My best answer is no if ABCD is a square and probably not otherwise.
By finding the angle of each connecting segment to the horizontal using arctan(slope) and the identity
arctan(x)arctan(y)=arctan((xy)/(1+xy))
we can get the angle of each segment. Then ∠BPA = ∠APC = ∠CPD becomes the system
(rh)/(r^2+sh+s^2) = (swrh)/(s^2+sh+r^2+rw) = (rh+hw)/(s^2+sh+r^2+2rw+w^2)
Cross multiplying the first by the third looks promising because there won't be an s^3 term and indeed a lot cancels leaving just
r^2hw+rhw^2=s^2hw+sh^2w
which reduces to
r^2 + rw = s^2 + sh
How to interpret this?
One implication is if ABCD is a square then r=s, the middle term of the equation above is zero and the implication rh=0=rh+hw which is impossible.
Solving for r gives r=(w +/sqrt(4s^2+4sh+w^2))/2
It's getting late and I really don't feel like substituting this into the above system. Will it lead to more contradictions or to an affirmative to part 1)? I'm not ready to answer that.
Edit: using this info to play with Geometer's Sketchpad leads me to believe there are no solutions for any (w,h)
The quadratic equation allows for making the first and third angles equal but to make them equal to the second seems to require letting r and s grow without bound along the curve implied. The difference goes to zero as (r,s) tends to a line with slope 1.
Edited on May 19, 2018, 11:21 pm

Posted by Jer
on 20180519 22:50:39 