【答案】
分析:(1)解直角三角形,求得點E到AC的距離等于
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193806737720390/SYS201311011938067377203026_DA/0.png)
a,這是一個定值;
(2)如答圖2所示,作輔助線,將四邊形MDEN分成一個等邊三角形和一個平行四邊形,求出其周長;
(3)可能存在三種情形,需要分類討論:
①若0<a≤
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193806737720390/SYS201311011938067377203026_DA/1.png)
,△DEF在△ABC內(nèi)部,如答圖3所示;
②若
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193806737720390/SYS201311011938067377203026_DA/2.png)
<a≤
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193806737720390/SYS201311011938067377203026_DA/3.png)
,點E在△ABC內(nèi)部,點F在△ABC外部,在如答圖4所示;
③若
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193806737720390/SYS201311011938067377203026_DA/4.png)
<a<3,點E、F均在△ABC外部,如答圖5所示.
解答:解:(1)由題意得:tanA=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193806737720390/SYS201311011938067377203026_DA/5.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193806737720390/SYS201311011938067377203026_DA/6.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193806737720390/SYS201311011938067377203026_DA/7.png)
,
∴∠A=60°.
∵DE∥AB,
∴∠CDE=∠A=60°.
如答圖1所示,過點E作EH⊥AC于點H,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193806737720390/SYS201311011938067377203026_DA/images8.png)
則EH=DE•sin∠CDE=a•
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193806737720390/SYS201311011938067377203026_DA/8.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193806737720390/SYS201311011938067377203026_DA/9.png)
a.
∴點E到AC的距離為一個常數(shù).
(2)若AD=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193806737720390/SYS201311011938067377203026_DA/10.png)
,當(dāng)a=2時,如答圖2所示.
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193806737720390/SYS201311011938067377203026_DA/images12.png)
設(shè)AB與DF、EF分別交于點M、N.
∵△DEF為等邊三角形,∴∠MDE=60°,
由(1)知∠CDE=60°,
∴∠ADM=180°-∠MDE-∠CDE=60°,
又∵∠A=60°,
∴△ADM為等邊三角形,
∴DM=AD=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193806737720390/SYS201311011938067377203026_DA/11.png)
.
過點M作MG∥AC,交DE于點G,則∠DMG=∠ADM=60°,
∴△DMG為等邊三角形,
∴DG=MG=DM=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193806737720390/SYS201311011938067377203026_DA/12.png)
.
∴GE=DE-DG=2-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193806737720390/SYS201311011938067377203026_DA/13.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193806737720390/SYS201311011938067377203026_DA/14.png)
.
∵∠MGD=∠E=60°,∴MG∥NE,
又∵DE∥AB,
∴四邊形MGEN為平行四邊形.
∴NE=MG=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193806737720390/SYS201311011938067377203026_DA/15.png)
,MN=GE=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193806737720390/SYS201311011938067377203026_DA/16.png)
.
∴T=DE+DM+MN+NE=2+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193806737720390/SYS201311011938067377203026_DA/17.png)
+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193806737720390/SYS201311011938067377203026_DA/18.png)
+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193806737720390/SYS201311011938067377203026_DA/19.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193806737720390/SYS201311011938067377203026_DA/20.png)
.
(3)若點D運動到AC的中點處,分情況討論如下:
①若0<a≤
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193806737720390/SYS201311011938067377203026_DA/21.png)
,△DEF在△ABC內(nèi)部,如答圖3所示:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193806737720390/SYS201311011938067377203026_DA/images24.png)
∴T=3a;
②若
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193806737720390/SYS201311011938067377203026_DA/22.png)
<a≤
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193806737720390/SYS201311011938067377203026_DA/23.png)
,點E在△ABC內(nèi)部,點F在△ABC外部,在如答圖4所示:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193806737720390/SYS201311011938067377203026_DA/images27.png)
設(shè)AB與DF、EF分別交于點M、N,過點M作MG∥AC交DE于點G.
與(2)同理,可知△ADM、△DMG均為等邊三角形,四邊形MGEN為平行四邊形.
∴DM=DG=NE=AD=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193806737720390/SYS201311011938067377203026_DA/24.png)
,MN=GE=DE-DG=a-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193806737720390/SYS201311011938067377203026_DA/25.png)
,
∴T=DE+DM+MN+NE=a+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193806737720390/SYS201311011938067377203026_DA/26.png)
+(a-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193806737720390/SYS201311011938067377203026_DA/27.png)
)+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193806737720390/SYS201311011938067377203026_DA/28.png)
=2a+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193806737720390/SYS201311011938067377203026_DA/29.png)
;
③若
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193806737720390/SYS201311011938067377203026_DA/30.png)
<a<3,點E、F均在△ABC外部,如答圖5所示:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193806737720390/SYS201311011938067377203026_DA/images35.png)
設(shè)AB與DF、EF分別交于點M、N,BC與DE、EF分別交于點P、Q.
在Rt△PCD中,CD=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193806737720390/SYS201311011938067377203026_DA/31.png)
,∠CDP=60°,∠DPC=30°,
∴PC=CD•tan60°=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193806737720390/SYS201311011938067377203026_DA/32.png)
×
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193806737720390/SYS201311011938067377203026_DA/33.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193806737720390/SYS201311011938067377203026_DA/34.png)
.
∵∠EPQ=∠DPC=30°,∠E=60°,∴∠PQE=90°.
由(1)知,點E到AC的距離為
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193806737720390/SYS201311011938067377203026_DA/35.png)
a,∴PQ=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193806737720390/SYS201311011938067377203026_DA/36.png)
a-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193806737720390/SYS201311011938067377203026_DA/37.png)
.
∴QE=PQ•tan30°=(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193806737720390/SYS201311011938067377203026_DA/38.png)
a-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193806737720390/SYS201311011938067377203026_DA/39.png)
)×
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193806737720390/SYS201311011938067377203026_DA/40.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193806737720390/SYS201311011938067377203026_DA/41.png)
a-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193806737720390/SYS201311011938067377203026_DA/42.png)
,PE=2QE=a-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193806737720390/SYS201311011938067377203026_DA/43.png)
.
由②可知,四邊形MDEN的周長為2a+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193806737720390/SYS201311011938067377203026_DA/44.png)
.
∴T=四邊形MDEN的周長-PE-QE+PQ=(2a+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193806737720390/SYS201311011938067377203026_DA/45.png)
)-(a-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193806737720390/SYS201311011938067377203026_DA/46.png)
)-(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193806737720390/SYS201311011938067377203026_DA/47.png)
a-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193806737720390/SYS201311011938067377203026_DA/48.png)
)+(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193806737720390/SYS201311011938067377203026_DA/49.png)
a-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193806737720390/SYS201311011938067377203026_DA/50.png)
)=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193806737720390/SYS201311011938067377203026_DA/51.png)
a+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193806737720390/SYS201311011938067377203026_DA/52.png)
-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193806737720390/SYS201311011938067377203026_DA/53.png)
.
綜上所述,若點D運動到AC的中點處,T的關(guān)系式為:
T=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193806737720390/SYS201311011938067377203026_DA/54.png)
.
點評:本題考查了運動型綜合題,新穎之處在于所求是重疊部分的周長而非面積.難點在于第(3)問,根據(jù)題意,可能的情形有三種,需要分類討論,避免漏解.