19. Let $x, y, z, w$ be different positive real numbers such that $x+\frac{1}{y}=y+\frac{1}{z}=z+\frac{1}{w}=w+\frac{1}{x}=t$. Find $t$.
(2 marks)
Let $x, y, z, w$ be different positive real numbers such that $x+\frac{1}{y}=y+\frac{1}{z}=z+\frac{1}{w}=w+\frac{1}{x}=t$. Find $t$.
(2 marks)
19. $\sqrt{2}$
19. Expressing $w, z$ and $y$ in terms of $t$ and $x$, we have
$$
w=\frac{x t-1}{x}, \quad z=\frac{x t^{2}-t-x}{x t-1} \text{ and } \quad y=\frac{x t^{3}-t^{2}-2 x t+1}{x t^{2}-t-x}.
$$

Substituting the last expression into $x+\frac{1}{y}=t$, we have
$$
\begin{aligned}
x+\frac{x t^{2}-t-x}{x t^{3}-t^{2}-2 x t+1} & =t \\
x t^{4}-x^{2} t^{3}-t^{3}-2 x t^{2}+2 x^{2} t+2 t & =0 \\
t\left(t^{2}-2\right)\left(x t-x^{2}-1\right) & =0
\end{aligned}
$$

Note that the last factor on the left hand side of the last row is non-zero, for otherwise we get $x+\frac{1}{x}=t=x+\frac{1}{y}$ which contradicts the fact that $x \neq y$. Since $t>0$, we must have $t=\sqrt{2}$.
Remark. Owing to some typo, $\sqrt{2},-\sqrt{2}$ and $\pm \sqrt{2}$ were all accepted as correct answers in the live paper.