| ■48891 / ) |
Re[3]: ベクトルについて。
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□投稿者/ muturajcp 一般人(14回)-(2018/11/14(Wed) 21:47:43)
 | 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)}
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