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40 BI TON HNH HC Oxy K2pi BI
Bi 1. Trong mt phng Oxy, cho hnh vung ABCD, bit nh B(3; 4) v phng trnh cnh ACl x y+ 2 = 0. Tm ta cc nh A, C, D, vi im A c honh b hn 3.Bi 2. Tm ta 3 nh ca tam gic ABC bit ng phn gic trong gc A, trung tuyn k tB, ng cao k t C ln lt l d1 : x 3y+ 1 = 0, d2 : 3x+ y+ 3 = 0, d3 : x+ y+ 3 = 0.Bi 3. Trong mt phng vi h trc ta Oxy, cho t gic ABCD ngoi tip ng trn (C) :(x+ 2)2 + (y 3)2 = 9, c phng trnh ng thng BD : 4x y+ 6 = 0. Cc cnh AB, AD lnlt tip xc vi (C) ti E(2; 6) v F
(22
5;65
). Tm ta cc nh ca t gic ABCD
Bi 4. Trong mt phng vi h trc ta Oxy, cho t gic ABCD ngoi tip ng trn (C) :(x+ 2)2 + (y 3)2 = 9, c nh A (8, 6) v phng trnh ng thng BD : 4x y+ 6 = 0.Tm ta cc nh cn li ca t gic ABCD
Bi 5. Trong mt phng Oxy, cho hnh thang cn ABCD c din tch bng452, y ln CD nm
trn ng thng x 3y 3 = 0. Bit hai ng cho AC, BD vung gc vi nhau ti I(2; 3). Vitphng trnh ng thng cha cnh BC, bit im C c honh dng.
Bi 6. Trong mt phng Oxy, cho tam gic ABC cn ti nh A(2; 2). ng thng i qua trungim ca cc cnh AB, AC c phng trnh x+ y 6 = 0 v im D(2; 4) nm trn ng caok t nh B. Xc nh ta cc nh B v C ca tam gic ABC.
Bi 7. Trong mt phng to Oxy, cho hnh thang cn ABCD (AB//CD v AB < CD), bitcc nh A(0; 2) v D(2;2). Giao im I ca hai ng cho AC v BD thuc ng thngd : x+ y 4 = 0. Tm to cc nh B v C, bit AID = 45o.Bi 8. Trongmt phng to Oxy, cho hnh ch nht ABCD c im A(3; 1) v C thuc ngthng d : x 2y 5 = 0. im E l giao im th hai ca DC v ng trn tm B bn knh BD(E khc D). Hnh chiu vung gc ca D xung BE l im N (6;2). Tm ta im B v C.Bi 9. Trong mt phng Oxy cho tam gic ABC nhn c M(1;2),N(2; 2) v P(1; 2), ln ltchn ng cao k t 3 nh. Tm ta 3 nh trn
Bi 10. Trong mt phng Oxy cho hnh vung ABCD c tm I(12;92
), cc nh A, B ln lt
thuc cc ng thng d1 : 3x+ 4y 8 = 0; d2 : 3x+ 4y 1 = 0. Tm ta cc nh ca hnhvung.
Bi 11. Cho hnh vung ABCD. M l trung im ca BC, phng trnh ng thng DM :x y 2 = 0, im C(3;3) v im A thuc d : 3x+ y 2 = 0. Xc nh to A, B,D.Bi 12. Trong mt phng ta Oxy, cho ng trn T : (x 4)2 + y2 = 40. Vit phng trnhng thng i qua gc ta v ct ng trn (T) ti hai im A, B sao cho AB = 4BO
Bi 13. Trong mt phng ta Oxy, cho im A(0;2) v ng thng d : x y+ 2 = 0. Tmta im M thuc d sao cho ng cao AH v ng trung tuyn ON ca tam gic OAM c di bng nhau.
Bi 14. Trong mt phng ta Oxy, cho tam gic ABC vung ti A. Bit A(1; 4), B(1;4) vng thng BC i qua im M(2;
12). Tm ta nh C.
1
Bi 15. Trong mt phng Oxy, cho tam gic ABC c A(1; 4), trc tm H(3;1), tm ng trnngoi tip tam gic I(2; 0). Xc nh im B,C bit honh ca C dng.Bi 16. Trong mt phng ta Oxy, cho ng trn (C) : (x 4)2 + (y + 3)2 = 4 v ngthng d : x+ y 1 = 0. Xc nh ta cc nh ca hnh vung ABCD ngoi tip ng trn(C), bit A thuc ng thng d.
Bi 17. Cho hnh vung ABCD .Gi M l trung im BC, N l im trn cnh CD sao cho CN =
2ND. Gi s M(112;12
)v ng thng AN c phng trnh : 2x y 3 = 0. Gi P l giao
im ca AN vi ng cho BD. Tm ta im P.
Bi 18. Trong mt phng ta Oxy, Cho tam gic ABC cn ti A, c 2AB =5BC. H(3; 3) l
hnh chiu ca A ln BC. Phng trnh ng thng AC : x 3y+ 2 = 0. Tm ta im B.Bi 19. Trong mt phng Oxy cho bn im A(1; 0), B(2; 4), C(1; 4),D(3; 5) v ng thngd : 3x y 5 = 0. Tm M trn d sao cho hai tam gic MAB,MCD c din tch bng nhau.Bi 20. Trong mt phng ta Oxy, cho hnh ch nht ABCD c din tch bng 6. ng choAC c phng trnh x+ 2y 9 = 0, ng thng AB i qua im M(5; 5), ng thng AD iqua im N(5; 1). Tm ta cc nh ca hnh ch nht, bit A c tung ln hn
32v B c
honh ln hn 3.
Bi 21. Trong mt phng ta Oxy cho ng trn (C) : x2 + y2 + 2x 4y 20 = 0 v imA(3; 0). Vit phng trnh ng thng i qua A v ct (C) theo dy cung MN sao cho di:a) MN ln nhtb) MN nh nht
Bi 22. Trong mt phng ta Oxy cho Elip (E) :x2
16+
y2
9= 1 v ng thng : 3x+ 4y
12 = 0 ct (E) ti hai im A v B. Tm im C (E) sao cho tam gic ABC c din tch ln nht.Bi 23. Trong mt phng ta Oxy cho hnh vung ABCD c cnh bng 1. Gi E, F ln lt l
cc im trn cnh AB v AD, sao cho SAEF =16v SCEF =
512
. Xc nh ta im F, bit
im E(1; 1
3
)v phng trnh ng thng CF : y+ 2x = 0.
Bi 24. Trong mt phng ta Oxy cho hnh vung ABCD . Gi E, F ln lt l cc im trn
cnh AB v AD, sao cho SAEF =16v tan
(ECB
)=
13. Xc nh ta im F, bit im
E(1; 1
3
), phng trnh ng thng CF : y+ 2x = 0 v im C c honh khng m.
Bi 25. Trongmt phng ta Oxy cho imM (1,1) v hai ng thng d1 : x y 1 = 0 vd2 : 2x+ y 5 = 0. Gi A l giao im ca hai ng thng trn. Vit phng trnh ng thngd i qua M, ct hai ng thng trn ln lt ti B v C sao cho tam gic ABC c BC = 3AB.
Bi 26. Trong mt phng to Oxy, cho hai ng thng d1 :3x+ y = 0 v d2 :
3x y = 0.
Gi (C) l ng trn tip xc vi d1 ti A v ct d2 ti hai im B,C sao cho tam gic ABC vung
ti B. Vit phng trnh ng trn (C), bit din tch tam gic ABC bng32
v im A chonh dng.
Bi 27. Trong mt phng ta Oxy, cho ng trn (C1) : x2 + y2 = 64 v A(3; 4). n trn(C2) c tm I2, tip xc (C1) v i qua trung im I2A. Vit phng trnh ng trn (C2) saocho bn knh ca ng trn ny nh nht.
2
Bi 28. Trong mt phng ta Oxy, cho hnh ch nht ABCD. nh B thuc ng thngd1 : 2x y + 2 = 0, nh C thuc ng thng d2 : x y 5 = 0. Gi H l hnh chiu ca Bxung ng cho AC, bit M
(95;25
),K(9; 2) ln lt l trung im ca AH v CD. Tm ta
cc nh ca hnh ch nht ABCD bit honh im C ln hn 4.
Bi 29. Trong mt phng ta Oxy cho4ABC ni tip ng trn (C) c tm I (d) : x y3 = 0. Bit rng trc tm H(3; 2); trng tm G(4;
53). Chn ng cao h t B;C ln lt l E(5; 3)
v Fsao cho EF = 22 ;SAEIF = 5. Tm ta cc nh ca4ABC
Bi 30. Trong mt phng ta Oxy, cho tam gic ABC c tm ng trn ngoi tip, ni tip lnlt l I(2;2),K(0, 1). ng thng AK ct ng trn ngoi tip tam gic ABC ti D(1, 2) (D khc A). Tm ta cc nh ca tam gic ABC.
Bi 31. Trong mt phng to Oxy, cho tam gic ABC cn ti nh A(0; 2). Gi D l imthuc on thng AB sao cho AB = 3AD v H l hnh chiu vung gc ca B trn CD. im
M(32;5
2
)l trung im ca CH. Xc nh to B v C.
Bi 32. Trong mt phng ta Oxy, cho tam gic ABC c tm ng trn ngoi tip I(1;2),tm ng trn ni tip K thuc ng thng 2x+ y 5 = 0. Phng trnh ng trn ngoitip tam gic KBC l (x 3)2 + (y 4)2 = 5. Tnh din tch tam gic ABC.Bi 33. Trong mt phng ta Oxy cho hnh vung ABCD. Trn cc cnh AD, AB ly 2 im Ev F sao cho AE = AF .Gi H l hnh chiu vung gc ca A ln BE. Tm ta im C, bit Cthuc ng thng d : x 2y+ 1 = 0 v ta F(2; 0),H(1;1).Bi 34. Trong mt phng ta Oxy cho 3 im A(2; 3); B(5; 2); (8; 6). Tm im D (d) :x y + 3 sao cho hnh vung MNPQ c cc cnh MN;NP; PQ;QM ln lt i qua cc imA; BC;D sao cho SMNPQ t gi tr ln nht
Bi 35. Trongmt phngOxy cho tam gic ABC c trng tmG. Gi cc im E(6; 4) ; F(387;327
)ln lt thuc cnh BC sao cho 2CE = 3BE v 5FB = 2FC. Tm ta im G.
Bi 36. DDDDDDDDDDDDDDDDDDD
Bi 37. DDDDDDDDDDDDDDDDDDD
Bi 38. DDDDDDDDDDDDDDDDDDD
Bi 39. DDDDDDDDDDDDDDDDDDD
Bi 40. DDDDDDDDDDDDDDDDDDD
3
LI GIIBi 1 Li Ca
Trong mt phng Oxy, cho hnh vung ABCD, bit nh B(3; 4) v phng trnh cnh AC lx y+ 2 = 0. Tm ta cc nh A, C, D, vi im A c honh b hn 3.Li gii(hungchng)
BI
D C
A
ng thng qua B vung gc AC : x y+ 2 = 0 c phng trnh(x 3) + (y 4) = 0 x+ y 7 = 0
Gi I l tm hnh vung ABCD ta c ta I l nghim ca h{x y+ 2 = 0x+ y 7 = 0 = I
(52;92
)I l trung im BD nn D(2; 5)
A,C l giao im ca ng thng AC v ng trn tm I bn knh IB =22
nn ta ca chng l nghim ca hx y+ 2 = 0(x 5
2
)2+
(y 9
2
)2=
12
{
x = 2y = 4 hay
{x = 3y = 5
= A (2; 4) ,C (3; 5) do A c honh b hn 3. Bi 2 catbuilata
Tm ta 3 nh ca tam gic ABC bit ng phn gic trong gc A, trung tuyn k t B,ng cao k t C ln lt l d1 : x 3y+ 1 = 0, d2 : 3x+ y+ 3 = 0, d3 : x+ y+ 3 = 0.Li gii(Li Ca)
d1 : x 3y = 1
d2 : 3x+ y = 3d3 : x+ y = 3
C
A
M
C
B
Tham s im C thuc ng cao d3: C(c;c 3).Tham s im A thuc ng phn gic trong d1: A(3a 1; a)Khi , trung im ca AC l: M
(c+ 3a 1
2;a c 3
2
). M thuc trung tuyn d2 nn
3c+ 3a 1
2+
a c 32
+ 3 = 0 c+ 5a = 0 c = 5aGi C l im i xng ca C qua d1 th C thuc ng thng AB.
Ta C l nghim ca h
{3(x c) + (y+ c+ 3) = 0x+ c 3(y c 3) + 2 = 0
Tm c C(c5 2; 7c
5+ 3)tc l C (a 2;7a+ 3) nn AC = (4a 1;8a+ 3)
Do ACd3 ta c iu kinAC.u3 = 0 vi u3 = (1;1) l VTCP ca d3.
iu kin thnh 1(4a 1) 1(8a+ 3) = 0 a = 1. Th th c = 5. Coi nh xong!S: C(5; 2), A(2; 1), B
(12;3
2
)
Bi 3 Letrungtin
Trong mt phng vi h trc ta Oxy, cho t gic ABCD ngoi tip ng trn (C) : (x+2)2 + (y 3)2 = 9, c phng trnh ng thng BD : 4x y+ 6 = 0. Cc cnh AB, AD lnlt tip xc vi (C) ti E(2; 6) v F
(22
5;65
). Tm ta cc nh ca t gic ABCD
Li gii(hungchng)
I
A B
D
C
E
F
G
H
ng trn (C) c tm I(2; 3) v bn knh r = 3 nn d thy ng thng AB tip xc (C) tiE(2; 6) c phng trnh y = 6 s ct ng thng BD : 4x y+ 6 = 0 ti B(0; 6)ng thng AI l trung trc ca EF c phng trnh(22
5+ 2)(x+ 2) +
(65 6)(y 3) = 0 x+ 2y 4 = 0
ta A l nghim ca h
{y = 6x+ 2y 4 = 0
{x = 8y = 6
= A(8; 6)
5
Phng trnh tip tuyn AF l(65 6)(x+ 8)
(22
5+ 8)(y 6) = 0 4x+ 3y+ 14 = 0
Nn ta D l nghim ca h
{4x y+ 6 = 04x+ 3y+ 14 = 0
{x = 2y = 2 = D(2;2)
Tip tuyn th hai ca (C) qua D tip xc ti H nn H v F i xng qua DI do ta H
l nghim ca h
(2+ 2)
(x+
225
)+ (3+ 2)
(y 6
5
)= 0
12
(x 22
5
)= 2
x =
25
y =65
nn phng
trnh tip tuyn DH l(65+ 2)(x+ 2)
(25+ 2)(y+ 2) = 0 4x 3y+ 2 = 0
ng thng BI c phng trnh (6 3)(x+ 2) (0+ 2)(y 3) = 0 3x 2y+ 12 = 0Tip tuyn th hai ca (C) qua B tip xc ti G nn G v E i xng qua BI do ta G l
nghim ca h
(0+ 2)(x+ 2) + (6 3)(y 6) = 0312(x 2) 21
2(y+ 6) + 12 = 0
x =
1013
y =5413
nn phng trnh tip
tuyn BG l(5413 6)(x 0)
(1013 0)(y 6) = 0 12x+ 5y 30 = 0
Do ta C l nghim ca h
{12x+ 5y 30 = 04x 3y+ 2 = 0
x =
107
y =187
= C(107;187
)
Bi 4 etrungtin
Trong mt phng vi h trc ta Oxy, cho t gic ABCD ngoi tip ng trn (C) : (x+2)2 + (y 3)2 = 9, c nh A (8, 6) v phng trnh ng thng BD : 4x y+ 6 = 0. Tmta cc nh cn li ca t gic ABCD
Li gii(hungchng)
I
A B
D
C
E
F
G
H
ng trn (C) c tm I(2; 3) v bn knh r = 3 nn d thy ng thng y = 6 i qua A vtip xc (C) ti E(2; 6) s ct ng thng BD : 4x y+ 6 = 0 (gi s) ti B(0; 6)ng thng AI c phng trnh (6 3)(x+ 2) (8+ 2)(y 3) = 0 x+ 2y 4 = 0
6
Tip tuyn th hai ca (C) qua A tip xc ti F nn F v E i xng qua AI do ta F l
nghim ca h
(8+ 2)(x+ 2) + (6 3)(y 6) = 012(x 2) + 21
2(y+ 6) 4 = 0
x = 22
5y =
65
nn phng trnh
tip tuyn AF l(65 6)(x+ 8)
(22
5+ 8)(y 6) = 0 4x+ 3y+ 14 = 0
Nn ta D l nghim ca h
{4x y+ 6 = 04x+ 3y+ 14 = 0
{x = 2y = 2 = D(2;2)
Tip tuyn th hai ca (C) qua D tip xc ti H nn H v F i xng qua DI do ta H
l nghim ca h
(2+ 2)
(x+
225
)+ (3+ 2)
(y 6
5
)= 0
12
(x 22
5
)= 2
x =
25
y =65
nn phng
trnh tip tuyn DH l(65+ 2)(x+ 2)
(25+ 2)(y+ 2) = 0 4x 3y+ 2 = 0
ng thng BI c phng trnh (6 3)(x+ 2) (0+ 2)(y 3) = 0 3x 2y+ 12 = 0Tip tuyn th hai ca (C) qua B tip xc ti G nn G v E i xng qua BI do ta G l
nghim ca h
(0+ 2)(x+ 2) + (6 3)(y 6) = 0312(x 2) 21
2(y+ 6) + 12 = 0
x =
1013
y =5413
nn phng trnh tip
tuyn BG l(5413 6)(x 0)
(1013 0)(y 6) = 0 12x+ 5y 30 = 0
Do ta C l nghim ca h
{12x+ 5y 30 = 04x 3y+ 2 = 0
x =
107
y =187
= C(107;187
)
Bi 5 letrungtin
Trong mt phng Oxy, cho hnh thang cn ABCD c din tch bng452, y ln CD nm trn
ng thng x 3y 3 = 0. Bit hai ng cho AC, BD vung gc vi nhau ti I(2; 3). Vitphng trnh ng thng cha cnh BC, bit im C c honh dng.
Li gii(Mai Tun Long)Gi M;N ln lt l trung im ca CD v ABTa c: AC BD CDI;ABI l cc tam gic vung cn ti I IN = NA = NB; IM = MC = MD. IM = d(I;CD) =
10 CD = 2
10
ng thng MN CD MN c PT:3x+ y 9 = 0{x 3y 3 = 03x+ y 9 = 0
{x = 3y = 0
M(3; 0)
C;D thuc ng trn tm M bn knh R =10 c PT:(x 3)2 + y2 = 10{
x 3y 3 = 0(x 3)2 + y2 = 10
{x = 6y = 1
hay
{x = 0y = 1 C(6; 1);D(0;1)
t IN = a, (a > 0) MN = a+10; AB = 2a
SABCD =452 (a+
10)2 =
452 a =
102
IDIB
=IMIN
= 2 DI = 2IB B(3; 5) BC = (3;4) ng thng BC c PT: 4x+ 3y 27 = 0
7
Li gii(hungchng)
x
y
x 3y 3 = 0
I
K
C
D
B
A
H
ng thng qua I vung gc CD : x 3y 3 = 0 c phng trnh3(x 2) + (y 3) = 0 3x+ y 9 = 0
Gi K l trung im CD ta c ta K l nghim ca h
{x 3y 3 = 03x+ y 9 = 0 = K (3; 0)
M KI = KC = KD nn A,C l giao im ca ng thng CD v ng trn tm K bn knhKI =
10. Do ta ca chng l nghim ca h{
x 3y 3 = 0(x 3)2 + (y2 = 10 = C (6; 1) ,D (0;1) do C c honh dng.
Gi H l trung im AB ta c452
= SABCD =12(AB+ CD)HK = (IH + IK)HK =
(IH +
10)2
= IH =102
mIDIB
=IKIH
= 2 DI = 2IB B(3; 5) BC = (3;4)Vy ng thng BC c phng trnh: 4(x 3) + 3(y 5) = 0 4x+ 3y 27 = 0
Bi 6 Li CaTrong mt phng Oxy, cho tam gic ABC cn ti nh A(2; 2). ng thng i qua trung imca cc cnh AB, AC c phng trnh x+ y 6 = 0 v im D(2; 4) nm trn ng cao k tnh B. Xc nh ta cc nh B v C ca tam gic ABC.
Li gii(tutuhtoi)d : x+ y 6 = 0. Gi AE l ng cao h t A xung cnh BC.Ta c AEd nn phng trnh AE : x y = 0. Gi I = d AE, ta c ta I(3; 3)Ta nhn thy I l trung im ca AE nn ta E(4; 4)ng thng BC i qua E v song song vi d nn phng trnh BC l: x+ y 8 = 0Gi B(t; 8 t). Do E l trung im BC nn ta c ta C(8 t; t)BD = (2 t; t 4); AC = (6 t; t 2). Do ACBD AC.BD = 0 t = 2 hoc t = 5.T ta c ta B,C.
Bi 7 Li CaTrong mt phng to Oxy, cho hnh thang cn ABCD (AB//CD v AB < CD), bit ccnh A(0; 2) v D(2;2). Giao im I ca hai ng cho AC v BD thuc ng thngd : x+ y 4 = 0. Tm to cc nh B v C, bit AID = 45o.Li gii(Phm Kim Chung)
8
Thc ra mu cht bi ton nm vn tm ta im I.
AID = 450 IA.IDIA . ID =
22
2m2 3m+ 6 = 2.(m2 2m+ 2) (m2 4m+ 20) [ m = 2m = 4
Sau khi tm c im I, ta vit phng trnh cc ng thng IA, ID ,ng thi tnh c IA, ID
v l :IA = IA
ICIC = IA
IDIC
Bi 8 theoanmTrong mt phng to Oxy, cho hnh ch nht ABCD c im A(3; 1) v C thuc ngthng d : x 2y 5 = 0. im E l giao im th hai ca DC v ng trn tm B bn knhBD (E khc D). Hnh chiu vung gc ca D xung BE l im N (6;2). Tm ta im Bv C.
Li gii(Monkey D.Luffy)Nt tht ca bi ton l chng minh c ANC = 90o.V ABCD nm trn ng trn ng knh BD. Mt khc DNB = 90o nn N cng thuc ngtrn ng knh BD. ng trn ny cng c ng knh AC. Suy ra ANC = 90o hay AN NC.Ta c:
AN = (9;3). V C (d) C(2t+ 5; t)NC = (2t 1; t+ 2).
Ta c: AN NCAN.NC = 0 9(2t 1) 3(t+ 2) = 0 t = 1 C(7; 1)To tm I ca hnh ch nht ABCD l trung im ca AC I(2; 1)ng trn tm I bn knh R =
AC2
= 5 ngoi tip hnh ch nht ABCD c phng trnh l:
(C) : (x 2)2 + (y 1)2 = 25Trong tam gic vung ANC v DNE ta c:AC2 = AN2 + NC2 = AN2 + CD2 = AN2 + AB2 AB =
10
To im B l nghim ca h phng trnh:{(x 2)2 + (y 1)2 = 25(x+ 3)2 + (y 1)2 = 10
[x = 2; y = 2x = 2; y = 4
[B(2;2)B(2; 4)
Phng trnh ng thng AC c dng: y 1 = 0V B v N cng pha vi ng thng AC nn B(2;2)Vy B(2;2),C(7; 1) l cc im cn tm. Li gii(Huynh)
Gi F i xng vi E qua B. Theo bi ra ta lun c BCAD (1).
9
Trong tam gic EDF cEBEF
=ECED
= BCFD (2).Theo tin Euclide, (1) ; (2) suy ra AD FD A,D, F thng hng.M theo Thales trong tam gic FDE c
ECED
=BCFD
=ADFD
=12
suy ra A l trung im ca FD suy ra ACEF. t AB = x ; AD = y.FDN : AN = AD = y
ADC : AC =x2 + y2
DNE : NC = DC = x= AC2 = AN2 + NC2 = ANNC .
Phng trnh NC l :3x y 20 = 0.Ta im C tha
{3x y 20 = 0x 2y 5 = 0
{x = 7y = 1 = C (7; 1) = I(2; 1).
Phng trnh EF l : y+ 2 = 0. Tham s ha B(b;2). Ta c AB = (b+ 3;3) ;BC = (b 7;3).Suy ra
AB.BC = 0 b2 4b 12 = 0
[b = 2b = 6 (Trng im N) = B (2;2).
Kt lun: Ta cc nh B(2;2) ; C(7; 1).
Bi 9 Tit Khnh Duy
Trong mt phng Oxy cho tam gic ABC nhn c M(1;2),N(2; 2) v P(1; 2), ln lt chnng cao k t 3 nh. Tm ta 3 nh trn
Li gii(tutuhtoi)Gi cc ng cao l AM, BN,CP. V H l trc tm tam gic ABC. Ta s chng minh kt qu sau:BN l phn gic ca gc MNP.Tht vy: Ta c t gic ANHP ni tip ng trn nn HNP = HAP.T gic NHMC ni tip ng trn nn HNM = HCM. V HNM = HAP (cng ph vi gcABC).Do HNM = HNP nn BN l phn gic ca gc MNP.Tng t ta cng chng minh c: AM l phn gic ca gc NMP, CP l phn gic ca gcNPM.S dng kt qu trn ta i gii quyt bi ton nh sau:- Vit phng trnh 2 ng phn gic trong: BN l phn gic trong ca gc MNP, AM l phngic trong ca gc NMP.- Tm ta H l giao ca BN v AM.- Vit phng trnh cnh BC (BC i qua M v c VTPT l
HM).
Tng t vit phng trnh cnh AC, BC. Sau tm ta ba nh tam gic.
Bi 10 PhHPhuong
Trong mt phng Oxy cho hnh vung ABCD c tm I(12;92
), cc nh A, B ln lt thuc
cc ng thng d1 : 3x+ 4y 8 = 0; d2 : 3x+ 4y 1 = 0. Tm ta cc nh ca hnh vung.Li gii(Li Ca)
Bi ny s dng tnh cht: tam gic AIB vung cn ti I.
{IA.IB = 0
IA2 = IB2
Tham s ha cc im A thuc d1 v B thuc d2: A(t;8 3t4
), B(k;1 3k
4
)
10
Ta c h phng trnh(t+
12
)(k+
12
)+
(3t+ 10
4
)(3k+ 17
4
)= 0 (1)(
t+12
)2+
(3t+ 10
4
)2=
(k+
12
)2+
(3k+ 17
4
)2(2)
T PT(1) ta c t = 38k+ 17425k+ 59
Thay vo PT(2) thu c k = 1 hoc k = 9325
tng ng t = 4 hoc t = 245
Bi 11 PhHPhuong
Cho hnh vung ABCD. M l trung im ca BC, phng trnh ng thng DM : x y 2 =0, im C(3;3) v im A thuc d : 3x+ y 2 = 0. Xc nh to A, B,D.Li gii(Li Ca)Mu cht ca bi ton ny l vit phng trnh ng thng CD. Ngoi ra, gi N l trung imca CD th AN vung gc vi DM.
Ta c tan MAC =MCDC
=12 cos MDC = 2
5ng thng CD c VTPT l n1 = (a; b). iu kin: a2 + b2 > 0.ng thng DM c VTPT l n2 = (1;1) Ta c phng trnh 2
5=
|a b|2.a2 + b2
Ta c hai trng hp:TH1. a = 3b Khi , phng trnh ca CD l 3x y 12 = 0 T y, D(5; 3). Suy ra, N(4; 0).Phng trnh ng thng AN: x + y 4 = 0. Suy ra A(1; 5) Gi I l tm ca hnh vungABCD. Khi , I(1; 1). Do , B(3;1)TH2. b = 3a. Gii quyt tng t.Bi ton hnh hc Oxy thng gii quyt theo iu kin cn, nn sau khi c p s ta cn kimtra li bng hnh v trn h trc.
Bi 12 Li CaTrong mt phng ta Oxy, cho ng trn T : (x 4)2 + y2 = 40. Vit phng trnh ngthng i qua gc ta v ct ng trn (T) ti hai im A, B sao cho AB = 4BO
Li gii(PhHPhuong)Dng ton ny nt tht bao gi cng l phi tnh cho c khong cch t tm ng trn Iti , sau gii bi ton quen thuc vit phng trnh ng thng i qua 1 im cho trcv cch 1 im cho trc bng 1 khong no Hng gii bi ny nh sau:- ng trn (T) c tm I(4; 0), bn knh r =
40 > IO = 4 nn O l im nm trong ng
trn- Gi H l hnh chiu vung gc ca I ln ng thng ,theo gi thit ta c O l trung im ca HB nn IO l trung tuyn ca tam gic IHB,
vy nn: IO2 =2(IH2 + IB2) HB2
4 64 = 2(IH2 + 40) (40 IH2) IH =
8
Bi ton coi nh c gii quyt.
Bi 13 bongbong
Trong mt phng ta Oxy, cho im A(0;2) v ng thng d : x y+ 2 = 0. Tm ta im M thuc d sao cho ng cao AH v ng trung tuyn ON ca tam gic OAM c dibng nhau.
11
Li gii(Li Ca)
Tham s im M thuc ng thng d l M(t; t+ 2). Ta cOM = (t; t+ 2) v N
(t2;t2
)Phng trnh ng thng OM l (t+ 2)x ty = 0Theo gi thit ta c
2|t|2t2 + 4t+ 4
=|t|2
Gii ra c t = 0 hoc t = 1+3 hoc t = 1
3
Bi 14 Li CaTrong mt phng ta Oxy, cho tam gic ABC vung ti A. Bit A(1; 4), B(1;4) v ngthng BC i qua im M(2;
12). Tm ta nh C.
Li gii(PhHPhuong)Bi ny tng i nh nhng:
- Vit pt ng BC i qua 2 im B v M bit:{
x = 1+ 2ty = 4+ 9t C (1+ 2t;4+ 9t) T
AB.AC = 0 tm ra C.
Bi 15 PhHPhuong
Trong mt phng Oxy, cho tam gic ABC c A(1; 4), trc tm H(3;1), tm ng trn ngoitip tam gic I(2; 0). Xc nh im B,C bit honh ca C dng.Li gii(tutuhtoi)- Gi A l im i xng vi A qua I. T c ta A.- Chng minh c tc gic BHCA l hnh bnh hnh. Gi M l giao ca 2 ng cho BC vHA. Khi M l trung im ca BC v HA. T tm c M.- Vit phng trnh ng thng BC (i qua M v c VTPT l
AH).
- Vit phng trnh ng trn ngoi tip tam gic ABC (c tm I bn knh IA).- Gii h giao im ca ng thng BC v ng trn ngoi tip tam gic ABC ta c ta B,C.
Bi 16 Li CaTrong mt phng ta Oxy, cho ng trn (C) : (x 4)2 + (y+ 3)2 = 4 v ng thngd : x+ y 1 = 0. Xc nh ta cc nh ca hnh vung ABCD ngoi tip ng trn (C),bit A thuc ng thng d.
Li gii(PhHPhuong)Bi ny cng kh nh nhng: ABCD ngoi tip ng trn C c tm I(4; -3), bn knh bng r =2 nn tnh c IA = 2
2 Vn cn li ch l gii bi ton quen thuc: " Tm A d sao cho
IA = 22
Li gii(tutuhtoi)Hnh vung ABCD ngoi tip ng trn ngha l cc cnh ca hnh vung tip xc vi ngtrn (C). Tm ng trn I(4; 3), bn knh R = 2. - Tham s ha im A(1 t; t). - K IHABti H. Ta c tam gic AIH vung cn ti H. Do IH = AH = R nn AI = 2
2. T tnh ra t. Suy
ra ta A. Do c ta C. - Vit phng trnh tip tuyn i qua A ca ng trn (C). S ra2 tip tuyn. Xt tng phng trnh tip tuyn v gi l AB. - Tm giao ca AB v (C) l ta H. T suy ra ta B v H l trung im AB.
Bi 17 tutuhtoiCho hnh vung ABCD .Gi M l trung im BC, N l im trn cnh CD sao cho CN = 2ND.
Gi s M(112;12
)v ng thng AN c phng trnh : 2x y 3 = 0. Gi P l giao im
ca AN vi ng cho BD. Tm ta im P.
12
Li gii(Li Ca)
Gi J l trung im ca AN. Khi , gc gia hai ng thng AN v MJ l cos AND =110
Phng trnh ng thng MJ l x+ y 6 = 0 hoc x+ 7y 9 = 0TH1. J(3; 3). Gi K l trung im ca AD. Ta c
MJ = 5 ~JK Do , K
(52;72
)Phng trnh AD l x y+ 1 = 0 Suy ra, A(4; 5).Li c P l trung im ca JN nn t P
(52; 2)
TH2. Tng t nh!
Bi 18 Phm Vn Lnh
Trong mt phng ta Oxy, Cho tam gic ABC cn ti A, c 2AB =5BC. H(3; 3) l hnh
chiu ca A ln BC. Phng trnh ng thng AC : x 3y+ 2 = 0. Tm ta im B.Li gii(PhHPhuong)
* ABC l tam gic cn nn H l trung im ca BC, do : d (B; AC) = 2d (H; AC) =810
* Din tch tam gic ABC l: SABC =12AH.BC =
12d (B; AC) .AC
AH.BC = 810
52BC AH = 2
2
* Ta c: sin HAC =HEHA
=15 tan HAC = 1
2AHC vung ti H
HC = AH tan HAC =2
* n y ta ch cn gii quyt cng vic quen thuc Tm im C AC ( bit phng trnh)sao cho HC =
2 .Tm c C suy ra B
ps: Tnh ton cn kim tra li cho chc, v mt phng php th chun Li gii(tutuhtoi)Do tam gic ABC cn ti A nn AB = AC v H l trung im ca BC. V 2 gc y l gc nhn.
Ta c 2AB =5BC AC = AB =
52BC =
5HC Tam gic AHC vung ti H nn AH2 +
HC2 = AC2 AH2 + HC2 = 5HC2 AH2 = 4HC2 AH = 2HC Trong tam gic AHC ta ctan ACH =
AHHC
= 2 Do cosACH =15(p dng cng thc
1cos2
= 1+ tan2) Gi VTPT
ca ng thng BC l n = (a, b). Khi phng trnh ca BC l: a(x 3) + b(y 3) = 0. Khi p dng cng thc tnh gc cos(AC, BC) = cosACH =
15suy ra mi quan h a, b. T vit
c phng trnh BC. Khi tm c C l giao ca AC v BC. V tm c B do H l trungim BC. Vit phng trnh ng cao AH (i qua H v vung gc vi BC). Tm c A l giaoim ca AH v AC.
Bi 19 Trng Nhc
Trong mt phng Oxy cho bn im A(1; 0), B(2; 4), C(1; 4),D(3; 5) v ng thng d :3x y 5 = 0. Tm M trn d sao cho hai tam gic MAB,MCD c din tch bng nhau.Li gii(Li Ca)Tham s im M thuc ng thng d nn M(t; 3t 5)Phng trnh AB : 3x 4y 3 = 0. Phng trnh CD : 4x+ y = 0Ta c phng trnh
|15t 23|5
.5 =|7t 5|
17.17
Tm c t =94hoc t =
1411
13
Bi 20 Li CaTrong mt phng ta Oxy, cho hnh ch nht ABCD c din tch bng 6. ng cho AC cphng trnh x+ 2y 9 = 0, ng thng AB i qua im M(5; 5), ng thng AD i quaim N(5; 1). Tm ta cc nh ca hnh ch nht, bit A c tung ln hn
32v B c honh
ln hn 3.
Li gii(Trng Nhc)Tham s ho im A(9 2a; a) hai nt tht+Tm im A, da vo k
AM = (2a 4; a) ,AN = (2a 4; 1 a)
AM AM a = 3 A(3; 3)+phng trnh AB : x y = 0 tham s B(b; b) AD : x+ y 6 = 0 tham s D(t; 6 t)AB =
2 (b 3)2, AD =
2 (t 3)2 AB.AD = 6 (b 3) (t 3) = 9
Gi I = AC BD Thay to ca I vo phng trnh AC 3b t 6 = 0Tm c b, t tho mn k bi ton. nh th bi ton gii quyt. Li gii(tutuhtoi)- Tham s ha im A(9 2t; t).Tm c ta im A nh vo iu kin: AMAN AM.AN = 0.- Vit phng trnh ng thng AB, AD.- Gi I l giao im ca 2 ng cho. Tham s ha im I(9 2a; a).Ta c AB.AD = 6 2d(I, AB).2d(I, AD) = 6.S dng cng thc tnh khong cch tm c a, t c ta I. Tm c ta C.- Vit phng trnh ng thng d i qua I v vung gc vi AB. Gi giao ca d v AB l H,suy ra ta H. T tm c ta B (do H l trung im AB).
Bi 21 PhHPhuong
Trong mt phng ta Oxy cho ng trn (C) : x2 + y2 + 2x 4y 20 = 0 v im A(3; 0).Vit phng trnh ng thng i qua A v ct (C) theo dy cung MN sao cho di:a) MN ln nhtb) MN nh nht
Li gii(noname)GGGGGGGGGGGGGGGGGGGGGG
Bi 22 PhHPhuong
Trong mt phng ta Oxy cho Elip (E) :x2
16+
y2
9= 1 v ng thng : 3x+ 4y 12 = 0
ct (E) ti hai im A v B. Tm im C (E) sao cho tam gic ABC c din tch ln nht.Li gii(noname)GGGGGGGGGGGGGGGGGGGGGG
Bi 23 Phm Kim Chung
Trong mt phng ta Oxy cho hnh vung ABCD c cnh bng 1. Gi E, F ln lt l cc
im trn cnh AB v AD, sao cho SAEF =16v SCEF =
512
. Xc nh ta im F, bit
im E(1; 1
3
)v phng trnh ng thng CF : y+ 2x = 0.
Li gii(noname)Nt tht l tnh cho c on EF
Ta c: 2SCEF = d (E,CF) .CF CF =52
FD =CF2 CD2 = 1
2 AF = AD FD = 1
2
14
AE = 2SAEFAF
=23 EF =
AE2 + AF2 =
56
n y a v gii bi ton quen thuc .
Tm trn ng thng d (xc nh) im F sao cho EF =56
Bi 24 Phm Kim Chung
Trongmt phng ta Oxy cho hnh vung ABCD . Gi E, F ln lt l cc im trn cnh AB
v AD, sao cho SAEF =16v tan
(ECB
)=
13. Xc nh ta im F, bit im E
(1; 1
3
),
phng trnh ng thng CF : y+ 2x = 0 v im C c honh khng m.
Li gii(ng Thnh Nam)
Gi s cnh hnh vung c di a > 0. T tan ECB =13 AE = 2
3a.
t AF = b SAEF = 12AE.AF =ab3
=16 b = 1
2a.
Ta tnh din tch tam gic CEFtheo hai cch nh sau
SCEF =12d (E,CF) .CF =
56.a2 + (a b)2 =
5(a2 +
(a 12a
)2)6
(1).
SCEF = SABCD SCEB SAEF SCFD = a2 16 a2 1
6 a (a b)
2=
5a2 1 3a (a 12a)6
(2).
So snh (1) v (2) suy ra c a : C th 2a2 +12=
5(2a2 +
14a2 1)
{
u = a2
(4u+ 1)2 = 5(8u2 4u+ 1
) { u = a224u2 28u+ 4 = 0 .
a2 = 1a2 =
16 a = 1
a =16.
v gi F (x0;2x0)th EF2 = a2 + (a b)2 = a2 +(a 1
2a
)2= (x0 + 1)
2 +
(2x0 +
13
)2.
T y suy ra c x0 F (x0;2x0).
Bi 25 D B A,A1 2012
Trong mt phng ta Oxy cho im M (1,1) v hai ng thng d1 : x y 1 = 0 v d2 :2x+ y 5 = 0. Gi A l giao im ca hai ng thng trn. Vit phng trnh ng thng di qua M, ct hai ng thng trn ln lt ti B v C sao cho tam gic ABC c BC = 3AB.
Li gii(khvav)D thy A(2, 1) Ly E thuc d1, E(0,1). EA = 2
2. K EFBC
F thuc BC, F( f , 5 2 f ). BC = 3AB nn EF = 3EA = 62 Suy ra ta im F.
pt BC qua M v c vct ch phngEF (C l 2 p s)
Bi 26 Li CaTrong mt phng to Oxy, cho hai ng thng d1 :
3x+ y = 0 v d2 :
3x y = 0. Gi
(C) l ng trn tip xc vi d1 ti A v ct d2 ti hai im B,C sao cho tam gic ABC vung
ti B. Vit phng trnh ng trn (C), bit din tch tam gic ABC bng32
v im A chonh dng.
Li gii(noname)GGGGGGGGGGGGGGGGGGGGGG
Bi 27 nh Nam
15
Trong mt phng ta Oxy, cho ng trn (C1) : x2 + y2 = 64 v A(3; 4). n trn (C2)c tm I2, tip xc (C1) v i qua trung im I2A. Vit phng trnh ng trn (C2) sao chobn knh ca ng trn ny nh nht.
Li gii(noname)GGGGGGGGGGGGGGGGGGGGGG
Bi 28 nh NamTrong mt phng ta Oxy, cho hnh ch nht ABCD. nh B thuc ng thng d1 : 2xy + 2 = 0, nh C thuc ng thng d2 : x y 5 = 0. Gi H l hnh chiu ca B xungng cho AC, bit M
(95;25
),K(9; 2) ln lt l trung im ca AH v CD. Tm ta cc
nh ca hnh ch nht ABCD bit honh im C ln hn 4.
Li gii(Monkey D.Luffy)ng thng AK ct ng thng BC ti E ta c C l trung im BE.V M;K ln lt l trung im ca AH,CD nn MK l ng trung bnh ca AHE.V B (d1),C (d2) B(b; 2b+ 2),C(c; c 5). V C l trung im ca BE E(2c b; 2c 2b 12) Ta c: HE = 2MK H
(2c b 72
5; 2c 2b 76
5
)Suy ra:
CK = (9 c; 7 c), BC = (c b; c 2b 7), BH =
(2c 2b 72
5; 2c 4b 86
5
);
MC =
(c 9
5; c 27
5
). Ta c:
{CK BCMC BH
{ CK.BC = 0
MC.BH = 0
(9 c)(c b) + (7 c)(c 2b 7) = 0(2c 2b 72
5
)(c 9
5
)+
(2c 4b 86
5
)(c 27
5
)= 0
2c
2 + 3bc+ 23c 23b 49 = 04c2 6bc+ 126
5b 46c+ 594
5= 0[b = 1; c = 4b = 1; c = 9
[B(1; 4),C(4;1)B(1; 4),C(9; 4)
B(1; 4),C(4;1) D(14; 5) A(11; 10). Loi v M v K cng pha vi AB. B(1; 4),C(9; 4) D(9; 0) A(1; 0)Vy to cc nh hnh ch nht ABCD l: A(1; 0), B(1; 4),C(9; 4),D(9; 0).
Bi 29 Huynh
Trong mt phng ta Oxy cho4ABC ni tip ng trn (C) c tm I (d) : x y 3 = 0.Bit rng trc tm H(3; 2); trng tm G(4;
53). Chn ng cao h t B;C ln lt l E(5; 3) v
Fsao cho EF = 22 ;SAEIF = 5. Tm ta cc nh ca4ABC
Li gii(Monkey D.Luffy)Gi D,M,N ln lt l chn ng vung gc ca A ln BC v trung im ca BC v AC.Ta c: AD BC; IM BC; IN AC. Gi A l im i xng vi A qua I A (C)Ta c:
{BA CH( AB)CA BH( AC) BHCA
l hnh bnh hnh
Suy ra hai ng cho ct nhau ti trung im mi ng M l trung im HA.R rng, IM l ng trung bnh AHA AH = 2IMTa c:
AH = 2
IMIH IA = 2IM = IB + ICIH = IA + IB + IC = 3IG
{3 x = 3(4 x)2y = 3(
53 y)
x =
92
y =32
I(92;32
)
Thay to im I vo ng thng (d) thy ng, vy I(92;32
)
16
AEHF ni tip (AEH = AFH = 90o) AEF = AHF (gc ni tip cng chn cung AF)Mt khc: BDHF ni tip (BDH = BFH = 90o) AHF = ABD (cng b DHF)IAC cn ti I AIN = CIN = 1
2AIC = ABC
Hai tam gic vung AIN v ADB c: ABD = AINSuy ra: AIN ADB BAD = IAN AEF + IAE = ABD + BAD = 90o
ATE = 90o (T l giao im ca EF v AI) EF IA. SAEIF = 12AI.EF = 5 AI =52
2ng thng AC vung gc vi HE: 2x+ y 13 = 0. Gi A(a; 13 2a)Ta c: AI2 =
(a 9
2
)2+
(232 2a
)2=
252 a = 7; a = 4
A(7;1) AH = (4; 3) = 2IM = (2x 9; 2y 3) M(52; 3)
ng thng IN vung gc vi AC : 2x 4y 3 = 0. To im N l nghim ca h:{2x+ y 13 = 02x 4y 3 = 0
{x =
112
y = 2 N
(112; 2) C (4; 5) B (1; 1)
A(4; 5) C (7;1) B(2; 7). Phng trnh ng thng BC : 8x+ 9y 17 = 0Ta thy A v H nm khc pha i vi ng thng BC(). Mt khc, cos(AB;AC) = 1
2< 0.
Vy BAC t. Khi , trc tm H s nm trn mt phng b BC c cha im A.iu ny mu thun (). Vy trng hp ny khng tho mn.Vy to cc nh ca ABC l: A (7;1) , B (1; 1) ,C (4; 5)
Bi 30 Chuyn Phan Bi Chu ln 2
Trong mt phng ta Oxy, cho tam gic ABC c tm ng trn ngoi tip, ni tip ln ltl I(2;2),K(0, 1). ng thng AK ct ng trn ngoi tip tam gic ABC ti D(1, 2) ( Dkhc A). Tm ta cc nh ca tam gic ABC.
Li gii(Monkey D.Luffy)
Gi (C) l ng trn ngoi tip ABC. Khi , D l im chnh gia cung nh BCPhng trnh ng trn (C) c dng: (C) : (x+ 2)2 + (y+ 2)2 = 25Phng trnh ng thng AD c dng: (AD) : x y+ 1 = 0
17
To im A l nghim ca h:{
x y+ 1 = 0(x+ 2)2 + (y+ 2)2 = 25
A(6; 5)
DCK c DCK = DKC =12(A + C). Suy ra DCK cn ti D.
Chng minh tng t, ta c: DBK cn ti D. Suy ra DB = DC = DKVy DBKC ni tip ng trn tm D bn knh DKPhng trnh ng trn (DBKC) c dng: (DBKC) : (x 1)2 + (y 2)2 = 2To im B;C l nghim ca h phng trnh:{(x+ 2)2 + (y+ 2)2 = 25(x 1)2 + (y 2)2 = 2 B(2; 1),C
( 625
;6725
)hoc B
( 625
;6725
),C(2; 1)
Vy to cc nh ca tam gic ABC l:
A(6; 5), B(2; 1),C( 625
;6725
)hoc A(6; 5), B
( 625
;6725
),C(2; 1)
Bi 31 Li CaTrong mt phng to Oxy, cho tam gic ABC cn ti nh A(0; 2). Gi D l im thuc on
thng AB sao cho AB = 3AD v H l hnh chiu vung gc ca B trn CD. im M(32;5
2
)l trung im ca CH. Xc nh to B v C.
Li gii(noname)GGGGGGGGGGGGGGGGGGGGGG
Bi 32 huynhcashin
Trong mt phng ta Oxy, cho tam gic ABC c tm ng trn ngoi tip I(1;2), tmng trn ni tip K thuc ng thng 2x+ y 5 = 0. Phng trnh ng trn ngoi tiptam gic KBC l (x 3)2 + (y 4)2 = 5. Tnh din tch tam gic ABC.Li gii(Monkey D.Luffy)
Gi (C) l ng trn ngoi tip tam gic ABC; (C1) l ng trn bng tip A ca ABC; J ltm ca (C1) v A1 l giao im ca AJ v (C).
Ta c: A1CK = A1KC =12(A + C) nn A1CK cn ti A1
Chng minh tng t, ta c: A1BK cn ti A1. Suy ra: A1B = A1C = A1KMt khc, ta c: KB BJ;KC CJSuy ra cc tam gic KBJ;KCJ ln lt vung ti B;C v A1 l trung im KJ.
Vy t gic KBJC ni tip ng trn tm A1 bn knhKJ2. Khi , (C1) (KBC).
18
Gi K(x; 5 2x). Ta c: A1K2 = (x 3)2 + (1 2x)2 = 5 x = 1 K(1; 3)Phng trnh ng trn (C) c dng: (C) : (x+ 1)2 + (y+ 2)2 = 52Phng trnh ng thng KA1 c dng: x 2y+ 5 = 0To im A l nghim ca h:
{x 2y+ 5 = 0
(x+ 1)2 + (y+ 2)2 = 52 A
(41
5;8
5
)(loi A (3; 4) v
trng A1)
To hai im B;C l nghim ca h:{(x+ 1)2 + (y+ 2)2 = 52(x 3)2 + (y 4)2 = 5
Suy ra B
(146 31015
52;193+ 2
1015
52
), C
(146+ 3
1015
52;193 21015
52
).
BC =
(3101526
;2101526
). BC =
101552
Phng trnh ng thng (BC) c dng: (BC) : 8x+ 12y 67 = 0. d (A; (BC)) = 7592013
Vy SABC =12.d (A; (BC)) .BC =
12.759
2013
.
101552
=7591015
1040
Bi 33 nh NamTrong mt phng ta Oxy cho hnh vung ABCD. Trn cc cnh AD, AB ly 2 im E v Fsao cho AE = AF .Gi H l hnh chiu vung gc ca A ln BE. Tm ta im C, bit C thucng thng d : x 2y+ 1 = 0 v ta F(2; 0),H(1;1).Li gii(Huynh)
x
y
x 2y+ 1 = 0 F
H
CI
B
K
D
A
E
Mu cht l chng minh FH HCGi
{AH DC = KBK FC = I . D dng chng minh FBCK l hnh ch nht .
Xt BHK c HI l trung tuyn nn HI =BK2
=FC2
Xt FHC c trung tuyn HI =FC2
= FHC vung ti H = FH HCTham s ha C(2t 1; t) (d). Ta c :FH.HC = 0 t = 1
3= C
(13;13
)KL: Ta C
(13;13
)
Bi 34 Huynh
19
Trong mt phng ta Oxy cho 3 im A(2; 3); B(5; 2); (8; 6). Tm im D (d) : x y+ 3sao cho hnh vung MNPQ c cc cnh MN;NP; PQ;QM ln lt i qua cc im A; BC;Dsao cho SMNPQ t gi tr ln nht
Li gii(noname)GGGGGGGGGGGGGGGGGGGGGG
Bi 35 Huynh
Trong mt phng Oxy cho tam gic ABC c trng tm G. Gi cc im E(6; 4) ; F(387;327
)ln
lt thuc cnh BC sao cho 2CE = 3BE v 5FB = 2FC. Tm ta im G.
Li gii(noname)GGGGGGGGGGGGGGGGGGGGGG
Bi 36 tc giDDDDDDDDDDDDDDDDDDD
Li gii(noname)GGGGGGGGGGGGGGGGGGGGGG
Bi 37 tc giDDDDDDDDDDDDDDDDDDD
Li gii(noname)GGGGGGGGGGGGGGGGGGGGGG
Bi 38 tc giDDDDDDDDDDDDDDDDDDD
Li gii(noname)GGGGGGGGGGGGGGGGGGGGGG
Bi 39 tc giDDDDDDDDDDDDDDDDDDD
Li gii(noname)GGGGGGGGGGGGGGGGGGGGGG
Bi 40 tc giDDDDDDDDDDDDDDDDDDD
Li gii(noname)GGGGGGGGGGGGGGGGGGGGGG
20