25．解：

（1）过点O作OH⊥CD，垂足为点H，联结OC．

在Rt△POH中，∵sinP＝，PO?6，∴OH?2．········································ （1分） ∵AB＝6，∴OC＝3． ···························································································· （1分） 由勾股定理得 CH?5． ······················································································· （1分）

∵OH⊥DC，∴CD?2CH?25． ································································· （1分） （2）在Rt△POH中，∵sinP＝，PO ＝m，∴OH＝21313m． ··································· （1分） 3?m?在Rt△OCH中，CH2＝9???． ····································································· （1分）

?3?m??在Rt△O1CH中，CH＝36??n??． ···························································· （1分）

3??223n2?81m???m?可得 36??n??＝9???，解得m＝． ········································ （2分）

3?2n??3?（3）△POO1成为等腰三角形可分以下几种情况：

● 当圆心O1、O在弦CD异侧时

223n2?81①OP＝OO1，即m＝n，由n＝解得n＝9． ········································ （1分）

2n即圆心距等于⊙O、⊙O1的半径的和，就有⊙O、⊙O1外切不合题意舍去．（1分） ②O1P＝OO1，由(n?m2m2)?m2?()＝n， 332293n2?81m＝nnn＝15． ·，即，········································ （1分） ＝解得解得

3352n81?3n2● 当圆心O1、O在弦CD同侧时,同理可得 m＝．

2n981?3n2∵?POO1是钝角，∴只能是m?n，即n＝，解得n＝5． ··········· （2分）

52n995或15．综上所述，n的值为55

青浦区

25.（本题满分14分，第（1）小题4分，第（2）小题6分，第（3）小题4分）

如图9-1，已知扇形MON的半径为2，∠MON=90o，点B在弧MN上移动，联结BM，作OD?BM，垂足为点D，C为线段OD上一点，且OC=BM，联结BC并延长交半径OM于点A，设OA= x，∠COM的正切值为y．

（1）如图9-2，当AB?OM时，求证：AM =AC； （2）求y关于x的函数关系式，并写出定义域； （3）当△OAC为等腰三角形时，求x的值.

25．解：（1）∵OD⊥BM，AB⊥OM，∴∠ODM =∠BAM =90°． ······································· （1分）

∵∠ABM +∠M =∠DOM +∠M，∴∠ABM =∠DOM． ································· （1分） ∵∠OAC=∠BAM，OC =BM，

∴△OAC≌△ABM， ························································································· （1分） ∴AC =AM． ······································································································· （1分） （2）过点D作DE//AB，交OM于点E． ······························································· （1分）

∵OB＝OM，OD⊥BM，∴BD＝DM． ···························································· （1分） ∵DE//AB， ∴

OCADMOCDMOMNBNBN

A图9-1 图9-2

备用图

MDME，∴AE＝EM， ?DMAE1∵OM=2，∴AE＝······························································· （1分） 2?x． ·

2??∵DE//AB， ∴

OAOC2DM， ················································································ （1分） ??OEODOD∴

DMOA， ?OD2OEx．（0?x?2） ······································································ （2分）

x?2∴y?（3）（i） 当OA=OC时， ∵DM?111BM?OC?x， 222OM2?DM2?2?在Rt△ODM中，OD?DM12x．∵y?， 4OD1x14?2?14?2x2∴．解得x?，或x?（舍）．（2分） ?221x?22?x24（ii）当AO=AC时，则∠AOC =∠ACO，

∵∠ACO >∠COB，∠COB =∠AOC，∴∠ACO >∠AOC，

∴此种情况不存在． ························································································ （1分） （ⅲ）当CO=CA时，

则∠COA =∠CAO=?，

∵∠CAO >∠M，∠M=90???，∴?>90???，∴?>45?，

∴?BOA?2??90?，∵?BOA?90?，∴此种情况不存在． ··········· （1分）

松江区

25．（本题满分14分，第（1）小题4分，第（2）小题每个小题各5分）

如图，已知Rt△ABC 中，∠ACB=90°，BC=2，AC=3，以点C为圆心、CB为半径的圆交AB于点D，过点A作AE∥CD，交BC延长线于点E. （1）求CE的长；

（2）P是 CE延长线上一点，直线AP、CD交于点Q.

① 如果△ACQ ∽△CPQ，求CP的长；

② 如果以点A为圆心，AQ为半径的圆与⊙C相切，求CP的长.

D A D A B

C

E B

C

E

25．（本题满分14分，第（1）小题4分，第（2）小题每个小题各5分） 解：（1）∵AE∥CD

∴BCDCBE?AE…………………………………1分 ∵BC=DC

∴BE=AE …………………………………1分 设CE=x 则AE=BE=x+2 ∵ ∠ACB=90°， ∴AC2?CE2?AE2

即9?x2?(x?2)2………………………1分

∴x?54 即CE?54…………………………………1分

（2）①

∵△ACQ ∽△CPQ，∠QAC>∠P ∴∠ACQ=∠P…………………………………1分 又∵AE∥CD ∴∠ACQ=∠CAE

∴∠CAE=∠P………………………………1分 ∴△ACE ∽△PCA，…………………………1分 ∴AC2?CE?CP…………………………1分

即32?54?CP ∴CP?365 ……………………………1分

A D B

C

E (第25题图)

Q A D B

C E P