| AP'=2APとなる点P'をとると |△AP'Q|:|△PP'Q|=2:1 |△PP'Q|:|△PP'R|=1+t:1 だから |△AP'Q|:|△PP'R|=2(1+t):1 だから |△PP'R|=[1/{2(1+t)}]|△AP'Q| だから |□APRQ| =|△AP'Q|-|△PP'R| =|△AP'Q|-[1/{2(1+t)}]|△AP'Q| =[(2t+1)/{2(1+t)}]|△AP'Q|
|△AP'Q| =(1/2)|AP'||AQ|sin∠PAQ =|AP||AQ|sin∠PAQ =√[|AP|^2|AQ|^2-(AP,AQ)^2]
△OABは辺長1の正3角形で APはAからOBへの垂直2等分線だから |AP|=(√3)/2 |AP|^2=3/4
△OADは辺長1の正3角形で ∠AOQ=∠AOD=60° |OA|=1 |OQ|=t|OD| だから |AQ|^2 =|OA|^2+|OQ|^2-2|OA||OQ|cos∠AOQ =t^2-t+1 だから |AP|^2|AQ|^2=3(t^2-t+1)/4
(AP,AQ) =((1/2)OB-OA,tOD-OA) =(t/2)(OB,OD)-t(OA,OD)-(1/2)(OB,OA)+|OA|^2 =(t/2)|OB||OD|cos∠BOD-t|OA||OD|cos∠AOD-(1/2)|OB||OA|cos∠AOB+1 =(t/2)|OB||OD|cos90°-t|OA||OD|cos60°-(1/2)|OB||OA|cos60°+1 =-(t/2)-(1/4)+1 =3/4-t/2 =(3-2t)/4 だから (AP,AQ)^2=(3-2t)^2/16=(4t^2-12t+9)/16 だから |AP|^2|AQ|^2-(AP,AQ)^2 =3(t^2-t+1)/4-(4t^2-12t+9)/16 =(8t^2+3)/16
|△AP'Q| =√[|AP|^2|AQ|^2-(AP,AQ)^2] ={√(8t^2+3)}/4
|□APRQ| =[(2t+1)/{2(1+t)}]|△AP'Q| =[(2t+1)/{2(1+t)}]√[|AP|^2|AQ|^2-(AP,AQ)^2] =[(2t+1)√(8t^2+3)]/{8(1+t)} |