说明:未注明自变量的取值范围不扣分．
∵a??473<0 ∴当t??时， △OPQ的面积最大． ········································ 6分 ?36102?(?)104710 此时P的坐标为（（4） 当 t?9453，） ． ··································································································· 7分 15105295或t?时， OP与PQ相等． ········································································ 9分 31310.解：（1）正确． ··································································· （1分）
·· （2分） ?EC，连接ME． ·D A
?BM?BE．??BME?45°，??AME?135°．
F M CF是外角平分线，
??DCF?45°，
B E C G ??ECF?135°．
??AME??ECF．
?AEB??BAE?90°，?AEB??CEF?90°， ??BAE??CEF．
····························································································· （5分） ?△AME≌△BCF（ASA）． ·
··························································································································· （6分） ?AE?EF． ·
（2）正确． ··········································································· （7分） 证明：在BA的延长线上取一点N．
················································· （8分） ?CE，连接NE． ·N F ?BN?BE． D A ??N??PCE?45°． 四边形ABCD是正方形， ?AD∥BE．
B C E G
??DAE??BEA． ??NAE??CEF．
··························································································· （10分） ?△ANE≌△ECF（ASA）． ·
?AE?EF． （11分）
11.解（Ⅰ）如图①，折叠后点B与点A重合， 则△ACD≌△BCD. 设点C的坐标为使AN证明：在AB上取一点M，使AM?0，m??m?0?.
则BC?OB?OC?4?m. 于是AC?BC?4?m.
在Rt△AOC中，由勾股定理，得AC?OC?OA， 即?4?m?2222?m2?22，解得m?3. 2?3??点C的坐标为?0，?. ···················································································································· 4分
2??（Ⅱ）如图②，折叠后点B落在OA边上的点为B?, 则△B?CD≌△BCD.
由题设OB??x，OC?y， 则B?C?BC?OB?OC?4?y，
在Rt△B?OC中，由勾股定理，得B?C?OC?OB?.
222??4?y??y2?x2，
1······························································································································· 6分 ??x2?2 ·
8由点B?在边OA上，有0≤x≤2，
1? 解析式y??x2?2?0≤x≤2?为所求.
8即y2? 当0≤x≤2时，y随x的增大而减小，
3································································································ 7分 ?y的取值范围为≤y≤2. ·
2（Ⅲ）如图③，折叠后点B落在OA边上的点为B??，且B??D∥OB. 则?OCB????CB??D. 又?CBD??CB??D，??OCB????CBD，有CB??∥BA. ?Rt△COB??∽Rt△BOA. OB??OC有，得OC?2OB??. ····························································································· 9分 ?OAOB在Rt△B??OC中，
设OB???x0?x?0?，则OC?2x0.
1??x20?2，
8由（Ⅱ）的结论，得2x0解得x0??8?45．x0?0，?x0??8?45. ?点C的坐标为0，85?16. ······························································································· 10分
12解：方法一：如图（1-1），连接BM，EM，BE．
F M A D 由题设，得四边形ABNM和四边形FENM关于直线MN对称． ∴MN垂直平分BE．∴BM ······················································ 1分 ?EM，BN?EN．?90°,AB?BC?CD?DA?2． ∵四边形ABCD是正方形，∴?A??D??CE
CE1?，?CE?DE?1． ∵设BN?x，则NE?x， NC?2?x．CD2B C N
222（1-1CE 在Rt△CNE中，NE?CN?图．） ∴x2??55，即BN?． ···························································· 3分 44 在Rt△ABM和在Rt△DEM中，
AM2?AB2?BM2，
解得x???2?x??12．2DM2?DE2?EM2，
?AM2?AB2?DM2?DE2． ······················································································ 5分
2222 设AM?y，则DM?2?y，∴y?2??2?y??1．
11 解得y?，即AM?． ·································································································· 6分
44AM1 ∴························································································································ 7分 ?．BN55 方法二：同方法一，BN?． ···························································································· 3分
4 如图（1－2），过点N做NG∥CD，交AD于点G，连接BE．
F
G M ∵AD∥BC，∴四边形GDCN是平行四边形．A D
∴NG?CD?BC．
同理，四边形ABNG也是平行四边形．∴AGE ?BN ∵MN ?BE，??EBC??BNM?90°． 在△BCE与△NGM中B C
5 ?．4NG??EBC??M，?，G ?BC?N∴△BCE≌△NGM，EC?MG． ···································· ５分
??C??NGM9?0°．?∵AMN
图（1-2）
51 ··········································································· 6分 ?AG?MG，AM=?1?．44AM1∴ ······················································································································ 7分 ?．BN5类比归纳
249（或）；； ··························································································10分 251017n?1联系拓广
?n?1?2n2m2?2n?1 ························································································································12分 22nm?1解1：依题意,得AQ=t,BP=2t,QD=16-t。过点Q作QF⊥BP,又 ∵AQ‖BF, ∴∠ABP=90° ∴四边形AQFB是矩形
∴AQ=BF=t ∵BP=2t ∴FP=t, ∴在Rt△QFP中,QP=√(122+t2) 又∵QD=QP=PD ∴√(122+t2)=16-t ∴122+t2=162-2*16*t+t2 ∴解得:t=7/2不知道对不对,错了别怪我。 解2：如图所示，
:这P作PE垂直AD于E,垂足为E点,则ABPE为矩形.PE=AB=12;AE=BP (1).s=1/2×AB×DQ=1/2×12×(AD-AQ)=6×(16-t)=96-6t;
(2).当 BC-2t=21-2t=PC=DQ=AD-t=16-t,即t=5时,四边形PCDQO为平形四边
形.
(3).①QE=AE-AQ=BP-AQ=2t-t=t,而ED=AD-AE=16-BP=16-2t;当QE=ED时,PE为QD的垂直平分线时,PQ=PD,而此时t=16-2t; t=16/3;所以当t=16/3时,PD=PQ;
.②在
Rt△PEQ
中,PE=AB=12;
EQ=AE-AQ=PB-AQ=2t-t=t;
PQ2=QE2+PE2=t2+122;
QD2=(AD-AQ)2=(16-t)2; 所以当t2+122=(16-t)2,即:t=3.5时,DQ=PQ; 解：因为∠C=90°，∠CBA=30°，BC=20√3 所以可求出AB＝40
如图，圆心从A向B的方向运动时，共有三个位置能使此圆与直线AC或直线BC相切
当圆心在O1点时，设切点为P
显然PO1＝6，∠APO1＝90°，∠AO1P＝30° 所以AO1＝4√3
因为圆O以2个单位长度/秒的速度向右运动
所以当t1＝4√3/2＝2√3（秒）时，圆O与直线AC相切 当圆心在O2点时，设切点为Q
显然QO2＝6，∠BQO2＝90°，∠QBO2＝30° 所以BO2＝12，AO2＝40－12＝28
因为圆O以2个单位长度/秒的速度向右运动 所以当t2＝28/2＝14（秒）时，圆O与直线BC相切 当圆心在O3点时，设切点为R
显然RO3＝6，∠BRO3＝90°，∠RBO3＝30° 所以BO3＝12，AO3＝40＋12＝52
因为圆O以2个单位长度/秒的速度向右运动 所以当t3＝52/2＝26（秒）时，圆O与直线BC相切
综上所述，当圆O运动2√3秒、14秒、26秒时与△ABC的一边所在的直线相切.
初中数学几何的动点问题专题练习附答案版
