38
Cut along the red lines and move the pieces as indicated by the yellow arrows.
As is usual with this kind of dissection, it helps if you look at the area to work out the length of the side of the final square. Given the grid lines, you also know the orientation of the square, so you can try to place a square over the original picture at such a location that ...
14
Aha! The solution:
6
6
4
In that triangle $OAX$, angle $OXA$ is $\frac\pi2-\phi$; let's call it $\theta$. We want to make $\theta$ as large as we can. The side-lengths of that triangle are $R$, $r$, and the distance $AX$ which I'll call $t$. Then the cosine rule for angle $X$ of triangle $OAX$ says $r^2=R^2+t^2-2Rt\cos\theta$ so maximizing $\theta$ is the same as minimizing $\frac{R^...
4
Construction
Proof
Addendum
Only top voted, non community-wiki answers of a minimum length are eligible
