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證明:(1)因?yàn)锳坐標(biāo)為(1,
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),
所以O(shè)A=2,∠AOB=60°.
因?yàn)镺M=2-4t,ON=6-4t,
當(dāng)
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=
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時(shí),解得t=0,
即在甲、乙兩人到達(dá)O點(diǎn)前,只有當(dāng)t=0時(shí),△OMN∽△OAB,所以MN與AB不可能平行;
(2)因?yàn)榧走_(dá)到O點(diǎn)時(shí)間為t=
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,乙達(dá)到O點(diǎn)的時(shí)間為t=
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=
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,所以甲先到達(dá)O點(diǎn),所以t=
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或t=
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時(shí),O、M、N三點(diǎn)不能連接成三角形,
①當(dāng)t≤
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時(shí),如果△OMN∽△OBA,則有
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=
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,解得t=2>
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,所以,△OMN不可能相似△OBA;
②當(dāng)
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<t≤
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時(shí),∠MON>∠AOB,顯然△OMN不相似△OBA;
③當(dāng)t>
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時(shí),
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=
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,解得t=2>
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,所以當(dāng)t=2時(shí),△OMN∽△OBA;
(3)①當(dāng)t≤
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時(shí),如圖1,過點(diǎn)M作MH⊥x軸,垂足為H,
在Rt△MOH中,因?yàn)椤螦OB=60°,
所以MH=OMsin60°=(2-4t)×
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=
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(1-2t),
OH=0Mcos60°=(2-4t)×
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=1-2t,
所以NH=(6-4t)-(1-2t)=5-2t,
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所以s=[
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(1-2t)]
2+(5-2t)
2=16t
2-32t+28
②當(dāng)
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<t≤
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時(shí),如圖2,作MH⊥x軸,垂足為H,
在Rt△MOH中,MH=
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(4t-2)=
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(2t-1),NH=
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(4t-2)+(6-4t)=5-2t,
所以s=[
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(1-2t)]
2+(5-2t)
2=16t
2-32t+28
③當(dāng)t>
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時(shí),同理可得s=[
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(1-2t)]
2+(5-2t)
2=16t
2-32t+28,
綜上所述,s=[
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(1-2t)]
2+(5-2t)
2=16t
2-32t+28.
因?yàn)閟=16t
2-32t+28=16(t-1)
2+12,
所以當(dāng)t=1時(shí),s有最小值為12,所以甲、乙兩人距離最小值為2
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km.
分析:(1)用反證法說明.根據(jù)已知條件分別表示相關(guān)線段的長(zhǎng)度,根據(jù)三角形相似得比例式說明;
(2)根據(jù)兩個(gè)點(diǎn)到達(dá)O點(diǎn)的時(shí)間不同分段討論解答;
(3)在不同的時(shí)間段運(yùn)用相似三角形的判定和性質(zhì)分別求解析式,運(yùn)用函數(shù)性質(zhì)解答問題.
點(diǎn)評(píng):此題綜合考查了坐標(biāo)與圖形、相似三角形的判定與性質(zhì)、分類討論數(shù)學(xué)思想的應(yīng)用等知識(shí)點(diǎn),難度較大.