摘要：004](1)解:由得点坐标为 由得点坐标为∴ 由解得∴点的坐标为 ∴ (2)解:∵点在上且 ∴点坐标为又∵点在上且∴点坐标为 ∴ (3)解法一:当时.如图1.矩形与重叠部分为五边形(时.为四边形)．过作于.则 ∴即∴ ∴ 即 [005](1)如图1.过点作于点·················· 1分 ∵为的中点. ∴ 在中.∴············· 2分 ∴ 即点到的距离为··········································· 3分 (2)①当点在线段上运动时.的形状不发生改变． ∵∴ ∵∴. 同理······························································································· 4分 如图2.过点作于.∵ ∴ ∴ ∴ 则 在中. ∴的周长=············································· 6分 ②当点在线段上运动时.的形状发生改变.但恒为等边三角形． 当时.如图3.作于.则 类似①. ∴································································································ 7分 ∵是等边三角形.∴ 此时.········································· 8分 当时.如图4.这时 此时. 当时.如图5. 则又 ∴ 因此点与重合.为直角三角形． ∴ 此时. 综上所述.当或4或时.为等腰三角形．

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