初三数学几何的动点问题专题练习(3)

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()()00m m >，.

则4BC OB OC m =-=-.

于是4AC BC m ==-.

在Rt AOC △中，由勾股定理，得222AC OC OA =+，

即()22242m m -=+，解得32

m =. ∴点C 的坐标为302?? ???

，. ··················································································· 4分 （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 '==，， A D F C G E B M A D F G E B N

17 则4B C BC OB OC y '==-=-，

在Rt B OC '△中，由勾股定理，得222

B C OC OB ''=+. ()2

224y y x ∴-=+， 即2128

y x =-

+ ···························································································· 6分 由点B '在边OA 上，有02x ≤≤，

∴ 解析式2128y x =-+()02x ≤≤为所求. ∴ 当02x ≤≤时，y 随x 的增大而减小，

y ∴的取值范围为322

y ≤≤. ····································································· 7分 （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 Rt COB BOA ''∴△∽△. 有OB OC OA OB

''=，得2OC OB ''=. ·································································· 9分 在Rt B OC ''△中，

设()00OB x x ''=>，则02OC x =. 由（Ⅱ）的结论，得2001228

x x =-+，

解得000808x x x =-±>∴=-+，∴点C

的坐标为()

016. ··································································· 10分

12解：方法一：如图（1-1），连接BM EM BE ，，．

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 EM BN EN ==，． ···································· 1分 ∵四边形ABCD 是正方形，∴902A D C AB BC CD DA ∠=∠=∠=====°,． ∵112

CE CE DE CD =∴==，．设BN x =，则NE x =，2NC x =-． N 图（1-1） A B C E F M

18 在Rt CNE △中，222NE CN CE =+．

∴()22221x x =-+．解得54x =，即54

BN =． ········································· 3分 在Rt ABM △和在Rt DEM △中，

222AM AB BM +=，

222DM DE EM +=，

∴2222AM AB DM DE +=+．

····························································· 5分 设AM y =，则2DM y =-，∴()2222221y y +=-+．

解得14y =，即14

AM =． ····································································· 6分 ∴15

AM BN =． ····················································································· 7分 方法二：同方法一，54

BN =． ································································ 3分 如图（1－2），过点N 做NG CD ∥，交AD 于点G ，连接BE ．

∵AD BC ∥，∴四边形GDCN 是平行四边形．

∴NG CD BC ==．

同理，四边形ABNG 也是平行四边形．∴54

AG BN ==． ∵90MN BE EBC BNM ⊥∴∠+∠=，°．

90NG BC MNG BNM EBC MNG ⊥∴∠+∠=∴∠=∠ ，°，．

在BCE △与NGM △中

90EBC MNG BC NG C NGM ∠=∠??=??∠=∠=?

，，°．∴BCE NGM EC MG =△≌△，． ························· ５分 ∵114AM AG MG AM =--=5，=

．4 ····················································· 6分 ∴

15

AM BN =． ··················································································· 7分 类比归纳 25（或410）；917； ()2211

n n -+ ································································· 10分 联系拓广

N 图（1-2）

A B C D E F M G

19 2222211

n m n n m -++ ······················································································ 12分