解:(1)∵M(jìn)N∥BC,
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∴△AMN∽△ABC,
∴
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=
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,即
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=
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,解得AN=
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x,
∴△AMN的面積=
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•x•
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x=
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x
2,
∵四邊形AMPN是矩形,
∴S=
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•x•
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x=
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x
2(0<x≤8);
(2)若P點在BC上時,
∵四邊形AMPN是矩形,
∴O點為AP的中點,
而MN∥BC,
∴MN為△ABC的中位線,此時AM=4,
當(dāng)0<x≤4時,y=S=
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•x•
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x=
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x
2,此時x=4,y的最大值為6;
當(dāng)4<x≤8時,PM與PN分別交BC于E、F,如圖,
y=S
梯形MEFN=S
△PMN-S
△PEF,
∵四邊形AMPN是矩形,
∴PN=AM=x,
∵M(jìn)N∥BC,
∴四邊形BFNM是平行四邊形,
∴FN=BM=8-x,PF=PN-FN=x-(8-x)=2x-8,
∵Rt△PEF∽Rt△ACB,
∴
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=(
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)
2=(
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)
2,
而S
△ABC=
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×8×6=24,
∴S
△PEF=
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(x-4)
2,
∴y=
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x
2-
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(x-4)
2=-
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x
2+12x-24,
=-
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(x-
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)
2+8(4<x≤8),
∵a=-
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<0,
∴當(dāng)x=
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時,y有最大值,最大值為8,
綜上所述,當(dāng)x=
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時,y有最大值,最大值為8.
分析:(1)先證明△AMN∽△ABC,則可根據(jù)相似三角形的對應(yīng)邊成比例求AN,然后由三角形的面積公式求得用x的代數(shù)式表示的△AMN的面積S;
(3)先求出P點在BC上時AM的值,然后進(jìn)行討論:當(dāng)0<x≤4時,y=S=
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•x•
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x=
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x
2,根據(jù)二次函數(shù)的性質(zhì)得到x=4,y的最大值為6;當(dāng)4<x≤8時,PM與PN分別交BC于E、F,y=S
梯形MEFN=S
△PMN-S
△PEF,利用矩形的性質(zhì)可表示出PN=AM=x;再由平行四邊形BFNM的性質(zhì)解得FN=8-x,PF=2x-8,則可利用相似三角形Rt△PEF∽Rt△ABC的性質(zhì)求得S
△PEF值;然后寫出y與x的解析式,再根據(jù)二次函數(shù)的性質(zhì)求出y的最大值,最后綜合兩種情況即可.
點評:本題考查了圓的綜合題:掌握圓周角定理及其推論;熟練運用相似三角形的有關(guān)知識進(jìn)行幾何計算和二次函數(shù)的性質(zhì)解決最值問題.