Hi Jim.
I labeled some points on your diagram.
Unfortunately I don't have a nice solution to your problem. Here is one solution.
Let y = PA and x = BQ then using similar triangles
y/4 = 4/x or x = 16/y
Also triangle POQ is a right triangle and hence Pythagoras Theorem gives
(y + 4)^{2} + (x + 4)^{2} = 20^{2}
Substitute x = 16/y and simplify. I got
y^{4} + 8 y^{3}  368 y2 + 128y + 256 = 0
I then used Mathematica to solve the quartic equation and got four solutions which numerically are
y = 1.04, y = 15.35, y = 23.72 and y = 0.67
From the diagram the result you are looking for is y = 15.34 which symbolically is
y = 2(1 + Sqrt[26] + Sqrt[23  2 Sqrt[26]]) where Sqrt is the square root.
Thus the length you want is
4 + 2(1 + Sqrt[26] + Sqrt[23  2 Sqrt[26]])
which is approximately 19.35 meters.
Penny
