The black outlined, inscribed square has a side length of $1/\sqrt{2}$ and a diagonal of $1$. You then might aim to calculate the position of the pink dot by means of Pythagoras according to
$$(\frac{1}{\sqrt{2}}+a)^2+a^2=1$$
But you do not even have to!
The lower bond of that dotted line is a shifted copy of the opposite copy wrt. that small pink square. The other bond in the very same way wrt. the upper pink square. Therefore the searched for dotted distance is exactly the same as the diagonal of the black outlined square. And that one already is mentioned to be $1$.
\--- rk