【答案】
分析:(1)根據(jù)B點(diǎn)的坐標(biāo)即可求出A、C的坐標(biāo).
(2)當(dāng)MN=
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AC時(shí),有兩種情況,①M(fèi)N是△OAC的中位線,此時(shí)OM=
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OA=2,因此t=2;
②當(dāng)MN是△ABC的中位線時(shí),OM=

OA=6,因此t=6;
(3)本題要分類(lèi)進(jìn)行討論:
①當(dāng)直線m在AC下方或與AC重合時(shí),即當(dāng)0<t≤4時(shí),可根據(jù)△OMN∽△OAC,用兩三角形的相似比求出面積比,即可得出S與t的函數(shù)關(guān)系式.
②當(dāng)直線m在AC上方時(shí),即當(dāng)4<t<8時(shí),可用矩形OABC的面積-三角形BMN的面積-三角形OCN的面積-三角形OAM的面積來(lái)求得.(也可過(guò)O作直線m的垂線設(shè)垂足為F,那么在直角三角形OMF中,可根據(jù)OD的長(zhǎng)和∠ODE的正弦值求出OF的長(zhǎng),求MN的方法一樣).
(4)根據(jù)(3)得出的函數(shù)的性質(zhì)和自變量的取值范圍即可求出面積S的最大值及對(duì)應(yīng)的t的值.
解答:
解:(1)(4,0),(0,3);
(2)當(dāng)MN=

AC時(shí),有兩種情況,
①M(fèi)N是△OAC的中位線,此時(shí)OM=

OA=2,因此t=2;
②當(dāng)MN是△ABC的中位線時(shí),
∴AM=

AB=

,OA=4,
∴AD=

=

=2
∴OD=OA+AD=4+2=6,因此t=6;
(3)當(dāng)0<t≤4時(shí),OM=t
∵由△OMN∽△OAC,得

=

,
∴ON=

,S=

t
2當(dāng)4<t<8時(shí),
如圖,∵OD=t,

∴AD=t-4
方法一:
由△DAM∽△AOC,可得AM=
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(t-4)
∴BM=6-

由△BMN∽△BAC,可得BN=

BM=8-t
∴CN=t-4
S=矩形OABC的面積-Rt△OAM的面積-Rt△MBN的面積-Rt△NCO的面積
=12-

(t-4)-
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(8-t)(6-
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)-

=
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t
2+3t
方法二:
易知四邊形ADNC是平行四邊形,
∴CN=AD=t-4,BN=8-t.
由△BMN∽△BAC,可得BM=

BN=6-

,
∴AM=

(t-4)
以下同方法一.
(4)有最大值.
方法一:
當(dāng)0<t≤4時(shí),
∵拋物線S=

t
2的開(kāi)口向上,在對(duì)稱(chēng)軸t=0的右邊,S隨t的增大而增大
∴當(dāng)t=4時(shí),S可取到最大值

×4
2=6;(11分)
當(dāng)4<t<8時(shí),
∵拋物線S=

t
2+3t的開(kāi)口向下,它的頂點(diǎn)是(4,6),
∴S<6.
綜上,當(dāng)t=4時(shí),S有最大值6.
方法二:
∵S=

∴當(dāng)0<t<8時(shí),畫(huà)出S與t的函數(shù)關(guān)系圖象
如圖所示.
顯然,當(dāng)t=4時(shí),S有最大值6.
點(diǎn)評(píng):本題考查了矩形的性質(zhì),二次函數(shù)的應(yīng)用、圖形的面積求法等知識(shí)及綜合應(yīng)用知識(shí)、解決問(wèn)題的能力.