【答案】
分析:(1)因?yàn)椤螩AO=30°,由折疊可知∠OCE=∠ECD=
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∠OCA=30°,
在Rt△COE中,利用三角函數(shù)可求OE=OC•tan∠OCE=
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×
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=1,從而可求點(diǎn)E的坐標(biāo)是(1,0).
因?yàn)镺C=
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,所以C(0,
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).
可設(shè)直線CE的解析式為y=kx+b,將C、E的坐標(biāo)代入,可得到關(guān)于k、b的方程組,解之即可;
(2)在Rt△AOC中,利用三角函數(shù)可求出AC、AO的值,因?yàn)镃D=OC=
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,可求出AD=AC-CD=2
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-
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=
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.
要求D的坐標(biāo),需過(guò)點(diǎn)D作DF⊥OA于點(diǎn)F.
在Rt△AFD中,利用三角函數(shù)可求DF=AD•sin∠CAO=
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,AF=AD•cos∠CAO=
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,所以O(shè)F=AO-AF=
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,從而點(diǎn)D的坐標(biāo)是(
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,
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);
(3)需分情況討論:
第一種情況:若此點(diǎn)在第四象限內(nèi),可設(shè)其為M
1,延長(zhǎng)DF交直線CE于M
1,連接M
1O,則有DM
1∥y軸.
因?yàn)镺F=
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,所以可設(shè)點(diǎn)M
1的坐標(biāo)為(
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,y
1),利用點(diǎn)M
1在直線CE上,可得y
1的值,即可求出點(diǎn)M
1的坐標(biāo)是(
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,-
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),所以有DM
1=DF+FM
1=
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+
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=
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,OC=
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,所以DM
1=OC.
利用一組對(duì)邊平行且相等的四邊形是平行四邊形可知四邊形CDM
1O為平行四邊形.而點(diǎn)O在y軸上,所以點(diǎn)M
1是符合條件的點(diǎn).
第二種情況:此點(diǎn)在第二象限內(nèi),設(shè)為M
2.可過(guò)點(diǎn)D作DN∥CE交y軸于N,過(guò)N點(diǎn)作NM
2∥CD交直線CE于點(diǎn)M
2,則四邊形M
2NDC為平行四邊形.
利用平行四邊形的對(duì)邊分別相等,可知M
2N=CD=
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,
又因M
2N∥CD,DN∥CE,所以∠NM
2C=∠ACE,∠OCE=∠M
2CN,CN=M
2N.
又因M
2N=CD=
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,所以CN=
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.
接著可作M
2H⊥y軸于點(diǎn)H,利用兩直線平行,內(nèi)錯(cuò)角相等可得∠M
2NC=∠NCD,∴∠M
2NH=∠OCA=60°.
在Rt△M
2NH中,利用三角函數(shù)可求出M
2H,NH的值,利用HO=HN+CN+OC=
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可得M
2(-
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,
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).
解答:解:(1)由題意知∠CAO=30°,
∴∠OCE=∠ECD=
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∠OCA=30°,
∴在Rt△COE中,OE=OC•tan∠OCE=
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×
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=1,
∴點(diǎn)E的坐標(biāo)是(1,0),
設(shè)直線CE的解析式為y=kx+b.
把點(diǎn)C(0,
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),E(1,0)代入得
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,
∴
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,
∴直線CE的解析式為y=-
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x+
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.
(2)在Rt△AOC中,AC=
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=2
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,
AO=
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=3,
∵CD=OC=
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,
∴AD=AC-CD=2
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-
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=
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,
過(guò)點(diǎn)D作DF⊥OA于點(diǎn)F,
在Rt△AFD中,DF=AD•sin∠CAO=
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,
AF=AD•cos∠CAO=
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,
∴OF=AO-AF=
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.
∴點(diǎn)D的坐標(biāo)是(
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,
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).
(3)存在兩個(gè)符合條件的M點(diǎn),
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第一種情況:此點(diǎn)在第四象限內(nèi),設(shè)為M
1,延長(zhǎng)DF交直線CE于M
1,
連接M
1O,M
1O∥AC,
則有DM
1∥y軸,
∵OF=
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,
∴設(shè)點(diǎn)M
1的坐標(biāo)為(
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,y
1),
又∵點(diǎn)M
1在直線CE上,
∴將點(diǎn)M
1的坐標(biāo)代入y=-
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x+
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中,
得y
1=-
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×
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+
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=-
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,即FM
1=
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.
∴點(diǎn)M
1的坐標(biāo)是(
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,-
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),
又∵DM
1=DF+FM
1=
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+
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=
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,OC=
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,
∴DM
1=OC,
又∵DM
1∥OC,
∴四邊形CDM
1O為平行四邊形,
又∵點(diǎn)O在y軸上,
∴點(diǎn)M
1是符合條件的點(diǎn).
第二種情況:此點(diǎn)在第二象限內(nèi),設(shè)為M
2,
過(guò)點(diǎn)D作DN∥CE交y軸于N,過(guò)N點(diǎn)作NM
2∥CD交直線CE于點(diǎn)M
2,
則四邊形M
2NDC為平行四邊形,
∴M
2N=CD=
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,
∵M(jìn)
2N∥CD,DN∥CE,
∴∠NM
2C=∠ACE,∠OCE=∠M
2CN,
∴CN=M
2N,
∵M(jìn)
2N=CD=
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,
∴CN=
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,
作M
2H⊥y軸于點(diǎn)H,
∵M(jìn)
2N∥CD,
∴∠M
2NC=∠NCD,
∴∠M
2NH=∠OCA=60°,
在Rt△M
2NH中,
M
2H=M
2N•sin60°=
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×
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=
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,
NH=M
2N•cos60°=
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×
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=
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,
∴HO=HN+CN+OC=
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,
∴M
2(-
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,
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),
∴點(diǎn)M
2是符合條件的點(diǎn),
綜上所述,符合條件的兩個(gè)點(diǎn)的坐標(biāo)分別為M
1(
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,-
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),M
2(-
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,
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).
點(diǎn)評(píng):本題的解決需要綜合運(yùn)用待定系數(shù)法、三角函數(shù)等知識(shí),另外解決這類問(wèn)題常用到分類討論、數(shù)形結(jié)合、方程和轉(zhuǎn)化等數(shù)學(xué)思想方法.