∴ OG＝OC. 即OG是⊙O的半径.
∴ 点G在⊙O上. 即直线CF与圆有两个交点. ······················ 9分 ∴ 此时CF不是⊙O的切线.
解3：不正确. ················································································ 5分
连结OC.
当 ∠CBA＝70°时， ································································ 6分 ∴ ∠OCB＝70°. ··································································· 7分 又∵∠BCF＝30°， ∴∠FCO＝100°， ··································································· 8分 ∴ CO与FC不垂直. ······························································ 9分 ∴ 此时CF不是⊙O的切线.
24．（本题满分10分）
75
（1）解：点Ｃ(，) 是线段AB的“邻近点”. ······································ 1分
22
7575
∵－1＝， ∴点Ｃ(，)在直线y＝x－1上. ······················· 2分 2222∵点A的纵坐标与点B的纵坐标相同， ∴ AB∥x轴. ·········································································· 3分 755∴Ｃ(，) 到线段AB的距离是3－，
222
51
∵3－＝＜1， ········································································ 4分
22
75
∴Ｃ(，)是线段AB的“邻近点”.
22
（2）解1：∵点Q(m，n)是线段AB的“邻近点”，
∴ 点Q(m，n)在直线y＝x－1上，
∴ n＝m－1. ··········································································· 5分 ① 当m≥4时，········································································ 6分 有n＝m－1≥3. 又AB∥x轴，
∴ 此时点Q(m，n)到线段AB的距离是n－3. ·························· 7分 ∴0≤n－3＜1.
∴ 4≤m＜5. ········································································· 8分 ② 当m≤4时，········································································ 9分 有n＝m－1≤3. 又AB∥x轴，
∴ 此时点Q(m，n)到线段AB的距离是3－n.
∴0≤3－n＜1. ∴ 3＜m≤4. ········································································10分
综上所述， 3＜m＜5.
解2：∵点Q(m，n)是线段AB的“邻近点”，
∴ 点Q(m，n)在直线y＝x－1上，
∴ n＝m－1. ··········································································· 5分 又AB∥x轴，
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∴ Q(m，n)到直线AB的距离是n－3或3－n， ························· 6分 ① 当0≤n－3＜1时， ······························································ 7分 即 当0≤m－1－3＜1时， 得 4≤m＜5. ··········································································· 8分 ② 当0≤3－n＜1时， ······························································ 9分 有0≤3－(m－1)＜1时， 得 3＜m≤4. ········································································10分
综上所述，3＜m＜5.
25．（本题满分10分） （1）解1：连结PO ,
∵ PE＝PF，PO＝PO， PE⊥AC、PF⊥BD， ∴ Rt△PEO≌Rt△PFO.
∴ ∠EPO＝∠FPO. 在Rt△PEO中，
??1分 ??2分
AEPFDOCBEO3tan∠EPO＝＝， ??3分
PE3
∴ ∠EPO＝30°. ∴ ∠EPF＝60°. ···································································· 4分
解2：连结PO ,
在Rt△PEO中， ······································································· 1分
PO＝3＋1 ＝2.
EO1
∴ sin∠EPO＝＝. ····························································· 2分
PO2
∴ ∠EPO＝30°. ··································································· 3分
PF3在Rt△PFO中，cos∠FPO＝＝，∴∠FPO＝30°.
PO2
∴ ∠EPF＝60°. ···································································· 4分
解3：连结PO ,
∵ PE＝PF，PE⊥AC、PF⊥BD，垂足分别为E、F，
∴ OP是∠EOF的平分线. ∴ ∠EOP＝∠FOP. ································································ 1分 在Rt△PEO中， ······································································· 2分 PE
tan∠EOP＝＝3 ·································································· 3分
EO∴ ∠EOP＝60°，∴ ∠EOF＝120°. 又∵∠PEO＝∠PFO＝90°，
∴ ∠EPF＝60°. ···································································· 4分
（2）解1：∵点P是AD的中点，∴ AP＝DP.
又∵ PE＝PF，∴ Rt△PEA≌Rt△PFD. ∴ ∠OAD＝∠ODA.
∴ OA＝OD. ··········································································· 5分 ∴ AC＝2OA＝2OD＝BD. ∴□ABCD是矩形. ·································································· 6分
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∵ 点P是AD的中点，点F是DO的中点，
∴ AO∥PF. ············································································ 7分 ∵ PF⊥BD，∴ AC⊥BD. ∴□ABCD是菱形. ·································································· 8分 ∴□ABCD是正方形. ······························································ 9分 ∴ BD＝2BC.
332∵ BF＝BD，∴BC＋32－4＝BC.
44
解得，BC＝4. ········································································10分
解2：∵ 点P是AD的中点，点F是DO的中点，
∴ AO∥PF. ············································································ 5分 ∵ PF⊥BD，∴ AC⊥BD.
∴□ABCD是菱形. ·································································· 6分 ∵ PE⊥AC，∴ PE∥OD.
PAD∴ △AEP∽△AOD. ∴
EPAP1＝＝. ODAD2
EOF∴ DO＝2PE.
∵ PF是△DAO的中位线， ∴ AO＝2PF. ∵ PF＝PE，
BC∴ AO＝OD. ············································································ 7分 ∴ AC＝2OA＝2OD＝BD. ∴ □ABCD是矩形. ································································ 8分 ∴ □ABCD是正方形. ····························································· 9分 ∴ BD＝2BC.
332∵ BF＝BD，∴BC＋32－4＝BC.
44
解得，BC＝4. ········································································10分
解3：∵点P是AD的中点，∴ AP＝DP.
又∵ PE＝PF， ∴ Rt△PEA≌Rt△PFD. ∴ ∠OAD＝∠ODA.
∴ OA＝OD. ··········································································· 5分 ∴ AC＝2OA＝2OD＝BD.
∴□ABCD是矩形. ·································································· 6分 ∵点P是AD的中点，点O是BD的中点，连结PO. ∴PO是△ABD的中位线，
∴ AB＝2PO. ·········································································· 7分 ∵ PF⊥OD，点F是OD的中点， ∴ PO＝PD. ∴ AD＝2PO.
∴ AB＝AD. ············································································ 8分 ∴□ABCD是正方形. ······························································ 9分 ∴ BD＝2BC.
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332∵ BF＝BD，∴BC＋32－4＝BC.
44
解得，BC＝4. ········································································10分
解4：∵点P是AD的中点，∴ AP＝DP.
又∵ PE＝PF， ∴ Rt△PEA≌Rt△PFD.
∴ ∠OAD＝∠ODA. ∴ OA＝OD. ··········································································· 5分 ∴ AC＝2OA＝2OD＝BD. ∴□ABCD是矩形. ·································································· 6分 ∵PF⊥OD，点F是OD的中点，连结PO. ∴PF是线段OD的中垂线， 又∵点P是AD的中点，
1
∴PO＝PD＝BD······································································· 7分
2∴△AOD 是直角三角形, ∠AOD＝90°. ································ 8分 ∴□ABCD是正方形. ······························································ 9分 ∴ BD＝2BC.
332∵ BF＝BD，∴BC＋32－4＝BC.
44
解得，BC＝4. ········································································10分
26．（本题满分12分） k
（1）解：∵点A(1，c)和点B (3，d )在双曲线y＝2（k2＞0）上，
x
∴ c＝k2＝3d ··········································································· 1分 ∵ k2＞0， ∴ c＞0，d＞0.
A(1，c)和点B (3，d )都在第一象限. ∴ AM＝3d. ············································································ 2分 过点B作BT⊥AM，垂足为T. ∴ BT＝2. ··············································································· 3分 TM＝d.
∵ AM＝BM， ∴ BM＝3d.
在Rt△BTM中，TM 2＋BT2＝BM2， ∴ d2＋4＝9d2， ∴ d＝点B(3，
2. 2
2) . ········································································· 4分 2
k
（2）解1：∵ 点A(1，c)、B(3，d)是直线y＝k1x＋b与双曲线y＝2（k2＞0）的交点，
x
∴ c＝k2,，3d＝k2，c＝k1＋b，d＝3k1＋b. ······························ 5分
14
∴ k1＝－k2，b＝k2.
33
∵ A(1，c)和点B (3，d )都在第一象限，∴ 点P在第一象限.
- 14 -
∴
PEk1x＋b＝ NEk2x kb＝1x2＋x k2k2
14
＝－x2＋x. ································································· 6分
33
PE
∵ 当x＝1，3时，＝1；
NE又∵当x＝2时， ∴ 1≤
PE4的最大值是. NE3
PE4≤. ······································································· 7分 NE3
∴ PE≥NE. ············································································ 8分 ∴
PNPE14
＝－1＝－x2＋x－1. ··············································· 9分 NENE33
∴ 当x＝2时，
PN1的最大值是. ······································································10分 NE31由题意，此时PN＝，
2
3
∴ NE＝. ··············································································· 11分
23
∴ 点N(2，) . ∴ k2＝3.
2
3∴ y＝. ··················································································12分
x
解2：∵ A(1，c)和点B (3，d )都在第一象限，∴ 点P在第一象限.
∵
PEk1x＋bk12b＝ ＝x＋x， NEk2k2k2
x
PE
当点P与点A、B重合时，＝1，
NEPE
即当x＝1或3时，＝1.
NE
kb9k3b
∴ 有 1＋＝－1， 1＋＝－1. ···································· 5分
k2k2k2k214
解得，k1＝－k2，b＝k2.
33∴
PE14
＝－x2＋x. ································································· 6分 NE33
∵ k2＝－3k1，k2＞0，∴ k1＜0. k23k1∵ PE－NE＝k1x＋b－＝k1x－4k1＋ x x
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