Point P is located inside a square ABCD of side length 10. Let O1, O2, O3, O4 be the circumcenters of PAB, PBC, PCD, and PDA, respectively. Given that PA+PB+PC+PD=232 and the area of O1O2O3O4 is 50, the second largest of the lengths O1O2, O2O3, O3O4, O4O1 can be written as ba, where a and b are relatively prime positive integers. Compute 100a+b.