∴E坐标为（m, ∵E在反比例y=∴3m=23． m． ··········· 6分 3m）23图像上， x∴m1=2， m2=-2(舍去)．
∴OE=22，EA=4?22，EG=2 ··················································· 7分 ∵4?22＜2，
∴EA＜EG．
∴以E为圆心，EA垂为半径的圆与y轴相离． ··································· 8分
（3） 存在．······························································································ 9分 方法一：假设存在点F，使AE⊥FE．过点F作FC⊥OB于点 C，过E点作EH⊥OB于点H．
设BF= x.
∵△AOB是等边三角形，
∴AB=OA=OB=4，∠AOB=∠ABO=∠A=60?．
1∴BC=FB·cos∠FBC＝x
2FC=FB·sin∠FBC=3x 21∴AF=4－x，OC=OB －BC=4－x
2∵AE⊥FE
1∴AE=AF·cos∠A=2－x
21∴OE=O A－AE=x+2
21∴OH=OE·cos∠AOB=x?1，
4EH=OE·sin∠AOB=3x?3 43311x?3)，F(4－x，x) ·∴E(x?1,··········································· 11分 4242
∵E、F都在双曲线y=
k的图象上， x3311x?3）＝（4－x）x ∴（x?1）（42424解得 x1=4，x2=． ································································· 12分
5[来源学+科+网]
当BF=4时，AF=0，当BF=
BF不存在，舍去． AF416BF1时，AF=，······················································ 13分 ?． ·55AF4方法二：假设存在点F，使AE⊥FE．过E点作EH⊥OB于 H.
∵△AOB是等边三角形，设E(m,
3m)，则OE=2m, AE=4－2m．
∴AB=OA=AB=4，∠AOB=∠ABO=∠A=60?． ∵Cos?A?AE1?， AF2∴AF=2AE=8－4m，FB=4m－4．
∴FC=FB·sin∠FBC=23m－23， BC=FB·cos∠FBC=2m－2． ∴OC=6－2m
∴F(6－2m, 23m－23)． ······························································· 11分 ∵E、F都在双曲线y=
k上， x∴m·3m=(6－2m)(23m－23) 化简得：5m2-16m+12=0 解得： m1=2，m2=
6． ····································································· 12分 5当m＝2时，AF=8－4m=0，BF=4，F与B重合，不合题意，舍去．
616164当m＝时，AF=8－4m＝BF=4－=．
55，55 ∴BF:FA?1:4． ·········································································· 13分
1125. (1)在菱形ABCD中， OA?AC?40,OD?BD?3022∵AC⊥BD
∴AD=302?402=50.
∴菱形ABCD的周长为200. ······················4分 (2) 过点M作MP⊥AD，垂足为点P. ①当0＜t≤40 ∵Sin?OAD?3∴MP=t
51∴S??DN?MP
232t ··························································································· 6分 10②当40 MPOD3?? AMAD5 = ∵Sin?ADO=MPAO? MDAD4∴MP=(70?t) 5∴S?DMN?1DN?MP 222··································································· 8分 ??t2?28t??(t?35)2?490 · 55?32t,0?t?40??10∴S?? 2??(t?35)2?490,40?t?50??5当0＜t≤40时，S随t的增大而增大，当t＝40时，最大值为480. 当40＜t≤50时，S随t的增大而减小，当t＝40时，最大值为480. 综上所述，S的最大值为480. ······································································ 9分 （3）存在2个点P，使得∠DPO=∠DON.····················································· 10分 方法一：过点N作NF⊥OD于点F， 401203090??24DF=ND?Cos?ODA?30???18.则NF?ND?Sin?ODA?30?505505， ∴OF=12，∴tan?NOD?NF24??2 ·················································· 11分 OF12作?NOD的平分线交NF于点G，过点G作GH⊥ON于点H. 1111∴S?ONF?OF?NF?S?OGN?S?OFG?OF?FG?ON?GH?(OF?ON)?FG 2222OF?NF12?2424??∴FG=OF?ON12?1251?5 24GF1?52tan?GOF???∴ OF121?5设OD中垂线与OD的交点为K，由对称性可知： 11∴?DPK??DPO??DON??FOG ·················································· 12分 22DK152tan?DPK???∴PKPK1?5 15(5?1)∴PK= ················································································ 13分 2根据菱形的对称性可知，在线段OD的下方存在与点P关于OD轴对称的点P'. 15(5?1). ·········································· 14分 2方法二：如图，作ON的垂直平分线，交EF于点I，连结OI，IN. ∴存在两个点P到OD的距离都是过点N作NG⊥OD，NH⊥EF,垂足分别为G，H. 当t=30时，DN=OD=30，易知△DNG∽△DAO， ∴即 DNNGDG??． DAAOOD30NGDG??．504030[来源学科网ZXXK] ∴NG=24,DG=18． ············································································ 10分 ∵EF垂直平分OD， ∴OE= ED＝15，EG=NH=3． ······························································ 11分 设OI=R，EI=x,则 在Rt△OEI中，有R2=152+x2 ① 在Rt△NIH中，有R2=32+(24-x)2 ② 15?x??2?由①、②可得：? 155?R???2∴PE=PI+IE=15?155． ·································································· 13分 2