Oxford University MAT 2019 Solved Papers

The hypotenuse is the longest side, so write it as $\mathit{a}{\mathit{r}}^2$, where $\mathit{a}>0$ is the shortest side and $\mathit{r}>1$. The remaining side must be ar for the sides to be in geometric progression. Then Pythagoras' theorem gives $1+{\mathit{r}}^2={\mathit{r}}^4$. Solve the quadratic for ${\mathit{r}}^2$ to get ${\mathit{r}}^2=\left(1+\sqrt{5}\right)∕2$, taking the positive root since ${\mathit{r}}^2>0$. Now $\mathit{r}>0$ so $\mathit{r}=\sqrt{\left(1+\sqrt{5}\right)∕2}\cdot \tan \angle \mathit{B}\mathit{A}\mathit{C}$ is either $\mathit{r}$ or $1∕\mathit{r}$ depending on which way round $\mathit{A}$ and $\mathit{C}$ are,