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Difference between revisions of "Mock AIME I 2015 Problems/Problem 11"
 
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Our solutions are <math>x = \frac{113}{7}</math> and <math>y = \frac{69}{7}</math>.
 
Our solutions are <math>x = \frac{113}{7}</math> and <math>y = \frac{69}{7}</math>.
  
−
Then <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>.
+
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 $\alpha$, $\beta$, and $\gamma$ are complex numbers that satisfy the system of equations \begin{align*}\alpha+\beta+\gamma&=6,\\\alpha^3+\beta^3+\gamma^3&=87,\\(\alpha+1)(\beta+1)(\gamma+1)&=33.\end{align*} If $\frac1\alpha+\frac1\beta+\frac1\gamma=\tfrac mn$ for positive relatively prime integers $m$ and $n$, find $m+n$.

Solution 1

For convenience, let's use $a, b, c$ instead of $\alpha, \beta, \gamma$. Define a polynomial $P(x)$ such that $P(x) = (x-a)(x-b)(x-c) = x^3 - (a+b+c)x^2 + (ab+ac+bc)x - abc$. Let $j = ab + ac + bc$ and $k = -abc$. Then, our polynomial becomes $P(x) = x^3 - (a+b+c)x^2 + jx + k$. Note that we want to compute $-\frac{j}{k}$.


From the given information, we know that the coefficient of the $x^2$ term is $6$, and we also know that $P(-1) = -33$, or in other words, $-j + k = -26$. By Newton's Sums (since we are given $a^3 + b^3 + c^3$), we also find that $6j + k = 43$. Solving this system, we find that $(j, k) \in (\frac{69}{7}, -\frac{113}{7})$. Thus, $\frac{j}{-k} = \frac{69}{113}$, so our final answer is $69 + 113 = \boxed{182}$.

Solution 2

Let $\alpha = a$, $\beta = b$, and $\gamma = c$. Then our system becomes \[a + b + c = 6\] \[a^3 + b^3 + c^3 = 87\] \[(a + 1)(b + 1)(c + 1) = 33\].

\[a^3 + b^3 + c^3 = 87\] \[a^3 + b^3 + c^3 - 3abc = 87 - 3abc\] \[(a + b + c)(a^2 + b^2 + c^2 - ab - ac - bc) = 87 - 3abc\] \[(a + b + c)((a + b + c)^2 - 3(ab + ac + bc)) = 87 - 3abc\] Since $a + b + c = 6$, this equation becomes $6(6^2 - 3(ab + ac + bc)) = 87 - 3abc$.

\[33 = (a + 1)(b + 1)(c + 1) = abc + ab + ac + bc + a + b + c + 1\]. Since $a + b + c = 6$, this equation becomes $26 = abc + ab + ac + bc$.

We will now use these $2$ equations to solve the problem. Let $x = abc$, and $y = ab + ac + bc$. Then we have \[6(6^2 - 3y) = 87 - 3x\] \[26 = x + y\]. Our solutions are $x = \frac{113}{7}$ and $y = \frac{69}{7}$.

Therefore, $\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{ab + ac + bc}{abc} = \frac{69/7}{113/7} = \frac{69}{113}$. So, $m + n = 69 + 113 = \boxed{182}$.

<baker77>

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