HongKong 2022-23 IMO Selection Tests

Answer: $\frac{55}{2}$ Let $x=\frac{1}{\sqrt{22}}\sec \theta$ and $y=\frac{1}{\sqrt{22}}\tan \theta$. Then $x^2-y^2=\frac{1}{22}$ and $$\begin{array}{rl}\frac{1-22xy}{x^2}& =\frac{1-\sec \theta \tan \theta }{\frac{1}{22}{\sec }^2\theta }\\ & =22({\cos }^2\theta -\sin \theta )\\ & =22(-{\sin }^2\theta -\sin \theta +1)\\ & =22\left[\frac{5}{4}-{\left(\sin \theta +\frac{1}{2}\right)}^2\right].\end{array}$$ The greatest possible value is thus $22\cdot \frac{5}{4}=\frac{55}{2}$, attained when $\sin \theta =-\frac{1}{2}$ (which is feasible, e.g. with $\theta =-30^{\circ }$, $x=\frac{2}{\sqrt{66}}$ and $y=-\frac{1}{\sqrt{66}}$).