John Hughes’s Garden Problem
Solution by Robin Corbett

We calculate the area PQRS as a function of θ₁, so that we can plot a graph of area against θ₁ (or better tan θ₁) and read off the angles needed to mark out the equal areas.

      Area PQRS = §PO₁S – §QO₁R
                           = §PO₁S – §QO₂R + ΔQO₂O₁
                           = ½ r₁² θ₁ – ½ r₂² θ₂ + ½ r₂ x sin θ₂

We need θ₂ as a function of θ₁. Using the sine rule in the triangle QO₂O₁, we get:

sin (θ1 – θ2) x

=

sin θ1 r2

      \   θ₂   =   θ₁   –   sin^-1[(x sin θ₁) / r₂]

So we now have the functions to produce the desired graph.