Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "Mock AIME I 2015 Problems/Problem 11"
Line 9: Line 9:
 
==Solution 2==
 
==Solution 2==
  
−
Let <math>\alpha = a</math>, <math>\beta = b</math>, and <math>\beta = c</math>. Then our system becomes
+
Let <math>\alpha = a</math>, <math>\beta = b</math>, and <math>\gamma = c</math>. Then our system becomes
 
<cmath>a + b + c = 6</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>

Revision as of 10:55, 12 October 2019

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 $y = \frac{69}{7}$.

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

<baker77>

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=Mock_AIME_I_2015_Problems/Problem_11&oldid=110245"