解:(1)當(dāng)x=0時(shí),y=4;當(dāng)y=0時(shí),-
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x+4=0,x=3.
∴A(3,0),B(0,4).
(2)設(shè)點(diǎn)C的橫坐標(biāo)為n.由(1)知AB=
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=5,
∴sin∠OBA=
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.
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過C作CE⊥x軸于E,過P作PG⊥x軸于G,PF⊥CE于F,
則∠FCP=∠OBA,PF=m-n.
①當(dāng)m<3時(shí),∵PC=PG=-
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m+4,
∴PF=PC•sin∠FCP=PC•sin∠OBA,
∴m-n=(-
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m+4)×
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.
解得n=
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m-
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.
②當(dāng)m>3時(shí),PC=PG=
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,PF=PC•sin∠FCP=PC•sin∠OBA,
∴m-n=(
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m-4)×
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.
解得n=
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m+
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.
(3)當(dāng)點(diǎn)C在線段AB上時(shí),由(2)知,C點(diǎn)的橫坐標(biāo)n=
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m-
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,
以下兩種情況△BOC為等腰三角形.
①當(dāng)CB=CO時(shí),
∵△OBA是直角三角形,∠BOA=90度.
∴此時(shí)C為AB的中點(diǎn),
∴C點(diǎn)的橫坐標(biāo)為
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.
∴
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,解得m=
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.
②當(dāng)CB=OB時(shí),
∵AB=5,
∴AC=AB-CB=1,
∴AE=AC•cos∠OAB=
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.
∵OE+AE=OA,
∴
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,解得m=
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.
∵OB>OA,
∴在線段AB上不存在點(diǎn)C,使OC=OB.
所以,當(dāng)m=
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或m=
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時(shí),△BOC為等腰三角形.
分析:(1)因?yàn)橹本€y=-
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x+4分別交x軸、y軸于A、B兩點(diǎn),所以分別令x=0、y=0,即可求出A、B的坐標(biāo);
(2)設(shè)點(diǎn)C的橫坐標(biāo)為n.由(1)知AB=
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=5,所以sin∠OBA=
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,要求點(diǎn)C的橫坐標(biāo),可過C作CE⊥x軸于E,過P作PG⊥x軸于G,PF⊥CE于F,則∠FCP=∠OBA,PF=m-n.
①若m<3時(shí),因?yàn)镻點(diǎn)的橫坐標(biāo)為m,P在直線y=-
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x+4上,所以PC=PG=-
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m+4,利用三角函數(shù)可得PF=PC•sin∠FCP=PC•sin∠OBA,即可得到關(guān)于m、m的關(guān)系式,整理即可;
②當(dāng)m>3時(shí),P在x軸的下方,所以PC=PG=
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,PF=PC•sin∠FCP=PC•sin∠OBA,整理即可得到另一個(gè)m、n的關(guān)系式;
(3)當(dāng)點(diǎn)C在線段AB上時(shí),由(2)知,C點(diǎn)的橫坐標(biāo)n=
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m-
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,因?yàn)椤鰾OC為等腰三角形,所以需要分情況討論:
①當(dāng)CB=CO時(shí),因?yàn)椤鱋BA是直角三角形,∠BOA=90°,所以此時(shí)C為AB的中點(diǎn),C點(diǎn)的橫坐標(biāo)為
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,即n=
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,即
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,解之即可;
②當(dāng)CB=OB=4時(shí),因?yàn)锳B=5,可得AC=AB-CB=1,利用三角函數(shù)可得AE=AC•cos∠OAB=
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,又因OE+AE=OA,就可得到關(guān)于m的方程,解之即可;
③當(dāng)OC=OB時(shí),因?yàn)镺B>OA,所以C在線段BA的延長線上,即在線段AB上不存在點(diǎn)C,使OC=OB.
點(diǎn)評:本題的解決需要用到分類討論、數(shù)形結(jié)合、方程和轉(zhuǎn)化等數(shù)學(xué)思想方法.