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 "2025 SSMO Accuracy Round Problems/Problem 4"

(Created page with "==Problem== Let <math>a,</math> <math>b,</math> and <math>c</math> be the roots of the polynomial <math>x^3+4x^2-3x-4</math>. Suppose that <cmath>\left|\frac{a}{b+c}+\frac{b...")
 
(→‎Solution)
 
Line 4: Line 4:
  
 
==Solution==
 
==Solution==
 +
 +
Let <math>f(x) = x^3 + 4x^2 - 3x - 4  =(x-a)(x-b)(x-c)</math>. By Vieta, <math>a+b+c = -4</math>. We can use this fact to manipulate the expression whose absolute value we seek. We have
 +
\begin{align*} \frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b}  &= \frac{a(c+a)(a+b) + b(a+b)(b+c) + c(b+c)(c+a)}{(a+b)(b+c)(c+a)} \\
 +
&= \frac{a(-4-b)(-4-c) + b(-4-c)(-4-a) + c(-4-a)(-4-b)}{(-4-a)(-4-b)(-4-c)} \\
 +
&= \frac{16(a+b+c) + 8(ab+bc+ca) + 3abc}{f(-4)}.
 +
\end{align*}
 +
The rest is computation; we have <math>f(-4) = 8</math>, and by Vieta, we know that <math>a+b+c = -4</math>, <math>ab+bc+ca = -3</math>, and <math>abc = 4</math>. Plugging everything in, we find that
 +
\begin{align*}
 +
\left | \frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} \right | &= \left| \frac{16(-4) + 8(-3) + 3(4)}{8} \right | \\
 +
&= \frac{19}{2}. \end{align*}
 +
We extract <math>19+2 = \boxed{21}</math>.
 +
 +
~Sedro

Latest revision as of 20:09, 12 September 2025

Problem

Let $a,$ $b,$ and $c$ be the roots of the polynomial $x^3+4x^2-3x-4$. Suppose that \[\left|\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right|=\frac{m}{n},\] for relatively prime positive integers $m$ and $n$. Find $m+n$.

Solution

Let $f(x) = x^3 + 4x^2 - 3x - 4 =(x-a)(x-b)(x-c)$. By Vieta, $a+b+c = -4$. We can use this fact to manipulate the expression whose absolute value we seek. We have \begin{align*} \frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} &= \frac{a(c+a)(a+b) + b(a+b)(b+c) + c(b+c)(c+a)}{(a+b)(b+c)(c+a)} \\ &= \frac{a(-4-b)(-4-c) + b(-4-c)(-4-a) + c(-4-a)(-4-b)}{(-4-a)(-4-b)(-4-c)} \\ &= \frac{16(a+b+c) + 8(ab+bc+ca) + 3abc}{f(-4)}. \end{align*} The rest is computation; we have $f(-4) = 8$, and by Vieta, we know that $a+b+c = -4$, $ab+bc+ca = -3$, and $abc = 4$. Plugging everything in, we find that \begin{align*} \left | \frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} \right | &= \left| \frac{16(-4) + 8(-3) + 3(4)}{8} \right | \\ &= \frac{19}{2}. \end{align*} We extract $19+2 = \boxed{21}$.

~Sedro

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2025_SSMO_Accuracy_Round_Problems/Problem_4&oldid=260474"