cmath> | <cmath>6(6^2 - 3y) = 87 - 3x</cmath> | ||
<cmath>26 = x + y</cmath>. | <cmath>26 = x + y</cmath>. | ||
− | + | Our solutions are <math>x = \frac{113}{7}</math> and <math>y = \frac{69}{7}</math>. | |
− | + | Therefore, <math>\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{ab + ac + bc}{abc} = \frac{69/7}{113/7} = \frac{69}{113}</math>. So, <math>m + n = 69 + 113 = \boxed{182}</math>. | |
<baker77> | <baker77> |
Latest revision as of 10:29, 29 October 2019
Problem
Suppose , , and are complex numbers that satisfy the system of equations If for positive relatively prime integers and , find .
Solution 1
For convenience, let's use instead of . Define a polynomial such that . Let and . Then, our polynomial becomes . Note that we want to compute .
From the given information, we know that the coefficient of the term is , and we also know that , or in other words, . By Newton's Sums (since we are given ), we also find that . Solving this system, we find that . Thus, , so our final answer is .
Solution 2
Let , , and . Then our system becomes .
Since , this equation becomes .
. Since , this equation becomes .
We will now use these equations to solve the problem. Let , and . Then we have . Our solutions are and .
Therefore, . So, .
<baker77>