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Difference between revisions of "Mock AIME I 2015 Problems/Problem 11"

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==Solution==
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For convenience, let's use <math>a, b, c</math> instead of <math>\alpha, \beta, \gamma</math>. Define a polynomial <math>P(x)</math> such that <math>P(x) = (x-a)(x-b)(x-c) = x^3 - (a+b+c)x^2 + (ab+ac+bc)x - abc</math>. Let <math>j = ab + ac + bc</math> and <math>k = -abc</math>. Then, our polynomial becomes <math>P(x) = x^3 - (a+b+c)x^2 + jx + k</math>.  
 
For convenience, let's use <math>a, b, c</math> instead of <math>\alpha, \beta, \gamma</math>. Define a polynomial <math>P(x)</math> such that <math>P(x) = (x-a)(x-b)(x-c) = x^3 - (a+b+c)x^2 + (ab+ac+bc)x - abc</math>. Let <math>j = ab + ac + bc</math> and <math>k = -abc</math>. Then, our polynomial becomes <math>P(x) = x^3 - (a+b+c)x^2 + jx + k</math>.  
 
Note that we want to compute <math>-\frac{j}{k}</math>.  
 
Note that we want to compute <math>-\frac{j}{k}</math>.  
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From the given information, we know that the coefficient of the <math>x^2</math> term is <math>6</math>, and we also know that <math>P(-1) = -33</math>, or in other words, <math>-j + k = -26</math>. By Newton's Sums (since we are given <math>a^3 + b^3 + c^3</math>), we also find that <math>6j + k = 43</math>. Solving this system, we find that <math>(j, k) \in (\frac{69}{7}, -\frac{113}{7})</math>. Thus, <math>\frac{j}{-k} = \frac{69}{113}</math>, so our final answer is <math>69 + 113 = \boxed{182}</math>.
 
From the given information, we know that the coefficient of the <math>x^2</math> term is <math>6</math>, and we also know that <math>P(-1) = -33</math>, or in other words, <math>-j + k = -26</math>. By Newton's Sums (since we are given <math>a^3 + b^3 + c^3</math>), we also find that <math>6j + k = 43</math>. Solving this system, we find that <math>(j, k) \in (\frac{69}{7}, -\frac{113}{7})</math>. Thus, <math>\frac{j}{-k} = \frac{69}{113}</math>, so our final answer is <math>69 + 113 = \boxed{182}</math>.

Revision as of 14:52, 12 February 2017

Solution

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 $69 + 113 = \boxed{182}$.

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