2025 SSMO Accuracy Round Problems/Problem 4

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

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2025 SSMO Accuracy Round Problems/Problem 4

Problem

Solution