2014/15 AHSMC Part 1 Problem 11

Problem 11 of 16

Let $a$ , $b$ and $c$ be real numbers such that $a + b + c = 5$ and $\dfrac{1}{b+c} + \dfrac{1}{c+a} + \dfrac{1}{a+b} = 6$ . The value of $\dfrac{a}{b+c} + \dfrac{b}{c+a} + \dfrac{c}{a+b}$ is

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Note that $\begin{align*} \frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} &= \frac{5-(b+c)}{b+c} + \frac{5-(c+a)}{c+a} + \frac{5-(a+b)}{a+b}\\ &= \frac{5}{b+c} + \frac{5}{c+a} + \frac{5}{a+b} - 3. \end{align*}$ Hence $\dfrac{a}{b+c} + \dfrac{b}{c+a} + \dfrac{c}{a+b} = 5 \times 6 - 3 = 27$ . The answer is (D).

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