A piece of wooden board in the shape of an isosceles right triangle, with sides $1$,$1$, $\sqrt{2}$ is to be sawn into two pieces. Find the length and location of the shortest straight cut which divides the board into two parts of equal area.
I think the answer would be a cut joining the midpoint of the hypotenuse and the opposite vertex. But, how can I write it out formally i.e. how to prove that this indeed is the shortest possible cut?
In the above configuration, $EF$ is the cut, where $CE=\lambda CD$ and $CF=\frac{1}{\lambda}CB$.
By the cosine theorem, $$ EF^2 = \lambda^2 CD^2 + \frac{1}{\lambda^2} CB^2 - \sqrt{2}\, CD\cdot CB $$
hence the shortest cut is achieved by the value of $\lambda\in[1,2]$ such that $\frac{\lambda^2}{2}+\frac{1}{\lambda^2}$ is minimal. By the AM-GM inequality, the minimum is attained when $\frac{\lambda^2}{2}=\frac{1}{\lambda^2}$, hence for: $$ \color{red}{\lambda = \sqrt[4]{2}}. $$ For such a configuration, the minimum cut length is so $\color{red}{\sqrt{\sqrt{2}-1}}<\frac{\sqrt{2}}{2}$, with $CEF$ being an isosceles triangle. If the cut joins two points on two equal sides, its minimal length is $1$.