∴
33AOPE4?,即=2， ABPB45PB315, 2315， 2∴PB?∴PO?BO?PB?8?∴P(0,∴k?315?8)， 2315?8. 2315－8)， 2当圆心P在线段OB延长线上时,同理可得P(0,－∴k=－315－8， 2315315－8或k=－－8时，以⊙P与直线l的两个交点和圆心P为顶点的三角形22∴当k=是正三角形.
4.
85.解：（1）1，；
5（2）作QF⊥AC于点F，如图3， AQ = CP= t，∴AP?3?t． 由△AQF∽△ABC，BC?52?32?4， 得
QFt4?．∴QF?t． 45514(3?t)?t， 25B ∴S?26即S??t2?t．
55（3）能．
①当DE∥QB时，如图4．
∵DE⊥PQ，∴PQ⊥QB，四边形QBED是直角梯形． 此时∠AQP=90°．
A Q D P E C AQAP由△APQ?∽△ABC，得， ?ACAB即
图4
B t3?t9． 解得t?． ?358Q D A P E C ②如图5，当PQ∥BC时，DE⊥BC，四边形QBED是直角梯形． 此时∠APQ =90°． 由△AQP?∽△ABC，得
AQAP， ?ABAC图5
B Q G 即
t3?t15
． 解得t?． ?538
545或t?． 214
（4）t?①点P由C向A运动，DE经过点C．
连接QC，作QG⊥BC于点G，如图6．
34PC?t，QC2?QG2?CG2?[(5?t)]2?[4?(5?t)]2．
55534由PC2?QC2，得t2?[(5?t)]2?[4?(5?t)]2，解得t?．
255②点P由A向C运动，DE经过点C，如图7．
34(6?t)2?[(5?t)]2?[4?(5?t)]2，t?45】
55146.解（1）①30，1；②60，1.5； ……………………4分
（2）当∠α=90时，四边形EDBC是菱形. ∵∠α=∠ACB=90，∴BC//ED.
∵CE//AB, ∴四边形EDBC是平行四边形. ……………………6分 在Rt△ABC中，∠ACB=90，∠B=60,BC=2,
∴∠A=30.
0
0
0
0
0
∴AB=4,AC=23. ∴AO=
1AC=3 . ……………………8分 20
在Rt△AOD中，∠A=30，∴AD=2. ∴BD=2. ∴BD=BC.
又∵四边形EDBC是平行四边形，
∴四边形EDBC是菱形 ……………………10分 7.解：（1）如图①，过是矩形
∴KH ················································································································ 1分 ?AD?3．则四边形ADHKA、D分别作AK?BC于K，DH?BC于H，
在Rt△ABK中，AK?ABsin45??42．2?4 2BK?ABcos45??422?4 ················································································· 2分 2在Rt△CDH中，由勾股定理得，HC?52?42?3
∴BC?BK?KH?HC?4?3?3?10····································································· 3分
A
D
A
D
N
B
C
K
H
B
G
M
C
（2）如图②，过D作DG∥AB交BC于G点，则四边形ADGB是平行四边形 ∵MN∥AB ∴MN∥DG ∴BG?AD?3
∴GC?10?3?7 ············································································································ 4分 由题意知，当M、N运动到t秒时，CN?t，CM?10?2t． ∵DG∥MN
∴∠NMC?∠DGC 又∠C?∠C
∴△MNC∽△GDC CNCM∴ ····················································································································· 5分 ?CDCGt10?2t即? 5750解得，t? ······················································································································ 6分
17（3）分三种情况讨论： ①当NC∴t ②当MN解法一： ?MC时，如图③，即t?10?2t
?10 ······························································································································ 7分 3D
?NC时，如图④，过N作NE?MC于E
N
A
A
D N
11MC??10?2t??5?t 22B C B C E M H M EC5?t在Rt△CEN中，cosc? ?NCt（图④） （图③） CH3又在Rt△DHC中，cosc??
CD55?t3∴?
t525解得t? ·························································································································· 8分
8由等腰三角形三线合一性质得EC?解法二：
∵∠C?∠C，?DHC??NEC?90? ∴△NEC∽△DHC
NCEC ?DCHCt5?t即? 5325∴t? ······························································································································ 8分
811③当MN?MC时，如图⑤，过M作MF?CN于F点.FC?NC?t
22∴
解法一：（方法同②中解法一）
1tFC32cosC??? MC10?2t560解得t?
17解法二：
∵∠C?∠C，?MFC??DHC?90? ∴△MFC∽△DHC ∴
B
A
D
N F
H M
C
（图⑤）
FCMC ?HCDC1t10?2t即2?
3560∴t?
17102560综上所述，当t?、t?或t?时，△MNC为等腰三角形 ························· 9分
38178.解（1）如图1，过点E作EG?BC于点G． ······························ 1分
∵E为AB的中点，
A D
1∴BE?AB?2．
2F E
在Rt△EBG中，∠B?60?，∴∠BEG?30?． ················· 2分
1∴BG?BE?1 ，EG?22?12?3．B C 2G
即点E到BC的距离为3． ····················································· 3分 （2）①当点N在线段AD上运动时，△PMN的形状不发生改变． ∵PM∴PM∥EG． ?EF，EG?EF，图1
∵EF∥BC，∴EP?GM，PM?EG?同理MN3．
·················································································································· 4分 ?AB?4．如图2，过点P作PH?MN于H，∵MN∥AB， ∴∠NMC?∠B?60?，∠PMH?30?． N
A
D
∴PH?13PM?． 22B
E
P H F
3∴MH?PMcos30??．
235则NH?MN?MH?4??．
222G M 图2
C
?5??3?22在Rt△PNH中，PN?NH?PH????? ?7．???22????∴△PMN的周长=PM?PN?MN?23?7?4． ······················································· 6分
②当点N在线段DC上运动时，△PMN的形状发生改变，但△MNC恒为等边三角形．
当PM ?PN时，如图3，作PR?MN于R，则MR?NR．3类似①，MR?．
2∴MN?2MR?3 ···················································································································· 7分 ．∵△MNC是等边三角形，∴MC?MN?3 ．此时，x?EP?GM?BC?BG?MC?6?1?3?2． ·················································· 8分 A E
P R
B
G
M
图3
C
B
G
图4
M
D N F
E A
P D F N C
B E A
D F（P） N
G
图5
M
C
当MP?MN时，如图4，这时MC?MN?MP?此时，x?EP?GM?6?1?3?5?3． 当NP?NM时，如图5，∠NPM3．
?∠PMN?30?．则∠PMN?120?，又∠MNC?60?， ∴∠PNM?∠MNC?180?．
因此点P与F重合，△PMC为直角三角形． ∴MC?PMtan30??1 ．此时，x?EP?GM?6?1?1?4． 综上所述，当x?2或4或
?5?3?时，△PMN为等腰三角形． ································ 10分
9解：（1）Q（1，0） ······················································································································ 1分 点P运动速度每秒钟1个单位长度． ····························································································· 2分 （2） 过点B作BF⊥y轴于点F，BE⊥x轴于点E，则BF＝8，OF?BE?4． ∴AF?10?4?6．
在Rt△AFB中，AB?82?62?10 3分 过点C作CG⊥x轴于点G，与FB的延长线交于点H． ∵?ABC?90?,AB?BC ∴△ABF≌△BCH． ∴BH?AF?6,CH?BF?8． ∴OG?FH?8?6?14,CG?8?4?12．
∴所求C点的坐标为（14，12）． 4分 （3） 过点P作PM⊥y轴于点M，PN⊥x轴于点N， 则△APM∽△ABF． ∴
APAMMPtAMMP??． ???． ABAFBF1068ONQyDCAMFPHGxBE3434 ∴AM?t，PM?t． ∴PN?OM?10?t,ON?PM?t．
5555设△OPQ的面积为S（平方单位）
13473∴S??(10?t)(1?t)?5?t?t2（0≤t≤10） ····································································· 5分
251010
初中数学几何的动点问题专题练习附答案版
