The triangles BOC, OQC, on the same base BQ, have the same "height" (namely the perpendicular distance from C to BQ). Since the area of BOC is twice the area of OQC, it follows that . By a similar argument from the triangles COB, OBP, it follows that .

Now let x be the area of triangle PAO, and let y be the area of triangle OAQ. Since , it follows (by the same reasoning as in the previous paragraph) that the area of triangle BAO is twice the area of triangle OAQ. In other words, . A similar argument in triangles PAO and OAC shows that .

That gives you two simultaneous equations for x and y. The answer to the problem is the sum x+y.