Geometry

Suppose $\overline{AC} = x$ and $\overline{BC} = y$. From $\overline{AM} = \overline{BN} = 1$, we have

$\begin{cases} \left(\dfrac x2\right)^2 + y^2 = 1\\ x^2 + \left(\dfrac y2\right)^2 = 1 \end{cases}$

Adding, we have 

$\dfrac 54 \left(x^2 + y^2\right) = 2\\ x^2 + y^2 = \dfrac 85\\ \sqrt{x^2 + y^2} = \dfrac{2 \sqrt {10}}5$

Hence, length of the hypotenuse is $\dfrac{2 \sqrt {10}}5$