∠∠AMB is an ordinary angle ∴ △ABM is similar to △BPM, △BPM is similar to △BNC ∴ S△BMP=(BM/BN)? S△BMA

Let MB=X, and the intersection point A is perpendicular to BC, and get AM=BN=√[27+(3-x)? ]

∴ S△BMP=x？ /[27+(3-x)？ ]s△BMA ∴s△apb= 1/2*3√3x*(36-6x)/[27+(3-x)？ ]

Let S△APB=S

(S+9√3)x？ -6x(9√3+S)+36=0，get△& gt； =0，36(S+9√3)？ -4*S*36(S+9√3)>=0，-9√3 & lt； = S & lt=3√3

∫0 & lt； = x< when X=3 =6 ∴, s max =3√3.

The line segment (chord) opposite the right angle is the diameter and the midpoint of the line segment is the center of the circle; Or the midpoint of a line segment can be regarded as the center of the circle; Two equal angles of the same line segment can be regarded as the circumferential angle of the same chord.