Oxford University MAT 2018 Solved Papers

H. The triangle $\mathit{T}$ could be any size; for instance it could be a right-angled triangle in a circle of arbitrary radius. Given such a triangle $\mathit{T}$, the triangle $\mathit{S}$ could be any size smaller than $\mathit{T}$; we can change the size of triangle $\mathit{S}$ from zero to$\mathit{t}$ by moving the other vertex of triangle $\mathit{T}$ around the circle.
So we have , but otherwise $\mathit{s}$ and $\mathit{t}$ could be any positive numbers. We want to minimise
$\frac{4{\mathit{s}}^2+{\mathit{t}}^2}{5\mathit{s}\mathit{t}}=\frac{1}{5}\left(4\left(\frac{\mathit{s}}{\mathit{t}}\right)+{\left(\frac{\mathit{s}}{\mathit{t}}\right)}^{-1}\right).$
Write $\mathit{u}=\mathit{s}∕\mathit{t}$, so that .
The expression $4\mathit{u}+{\mathit{u}}^{-1}$ is minimised when $4-{\mathit{u}}^{-2}=0$, which is when $\mathit{u}=1∕2$ (which is between 0 and 1 ).
The value at $\mathit{s}∕\mathit{t}=1∕2$ is $4∕5$. The answer is (c).