设CP=x,PQ =y=>x+y+√(x^2+y^2)=2 =>(x+y-2)^2=(x^2+y^2)=>-4(x+y)+4+2xy=0=>x+y=1+xy/2x+y>=2√(xy)=>1+xy/2>=2√(xy)设√(xy)=t>=0=>1+t^2/2>=2t=>2+t^2-4t>=0=>0三角形PAQ面积=正方形面积-三角形PBA面积-三角形PD...