(In reply to
Full solution by Jer)
here is how I analized your solution. First some labeling. In the first drawing, let the length of the horizontal (top/bottom) side by m and let the length of the vertical (left/right) side be n. Also, I assumed that m<n, if m=n then any cut works as it is already a square. Now for the diagonal cut, let end of the cut be x from the top. Also, let the horizontal cut be y from the top. Finally, let the length of the horizontal cut be z. Now from similar triangles we have x/m=y/z thus z=my/x.
Now in the bottom drawing we have the left side as x and the right side as ny. Also both the top and bottom are given by m+z=m+my/x. So for this to be a square we need both
x=ny
x=m+my/x
substituting the first in the second we have
ny=m+my/(ny) since y<n (otherwise there can't be a cut) we can multiply through by ny to get
(ny)^2=mn
y=n+sqrt(mn)
nowhever since m<n then sqrt(mn)<n and since y<n we have
y=nsqrt(mn)
which gives x=sqrt(mn)<n thus the diagonal cut can happen as well.
So from this analysis I conclude that your solution works for all rectangles.

Posted by Daniel
on 20130816 10:34:57 