b Duong Thang.doc

5/28/2018 b Duong Thang.doc

1/38

Nguyn Ph Khnh

549

Trong mt phng to cc vung gc O.

. T C hc g hg 2 0 + = a ch ABC g C , b

( ) ( )A 1; 2 ,B 1; 3 .

Cho tam gic ABC c ( )A 3;2 v phng trnh hai ng trung tuyn

+ = =BM : 3 x 4y 3 0, CN : 3x 10y 17 0. Tnh ta cc im B, C.

Cho tam gic ABC c ( )A 3; 0 v phng trnh hai ng phn gic trong

= + + =BD : x y 1 0,CE : x 2y 17 0. Tnh ta cc im B, C.

Trong mt phng Oxy, cho tam gic ABC vung cn ti A . Xc nh ta 3 nh ca tam gic ng thng AC i qua im ( )N 7;7 , ( )M 2; 3 thuc

AB v nm ngoi AB , phng trnh BC : + =x 7 y 3 1 0.

Trong mt phng Oxy, cho hnh bnh hnh ABCD c ( )B 1;5 , ng cao

+ =AH : x 2y 2 0,phn gic ACB c phng trnh =x y 1 0 . Tm ta

im A.

Trong mt phng to cc vung gc O, cho ( )A 1;3

g hg ( ) : 2 2 0 + = .Ng a dg hh g ABCD a ch 2 B C

g hg ( ) cc a ca h C dg. T a cc h B,C,D ;

T ch d ch hh g ABCD.

Trong mt phng to cc vung gc O,

Cho tam gic MNP c ( )N 2; 1 , ng cao h t M xung NP c phng

trnh: 3x 4y 27 0 + = v ng phn gic trong nh P c phng trnh:

x 2y 5 0+ = . Vit phng trnh cc cnh cha cc cnh tam gic.

Cho tam gic ABC c ( )C 5 ; 3 v phng trnh ng cao + =AA' : x y 2 0,

ng trung tuyn + =BM : 2x 5y 13 0 .Tnh ta cc im A, B.

Cho tam gic ABC c ( )B 1; 3 v phng trnh ng cao + =AD : 2x y 1 0 ,

ng phn gic + =CE :x y 2 0 .Tnh ta cc im A, C.

www.VNMATH.com


2/38

?

Nguyn Ph Khnh

550

Trong mt phng Oxy, cho im ( )E 1; 1 l tm ca mt hnh vung, mt

trong cc cnh ca n c phng trnh + =x 2y 12 0 . Vit phng trnh cc cnh

cn li ca hnh vung.Trong mt phng Oxy, cho hnh vung ABCD c chu vi bng 6 2, nh A

thuc trc Ox ( A c honh dng) v hai nh B,C thuc ng thng + =d : x y 1 0 . Vit phng trnh ng thng BD


Cho tam gic ABC vung cn ti A c trng tm2

G 0;3

. Vit phng trnh

cha cc cnh tam gic 1 1

I ;2 2

l trung im cnh BC.

Cho tam gic ABC c ( )M 2; 0 l trung im ca cnh AB . ng trung tuyn

v ng cao qua nh A ln lt c phng trnh l =7x 2y 3 0 v

=6x y 4 0 . Vit phng trnh ng thng AC .

cho im ( )C 2 ; 5 v ng thng + =: 3x 4y 4 0 .Tm trn hai im A

v B i xng nhau qua5

I 2;2

sao cho din tch tam gic ABC bng15.

Trong mt phng Oxy cho hnh ch nht ABCD c din tch bng 12 , tm I

l giao im ca ng thng ( ) ( ) = + =1 2d : x y 3 0, d : x y 6 0 . Trung im

ca mt cnh l giao im ca ( )1d vi trc Ox .Tm ta cc nh ca hnh ch

nht ABCD .Trong mt phng ta Oxy, cho hnh ch nht ABCD c phng trnh cnh

=A B : x 2 y 1 0, ng cho + =BD : x 7y 14 0 v ng cho AC i qua

im ( )E 2;1 . Tm ta cc nh ca hnh ch nht.


Cho a gc ABC c 3 ch he h 3 g hg :

( ) + =1d : 6 0 , ( ) + =2d : 4 14 0, ( ) =3d :4 19 0 . H hh dg ca

a gc.

Cho im ( )A 2; 2 v hai ng thng: + =1d : x y 2 0, + =2d : x y 8 0 . Tm

ta im B,C ln lt thuc 1 2d , d sao cho tam gic ABC vung cn ti A.

www.VNMATH.com


3/38

Nguyn Ph Khnh

551

Trong mt phng ta Oxy, cho tam gic ABC cn ti A c phng trnh 2

cnh AB, AC ln lt l: + =x 2y 2 0 v + + =2x y 1 0, im ( )M 1;2 thuc on

BC . Tm ta im D sao cho DB.DC

c gi tr nh nht.Trong mt phng ta O,cho ng trn ( )C : + + =2 2x y 2x 2y 14 0 c

tm I v ng thng ( )d : + + =x y m 0 . Tm d ct ( )C ti hai im phn

bit A, B ng thi din tch tam gic IAB ln nht.Trong mt phng ta Oxy, cho hnh ch nht ABCD c phng trnh

ng thng AB, BDln lt l: + =x 2y 1 0 v + =x 7y 14 0, ng thng

ACi qua ( )M 2; 1 .Tm to im N thuc BDsao cho +NA NC nh nht.


Cho tam gic ABC c ( )A 4 ; 1 v phng trnh hai ng trung tuyn 1BB :

8x y 3 0, = 1CC : 14x 13y 9 0 = . Tnh ta cc im B, C.

Cho hnh ch nht ABCD,vi to cc nh ( )A 1;1 .Gi4

G 2;3

l trng

tm tam gic ABD. Tm ta cc nh cn li ca hnh ch nht bit D nm trnng thng c phng trnh: x y 2 0. =

. Trong mt phng ta Oxy, cho hnh ch nht ABCD c din tch bng 22 .

ng thng AB c phng trnh 3x 4y 1 0,+ + = ng thng BD c phng

trnh x y 2 0.+ = Tm ta cc nh A ,B,C, D? .

Trong mt phng ta Oxy, cho hnh ch nht ABCD c ( )M 4;6 l trung

im ca AB .Giao im I ca hai ng cho nm trn ng thng ( )d cphng trnh 3x 5y 6 0,+ = im ( )N 6; 2 thuc cnh CD . Hy vit phng

trnh cnh CD bit tung im I ln hn 4 .


Cho tam gic ABC c ( )A 4 ; 1 , phng trnh hai ng phn gic

B E : x 1 0 , C F : x y 1 0 = = . Tnh ta cc im B, C.

Cho tam gic ABC vung ti C , bit ( )A 3;0 , nh C thuc trc tung, im B

nm trn ng thng : 4x 3y 12 0. + = Tm ta trng tm tam gic ABC ,

bit din tch tam gic ABC bng 6.

www.VNMATH.com


4/38

3

Nguyn Ph Khnh

552

Cho hnh bnh hnh ABCD c ( )B 1;5 v ng cao AH c phng trnh

x 2y 2 0+ = , vi H thuc BC, ng phn gic trong ca gc ACB c phng

trnh l x y 1 0 = . Tm ta nh A,C,D. Cho tam gic ABC vi hai im ( )A 2; 1 , ( )B 1; 2 v trng tm G nm trn

ng thng d : x y 2 0.+ = Tm ta im C,bit din tch tam gic ABC

bng3

.2

Cho hnh bnh hnh ABCD c ( )D 6; 6 . ng trung trc ca on DC c

phng trnh ( )d : 2x 3y 17 0+ + = v ng phn gic gc BAC c phng

trnh ( )d' : 5x y 3 0+ = .Xc nh to cc nh cn li ca hnh bnh hnh.

Trong mt phng to cc vung gc O,Cho tam gic ABC c ( )C 4; 5 v phng trnh ng cao AD : x 2y 2 0+ = ,

ng trung tuyn 1BB : 8x y 3 0 = .Tm ta cc im A, B.

Cho hnh thang vung ABCD, vung ti A v D. Phng trnh AD

= 2 0 . Trung im M ca BC c ta ( )M 1;0 . Bit = =BC CD 2AB. Tm ta

ca im A.

Cho ABC ,bit ta im ( )A 2 ; 3 v ( )B 3; 2 , din tch tam gic ABC l

32

v trng tm G ca tam gic thuc ng thng : 3x y 8 0 = .Tm ta

im C .Cho tam gic ABC vung ti A, bit B v C i xng nhau qua gc ta .

ng phn gic trong ca gc ABC c phng trnh l: x 2y 5 0.+ = Tm ta

cc nh ca tam gic bit ng thng AC i qua im ( )K 6;2 .

Cho tam gic ABC cn ti C c phng trnh cnh AB l : x 2y 0 = , im

( )I 4; 2 l trung im ca AB , im9

M 4;2

thuc cnh BC , din tch tam gic

ABC bng 10. Tm ta cc nh ca tam gic ABCbit tung im B lnhn hoc bng 3.

Trong mt phng to cc vung gc O, Cho tam gic ABC c ( )B 1;5 v phng trnh ng cao AD : x 2y 2 0+ = ,

ng phn gic trong 1CC : x y 1 0 = . Tnh ta cc im A, C.

www.VNMATH.com


5/38

3

Nguyn Ph Khnh

553

Cho tam gic ABC vung cn ti A, phng trnh BC : 2x y 7 0, = ng

thng AC i qua im ( )M 1; 1 , im A nm trn ng thng

: x 4y 6 0. + = Tm ta cc nh ca tam gic ABC bit rng nh A chonh dng.

Cho 2 ng thng ln lt c phng trnh l ( )1d : 2x 3y 3 0 = v

( )2d : 5x 2y 17 0+ = . Vit phng trnh ng thng i qua giao im ca ( )1d

, ( )2d ln lt ct cc tia Ox, Oy ti A v B sao cho2

OAB

ABS

t gi tr nh

nht.

Cho parabol ( )P : 2y x 2x 3= + . Xt hnh bnh hnh ABCD

( ) ( )A 1; 4 , B 2;5 thuc ( )P v tm I ca hnh bnh hnh thuc cung AB ca( )P sao cho tam gic IAB c din tch ln nht. Hy xc nh ta hai imC, D.

Cho tam gic ABC cn ti A, c nh B v C thuc ng thng

1:d x y 1 0+ + = ng cao i qua nh B l 2d :x 2y 2 0 = im ( )M 2;1

thuc ng cao i qua nh C.Vit phng trnh cc cnh bn ca tam gicABC

f. Cho tam gic ABC c A nm trn Ox vi A5

0 x2

< < . Hai ng cao xut

pht t B v C ln lt c phng trnh: 1d : x y 1 0, + = 2d : 2x y 4 0.+ = Tm

ta A, B, C sao cho din tch tam gic ABCl ln nht.


Cho tam gic ABC c phng trnh cc ng caoAD : 2x y 1 0, BE : x y 2 0 + = + = , C thuc ng thng d : x y 6 0+ = v BC

i qua ( )M 0;3 . Tm ta cc nh ca tam gic.

Cho hnh vung ABCD c phng trnh ng thng AB : 2x y 1 0,+ = v

C, D ln lt thuc 2 ng thng 1d : 3x y 4 0, = 2d : x y 6 0.+ = Tnh din

tch hnh vung

Cho hnh bnh hnh ABCD c ( )A 2;1 , ng cho BD c phng trnh

x 2 y 1 0 .+ + = im M nm trn ng thng AD sao cho AM AC.= ngthng MC c phng trnh x y 1 0.+ = Tm ta cc nh cn li ca hnh

bnh hnh ABCD .

www.VNMATH.com


6/38

3

Nguyn Ph Khnh

554

Cho tam gic ABC vung cn ti A. Bit phng trnh cnh BC l

( )d : x 7y 31 0+ = , im ( )N 7; 7 thuc ng thng AC, im ( )M 2; 3 thuc

AB v nm ngoi on AB . Tm ta cc nh ca tam gic ABC .


Cho tam gic ABC cn ti A c trng tm4 1

G ;3 3

, phng trnh ng

thng BC : x 2y 4 0 = v phng trnh ng thng BG : 7x 4y 8 0 = . Tm

ta cc nh A,B,C.

Cho hnh thang ( )ABCD AB CD . Bit hai nh ( )B 3;3 v ( )C 5; 3 . Giao

im I ca hai ng cho nm trn ng thng : 2x y 3 0.+ = Xc nh ta

cc nh cn li ca hnh thang ABCD CI 2BI,= tam gic ACB c din

tch bng 12 , im I c honh dng v im A c honh m .


Cho tam gic ABC c nh A thuc ng thng d :x 4y 2 0 = , cnh BC

song song vi d, phng trnh ng cao BH : x y 3 0+ + = v trung im cnh

AC l ( )M 1;1 . Tm ta cc nh A,B,C.

Ch hh h ABCD c hg h ha ch AB AD he h

x 2y 2 0+ = 2x y 1 0+ + = . Ch BD cha ( )M 1;2 . T a cc hca hh h


Cho tam gic ABC cn ti A c nh ( )A 6;6 , ng thng i qua trung im

ca cc cnh AB v AC c phng trnh x y 4 0+ = . Tm ta cc nh B v

C , bit im ( )E 1; 3 nm trn ng cao i qua nh C ca tam gic cho.

Cho hai ng thng 1d : x y 2 0, = 2d : 2x y 5 0+ = . Vit phng trnh

ng thng i qua gc ta O ct 1d , 2d ln lt ti A , B sao cho

OA.OB 10= .

14. Trong mt phng to cc vung gc O,cho tam gic ABC , bit

( )C 4;3 v cc ng phn gic trong, trung tuyn k t A ln lt cphng trnh x 2y 5 0, 4x 13y 10 0+ = + = .Tm ta im A,B .

www.VNMATH.com


7/38

3

Nguyn Ph Khnh

555

Trong mt phng to cc vung gc O,cho tam gic ABCcn tiA c nh A( 1;4) v cc nh B, C thuc ng thng : x y 4 0 = .

Xc nh to cc im B v C , bit din tch tam gic ABCbng 18.

Trong mt phng to cc vung gc O,cho tam gic ABC vi

( ) ( )A 2 ; 4 ,B 0 ; 2 v trng tm G thuc ng thng d : 3x y 1 0 + = . Hy tm

ta ca C , bit rng din tch tam gic ABC bng 3 .

Trong mt phng to cc vung gc O,cho tam gic ABC vungti A, c nh C( 4;1) , phn gic trong gc A c phng trnh x y 5 0+ = . Vit

phng trnh ng thng BC, bit din tch tam gic ABC bng 24 v nh A c honh dng.

Trong mt phng to cc vung gc O,cho tam gic ABC c( ) ( )M 1;0 ,N 4; 3 ln lt l trung im ca AB,AC ; ( )D 2;6 l chn ng cao

h t A ln BC . Tm ta cc nh ca tam gic ABC .

Trong mt phng to cc vung gc O,cho a gc ABC c

( )M 2;2 g BC hg h ch ( ) =AB : 2 2 0,

( ) + + =AC :2 5 3 0 . H c h a h ca a gc.

Trong mt phng to cc vung gc O,cho a gc ABC cg ( ) G 2; 1 hg h cc ch ( ) + + =AB : 4 15 0,

( ) + + =AC :2 5 3 0 . T a h A , g M ca BC , h B,C.

Trong mt phng to cc vung gc O,cho ABC c ( )B 3;5 ,g A c hg h : + =2 5 3 0 g C c hg

h : + = 5 0 . T a h A , g M AB .

Vit phng trnh cc cnh hnh vung ABCD, bit rng ( )M 0; 2 AB,

( )N 5; 3 BC, ( )P 2; 2 CD, ( )Q 2; 4 DA .

Trong mt phng to cc vung gc O,cho 2

( ) ( )A 1;1 ,B 2;2 . T C g hg ( ) =d : 3 , a ch =ABCS 2 (d).

Kh h b h R ca g g ABC .

www.VNMATH.com


8/38

3

Nguyn Ph Khnh

556

Trong mt phng to cc vung gc O, cho hnh bnh hnh

ABCD c ( )B 1;5 , ng cao AH : 2 2 0+ = , ng phn gic trong d ca gc

ACB c phng trnh 1 0 = . Tm ta cc nh cn li ca hnh bnhhnh.

Trong mt phng to cc vung gc O, cho tam gic ABCvung cn ti A , cc nh A ,B,C ln lt nm trn cc ng thng d :

x y 5 0,+ = 1d : x 1 0 ,+ = 2d : y 2 0+ = v BC 5 2= . Tm ta nh A,B,C

ca tam gic.

Trong mt phng to cc vung gc O, cho ABC c ( )C 1;1 v

AB 5= v AB: x 2y 3 0+ = , trng tm ABC thuc ng thng

x y 2 0+ = . Tm ta im A, B.

Trong mt phng to cc vung gc O, cho hnh vung ABCD

c ( )A 2; 6 , nh B thuc ng thng d :x 2y 6 0 + = . Gi M, N ln lt l

hai im trn 2 cnh BC,CD sao cho BM CN= . Xc nh ta nh C , bit

rng AM ct BN ti2 14

I ;5 5

.

Trong mt phng to cc vung gc O, cho ABC c ( )A 2;7 ,

ng thng AB ct trc Oyti E sao cho AE 2EB= , ng thi AEC cn ti

A v c trng tm13

G 2;3

. Vit phng trnh cha cnh BC.

Trong mt phng to cc vung gc O, cho hnh ch nht

ABCD c din tch bng 12, tm I l giao im ca ng thng 1d : 3 0 =

v 2d : 6 0+ = . Trung im ca AB l giao im ca 1d vi trc Ox.

Tm to cc nh ca hnh ch nht.

Trong mt phng to cc vung gc O,cho hnh vung ABCD

bit ( )M 2;1 , ( )N 4; 2 ; ( ) ( )P 2;0 ; Q 1;2 ln lt thuc cnh AB, BC, CD, AD . Hy lp

phng trnh cc cnh ca hnh vung.

www.VNMATH.com


9/38

3

Nguyn Ph Khnh

557

31. Trong mt phng to cc vung gc O, cho hnh thoi ABCD c tm

I(2; 1) v AC = 2BD. im 1M 0; 3 thuc ng thng AB; im ( )N 0; 7 thuc

ng thng CD. Tm ta nh B bit B c honh dng.

. Trong mt phng to cc vung gc O, cho im ( )A 2;0 v 2

ng thng 1d : x y 0, = 2d : x 2y 1 0+ + = . Tm cc im 1B d , 2C d

tam gic ABC vung cn ti A .

Trong mt phng to cc vung gc Oxy cho hai ng thng

1 2d : x 2y 1 0,d : 2x 3y 0 + = + = . Xc nh ta cc nh ca hnh vung

ABCD, bit A thuc ng thng d1

, C thuc ng thng d2

v hai im B, Dthuc trc Ox .

Cho hnh bnh hnh ABCD . Bit7 5

I ;2 2

l trung im ca cnh CD ,

3D 3;

2

v

ng phn gic gc BAC c phng trnh l : x y 1 0 + = . Xc nh ta nh

B .

. Trong mt phng to cc vung gc O,cho ba im ( )I 1; 1 ,( ) ( ) J 2; 2 , K 2; 2 . Tm ta cc nh ca hnh vung ABCD sao cho I l tm

hnh vung, J thuc cnh ABv K thuc cnh CD.

Trong mt phng to cc vung gc O,cho ba ng thng

1d : 4 9 0,+ = 2d : 2 6 0, + = 3d : 2 0 + = . Tm ta cc nh ca hnh

thoi ABCD, bit hnh thoi ABCD c din tch bng 15, cc nh A,C thuc 3d , B

thuc 1d v D thuc 2d .

Trong mt phng to cc vung gc O,cho ( ) ( )A 2;2 ,B 7;2 g hg ( ) + =: 3 3 0 . H ( ) cc C D a ch : ABC c A ;

www.VNMATH.com


10/38

3

Nguyn Ph Khnh

558

( )+AD BD g h.

Trong mt phng vi h ta Oxy, cho ba im ( )A 1; 1 , ( )B 0;2 ,

( )C 0;1 . Vit phng trnh ng thng i qua A sao cho tng khong cch t

B v C ti l ln nht.

( ) 2

d : 2 0

= + =

= R ; ( )C d ( )C 2;

( ) ( )AC 3; 2 ,BC 1; 3= + = +

ABC g C h ( ) ( ) = + =AC BC AC.BC 0 3 2 3 0

( ) = 1 1 3 C 1;3 hc

=

2 23 7 3

C ;2 2 2

Gi G l trong tm ca tam gic, suy ra ta ca G l nghim ca h

+ = = = =

73 x 4 y 3 0 x 7G ; 13

3x 10y 17 0 3y 1.

Gi E l trung im ca BC, suy ra3 5

EA GA E 2;2 2

=

.

Gi s ( )B a;b , suy ra ( )C 4 a; 5 b . T ta c h:

( ) ( )

3a 4b 3 0 3a 4b 3 0 a 53 4 a 10 5 b 17 0 3a 10b 45 0 b 3

+ = + = =

= + + = =

.

Vy, ( ) ( )B 5; 3 , C 1; 2 .

Gi 1A i xng vi A qua BD, suy ra 1A BC v ( )1A 1; 4

2A i xng vi A qua CE, suy ra 2A BC v 243 56

A ;5 5

.

Suy ra phng trnh =BC : 3x 4y 19 0.

Ta B l nghim ca h: ( )x y 1 0 x 15

B 15; 163x 4y 19 0 y 16

= =

= = .

Ta C l nghim ca h: ( )x 2y 17 0 x 3 C 3; 73x 4y 19 0 y 7

+ + = = = =

.

www.VNMATH.com


11/38

3

Nguyn Ph Khnh

559

Vy, ( ) ( )B 15; 16 , C 3; 7 .

AB i qua ( )M 2; 3 c phng trnh: ( ) ( ) + + =x 2 b y 3 0a

ABC vung cn ti A =2 212a 7ab 12b 0Gi s ( )+ + = + = AB : 4x 3y 1 0 AC : 3x 4y 7 0 A 1;1 , ( )B 4; 5 , ( )C 3;4 .

+ =B C : 2 x y 3 0 Gi A' l im i xng ca B qua phn gic ACB, ta tm

c ( )A' 6;0 . Ta im A l giao im A'C v AH ( ) A 4; 1

Dg g hg ( )d a ( )A 1;3

( ) ( ) ( ) ( ) ( )d d :2 1 1 3 0 + + =

ha ( )d :2 1 0+ =

( ) ( ) ( ){ }

d B 0;1 AB 5 = =

G ( )C C C CC ; , 0; 0> > . Ta c( )

( )C C

22C C

2 2 0C 2;2

1 5

+ =

+ =

Hh ABCD hh g : BA CD=

; a c :

( )D D

D D

1 2 1D 1;4

2 2 4

= =

= =

Ch hh g : 2P 4.AB 4 5 S AB 5= = = =

NP i qua N v vung gc vi ng cao h t M , nn c phng

trnh: 4x 3y 5 0+ = . ( )P 1; 3 l ta giao im ca NP v phn gic tronggc P . Gi s PI l phn gic trong P th MPI IPN= .PM : y 3 0, = MN : 4x 7y 1 0+ =

:

MH : 3x 4y 27 0, + = phn gic PI : x 2y 5 0+ =

Ly N' i xng vi N qua PI .

Vit NP qua N v vung gc MH

Vit PM qua P c PMu PN'=

Ta c phng trnh BC : + =x y 2 0

Suy ra ta ca B l nghim ca h: x y 2 0 x 12x 5y 13 0 y 3

+ = = + = =

( )B 1; 3 .

www.VNMATH.com


12/38

3

Nguyn Ph Khnh

560

Gi ( )A a; a 2+ , suy ra ta ca trung im AC l +

a 5 a 1M ;

2 2

M M BM nn ( )a 5 a 12 5 13 0 a 3 A 3;52 2+ + = = .

Vy, ( ) ( )A 3;5 ,B 1;3 .

Ta c phng trnh + + =BC : x 2y 5 0.

Ta im C l nghim ca h ( )x y 2 0 x 9

C 9 ; 7x 2y 5 0 y 7

+ = =

+ + = = .

Gi B' l im i xng vi B qua CE, suy ra ( )B' 5;1 v B' AC

Do , ta c phng trnh + =AC : 2x y 11 0.

Ta im A l nghim ca h: + = =

+ = =

52x y 1 0 x 5A ;62

2 x y 1 1 0 2y 6.

Vy, ( )5

A ;6 ,C 9; 72

.

Gi hnh vung cho l ABCD . ( )AB l + =x 2y 12 0.

Gi H l hnh chiu ca E ln ng thng AB . Suy ra ( )H 2; 5

A, B thuc ng trn tm H , bn knh =EH 45 c phng trnh:

( ) ( )+ + =

2 2

x 2 y 5 45

To hai im A, B l nghim ca h:( ) ( )

+ =

+ + =2 2

x 2y 12 0

x 2 y 5 45.

Gii h tm c ( ) ( )A 4;8 ,B 8;2 . Suy ra ( ) C 2; 10

+ =AD : 2x y 16 0, + + =BC : 2x y 14 0, =CD : x 2y 18 0.

Chu vi bng 6 2 =3 2

AB2

. A thuc Ox v ( ) ( )= 3 2

d A,Ox A 2;02

B l hnh chiu ca A trn d nn c to : + =

+ =

x y 1 0 1 3B ;

x y 2 0 2 2 .

BD hp vi d gc 045 v c VTPT ( )= n a; b 0 tho :

www.VNMATH.com


13/38

3

Nguyn Ph Khnh

561

= =

+

0

2 2

a.1 b.1cos 45 BD : 2x 1 0

2 a bhoc =2y 3 0

Gi ( )B BB x ; y , ( )C CC x ; y l ta cn tm.

G l trng tm tam gic nn c: ( )1

GI AI I 1; 33

=

I l trung im BC v tam gic ABC vung cn ti A nn c:

( ) ( )

( ) ( )

B C I

B C I

x x 2x 1

y y 2y 1 B 4;1 ,C 3;2

AB AC B 3;2 ,C 4;1

ABAC 0

+ = = + = = = =

Ta A tha mn h: ( ) =

=

7x 2y 3 0A 1; 2

6x y 4 0

V B i xng vi A qua M nn suy ra ( )B 3; 2 .

ng thng BCi qua B v vung gc vi ng thng: =6x y 4 0nn suy

ra phng trnh + + =BC : x 6y 9 0.

Ta trung im N ca BC tha mn h: =

+ + =

7x 2y 3 0 3N 0;

x 6y 9 0 2

Suy ra ( )= = AC 2.MN 4; 3 .

Phng trnh ng thng + =AC : 3x 4y 5 0

Gi 3a 4 16 3aA a; B 4 a;4 4

+

.

Khi din tch tam gic ABC l: ( )ABC1

S AB.d C, 3AB2

= =

Theo gi thit ta c: ( )2

2 6 3aAB 5 4 2a 25 a 0

2

= + = =

hoc a 4=

Vy hai im cn tm l ( )A 0;1 v ( )B 4;4 .

9 3I ;

2 2 . Gi s M l trung im ( )AD M 3; 0 . = =AB 2IM 3 2

=AD 2 2 ,

= =MA MD 2 . ( )

+ =AD : x y 3 0 .

www.VNMATH.com


14/38

3

Nguyn Ph Khnh

562

Ta A, D l nghim ca h:( )

2 2

x y 3 0

x 3 y 2

+ =

+ =

( ) ( ) ( ) ( )A 2;1 ,D 4; 1 ,C 7 ;2 ,B 5 ;4 Ta c: = B AB BD suy ra ta B l nghim h:

( ) = =

= + = =

x 2y 1 0 x 7B 7; 3

x 7y 14 0 y 3

Gi s ( ) ( )+ A 2a 1; a AB, D 7d 14; d BD

( ) ( ) ( ) = = = AB 6 2a; 3 a , BD 7d 21; d 3 , AD 7d 2a 15; d a

Do ( )( ) = = =AB AD AB.AD 0 3 a 15d 5a 30 0 3d a 6 0

( ) = = a 3d 6 AD d 3; 6 2d

.

Li c:( )

= C C

BC x 7; y 3

. M ABCD l hnh ch nht nn =AD BC

( ) = = +

= + = =

C C

C C

d 3 x 7 x d 4C d 4; 9 2d

6 2d y 3 y 9 2d .

( ) ( ) = = + EA 6d 13; 3d 7 , EC d 2; 8 2d

vi ( )=E 2;1

Mt khc im ( ) E 2; 1 AC EA, EC

cng phng

( )( ) ( )( ) = + + =26d 13 8 2d d 2 3d 7 d 5d 6 0 = =d 2 a 0

Vy ( ) ( ) ( ) ( )= = = =A 1; 0 , B 7; 3 , C 6; 5 , D 0; 0 l cc nh ca hnh ch nht cn

tm.

( )

( )

= = =

= =

1 1 2

1 2

2 1 3

3c c d ;d 34

3c c d ;d

34

V, ABC c c ch + = 6 0 .

V ( ) ( )1 2B d B b; 2 b ,C d C c; c 8 .

Ta c h:( )( )

( ) ( )

= =

= = 2 2

b 1 c 4 2AB.AC 0AB AC b 1 c 4 3

t = = x b 1;y c 4 ta c h: 2 2xy 2 x 2

y 1x y 3

= =

= =

hocx 2y 1

=

=

Vy, ( ) ( )B 3; 1 ,C 5;3 hoc ( ) ( )B 1;3 ,C 3;5 .

www.VNMATH.com


15/38

3

Nguyn Ph Khnh

563

Gi vect php tuyn AB, AC, BCln lt l: ( ) ( ) ( )1 2 3n 1; 2 , n 2; 1 , n a; b

Phng trnh BCc dng: ( ) ( ) + = + >2 2a x 1 b y 2 0,a b 0

Tam gic ABC cn ti A nn: ( ) ( )= =1 3 2 3cos B cos C cos n , n cos n , n

+ + = = =+ +2 2 2 2

a 2b 2a b a b

a ba b . 5 a b . 5

Vi = a b , chn = =b 1 a 1 + =BC: x y 1 0 ( )

2 1B 0;1 ,C ;

3 3 . Khng

tha mn M thuc on BC.Vi =a b , chn = =a b 1 + =BC: x y 3 0 ( ) ( ) B 4; 1 ,C 4;7 . Tha mn M

thuc on BC. Gi trung im ca BC l

( )K 0;3 .

Ta c: ( )( )= + + = 2 2

2 BC BCDB.DC DK KB . DK KC DK4 4

Du bng xy ra khi D K . Vy ( )D 0;3

Ta c ( ) ( )+ + = + + =2 22 2x y 2x 2y 14 0 x 1 y 1 16

Do vy ng trn ( )C c tm ( )I 1;1 v bn knh =R 4.

ng thng d ct ( )C ti hai im phn bit ( ) 2 2a b 0 ) . Khi ta c: ( ) ( )=AB BD AC ABcos n ,n cos n ,n

www.VNMATH.com


16/38

3

Nguyn Ph Khnh

564

= = + + + = =

2 2 2 2a b

3a 2b a b 7a 8ab b 0 b

a27

Vi = a b , chn = = =a 1 b 1 AC : x y 1 0

Vi = b

a7

, chn = = + =a 1 b 7 AC : x 7y 5 0 ( khng tha v AC

khng ct BD )Gi I l tm hnh ch nht th = I AC BD nn to I l nghim ca h:

= =

+ = =

7xx y 1 0 7 52 I ;

x 7y 14 0 5 2 2y

2

Hn na A, C khc pha so vi BD nn: + NA NC AC

ng thc xy ra khi = N AC BD N I . Vy

7 5N ;2 2 .

1B BB ( )B b; 8b 3 , 1C l trung im ca AB nn c

1b 4

C ; 4b 22

+

Mt khc: 1 1C CC nn suy ra

( ) ( )7 b 4 13 4b 2 9 0+ =

( )b 1 B 1; 11 = Gi G l trng tm tam gic ABC, suy ra

ta ca G l nghim ca h :

1x8x y 3 0 1 13 G ;

14x 13y 9 0 1 3 3y

3

= =

= =

B1C1

B

C

Suy ra ( )C G A BC G A B

x 3x x x 2C 2; 11

y 3y y y 11

= =

= =.

Gi I l giao im 2 ng cho hnh ch nht ABCD. V G l trngtm tam gic ABD nn A,G,I thng hng. Theo tnh cht trng tm tam gic ta

www.VNMATH.com


17/38

3

Nguyn Ph Khnh

565

d dng tm ra ta im5 3

I ; .2 2

V I l trung im AC nn bit ta A, I

ta s tm ra ta ( )

C 4;2 .

V D thc ng thng x y 2 0 = m C tha mn phng trnh ny . Do

DC: x y 2 0 = .

Bit phng trnh DC s vit c phng trnh AB m ABCD l hnh chnht nn bit php tuyn AB ta s bit php tuyn AD t vit c phngtrnh AD . Ta D l giao im ca AD v DC . Ta tm c D . V I l trungim BD nn ta tm c nt B

Gi I l trung im ca BD . Theo tnh cht trng tm ta c :

( )

( )

IG A I G

G A I GI

3xx x 2 x x 2AG 2GI

5y y 2 y yy 2

= =

= =

=

Do ABCD l hnh ch nht nn ta c I l trung im ca AC . T :

C M A

C M A

x 2x x 4

x 2y y 2

= =

= = ( )C 4;2

AD DC DA.DC 0 =

( ){ }AB BD B 9; 7 =

Gi I l giao im AC v BD , suy ra ( ) ( )I a; 2 a , D 9 2a; 11 2a +

V BC AB BC : 4x 3y m 0 + = . BCqua im B nn ta tm c m.

Theo gi thit din tch hnh ch nht l 22 nn ta c: AB BC 22. = Hn na

BC AB BC.AB 0 =

d dng tm ra ta C,D.

Gi ( )P PP x ;y i xng vi ( )M 4;6 qua I nnP I

P I

4 x 2x

6 y 2y

+ =

+ =

I thuc ( )d nn ( ) ( )P P

P P3 4 x 5 6 y

6 0 3x 5y 6 02 2

+ + + = = ( )1

Li c PM PN ( ) ( ) ( ) ( )P P P PPM.PN 0 x 4 x 6 y 6 y 2 0 = + =

( )2

T ( )1 v ( )2 , suy ra: 2P P P34y 162y 180 0 y 3 + = = hoc P30

y17

=

Gi M l im i xng vi A qua CF, suy ra M BC . V AM CF

nn AM :x y 3 0+ = .

( )D d :x y 2 0 D x; x 2 . =

www.VNMATH.com


18/38

3

Nguyn Ph Khnh

566

Do AM CF ti ( )I 2;1 , M i xng vi A qua I ( )M 0;3 .

Tng t, gi N l im i xng

vi A qua BE , suy ra N BC v

( )N 2; 1 .

Suy ra ( )MN 2; 4=

phng

trnh BC : 2x y 3 0 + = .

x 1 0B BE BC B :

2x y 3 0 =

= + =

I EF

A

BC

( )x 1

B 1;5y 5

=

=

x y 1 0C CF BC C :

2x y 3 0 =

= + =

( )x 4

C 4; 5y 5

=

= .

Gi s rng: ( ) ( )B 3b; 4b 4 ,C 0;c . +

Ta c: ( ) ( )AC 3;c ,BC 3b; 4b c 4= = +

Gi thit tam gic ABC vung ti C ta c: AC.BC 0=

29b 4bc c 4c 0 + + =

( )1

( )ABC1

S AB.d C; AB2

= , trong : ( ) 3c 12

AB 5 b 1 , d C;AB5

= =

Theo bi ton, ta c: ( ) ( )3c 121 .5 b 1 6 b 1 c 4 42 5

= = ( )2

BC i qua ( )B 1;5 v vung gc AH nn BC: 2x y 3 0 + =

To C l nghim ca h: ( )2x y 3 0

C 4; 5x y 1 0

+ =

=

Gi A' l im i xng B qua ng phn gic ( )d : x y 1 0, =

( )BA d K = . ng thng KB i qua B v vung gc ( )d nn KB c phng

trnh : x y 6 0+ =

To im K l nghim ca h: ( )x y 6 0 7 5

K ; A' 6;0x y 1 0 2 2 + =

=

Phng trnh AC : x 2y 6 0= , ( )A CA' AH A 4; 1=

www.VNMATH.com


19/38

0

Nguyn Ph Khnh

567

Trung im ( )I 0 ; 3 ca AC, ng thi I l trung im BDnn ( )D 1; 11 .

AB : x y 3 0= Gi s ( ) ( )G m; 2 m d C 3m 3;9 3m .

( ) ( )ABC3 1 3 3S .AB.d C;AB d C;AB2 2 2 2

= = = 6m 15 m 23

m 3.2 2 = = =

Phng trnh DC qua D v vung gc ( )d l: 3x 2y 6 0 + = .

Giao im ca DC v ( )d l: ( )M 4; 3 v cng l trung im DC . Suy ra ta

( )C 2; 0 .

Gi C l im i xng ca C qua d' th C AB, phng trnh CC :

x 5y 2 0 + = . Giao im CC v d' l1 1

I ;2 2

.Suy ra ta ( )C' 3;1 .

Phng trnh AB qua C vung gc( )d l: 3x 2y 7 0. =

.V BC AD nn phng trnh BC : 2x y 3 0 + = .

18x y 3 0

B BC BB B :2x y 3 0

==

+ =

( )x 1

B 1;5y 5

=

=

Do A AD , suy ra ( )A 2 2a;a .

Do 1a 5

B a 1;2

.

D

1

C

M 1B BB nn ta c: ( ) ( )a 5

8 a 1 3 0 a 1 A 4; 12

= = .

Gi H l hnh chiu ca M ln AD ta c2 2

H ;3 3

. t AB x=

BC CD 2x = = 3x 1

MH2 3

= = . Vy,2

AD3

= .

www.VNMATH.com


20/38

0

Nguyn Ph Khnh

568

Gi ( )A 2a; a v H l trung im ca AD suy ra ( )D ; v 2AD 3= suy ra

2t 3=

2 2A ;3 3

Gi M l trung im AB 5 5M ; AB : x y 5 0

2 2

=

V G l trng tm ABG ABC1 1

ABC S S .3 2

= =

Gi s ( )0 0G x ; y , ta c:

( ) 0 0 ABGx y 5 2S 1

d G;ABAB2 2 = = = ( )0 0x y 5 1 1 =

V ( )0 0G : 3x y 8 0 3x y 8 0 2 = = T ( )1 v ( )2 suy ra ( ) ( )G 1; 5 C 2; 10 hoc ( ) ( )G 2; 2 C 1; 1

Gi ta im ( ) ( )B 2b 5; b C 2b 5; b + .

im O BC . Ly i xng O qua phn gic ca gc B ta c im

( )M 2; 4 AB ( )BM 7 2b; 4 b = +

( )CK 1 2b; 2 b= +

V tam gic ABC vung ti A nn c: BM.CK 0 b 3= =

hoc b 1= .

Gi ta im ( ) ( )B B B BB 2y ; y A 8 2y ; 4 y

Phng trnh ng thng CI : 2x y 10 0+ =

Gi ta im ( )C CC x ;10 2x CCI 5 4 x , =

BAB 20 y 2=

Din tch tam gic ABC l: ABC B C C B1S CI.AB 10 4y 2x x y 8 22= = + =

( )C B B Cx y 4y 2x 6 1 = hoc ( )C B B Cx y 4y 2x 10 2 =

V( )C B

C B

4 x k 2y 4M BC CM kMB 11 9

2x k y2 2

=

= + =

v By 3

C B B C2x y 6y 5x 16 0 + = ( )3

T ( )1 v ( )3 ta c h: C B B C BC B B C C

x y 4y 2x 6 y 1 2

2x y 6y 5x 16 0 x 1 2

= =

+ = = +

T ( )2 v ( )3 ta c h: C B B C BCC B B C

x y 4y 2x 10 y 3

x 22x y 6y 5x 16 0

= =

= + =

www.VNMATH.com


21/38

Nguyn Ph Khnh

569

Vy, ta cc nh ca tam gic ABC l: ( ) ( ) ( )A 2;1 , B 6;3 , C 2;6

Ta c phng trnh BC : 2x y 3 0 + = .

V ( )12x y 3 0 x 4

C CC BC C : C 4; 5x y 1 0 y 5

+ = = =

= = .

Gi N l im i xng vi B qua

1CC , ta c N AC v ( )N 6;0

NC (10;5) =

, phng trnh

AC : x 2y 6 0 = .

Ta ca A l nghim ca h

( )x 2y 6 0 x 4

A 4 ; 1x 2y 2 0 y 1

= =

+ = = .

I

C1

D

A

BC

Gi im ( )0 0A A 4y 6; y ( )AC 0 0n y 1; 5 4y =

Tam gic ABC vung cn ti A, nn c:

( )0

20 0

6y 7 1cosACB

25 17y 42y 26

= =

+

20 0 0 013y 42y 32 0 y 2 x 2 + = = = hoc 0 0

16 14y x

13 13

= = ( loi ).

Vy, ( ) ( ) ( )A 2;2 , B 3 ; 1 , C 5;3 l ta cn tm. Giao im ca ( )1d v ( )2d l ( )M 3;1 .

: OAB1

S AB.OH2

= vi H l chn ng cao h t O ln AB

2

2OAB

AB 4S OH

=

V OH OM OHmax OM = th

2

4

OHnh nht.

Khi AB nhn OM lm vc t php tuyn. Ta vit c phng trnh AB

:

Phng trnh ng thng d c dng: ( ) ( ) ( )a x 3 b y 1 0 , a,b 0 + = >

Theo bi ton, ta tm c:3a b

A ;0 ,a

+

3a b

B 0;b

+

www.VNMATH.com


22/38

Nguyn Ph Khnh

570

2 2b a

AB 3 1 3a b

= + + +

2

2 2OAB

AB 4 4S a b

1 3 3b a

= +

+ +

. ta

t , t 0b

= >

Xt hm s: ( )( )

2

2

t 1f t 4.

3t 1

+=

+vi t 0>

Gi tr nh nht ca ( )f t l25

t c khi t 3= hay a 3b=

Phng trnh ng thng cn tm l: 3x y 10 0+ =

I thuc cung AB ca ( )P sao cho din tch IAB ln nht I xa AB nht,

tc I l tip im ca tip tuyn ( )d AB ca ( )P .

Phng trnh ng thng ( )AB : y 3x 1 d : y 3x c= = +

( )d tip xc ( )P ti im I 1 7 17 1

I ; C 1; , D 2;2 4 2 2

( )1B BC d B 0; 1= ( )BM 2; 2 =

. Do BM

l mt vc t php tuyn

ca BC MB BC

K MN BC ct 2d ti N ,v tam gic ABCcn ti A nn t gic BCNM l

hnh ch nht.

Do ( )MN BCQua M 2;1 MN : x y 3 0 + = , 2 8 1N MN d N ;3 3 =

DoNC BC

8 1Qua N ;

3 3

7NC : x y 0

3 = , 1

2 5C NC d C ;

3 3

=

Hn na: ( )4 8

CM ; n 1;23 3

l mt vc t php tuyn ca AB nn phng

trnh A B : x 2 y 2 0+ + = v ( )8 4

BN ; u 2;1 AC : 6x 3y 1 03 3

+ + =

( ) ( ) ( )A a ; 0 ,B b ; b 1 ,C 4 2c ; c .+

www.VNMATH.com


23/38

Nguyn Ph Khnh

571

Ta c:d2

d1

2a 1bAB.n 0 2a 1 2a 2 2a 4 4 a3 B ; , C ;

4 a 3 3 3 3AC.n 0c 3

= = + +

= =

Gi H l trc tm tam gic ABC, suy ra ta ca H l nghim ca h

1x2x y 1 0 1 53 H ;

x y 2 0 5 3 3y

3

= + =

+ = =

.

V C d C(a; 6 a) . Do AC BE nn

phng trnh AC c dng:

x y 6 2a 0 + =

Tng t, phng trnh

BC : x 2y a 12 0 0+ + = = .

Suy rax 2y a 12 0

B :x y 2 0

+ + =

+ =

( )x a 8 B a 8; 10 ay 10 a =

=

d

D

E

H

A

B

C

2x y 1 0A :

x y 6 2a 0 + =

+ = ( )

x 5 2aA 5 2a;11 4a

y 11 4a =

=

Suy ra ( ) ( )MC a;3 a ,MB a 8;7 a= =

.

V B,C,M thng hng nna 8 a 7

a 6a a 3

= =

.

Vy , ( ) ( ) ( )A 7; 13 , B 2; 4 , C 6; 0 .

Vit BC qua M v vung gc vi AD

Vit CA qua C v vung gc vi BE

www.VNMATH.com


24/38

Nguyn Ph Khnh

572

C, D ln lt thuc 2 ng thng 1d : 3x y 4 0, = 2d : x y 6 0+ =

( )C a; 3a 4 , ( )D b; 6 b ( )CD b a;10 b 3a =

Gi ( )n 2;1=

l mt vect php tuyn ca AB.

V ABCD l hnh vung nn( )

CD n

d C; AB CD

=

( ) ( )

( ) ( )2 2

2 2

2 b a 1. 10 b 3a 0b 5a 10

2b 6 b 1a 1 4a 10b a 10 b 3a

2 1

+ = =

+ = = +

+

Vit phng trnh ng thng qua A song song vi CM

Tm cc giao im E, F ca AE,CM vi BD suy ra I l trung im ca EF.

Tnh c im C

Tm c trung im H ca CM chnh l hnh chiu vung gc ca A lnCM suy ra ta im M .

Vit phng trnh AD i qua A, M . Tm c im D suy ra B .

ng thng AB i qua M nn c phng trnh ( ) ( )a x 2 b y 3 0 + + =

( )2 2a b 0+ . ( ) 0AB;BC 45= nn 02 2

a 7b 3a 4bcos45

4a 3b50 a b

+ == = +

.

Nu 3a 4b,= chn a 4, b 3= = c ( )AB : 4x 3y 1 0+ + = . ( )AC : 3x 4y 7 0 + = .

T ( )A 1;1 v ( )B 4; 5 . Kim tra MB 2MA=

nn M nm ngoi on AB

Nu 4a 3b,= chn a 3, b 4= = c ( )AB : 3x 4y 18 0 = ,

( )AC : 4x 3y 49 0+ = ( )A 10; 3 , ( )B 10;3 ( khng tha )

Ta nh B l nghim ca h ( )x 2y 4 0

B 0; 27x 4y 8 0

=

=

www.VNMATH.com


25/38

Nguyn Ph Khnh

573

V ABC cn ti A nn AG l ng cao

ca ABC , suy ra phng trnh AG c

dng: 4 12 x 1 y 03 3

+ =

2x y 3 0 + =

Gi H AG BC= th ta im H l

nghim ca h : ( )2x y 3 0

H 2 ; 1x 2y 4 0

+ =

=

HB C

A

G

V H l trung im ca ( )C H BC H B

x 2x xBC C 4; 0

y 2y y

=

= .

Ta c ( ) ( ) ( )G A B C G A B C1 1x x x x ,y y y y A 0; 33 3= + + = + + .

Vy, ( ) ( ) ( )A 0;3 ,B 0; 2 ,C 4 ;0 .

( ) ( )I I t; 3 2t , t 0 >

2IC 2IB 15t 10t 25 0= + = 5

t3

= ( khng tha t 0> ) hoc t 1= ( )I 1;1 .

Phng trnh ng thng IC: x y 2 0+ =

( )ABC1

S AC.d B; AC AC 6 22

= = .

V ( )A IC A a; 2 a nn c: ( )2a 5 36 a 11 = = hoc a 1= ( )A 1; 3 Phng trnh ng thng CD : y 3 0+ =

Ta D l nghim ca h: ( )x y 0

D 3; 3y 3 0

=

+ =

Cnh AC nm trn ng thng i qua M v vung gc vi BH

Phng trnh cnh AC : x y 0 = .

Ta im A l nghim ca h:x 4y 2 0

x y 0

=

=

2

x y

3

= =

2 2A ;

3 3

.Suy ra ta im8 8

C ;3 3

.

www.VNMATH.com


26/38

Nguyn Ph Khnh

574

Cnh BCi qua C v song song vi ng thng d nn c phng trnhBC : x 4y 8 0 + =

Ta nh B l nghim ca h: ( )x y 3 0 x 4 B 4;1x 4y 8 0 y 1 + + = = + = = .

Ta h A ga ca AB AD ( )A x;y gh ca h

4xx 2y 2 0 4 53 A ;

2x y 1 0 5 3 3y

3

= + =

+ + = =

.

Phg h g h gc gc A

x 2y 2 2x y 1

5 5

+ + += ha

( )

( )1

1

d : x y 3 0

d : 3x 3y 1 0

+ =

+ =.

Tg h ( )1d : x y 3 0 + = .g hg ( )BD a M g gc ( )1d ( )BD : x y 3 0+ = .

S a ( )B AB BD B 4; 1= , ( )D AD BD D 4;7= .

G ( ) ( )1I BD d I 0; 3= . V C g A a I 4 13

C ;3 3

.

Tg h ( )2d : 3x 3y 1 0+ = . B c g .

Gi d : x y 4 0+ = .

V BC d

nn phng trnhBC

c dng: x y m 0+ + =

Ly ( )I 1; 3 d , ta c: ( ) ( )d I, BC d A,d 4 2= = m 4 8 + = m 12,m 4 = =

V A v I cng pha so vi BC nn ta c m 4 BC : x y 4 0= + + = .

ng cao h t nh A c phng trnh : x y 0 = .

Ta trung im P ca ( )x y 0

BC : P 2; 2x y 4 0

=

+ + =

Do ( )B BC B b; 4 b v P l trung im BCsuy ra ( )C 4 b; b

Mt khc AB CE nn ta c ( )( ) ( )( )b 6 b 4 b 10 b 3 0 + + + + =

b 0, b 6 = =

Vy c hai b im tha yu cu bi ton:

www.VNMATH.com


27/38

0

Nguyn Ph Khnh

575

( ) ( )B 0; 4 , C 4; 0 hoc ( ) ( )B 6; 2 , C 2; 6 .

Do qua O, nn c phng trnh dng : x 0= hoc y kx=

Nu phng trnh : x 0= , khi ( )1A d : x y 2 0 A 0; 2= =

( )2d : 2x y 5 0 B 0; 5 OA.OB 10 + = = ( tha mn )

Nu phng trnh : y kx=

Do 1A d= nn ta ca A l nghim ca h phng trnh:

x y 2 0

y kx =

=

2x

1 k2k

y1 k

=

=

2 2kA ;

1 k 1 k

Do 2B d= nn ta ca B l nghim ca h phng trnh:

2x y 5 0y kx

+ =

=

5x2 k

5ky

2 k

= + = +

5 5kB ;

2 k 2 k

+ +

Khi : OA.OB 10= ( ) ( )

2 22 2

2 2

4 4k 25 25kOA .OB 100 . 100

1 k 2 k

+ + = =

+

( ) ( ) 2 22 22 2

2 2

k 1 k k 2k 1 k k 2

k 1 k k 2

+ = + + = + + = +

k 3

1k 1,k

2

= = =

Phng trnh ca ng thng l y 3x, y x ,= =

1

y x2=

Gi C ' l im i xng ca C qua ng phn gic AD .

Khi C ' AB .

Gi ( ) ( )H AD CC' H 5 2t; t CH 1 2t; t 3= =

Mt khc ADc ( )u 2;1=

l VTCP v do CH u

nn ta c:

( ) ( )CH.u 0 2 1 2t 1 t 3 0 t 1= + = =

( )H 3;1 .

Do H l trung im ca CC ' , nn ( )C' 2 ; 1 .

www.VNMATH.com


28/38

Nguyn Ph Khnh

576

V A AD AM= ( M l trung im ca BC ) nn ta ca A l nghim ca h

phng trnh : ( )x 2y 5 0 x 9

A 9 ; 24x 13y 10 0 y 2

+ = =

+ = = .

Khi ng thng AB c phng trnh x 7 y 5 0+ + = nn ( )B 7t 5; t .

V13s 10

M AM M ; s4

+

.

Li v M l trung im ca BC nn ( )13s 10 14t 2

B 12;12s 3 t

+ =

= +.

Gi M l trung im cnh BC, do tam gic ABC cn ti A nn

AM BC .Suy ra phng trnh ca AM : x y 3 0+ = .

Ta im M l nghim ca h:x y 4 0

x y 3 0 =

+ =

7x

21

y2

=

=

7 1 9M ; AM

2 2 2

=

.

Ta c: ABC1 18

S AM.BC AM.BM 18 BM 2 22 AM

= = = = =

Mt khc: B , suy ra ( )B b; b 4 nn:

2 22 7 7BM 8 b b 8

2 2

= + =

27 11 3

b 4 b ,b2 2 2

= = =

.

Vi11 11 3 3 5

b B ; ,C ;2 2 2 2 2

=

.

Vi11 3 5 11 3

b B ; ,C ;2 2 2 2 2

=

.

Trung im I ca AB l ( )I 1; 3 , v G l trng tm tam gic ABC nn

suy ra : ( ) ( )AGB ABC1 1 1

S S 1 d G,AB .AB 1 d G,AB3 2 2

= = = =

www.VNMATH.com


29/38

Nguyn Ph Khnh

577

V G d nn suy ra ( )G a;3a 1+ .

Phng trnh ng thng AB :x y 2 0+ + =

nn ( )

4a 3

d G,AB 2

+=

Do ( ) 4a 31 1 1

d G,AB a 1,a22 2 2

+= = = =

1 1 1

a G ;2 2 2

=

, m

CA B C G

A B C GC

7xx x x 3x 2

y y y 3y 9y

2

= + + =

+ + = =

.

Do 7 9

C ;2 2

Tng t vi a 1= ta tm c ( )C 5; 0 .

Gi ( )M 1; 3 l trung im AB , ( )G d G t;1 3t +

G l trng tm tam gic ABC nn

2CM CG

32

AN AG3

=

=

Hn na Nl trung im BC v ( )ABC1

S AB.d C,AB2

=

Gi D l im i xng vi C qua ng thng d : x y 5 0+ = , ta tm

c ( )D 4;9 .

V Athuc ng trn ng knh CD nn A l giao im ca ng thng d v ng trn ng knh CD, suy ra ta ca A l nghim ca h:

( ) ( )22

x y 5 0A 4;1

x y 5 32

+ =

+ =v Ax 0>

www.VNMATH.com


30/38

Nguyn Ph Khnh

578

Suy ra ABC2S

AC 8 AB 6AC= = = .

V B thuc ng thng AD : x 4 0 = nn( )B 4;y .

T ( )2

AB 6 y 1 36 y 5,y 7= = = =

V AB

v AD

cng hng nn ta c

( )B 4;7 BC : 3x 4y 16 0 + = .

Gi ( )A a;b , suy ra ( )DA a 2; b 6 ,=

( )MN 3; 3=

.

V AD MN DA.MN 0 a 2 b 6 0 a b 4 = + = =

( )1 .Ly i xng im A qua M,N ta c: ( ) ( )B 2 a; b , C 8 a; 6 b

Suy ra ( ) ( )BD a; 6 b , CD a 6; b 12= + = +

V B,C,D thng hng nn ta c:a 6 b 12

a b 6a b 6 +

= + = +

( )2 .

T ( )1 v ( )2 ta suy ra a 5, b 1= = .

Vy, ( ) ( ) ( )A 5; 1 ,B 7 ;1 ,C 13; 5 .

MN : x y 1 0+ =

Vit AD qua D v vung gc MN nn c x y 4 0 + =

AD ct MN ti3 5

H ;2 2

. Sau s dng tnh cht trung im .

( ) ( ) =

4 4AB AC A ;

9 9 . ( )

( ) ( ) ( )

+ =

a M: :2 5 6 0

AC

( ) ( ) =

22 2AB N ;

9 9 . V Ng ( )AB =

AB 2AN

D

C A

B

www.VNMATH.com


31/38

Nguyn Ph Khnh

579

=

+ = +

4 22 4 2

9 9 9 40 11B ;

9 97 2 7

29 9 2

. V M g ( )BC

76 25C ;

9 9

( ) ( ) ( ) = AB AC A 4;1 : ( )= 1

GM AG M 1; 22

M g BC ( )

( )

+ = = + = =

+ + =

+ + =

B C M

B C M

B B

C C

2 2

B 3; 3 2 4B AB

C AC 4 15 0 C 1; 1

2 5 3 0

G ( ) ( )0 0 M MA ; ,M ; g ( )AB

+ += =0 0M M

3 5 ,

2 2 M hc g : + =M M 5 0

+ =0 0 2 0 , A hc g ca : + =0 02 5 3 0

Ta c( )

( )

+ =

+ =

0 0

0 0

A 1;1 2 0

2 5 3 0 M 2;3

( ) ( )AB : ax b y 2 0,+ = ( ) ( ) ( )AD : b x 2 a y 4 0 =

d P,AB d N,AD 3a b 0 = + = hoc a 7b 0+ =

( ) ( ) ( ) = = = +

0 0C ;3 3 AB 3;3 ,AC 1;2

( ) ( )= = + = =

0 01 1

S 2 de AB;AC 2 3.2 1 1 2 92 2

hc =0 1

( ) 1C 1;3 hc ( )2C 9;3

* ( ) ( ) ( ) = = =

C 1;3 AC 2;2 ,BC 1;1 AC.BC 0 AC BC

ABC g = =AB 10

C R2 2

* ( ) ( ) ( ) ( ) = = =

C 9;3 AC 10;2 , BC 7;1 , AB 3;1

( )= = = =

28 1cA c AB;AC A 1 c A65 65

The h h : = =BC 5

R 1302 A 2

www.VNMATH.com


32/38

Nguyn Ph Khnh

580

Phng trnhBC :2 3 0 + = ,

ta ca im C l nghim

ca h: + =

=

2 3 0

1 0

=

=

4

5

( ) C 4; 5 .

B'

HB C

Gi B' i xng vi B qua d, ta tm c ( )B' 6;0 v B' AC .

Suy ra phng trnh AC : 2 6 0 = .

Ta im A l nghim ca h: ( ) 2 6 0 4

A 4; 1 2 2 0 1

= =

+ = = .

V ( )AD BC D 1; 11=

.

V 1 2d d v ABC vung cn ti A nn A cch u 1 2d ,d , do A

l giao im ca d v phn gic hp bi 1 2d ,d .

Phng trnh phn gic hp bi 1 2d ,d l ( )x 1 y 2 = ( )1t : x y 1 0 =

hoc ( )2t : x y 3 0+ + = khng tha v ( ) ( )2t d .

Ta im A l nghim h: ( )x y 1 0

A 3;2x y 5 0

=

+ =

Gi ( ) 1B 1; b d , ( ) 2C c; 2 d

Theo bi ton ta c:( ) ( )

( ) ( )2B 1; 5 , C 0; 2AB.AC 0

B 1; 1 , C 6; 2BC 50

= =

Gi ( )A 3 2a; a , ( )B 3 2b; b l ta cn tm. G l trng tm ABC

v

thuc ng thng x y 2 0+ = nn suy ra a b 2 0+ = . Hn na AB 5= suy ra

( )2

a b 1 = . T y, tm c3

a ,

2

= 1

b

2

=

( )B d : x 2y 6 0 B 2y 6; y + =

www.VNMATH.com


33/38

Nguyn Ph Khnh

581

Ta thy AMB v BNC vung bng nhau AI BI IA.IB 0 =

y 4 =

( )B 2;4 . ( )BC : 2x y 0 C c; 2c , = AB 2 5,= ( ) ( )2 2BC c 2 2c 4= +

Theo bi ton, ( ) ( )AB BC c 2 2 C 0;0 ,C 4;8= =

V I nm trong hnh vung nn I, C cng pha vi ng thng

( )AB C 0; 0 .

Gi I l trung im ca EC . V G l trng tm AEC nn

2AG AI

3=

( )I 2;3 .Hn na E Oy nn ( )E o ;e

V AEC cn ti A nn AI EC AI.EC 0 e 3 = =

( )E 0;3 , ( )C 4;3 .

Mt khc, ( )AE 2EB B 1;1=

Ta c 1 2 3 0 9 3

d d I : I ; 6 0 2 2

= = + =

Gi M l giao ca ng thng 1d vi O, suy ra ( )M 3;0 .

V AB MI nn suy ra phng trnh AB: 3 0+ =

ABCDSAD 2MI 3 2 AB 2 2 AM 2AD

= = = = =

M ( ) ( )2

2A AB A a;3 a AM 2 a 3 1 = = a 2,a 4 = = , ta chn( ) ( )A 2;1 ,B 4; 1 .

Do I l tm ca hnh ch nht nn ( ) ( )C 7;2 , D 5;4 .

Vy, ta cc nh ca hnh ch nht l: ( ) ( )A 2;1 ,B 4; 1 , ( ) ( )C 7;2 , D 5;4 .

Trc ht ta chng minh tnh cht sau y:

Cho hnh vung ABCD, cc im M,N,P,Q ln lt nm trn cc ng thng

AB, BC, CD, DA. Khi MP NQ MP NQ= .

www.VNMATH.com


34/38

Nguyn Ph Khnh

582

Chng minh: V ME CD, E CD,

NF AD, F AD

Hai tam gic vung MEP v NFQ cNF ME= .

Do MP NQ MEP NFQ= =

0EPM FQN QIM 90 MP NQ = =

Tr li bi ton:

Ta c: ( )MP 0; 1 MP 1= =

. Gi d l

I

E

F Q

B C

D

M

P

N

ng thng i qua N v vung gc vi MP .

Suy ra phng trnh d : 4 0 = .Gi E l giao im ca d vi ng thng AD, p dng tnh cht trn ta suy ra

NE MP=

M ( )E 4; nn ( )2

NE MP 2 1 3, 1= = = = .

Vi 3= ( ) ( )E 4;3 QE 3;1 =

, suy ra phng trnh AD : 3 5 0 + =

Phng trnh AB :3 7 0, BC : 3 10 0, CD :3 6 0+ = = + = .

Vi 1= ( ) ( )E 4;1 QE 3; 1 =

, suy ra phng trnh AD : 3 7 0+ =

Phng trnh AB :3 5 0, BC : 3 2 0, CD :3 6 0 = + + = = .

Gi N l im i xng ca N qua tm I th ta c N(4; 5) v N thuccnh AB .Suy ra

16MN' 4;

3

=

nn phng trnh AB: 4x + 3y 1 = 0

V AC = 2BD nn AI = 2BI. Gi H l hnhchiu ca I ln AB, ta c:

( ) 8 3 1

IH d I,AB 25

+ = = = v

2 2 2 2

1 1 1 5 IH 5IB 5

2IH IA IB 4IB= + = = =

Mt khc ( )2

221 4b 4b 2B AB B b; ,b 0 IB b 2 5

3 3

+ > = + =

b 1 = . Vy ( )B 1; 1 cnh BC : 2x 5y 7 0 + = .

'

H

D

B

IA

C

www.VNMATH.com


35/38

Nguyn Ph Khnh

583

( ) ( )1 2B d ,C d B b; b , C 1 2c; c

4c 6AB AC AB.AC 0 bc 3b 4c 6 0 b

c 3

+ = + + = =

+

( )1

( )2 2AB AC 2 b 1 5c 12c 7= = + + ( )2

T ( )1 v ( )2 suy ra2

24c 62 1 5c 12c 73 c

+ = + + +

( )( )( )4 3 2 25c 42c 106c 114c 45 0 c 1 c 5 5c 12c 9 0 + + + + = + + + + = ( ) ( )c 1 B 1;1 , C 1; 1= hoc ( ) ( )c 5 B 7;7 , C 9; 5= .

V 1 2A d ,C d nn ( ) ( )A 2a 1;a ,C 3c; 2c , suy ra

2a 3c 1 a 2cI ;2 2 + l trung im AC

Do ABCD l hnh vung nn I l trung im ca BD, hay I Ox.

Do =a 2c.

Mt khc AC BD Ox nn suy ra 2a 1 3c c 1 = = .

T , ta tm c ( ) ( ) ( )A 3;2 , C 3; 2 , I 3;0 .

V ( )B Ox B b;0 , m IB IA 2 b 3 2 b 5,b 1= = = = = .

Vy ta cc nh ca hnh vung ABCD l:

( ) ( ) ( ) ( )A 3;2 , B 1;0 , C 3; 2 , D 5;0 hoc ( ) ( ) ( ) ( )A 3;2 , B 5;0 , C 3; 2 , D 1;0 .

: im I l trung im ca CD nnC I D

C I D

x 2x x 4

7y 2x y

2

= =

= =

7

C 4;2

V A nn ta im A c dng ( )A a; a 1+

Mt khc ABCD l hnh bnh hnh tng ng vi DA,DC

khng cng phng

v ( )B

B

BB

x a 4 3 x a 1AB DC B a 1;a 37 3 y a 3y a 1

2 2

= = + = + +

= + =

DA,DC

khng cng phng khi v ch khi

3a 1a 3 112 a

1 2 2

+

www.VNMATH.com


36/38

Nguyn Ph Khnh

584

ng thng l phn gic gc BAC nhn vect ( )u 1;1=

lm vec t ch

phng nn

( ) ( )

AB.u AC.ucos AB;u cos AC;u

AB u AC u= =

( )

C ( ) 5

AB 1; 2 , AC 4 a; a2

nn ( )

( )2

2

132a3 2

5 54 a a

2

=

+

22a 13a 11 0 a 1 + = = hoc11

a2

= ( loi )

Vy, ta im ( )B 2;4

: Ta c7

C 4;

2

.

ng thng d i qua C vung gc vi nhn ( )u 1;1

lm vect php

tuyn nn c phng trnh l ( ) 7

1. x 4 1. y 02

+ =

hay 2x 2y 15 0+ =

Ta giao im H ca v d l nghim ca h:

13xx y 1 0 13 174 H ;

2x 2y 15 0 17 4 4y

4

= + =

+ = =

Gi C' l im i xng vi C qua th khi C' thuc ng thng cha cnh

AB v H l trung im ca CC' do C' H C

C' H C

x 2x x

y 2y y

=

= C'

C'

5x2

y 5

=

=

5

C' ; 52

Suy ra ng thng cha cnh AB i qua C' v nhn ( )DC 1;2

lm vect ch

phng nn c phng trnh l5

x t2

y 5 2t

= +

= +

Thay x, y t phng trnh ng thng cha cnh AB vo phng trnh ng

thng ta c5 3

t 5 2t 1 0 t

2 2

+ + = = suy ra ( )A 1;2

ABCD l hnh bnh hnh nn B B

B B

x 1 1 x 2AB DC

y 2 2 y 4 = =

= = =

www.VNMATH.com


37/38

Nguyn Ph Khnh

585

Vy, ta im ( )B 2;4

Gi J i xng vi J qua I, ta c J(4; 0) v J CD.

Ta c: ( )KJ ' 2; 2=

, suy ra phng trnh

CD : x y 4 0 = .

V AB / /CD nn phng trnh:AB : x y 4 0 + =

Do ( )d I,AB 2 2= nn suy ra AB 4 2 IA 4= =

( )A AB A a; 4 a + , do ( ) ( )2 2

IA 4 a 1 a 3 16= + + =

2a 2a 3 0 a 1,a 3 + = = =

a = 1, ta c ( ) ( ) ( ) ( )A 1;3 , B 3;1 , C 1; 1 , D 5 ;1

a = 3, ta c ( ) ( ) ( ) ( )A 3;1 , B 1;3 , C 5 ;1 , D 1; 1 .

V BD AC nn phng trnh BD: = +

1B BD d= , suy ra 9 4 9

B B ;4 9 0 3 3

= + + =

Tng t2

6 2 6D BD d D ;

3 3

+ =

.

Suy ra ta trung im ca BD l1 2 1

I ;2 2

.

V1 2 1

I AC 2 0 32 2

+ = = . Suy ra ( ) ( )

1 5B 2;1 ,D 1;4 ,I ;

2 2

Ta c: 2BAD ABCD1 15 15 5 25

S S AI AI2 2 BD 22

= = = = =

M ( )2

23

1A d A a;a 2 AI 2 a

2

+ =

nn ta c:

21 25

a a 3,a 22 4

= = =

Vy ta cc nh ca hnh thoi l:

( ) ( ) ( ) ( )A 3;5 ,B 2;1 ,C 2;0 ,D 1;4 hoc ( ) ( ) ( ) ( )A 2;0 ,B 2;1 ,C 3;5 ,D 1;4 .

'

I

B C

D

K

www.VNMATH.com


38/38

Nguyn Ph Khnh

586

( ) ( ) x 3t

: x 3y 3 0 :

y 1 t

= + =

=

. V ( )C ( )C 3;1

ABC c =A AB AC AB,AC hg cg hg

( ) ( ) = + 2 2

5 3 2 1 2 AB,AC hg cg hg

1

2

= =

V AB,AC hg cg hg 2 = ( )AB 5;0=

hg cg hg

( )AC 4; 3=

( )C 6; 1

. G ( ) ( ) 2 2D D 3;1 AD BD 10 10 5 10 40 50 + = + + +

X

10 10

a 10 ;2 2

=

( )b 2 10 10; 10=

Ta c AB BD a b a b 3 3+ = + + =

V ( )AB BD 3 5+ = . Kh ( )

= =

1

2 2a b 1 D 3;01 12

38. Gi s i qua im A v c vect php tuyn l ( )n a; b 0=

, nn

c phng trnh: ( ) ( )a x 1 b y 1 0+ + + =

( )2 2

a 3bd B, ,

a b

+ =

+ ( )

2 2

a 2bd C,

a b

+ =

+

Gi ( ) ( )2 2 2 2

a 3b a 2bd d B, d C,

a b a b

+ += + = +

+ + ( )

2 2

1a 3b a 2b

a b= + + +

+

( ) ( )( )2 2 2 22 22 21 1

d 2 a 5 b 2 5 a b 29a ba b

+ + + =++

ng thc xy ra khi

ab 0

a a 2,b 5 :25b

>

= = =

2x 5y 7 0+ + =

www.VNMATH.com

b Duong Thang.doc

Documents