题目内容
(2010•重庆)已知:如图(1),在平面直角坐标xOy中,边长为2的等边△OAB的顶点B在第一象限,顶点A在x轴的正半轴上.另一等腰△OCA的顶点C在第四象限,OC=AC,∠C=120°.现有两动点P、Q分别从A、O两点同时出发,点Q以每秒1个单位的速度沿OC向点C运动,点P以每秒3个单位的速度沿A→O→B运动,当其中一个点到达终点时,另一个点也随即停止.(1)求在运动过程中形成的△OPQ的面积S与运动的时间t之间的函数关系,并写出自变量t的取值范围;
(2)在等边△OAB的边上(点A除外)存在点D,使得△OCD为等腰三角形,请直接写出所有符合条件的点D的坐标;
(3)如图(2),现有∠MCN=60°,其两边分别与OB、AB交于点M、N,连接MN.将∠MCN绕着C点旋转(0°<旋转角<60°),使得M、N始终在边OB和边AB上.试判断在这一过程中,△BMN的周长是否发生变化?若没有变化,请求出其周长;若发生变化,请说明理由.
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231047426185595/SYS201310212310474261855005_ST/images0.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231047426185595/SYS201310212310474261855005_ST/images1.png)
【答案】分析:(1)由于点Q从点O运动到点C需要
秒,点P从点A→O→B需要
秒,所以分两种情况讨论:①0<t<
;②
≤t<
.针对每一种情况,根据P点所在的位置,由三角形的面积公式得出△OPQ的面积S与运动的时间t之间的函数关系,并且得出自变量t的取值范围;
(2)如果△OCD为等腰三角形,那么分D在OA边或者OB边上或AB边上三种情形.每一种情形,都有可能O为顶点,C为顶点,D为顶点,分别讨论,得出结果;
(3)如果延长BA至点F,使AF=OM,连接CF,则由SAS可证△MOC≌△FAC,得出MC=CF,再由SAS证出△MCN≌△FCN,得出MN=NF,那么△BMN的周长=BA+BO=4.
解答:
解:(1)过点C作CD⊥OA于点D.(如图)
∵OC=AC,∠ACO=120°,
∴∠AOC=∠OAC=30°.
∵OC=AC,CD⊥OA,∴OD=DA=1.
在Rt△ODC中,OC=
=
=
(1分)
(i)当0<t<
时,OQ=t,AP=3t,OP=OA-AP=2-3t.
过点Q作QE⊥OA于点E.(如图)
在Rt△OEQ中,![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231047426185595/SYS201310212310474261855005_DA/images10.png)
∵∠AOC=30°,
∴QE=
OQ=
,
∴S△OPQ=
OP•EQ=
(2-3t)•
=-
+
t,
即S=-
+
t;(3分)
(ii)当
<t≤
时(如图)
OQ=t,OP=3t-2.
∴∠BOA=60°,∠AOC=30°,∴∠POQ=90°.
∴S△OPQ=
OQ•OP=
t•(3t-2)=
-t,
即S=
-t;
故当0<t<
时,S=-
+
t,当
<t≤
时,S=
-t(5分)
(2)D(
,1)或(
,0)或(
,0)或(
,
)(9分)
(3)△BMN的周长不发生变化.理由如下:
延长BA至点F,使AF=OM,连接CF.(如图)![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231047426185595/SYS201310212310474261855005_DA/images37.png)
又∵∠MOC=∠FAC=90°,OC=AC,
∴△MOC≌△FAC,
∴MC=CF,∠MCO=∠FCA.(10分)
∴∠FCN=∠FCA+∠NCA=∠MCO+∠NCA
=∠OCA-∠MCN
=60°,
∴∠FCN=∠MCN.
在△MCN和△FCN中,
,
∴△MCN≌△FCN,
∴MN=NF.(11分)
∴BM+MN+BN=BM+NF+BN=BO-OM+BA+AF=BA+BO=4.
∴△BMN的周长不变,其周长为4.
点评:本题综合考查了等腰三角形、等边三角形的性质,全等三角形的判定.难度很大.注意分类讨论时,做到不重复,不遗漏.
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231047426185595/SYS201310212310474261855005_DA/0.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231047426185595/SYS201310212310474261855005_DA/1.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231047426185595/SYS201310212310474261855005_DA/2.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231047426185595/SYS201310212310474261855005_DA/3.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231047426185595/SYS201310212310474261855005_DA/4.png)
(2)如果△OCD为等腰三角形,那么分D在OA边或者OB边上或AB边上三种情形.每一种情形,都有可能O为顶点,C为顶点,D为顶点,分别讨论,得出结果;
(3)如果延长BA至点F,使AF=OM,连接CF,则由SAS可证△MOC≌△FAC,得出MC=CF,再由SAS证出△MCN≌△FCN,得出MN=NF,那么△BMN的周长=BA+BO=4.
解答:
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231047426185595/SYS201310212310474261855005_DA/images5.png)
∵OC=AC,∠ACO=120°,
∴∠AOC=∠OAC=30°.
∵OC=AC,CD⊥OA,∴OD=DA=1.
在Rt△ODC中,OC=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231047426185595/SYS201310212310474261855005_DA/5.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231047426185595/SYS201310212310474261855005_DA/6.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231047426185595/SYS201310212310474261855005_DA/7.png)
(i)当0<t<
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231047426185595/SYS201310212310474261855005_DA/8.png)
过点Q作QE⊥OA于点E.(如图)
在Rt△OEQ中,
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231047426185595/SYS201310212310474261855005_DA/images10.png)
∵∠AOC=30°,
∴QE=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231047426185595/SYS201310212310474261855005_DA/9.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231047426185595/SYS201310212310474261855005_DA/10.png)
∴S△OPQ=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231047426185595/SYS201310212310474261855005_DA/11.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231047426185595/SYS201310212310474261855005_DA/12.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231047426185595/SYS201310212310474261855005_DA/13.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231047426185595/SYS201310212310474261855005_DA/14.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231047426185595/SYS201310212310474261855005_DA/15.png)
即S=-
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231047426185595/SYS201310212310474261855005_DA/16.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231047426185595/SYS201310212310474261855005_DA/17.png)
(ii)当
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231047426185595/SYS201310212310474261855005_DA/18.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231047426185595/SYS201310212310474261855005_DA/19.png)
OQ=t,OP=3t-2.
∴∠BOA=60°,∠AOC=30°,∴∠POQ=90°.
∴S△OPQ=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231047426185595/SYS201310212310474261855005_DA/20.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231047426185595/SYS201310212310474261855005_DA/21.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231047426185595/SYS201310212310474261855005_DA/22.png)
即S=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231047426185595/SYS201310212310474261855005_DA/23.png)
故当0<t<
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231047426185595/SYS201310212310474261855005_DA/24.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231047426185595/SYS201310212310474261855005_DA/25.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231047426185595/SYS201310212310474261855005_DA/26.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231047426185595/SYS201310212310474261855005_DA/27.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231047426185595/SYS201310212310474261855005_DA/28.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231047426185595/SYS201310212310474261855005_DA/29.png)
(2)D(
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231047426185595/SYS201310212310474261855005_DA/30.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231047426185595/SYS201310212310474261855005_DA/31.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231047426185595/SYS201310212310474261855005_DA/32.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231047426185595/SYS201310212310474261855005_DA/33.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231047426185595/SYS201310212310474261855005_DA/34.png)
(3)△BMN的周长不发生变化.理由如下:
延长BA至点F,使AF=OM,连接CF.(如图)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231047426185595/SYS201310212310474261855005_DA/images37.png)
又∵∠MOC=∠FAC=90°,OC=AC,
∴△MOC≌△FAC,
∴MC=CF,∠MCO=∠FCA.(10分)
∴∠FCN=∠FCA+∠NCA=∠MCO+∠NCA
=∠OCA-∠MCN
=60°,
∴∠FCN=∠MCN.
在△MCN和△FCN中,
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231047426185595/SYS201310212310474261855005_DA/35.png)
∴△MCN≌△FCN,
∴MN=NF.(11分)
∴BM+MN+BN=BM+NF+BN=BO-OM+BA+AF=BA+BO=4.
∴△BMN的周长不变,其周长为4.
点评:本题综合考查了等腰三角形、等边三角形的性质,全等三角形的判定.难度很大.注意分类讨论时,做到不重复,不遗漏.
![](http://thumb2018.1010pic.com/images/loading.gif)
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