【答案】
分析:(1)過點(diǎn)作CD⊥x軸于點(diǎn)D,先由等邊三角形的性質(zhì)求出P點(diǎn)坐標(biāo)及BP的長,故可得出PE的長,由圖形旋轉(zhuǎn)的性質(zhì)求出PC=PE及∠CPD的度數(shù),再由銳角三角函數(shù)的定義即可求出PD及CD的長,進(jìn)而可得出結(jié)論;
(2)過P作PD⊥OB于點(diǎn)D,過C作CF⊥PA于點(diǎn)F,在Rt△OPD中 PD=OP•sin60°=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191530455162510/SYS201311011915304551625022_DA/0.png)
,由相似三角形的判定定理得出△BPD∽△PCF,故可得出CF及PF的長,進(jìn)而可得出C點(diǎn)坐標(biāo);
(3)取OA的中點(diǎn)M,連接MC,由(2)得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191530455162510/SYS201311011915304551625022_DA/1.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191530455162510/SYS201311011915304551625022_DA/2.png)
,由銳角三角函數(shù)的定義得出∠CMF=30°,可知點(diǎn)C在直線MC上運(yùn)動(dòng).故當(dāng)點(diǎn)P在點(diǎn)O時(shí),點(diǎn)C與點(diǎn)M重合.
當(dāng)點(diǎn)P運(yùn)動(dòng)到點(diǎn)A時(shí),點(diǎn)C的坐標(biāo)為(5,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191530455162510/SYS201311011915304551625022_DA/3.png)
),由兩點(diǎn)間的距離公式即可得出結(jié)論.
解答:![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191530455162510/SYS201311011915304551625022_DA/images4.png)
解:(1)如圖1,過點(diǎn)作CD⊥x軸于點(diǎn)D,
∵△AOB是等邊三角形,P是OA的中點(diǎn),
∴P(2,0),BP=OB•sin60°=4×
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191530455162510/SYS201311011915304551625022_DA/4.png)
=2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191530455162510/SYS201311011915304551625022_DA/5.png)
,
∵E是BP的中點(diǎn),
∴PE=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191530455162510/SYS201311011915304551625022_DA/6.png)
,
∴PE=PC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191530455162510/SYS201311011915304551625022_DA/7.png)
,
∵∠BPC=60°,
∴∠CPA=30°,
∴PD=PC•cos30°=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191530455162510/SYS201311011915304551625022_DA/8.png)
×
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191530455162510/SYS201311011915304551625022_DA/9.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191530455162510/SYS201311011915304551625022_DA/10.png)
,CD=PC•sin30°=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191530455162510/SYS201311011915304551625022_DA/11.png)
×
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191530455162510/SYS201311011915304551625022_DA/12.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191530455162510/SYS201311011915304551625022_DA/13.png)
,
∴OD=OP+PD=2+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191530455162510/SYS201311011915304551625022_DA/14.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191530455162510/SYS201311011915304551625022_DA/15.png)
,
∴C(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191530455162510/SYS201311011915304551625022_DA/16.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191530455162510/SYS201311011915304551625022_DA/17.png)
);
(2)如圖2,過P作PD⊥OB于點(diǎn)D,過C作CF⊥PA于點(diǎn)F
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191530455162510/SYS201311011915304551625022_DA/images19.png)
在Rt△OPD中 PD=OP•sin60°=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191530455162510/SYS201311011915304551625022_DA/18.png)
,
∵∠OBP+∠OPB=∠CPF+∠OPB=120°
∴∠DBP=∠FPC,
∵∠PDB=∠CFP=90°
∴△BPD∽△PCF,
∴CF=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191530455162510/SYS201311011915304551625022_DA/19.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191530455162510/SYS201311011915304551625022_DA/20.png)
∴點(diǎn)C的坐標(biāo)是(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191530455162510/SYS201311011915304551625022_DA/21.png)
);
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191530455162510/SYS201311011915304551625022_DA/images24.png)
(3)取OA的中點(diǎn)M,連接MC,由(2)得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191530455162510/SYS201311011915304551625022_DA/22.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191530455162510/SYS201311011915304551625022_DA/23.png)
.
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191530455162510/SYS201311011915304551625022_DA/24.png)
∴∠CMF=30°.
∴點(diǎn)C在直線MC上運(yùn)動(dòng).
當(dāng)點(diǎn)P在點(diǎn)O時(shí),點(diǎn)C與點(diǎn)M重合.
當(dāng)點(diǎn)P運(yùn)動(dòng)到點(diǎn)A時(shí),點(diǎn)C的坐標(biāo)為
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191530455162510/SYS201311011915304551625022_DA/25.png)
∴點(diǎn)C所經(jīng)過的路徑長為
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191530455162510/SYS201311011915304551625022_DA/26.png)
.
點(diǎn)評:本題考查的是相似形綜合題及旋轉(zhuǎn)的性質(zhì)、等邊三角形的性質(zhì),根據(jù)題意作出輔助線,構(gòu)造出直角三角形是解答此題的關(guān)鍵.