解:(1)∵正方形OABC的邊長為2cm,
∴點A(0,-2),B(2,-2),
∴
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,
解得
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,
∴拋物線的表達式為y=
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x
2-
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x-2;
(2)移動t秒時,AP=2t,BP=2-2t,BQ=t,

①(i)OA與BP是對應(yīng)邊時,∵以O(shè)、A、P為頂點的三角形與△BPQ相似,
∴
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=
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,
即
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=
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,
解得t=
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,
(ii)OA與BQ是對應(yīng)邊時,∵以O(shè)、A、P為頂點的三角形與△BPQ相似,
∴
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=
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,
即
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=
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,
解得t=-1+

,t=-1-
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(舍去),
綜上所述,當(dāng)t=
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或-1+
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時,以O(shè)、A、P為頂點的三角形與△BPQ相似;
②根據(jù)勾股定理,S=PQ
2=BP
2+BQ
2=(2-2t)
2+t
2=5t
2-8t+4,
所以,當(dāng)t=-

=
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時,S有最小值,
此時BP=2-2t=2-2×
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=
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,BQ=t=
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,
(i)當(dāng)BP為對角線時,根據(jù)平行四邊形的對邊平行且相等,
點R的橫坐標為2t=
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,
縱坐標為-(2+
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)=-
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,
此時,
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×(
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)
2-
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×
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-2=
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-
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-2=-
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≠-
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,
點R不在拋物線上,所以,此時不成立,
(ii)BQ為對角線時,根據(jù)平行四邊形的對邊平行且相等,
點R的橫坐標為2+
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=
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,
縱坐標為-(2-
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)=-
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,
此時,
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×(
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)
2-
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×
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-2=
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-4-2=-
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,
點R在拋物線上,
所以,點R的坐標為(
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,-
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);
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(3)根據(jù)三角形三邊關(guān)系,|MA-MD|<DA,
所以,當(dāng)點M為直線AD與對稱軸交點時,M到D、A的距離之差最大,
此時,設(shè)直線AD的解析式為y=kx+b,
則
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,
解得
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,
所以,直線AD的解析式為y=

x-2,
∵拋物線y=

x
2-

x-2的對稱軸為x=-

=1,
∴y=

×1-2=-
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,
∴點M的坐標為(1,-
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).
分析:(1)根據(jù)正方形的四條邊都相等寫出點A、B的坐標,然后代入拋物線解析式得到關(guān)于b、c的方程組,解方程組求出b、c的值即可得解;
(2)表示出AP、BP、BQ的長,①然后分(i)OA與BP是對應(yīng)邊,(ii)OA與BQ是對應(yīng)邊兩種情況,根據(jù)相似三角形對應(yīng)邊成比例列出比例式求解即可;
②根據(jù)勾股定理表示出S,然后利用二次函數(shù)的最值問題確定出S取最小值時的t值,然后求出BP、BQ的值,再分(i)BP為對角線,(ii)BQ為對角線兩種情況,根據(jù)平行四邊形的對邊平行且相等求出點R的坐標,然后把點R的坐標代入拋物線,如果點R在拋物線上則,存在,否則不存在;
(3)根據(jù)三角形的任意兩邊之差小于第三邊判斷出當(dāng)點M為拋物線對稱軸與直線AD的交點時,M到D、A的距離之差最大,然后利用待定系數(shù)法求一次函數(shù)解析式求出直線AD的解析式,再求兩直線的交點即可.
點評:本題是二次函數(shù)的綜合題型,主要涉及正方形的性質(zhì),待定系數(shù)法求函數(shù)解析式(二次函數(shù)解析式與直線解析式),相似三角形對應(yīng)邊成比例的性質(zhì),平行四邊形的性質(zhì),三角形的三邊關(guān)系,分情況討論的思想,綜合性較強,難度較大,但只要仔細分析認真求解,也不難解答.