解:(1)由題意知:∠AMB=120°,
∴∠CMB=60°,∠OBM=30度.
∴OM=
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MB=1,
∴M(0,1).
(2)由A,B,C三點的特殊性與對稱性,知經過A,B,C三點的拋物線的解析式為y=ax
2+c.
∵OC=MC-MO=1,OB=
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,
∴C(0,-1),B(
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,0).
∴c=-1,a=
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.
∴y=
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x
2-1.
(3)∵S
四邊形ACBD=S
△ABC+S
△ABD,又S
△ABC與AB均為定值,
∴當△ABD邊AB上的高最大時,S
△ABD最大,此時點D為⊙M與y軸的交點.
∴S
四邊形ACBD=S
△ABC+S
△ABD=
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AB•OC+

AB•OD
=
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AB•CD
=4

cm
2.
(4)假設存在點P,如下圖所示:
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方法1:
∵△ABC為等腰三角形,∠ABC=30°,

,
∴△ABC∽△PAB等價于∠PAB=30°,PB=AB=2

,PA=

PB=6.
設P(x,y)且x>0,則x=PA•cos30°-AO=3

-

=2

,y=PA•sin30°=3.
又∵P(2

,3)的坐標滿足y=

x
2-1,
∴在拋物線y=

x
2-1上,存在點P(2

,3),
使△ABC∽△PAB.
由拋物線的對稱性,知點(-2

,3)也符合題意.
∴存在點P,它的坐標為(2

,3)或(-2

,3).
說明:只要求出(2

,3),(-2

,3),無最后一步不扣分.下面的方法相同.
方法2:
當△ABC∽△PAB時,∠PAB=∠BAC=30°,又由(1)知∠MAB=30°,
∴點P在直線AM上.
設直線AM的解析式為y=kx+b,
將A(-

,0),M(0,1)代入,
解得
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,
∴直線AM的解析式為y=

x+1.
解方程組
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,
得P(2

,3).
又∵

,
∴∠PBx=60度.
∴∠P=30°,
∴△ABC∽△PAB.
∴在拋物線y=
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x
2-1上,存在點(2

,3),使△ABC∽△PAB.
由拋物線的對稱性,知點(-2

,3)也符合題意.
∴存在點P,它的坐標為(2

,3)或(-2

,3).
方法3:
∵△ABC為等腰三角形,且

,
設P(x,y),則△ABC∽△PAB等價于PB=AB=2

,PA=

AB=6.
當x>0時,得
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,
解得P(2

,3).
又∵P(2

,3)的坐標滿足y=

x
2-1,
∴在拋物線y=

x
2-1上,存在點P(2

,3),使△ABC∽△PAB.
由拋物線的對稱性,知點(-2

,3)也符合題意.
∴存在點P,它的坐標為(2

,3)或(-2

,3).
分析:(1)在直角△AMO中,根據三角函數就可以求出OM,就可以得到M的坐標.
(2)根據三角函數就可以求出A,B的坐標,拋物線經過點A、B、C,因而M一定是拋物線的頂點.根據待定系數法就可以求出拋物線的解析式.
(3)四邊形ACBD的面積等于△ABC的面積+△ABP的面積,△ABC的面積一定,△ABP中底邊AB一定,P到AB的距離最大是三角形的面積最大,即當P是圓與y軸的交點時面積最大.
(4)△PAB和△ABC相似,根據相似三角形的對應邊的比相等,就可以求出P點的坐標.
點評:本題主要考查了二次函數的知識,其中涉及了待定系數法求函數的解析式、相似三角形的對應邊的比相等等知識,注意熟練掌握這些知識并靈活應用.