2006/07 AHSMC Part 1 Problem 10

Problem 10 of 16

The non-zero numbers $a$ , $b$ , $c$ , $x$ , $y$ and $z$ are such that $\dfrac{x}{a} = \dfrac{y}{b} = \dfrac{z}{c}$ . The value of $\dfrac{xyz(a + b)(b + c)(c + a)}{abc(x + y)(y + z)(z + x)}$ is

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Let $\dfrac{x}{a} = \dfrac{y}{b} = \dfrac{z}{c} = r$ . Then $\dfrac{xyz}{abc} = r^3$ , $x = ra$ , $y = rb$ and $z = rc$ . It follows that $x + y = r(a + b)$ , $y + z = r(b + c)$ and $z + x = r(c + a)$ , so that $\dfrac{x + y}{a + b} = \dfrac{y + z}{b + c} = \dfrac{z + x}{c + a} = r$ . The given expression is equal to $r^3 \left(\dfrac{1}{r}\right)^3 = 1$ . The answer is (c).

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