1.如图:画出A、B旋转的弧线,弧线所在的圆是以O为圆心的同心圆
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∠AOD=∠BOB'=θABE面积=△AOB面积-扇形AOA'面积-△A'OE面积BEB'面积=扇形BOB'面积-△B'OE面积△A'OE面积+△B'OE面积=△A'OB'面积△AOB面积=△A'OB'面积AB扫过面积=ABE面积+BEB'面积=△AOB面积-扇形AOA'面积-△A'OE面积+扇形BOB'面积-△B'OE面积=△AOB面积-扇形AOA'面积+扇形BOB'面积-△A'OB'面积=扇形BOB'面积-扇形AOA'面积=(θ/360°)πOB^2-(θ/360°)πOA^2=(θ/360°)π2.∠AOM=45°-θ AM=tan(45°-θ)BM=1-tan(45°-θ)=tan45°-tan(45°-θ)=sin45°/cos45°-sin(45°-θ)/cos(45°-θ)=(sin45°cos(45°-θ)-cos45°sin(45°-θ))/(cos45°cos(45°-θ))=2^0.5sinθ/cos(45°-θ)∠CON=θ CN=tanθBN=1-tanθ=tan45°-tanθ=sin45°/cos45°-sinθ/cosθ=(sin45°cosθ-sinθcos45°)/(cos45°cosθ)=2^0.5(sin45°-θ)/cosθBM+BN=2^0.5sinθ/cos(45°-θ)+2^0.5sin(sin45°-θ)/cosθ=2^0.5(sinθcosθ+sin(45°-θ)cos(45°-θ))/(cos(45°-θ)cosθ)=(1/2)2^0.5(sin2θ+sin(90°-2θ))/(cos(45°-θ)cosθ)=2^0.5sin45°cos(2θ-45°)/(cos(45°-θ)cosθ)=cos(2θ-45°)/(cos(45°-θ)cosθ)MN^2=BM^2+BN^2=2(((sinθ/cos(45°-θ))^2+(sin(45°-θ)/cosθ)^2)=2((sinθcosθ)^2+(sin(45°-θ)cos(45°-θ))^2)/(cosθcos(45°-θ))^2=(1/2)((sin2θ)^2+(sin(90°-2θ))^2)/(cosθcos(45°-θ))^2=(1/4)(1-cos4θ+1-cos(180°-4θ))/(cosθcos(45°-θ))^2=(1/4)(2-cos4θ+cos4θ)/(cosθcos(45°-θ))^2=1/2(cosθcos(45°-θ))^2MN=(1/2)2^0.5/cosθcos(45°-θ)=cos45°/cosθcos(45°-θ)BM+BN+MN=cos(2θ-45°)/(cos(45°-θ)cosθ)+cos45°/cosθcos(45°-θ)=(cos45°+cos(2θ-45°))/cosθcos(45°-θ)=2cosθcos(45°-θ)/cosθcos(45°-θ)=2p=2是定值3.OA/OM=cos(45°-θ) OM=1/cos(45°-θ)三角形高h=2^0.5/2cos(45°-θ)OC/ON=cosθ ON=1/cosθS△MON=h×ON/2=2^0.5/(4cosθcos(45°-θ))=(1/2)2^0.5/(cos45°+cos(2θ-45°)θ∈(0°,45°)θ=22.5°时,cos(2θ-45°)=1,分子最大,S取最小值=(2^0.5)-1m=MN=cos45°/(cos22.5°×cos(45°-22.5°))=cos45°/(cos22.5°)^2=2^0.5/(1+cos45°)=2(2^0.5-1)M点坐标(2^0.5/2cos22.5°,2^0.5/2cos22.5°)N点坐标(1/cos22.5°,0)△BMN是直角三角形,MN是斜边,也是外接圆的直径求外接圆的圆心点坐标cos22.5°=((cos45°+1)/2)^0.5=(1/2)(2+2^0.5)^0.5(2^0.5/2cos22.5°+1/cos22.5°)/2=(1/4)(2+2^0.5)/cos22.5°=(1/4)(2+2^0.5)/((1/2)(2+2^0.5)^0.5)=(1/2)(2+2^0.5)^0.5=cos22.5°圆心点坐标为(cos22.5°,2^0.5/4cos22.5°)半径=MN/2=2^0.5-1外接圆方程为(x-cos22.5°)^2+(y-2^0.5/4cos22.5°)^2=(2^0.5-1)^2