I'd still like to see a solution that's linear in the areas, rather than quadratic, but perhaps there's no such animal?? R.
What I thought of as the uninsightful version of the answer is quadratic in edge lengths -- will that do? You can un-shear the parallelogram and re-scale the axes, both with area-preserving linear maps -- and this point of view is certainly linear in area, so I don't think I'm cheating here. Anyway, this lets you canonicalize the diagram to leave rectangle ABCD with lengths AB=12, AQ=1. B--------C |\\__ | | \ \__ | | \ \P | \ /| | \ / | A-----Q--D Now there is only one free parameter, the length of QD. Say QD=x, so DP=34/x, and BC=x+1. So the length of PC is both 12-34/x and 58/(x+1), and we get x=(20+sqrt(502))/6. The area of the whole rectangle is 12(x+1), and then subtract off 6+17+29. I wouldn't say it's particularly insightful, but it does get the answer by solving for a quadratic in edge lengths, not in areas. --Michael Kleber -- It is very dark and after 2000. If you continue you are likely to be eaten by a bleen.