Chinh phục hình học tọa độ phẳng Oxy

CHINH PHUC

HNH HOCHNH HOCHNH HOCHNH HOC

ta phng Oxy (phan 1)

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MT S K THUT IN HNH KHI GII TON HNH HC TA PHNG

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Bi 1: ABC ni tip ng trn ng knh AD, M(3; 1) l trung im cnh BC. ng cao k t B ca ABC i qua im E( 1; 3) , im F(1;3) nm trn ng thng AC. Tm ta nh A v vit phng trnh cnh BC bit D(4; 2)

Hng dn tm li gii

+ Trc ht, khi gp loi bi tp m tam gic ni tip ng trn, d kin bi cho ng cao ca tam gic th ta thng ngh n vic to ra 1 hnh bnh hnh bng cch: - Nu tam gic c 2 ng cao th ta ch vic k 1 ng knh i qua nh cn li (khng cha 2 ng cao kia). - Nu tam gic c ng knh i qua nh v 1 ng cao th ta s k ng cao th 2 (bi ton ny ta s lm nh vy) + Vi bi ton ny ta s to ra im H l trc tm ABC ta chng minh c BHCD l hnh bnh hnh (ci ny qu quen ri phi khng - t lm nh) + Cng vic chun b xong, by gi ta s lm theo cc bc suy lun sau nh: - Thy ngay H l trung im AC H(2;0) - Lp c phng trnh BH (qua 2 im H v E) BH : x y 2 0 = - Lp c phng trnh DC (qua D v // BH) DC : x y 6 0 = - Lp c phng trnh AC (qua F v BH ) AC : x y 4 0 + = - Ta C AC DC= , gii h C(5; 1) - Lp phng trnh BC i qua 2 im M v C BC : y 1 0 + = - Lp phng trnh AH (qua H v BC ) AH : x 2 0 = - Ta A AH AC= , gii h A(2;2)

Bi 2: Cho ABC ni tip ng trn (C), ng phn gic trong v ngoi ca A ct ng trn (C) ln lt ti M(0; 3), N( 2;1) . Tm ta cc im B, C bit ng thng BC i qua E(2; 1) v C c honh dng.

Hng dn tm li gii + Trc ht ta thy ngay AN AM (t.c phn gic ca 2 gc k b) ng trn (C) s c tm I( 1; 1) l trung im MN, bn knh ( ) ( )2 2MNR 5 (C) : x 1 y 1 5

2= = + + + =

+ Nh vy n y thy rng tm ta B, C ta cn thit lp phng trnh ng thng BC ri cho giao vi ng trn (C).

F(1;3)H

M(3;-1)

E(-1;-3)

D(4;-2)

C

B

A ?



x

N(1;1)

M(-1;0)

O(0;0)

3x+y-1=0

CB

A

+ Quan st tip thy BC qua E(2;-1) ri, gi th ta cn tm VTCP hoc VTPT na l n ng khng !

Nu v hnh chnh xc th ta s d on c BC MN !!! (ta s chng minh nhanh nh: 1 2A A MB MC= = M l im chnh gia BC H l trung im BC ( H MN BC= ) BC MN (q. h gia ng knh v dy cung - hnh hc lp 9)) + Nh vy, tm li, ng thng BC qua E,

MN BC : x 2y 4 0 = + Cui cng, ta ch cn gii h phng trnh

gm 6 7(C) BC B( 2; 3),C ;5 5

Bi 3: Cho ABC ni tip ng trn tm O(0;0). Gi M(-1;0, N(1;1) ln lt l cc chn ng vung gc k t B, C ca ABC . Tm ta cc nh A, B, C ca ABC , bit im A nm trn ng thng c phng trnh : 3x + y - 1 = 0 Hng dn tm li gii + Ta thy A A(a;1 3a) , by gi cn thit lp 1 phng trnh tm a. + V hnh chnh xc cc ta s d on c AO MN (Tht vy: ta s c.minh nhanh nh sau: k tip tuyn Ax AOAx (*), c

sdACxAC ABC

2= = , m ABC AMN= (do

t gic MNBC ni tip) xAC AMN / /MNAx = (**). T (*) v

(**) AO MN ) Gii phng trnh : AO.MN 0 a 1 A(1; 2)= =

+ ng thng AB i qua A, N AB : x 1 0 =

+ ng thng AC i qua A, M AC : x y 1 0 + + =

+ ng cao BM i qua M v AC BM : x y 1 0 + =

+ Ta B AB BM B(1;2)= , tng t C( 2;1) Nh vy im quan trng nht i vi bi ny l pht hin ra AO MN

21

H

IE(2;-1)

N(-2;1)

M(0;-3)

C

B

A



By gi ta cng vn dng PP trn lm bi tng t sau nh: Bi 4 : Cho ABC ni tip ng trn tm I(1;2), bn knh R = 5. Chn ng cao k t B, C ln lt l H(3;3), K(0;-1). Vit phng trnh ng trn ngoi tip t gic BCHK, bit A c tung dng

Hng dn tm li gii

+ ng trn (C) tm I, bn knh R = 5 c phng trnh ( ) ( )2 2x 1 y 2 25 + = + Ta thy ngay ng trn ngoi tip t gic BCHK c tm M l trung im BC, ng knh BC (do 0BKC BHC 90= = ). Nh vy vn quyt nh ca bi ton ny l i tm ta B, C. + Theo bi ton gii thiu ln trc, do ta chng minh c AI KH AI l t qua I, AI KH AI c phng trnh: 3x 4y 11 0+ = + Ta A AI (C)= , gii h c A( 3;5) + ng thng AB i qua A, K AB : 2x y 1 0 + + = + Ta B AB (C)= , gii h c B(1; 3) , suy lun tng t c C(6;2)

Vy ng trn ngoi tip t gic BCHK c tm M l trung im BC, ng knh

BC c phng trnh: 2 27 1 25

x y2 2 4

+ + =

Bi 5: (KD-2014) Cho ABC ni tip ng trn, D(1;-1) l chn ng phn gic ca A , AB c phng trnh 3x 2y 9 0+ = , tip tuyn ti A c phng trnh : x 2y 7 0+ = . Hy vit phng trnh BC.

Hng dn tm li gii

+ Vi d kin bi cho, trc ht ta xc nh c ngay ta A AB A(1;3)= + ng thng BC i qua D(1;-1) nn lp phng trnh BC ta cn tm ta mt im na thuc BC. Gi E BC=

( )E E 7 2x;x

3x+2y-9=0

x+2y-7=0

E

A

CB D(1;-1)

1 1

2

1

K(0;-1)H(3;3)

I(1;2)

C

D

B

A



+ By gi cn thit lp 1 phng trnh tm x, v hnh chnh xc s cho ta d on EAD cn ti E gii phng trnh ED = EA s tm c x 1 E(5;1)= .

(chng minh EAD cn ti E nh sau: 1 1D C DAC= + (gc ngoi ADC ), m

1 1sdABC A

2= = ,

2 1 1 2DAC A D A A EAD EAD= = + = cn ti E)

+ ng thng BC i qua 2 im E v D BC : x 2y 3 0 =

By gi ta cng vn dng PP trn lm bi tng t sau nh: Bi 6 : Cho ABC c nh A(1;5) . Tm ng trn ni tip v ngoi tip ABC ln lt l 5I(2;2), K ;3

2

. Tm ta B, C

Hng dn tm li gii Mi bi hnh hc ta phng trong thi H u c mt nt tht ring, lm th no tm c nt tht v ci nt tht. Cu tr li l : Phi hc nhiu, lm nhiu, chu kh tng hp kin thc v t duy theo kinh nghim tch ly .

SAU Y TA S I TM NT THT CA BI TON LN TRC NH ! + Ta lp c ngay ng trn (C) ngoi tip ABC c tm K, bn knh AK.

( )2

25 25(C) : x y 32 4

+ =

+ ng thng AI qua A, I 5 1AI :3x y 8 0 D AI (C) D ;2 2

+ = =

+ By gi ta cn chng minh BD DI CD (*)= = B, C nm trn ng trn (T) tm D, bn knh DI ta B, C l giao ca 2 ng trn (C) v (T) (Tht vy, by gi ta chng minh (*) - y l yu t quyt nh ca bi ton ny !!! - Ta c 1 2A A DB DC DB DC (1)= = = - M 1 1 1I A B= + (gc ngoi ABI ), li c

1 2 2 3sdDCA A ,A B

2= = = ,

1 2B B=

1 2 3I B B IBD BDI = + = cn ti D DB DI (2) =

- T (1) v (2) BD DI CD (*) = = )

+ Nh vy ng trn (T) tm D, bn knh DI c phng trnh: 2 25 1 10

x y2 2 4

+ =

+ { } B(4;1),C(1;1)B,C (C) (T)B(1;1),C(4;1)

=

1

(C)2

1

32

1

K( 52 ;3)

I(2;2)

C

D

B

A(1;5)



Bi 7: Cho ABC c tm ng trn bng tip ca gc A l K(2; 9) , nh B( 3; 4),A(2;6) . Tm ta nh C Hng dn tm li gii

+ Ta thy C AC BC= , vy ta cn i tm phng trnh ng thng AC v BC * Bc 1: Tm phng trnh AC - ng thng AC i qua A v B (trong B(7;4) l im i xng ca B qua phn gic AK: x - 2 = 0)

AC : 2x 5y 34 0+ = (Trong qu trnh hc ta c c kinh nghim: khi gp ng phn gic v 1 im, ta s ly im i xng qua ng phn gic - hy vng bn cn nh) * Bc 2: Tm phng trnh BC Suy lun tng t ta cng c: ng thng BC i qua B v A (trong A l im i xng ca A qua phn gic BE) + Gii h C AC BC= . p s C(5;0)

Bi 8: ABC ni tip ng trn tm I(2;1), bn knh R = 5. Trc tm H(-1;-1), di BC = 8. Hy vit phng trnh BC

Hng dn tm li gii + y l 1 bi ton quen thuc tam gic ni tip ng trn, cho bit trc tm, vy ta s ngh ngay n vic to ra hnh bnh hnh bng cch k ng knh AD BHCD l hnh bnh hnh (bn hy t xem li cch chng minh nh) MI l ng trung bnh ca AHD

AH 2.MI = (mt kt qu rt quen thuc) + Vi cc suy lun trn, ta s tm c ta A trc tin. Tht vy, gi A(x;y) Ta c:

2 2 2 2AH 2.IM 2. CI BM 2 5 4 6AI 5

= = = =

=,

gii h ny x 1

A( 1;5) D(5; 3) M(2; 2)y 5

=

= (do I l trung im AD, M l trung im HD)

+ Nh vy, sau khi c im A, M ta thy ng thng BC i qua M, vung gc vi AH BC : y 2 0 + =

K(2;-9)

B(-3;-4)

A'C

B'

E

A(2;6)

I

C

H

B

A

D

M



Bi 9: ABC ni tip ng trn tm I(-2;0), A(3;-7), trc tm H(3;-1). Xc nh ta C bit C c honh dng. Hng dn tm li gii

+ Hon ton vi phng php lp lun nh bi trn, ta cng c c kt qu AH 2.MI AH 2.IM= =

, nu gi M(x;y) th gii

phng trnh AH 2.IM=

x 2, y 3 M( 2;3) = = + ng thng BC i qua im M, vung gc vi AH BC : y 3 0 = + ng trn (C) tm I, bn knh R = IA c phng trnh : ( )2 2x 2 y 74+ + = + Ta B, C l giao ca BC v (C), gii h ta s c ( )C 2 65;3 + (ch Cx 0> nh) Nh vy qua bi ton trn, cc bn cn ghi nh 1 kt qu quan trng sau: Nu H, I ln lt l trc

tm v tm ng trn ngoi tip ABC , M l trung im BC th ta c: AH 2.IM=

(y l im nt ca vn ). Tip theo mch t tng , ta nghin cu bi sau cng c cch khai thc tng t. Bi 10: Cho hnh ch nht ABCD, qua B k ng thng vung gc vi AC ti H. Gi E, F, G ln lt l trung im cc on thng CH, BH v AD. Bit

( )17 29 17 9E ; ;F ; ,G 1;55 5 5 5

. Tm ta tm ng trn ngoi tip ABE

Hng dn tm li gii + y l bi ton pht trin theo mch t duy ca dng bi trn + ABE c F l trc tm, vy nu gi I l tm ng trn ngoi tip ABE , M l trung i AB th ta chng minh c 2.IMEF =

(xem li bi

trn) Do ta E, F bit, vy c I ta cn tm ta M, m M l trung im AB nn ta cn tm ta A, B. (y l im nt ca bi ton ny) + Ta thy ngay EF l ng trung bnh ca

HCB AG FE =

. Nh vy nu gi A(x;y) th gii phng trnh AG FE x 1; y 1 A(1;1)= = =

+ Tip theo lp c phng trnh t AE i qua A, E AE : 2x y 1 0 + + = + ng thng AB qua A v vung gc vi EF AB : y 1 0 = + ng thng BH qua F v vung gc vi AE BH : x 2y 7 0 + =

B BH AB B(5;1) = M(3;1) + Gii phng trnh 2.IMEF =

I(3;3)

I

C

H

B

A

D

M

M

I

DC

A

G

H

B

F

E



Bi 11: Cho ABC c trc tm H, ng trn ngoi tip HBC c phng trnh ( )2 2x 1 y 9+ + = . Trng tm G ca ABC thuc Oy. Tm ta cc nh ca ABC bit BC c phng trnh x y 0 = v B c honh dng. Hng dn tm li gii

+ Trc ht ta c ta B, C l giao im ca ng trn ( )2 2x 1 y 9 (C)+ + = v ng thng BC : x y 0 = .

Gii h phng trnh 1 17 1 17B ; ;

2 2

1 17 1 17C ;2 2

+ +

+ By gi vic kh khn s l tm ta A(x;y) theo trnh t suy lun sau: - im G(0;a) thuc Oy l trng tm ABC , s dng cng thc trng tm A( 1; y) - Gi O v I ln lt l tm ng trn ngoi tip ABC v HBC I v O i xng nhau qua BC (*), t y ta lp c phng trnh OI qua I(-1;0) v vung gc BC

OI : x y 1 0 + + = . - Ta c, ta

1 1M OI BC M ; O(0; 1)2 2

=

- Mt khc OA = 3 (bng vi bn knh ng trn (C)) - do ng trn tm O v ng trn tm I i xng nhau qua BC nn bn knh bng nhau. Gii phng trnh

( )OA 3 A 1; 1 2 2= + hoc ( )A 1; 1 2 2 Chc bn s thc mc ch (*), by gi ta s cng gii thch nh: + Do t gic BHCD l hnh bnh hnh (vn ny chng minh hoi ri) M l trung im HA + Gi D l im i xng ca H qua BC 0ADA ' 90 = (do KM l ng trung bnh

HDA ' , m KM HD DA ' HD ) D (O) (O) ngoi tip BDC . + ng trn (I) ngoi tip BHC , m BHC i xng vi BDC qua BC ng trn tm I v ng trn tm O i xng nhau qua BC I v O i xng nhau qua BC (*)

K

A'

M

DI

OGH

CB

A



Bi 12: ABC cn ti A, gi D l trung im ca AB, D c tung dng, im 11 5I ;3 3

l tm ng trn ngoi tip ABC . im 13 5E ;3 3

l trng tm ADC . im

M(3; 1) DC, N( 3;0) AB . Tm ta A, B, C

Hng dn tm li gii

+ Trc tin ta vit phng trnh DC i qua M v vung gc vi EI DC : x 3 0 = (Ti s gii tch DC EI bn hiu: - Gi F, H, K ln lt l cc trung im BC, AC, AD

E DH CK = . - Gi G l trng tm ABC

G CDAF = - Ta c CE CG 2 GE / /ABCK CD 3

= = , m

AB DI GE ID - Li c DE / /BC

GI DE IGI BC

l trc

tm DGE .. ) + Tip theo ta tm ta D : do D DC D(3; x) , gii phng trnh DN.DI 0 x 3 D(3;3)= =

+ Ta s vit tip phng trnh AB (qua N, D)

AB : x 2y 3 0 + = + ng thng AF qua I v vung gc vi DE

: x y 2 0AF = + Gii h A AB A(7;5) B( 1;1)AF= (do D l trung im AB) + ng thng BC qua B v vung gc vi IA BC : x y 0 + = + Gii h C BC C(3; 3)CD=

(Lu l ng thng CD i qua M v D - bn t vit nh)

G

M(3;-1)

N(-3;0)

K

F

HE

I

CB

D

A



Bi 13: Cho ABC vung cn ti A. Gi M l trung im BC. G l trng tm ABM , im D(7;-2) l im nm trn on MC sao cho GA GD= . Tm ta im A, lp phng trnh AB, bit honh ca A nh hn 4 v AG c phng trnh 3x - y - 13 = 0

Hng dn tm li gii

Bc 1: Tm ta A + Ta tnh c ngay khong cch d(D;AG) 10= + A AG A(a;3a 13) + Ta c gi N l trung im AB, do BMA vung cn ti M nn NM l ng trung trc ca AB GA GB = , m GA GD(gt) GA GB GD= = = G l tm ng trn ngoi tip

0ABD AGD 2.ABD 90 = = (lin h gia gc tm v gc ni tip trong ng trn tm G ngoi tip ABD ) AGD vung cn ti G 2 2AD 2.DG 2.10 20 = = = (gii thch cht

xu: AGD vung ti G d(D;AG) DG 10 = = ). Gii phng trnh 2 a 5 4AD 20

a 3 A(3; 4)= >

= =

Bc 2: Lp phng trnh ng thng AB ng thng AB khng d g lp c nn trong TH ny ta s da vo gc gia 2 ng thng gii quyt. + Gi VTPT ca ng thng AB l ABn (a;b)=

, ng thng AG c VTPT l

AGn (3; 1)=

+ Ta c ( )AB AG 2 23a bc NAG c n ;n a b . 10os os

= =

+

+ Mt khc ( )22 2 21 1NG NM NA, AG NA NG 3.NG NG NG. 103 3

= = = + = + =

2 2

3a bNA 3 3c NAG

AG 10 10a b . 10os

= = =+

2 b 06ab 8b 03a 4b

= + =

=

- Vi b = 0, chn a = 1 AB : x 3 0 = - Vi 3a = -4b, chn a = 4, b = - 3 AB : 4x 3y 24 0 = * Nhn thy nu AB c phng trnh 4x 3y 24 0 = th d(A;AB) 10 G< nm ngoi

ABC (loi)

N

3x-y-13=0

D(7;-2)

G

C

M

B

A



Bi 14: Cho hnh ch nht ABCD c AB, AD tip xc vi ng trn (C) c phng trnh : 2 2x y 4x 6y 9 0+ + + = , ng thng AC ct (C) ti 16 23M ;

5 5

v N, vi

N Oy . Bit ANDS 10 = . Tm ta A, B, C, D bit A c honh m, D c honh dng.

Hng dn tm li gii

+ Cng vic chun b: theo bi ta th ng trn (C) c tm I( 2;3),R 2, N(0;3) Oy = + Lp c ngay phng trnh AC (i qua N v M) : x 2y 6 0+ = + ( )A AC A 6 2a;a , chng minh c APIQ l hnh vung (P, Q l tip im ca AD, AB vi (C))

2 2 2 2AI AQ QI 2 2 2 2 = + = + = .

Gii phng trnh ny A

a 5 A( 4;5)13 4 13

a A ; , x 05 5 5

= = >

+ Gi VTPT ca AD l n (m;n) AD : m(x 4) n(y 5) 0 mx ny 4m 5n 0= + + = + + =

M D

m 0 AD : y 5 0 D(d;5)d(I;AD) 2 ... 2mn 0

n 0 AD : x 4 0 x 4 0= =

= = = + = =



+ Mt khc ABCDBC AD 9S 9 .BK 9 .BK 9

2 2 2EF EF

+= = = =

+ im F F(x;2 x)EF + , gii phng trnh 9 17F ;4 49 9

x4 9 12 2 F ;

4 4

EF

= =

* TH1: 9 17F ;4 4

, ta lp c ng thng CD i qua 2 im F, P CD : 5x y 7 0 + + =

D CD AD = , gii HPT 7 7 11 27D ; C ;4 4 4 4

(do F l trung im CD)

* TH2: Cc bn t lm tng t nh. Bi 16: Cho hnh vung ABCD c tm I(1;-1) v im M thuc CD sao cho MC 2.MD= . ng thng AM c phng trnh 2x y 5 0 = . Tm ta nh A.

Hng dn tm li gii

+ Trc ht ta tnh c ngay 2IH d(I;AM) ...5

= = =

+ Do A AM A(x;2x 5) , vn by gi l phi thit lp 1 phng trnh tm x !!! + Ta thy AIH vung ti H, nu tnh c AI (hoc AH) th s c c phng trnh n x. Tht vy, em hy quan st suy lun sau y: - Em s chng minh c ( ) 0 1 21 2 1 2

1 2

tan A tan AA A 45 tan A A 1 11 tan A .tan A

(*)+

+ = + = =

- M 2DM 1

tan AAD 3

= = , thay vo (*) 1 1tan A 2 =

- Li c: AIH vung ti H 2 21IH 4

tan A AH AI AH IH 2AH 5

= = = + =

- By gi gii phng trnh AI = 2 13 13 1

x A ;5 5 5

x 1 A(1; 3)

=

=

I(1;-1)H

2x-y-5=0

12

MD C

BA



By gi chng ta cng xem li thi khi A-2012 c cch khai thc lm tng t (trong khi p n ca BGD rt kh hiu) Bi 17: (KA-2012) Cho hnh vung ABCD, M l trung im BC. N thuc CD sao cho

.CN 2 ND= . im 11 1M ; ,AN : 2x y 3 02 2

=

. Tm ta ca A.

Hng dn tm li gii + Do ( )A AN A x;2x 3 + Tnh c ngay khong cch

3 5AH d(M;AN)2

= =

+ By gi ta cn tnh on AM thit lp phng trnh tm x nh sau: - Ta c ( )0 01 2 3 2 1 3A A A 90 A 90 A A+ + = = +

( ) ( )02 1 3 1 3cot A cot 90 A A tan A A = + = +

01 3

2 21 3

1 1DN BMtan A tan A 3 2AD ABcot A 1 A 45DN BM 1 11 tan A .tan A 1 . 1 .

AD AB 3 2

+++ = = = = =

- Xt AHM vung ti H 0HM 5AM 3

sin 45 2 = =

- Gii phng trnh 5AM 3 x ? A?2

= =

Bi 18: (KA-2013) Cho hnh ch nht ABCD c M i xng vi B qua C. im N(5; 4) l hnh chiu vung gc ca B trn DM. im C nm trn ng thng 2x y 5 0,A( 4;8)+ + = . Tm ta ca B v C.

Hng dn tm li gii + im C d C(x; 2x 5) + Gi I l tm hnh ch nht ABCD I l trung im AC

x 4 2x 3I ;2 2 +

+ Ta d dng chng minh c IN IA= , gii phng trnh ny ( )x 1 C 1; 7 = + n y ta s lp c phng trnh AC (i qua 2 im A v C), im B l im i xng ca N qua AC

B( 4; 7)

2x-y-3=0

H

D NC

M( 112 ;12 )

BA3

2

1

I

D

N(5;-4)

M

C

BA(-4;8)

d:2x+y+5=0



Cch khc: + im C d C(x; 2x 5) , v hnh chnh xc, d on c ngay rng: AN NC AN.NC 0 =

,

gii phng trnh ny s x C

(Ta chng minh AN NC nh sau: Chng minh ADMC l hnh bnh hnh

AC NB . Trong ANM c C l trung

im BM, EC // NM E l trung im BN 0ABC ANC ANC 90 = = ) + tm ta B ta gii h B BN

BC CN

=

(trong BN l ng thng qua N v vung gc vi AC) Bi 19: Cho hnh ch nht ABCD, A(5; 7),C d : x y 4 0 + = . ng thng i qua D v trung im M ca AB c phng trnh : 3x 4y 23 0 = . Tm ta B, C bit Bx 0>

Hng dn tm li gii

+ C d C(x;x 4) + + Do M l trung im AB

d(C; ) 2.d(A; ) = , gii phng trnh ny

x 1 C(1;5)x 79 0

=

=



Bi 20: Cho ng trn ( )2 2(C) : x 4 y 4 + = . Tm M thuc trc tung sao cho qua M k c 2 tip tuyn MA, MB ti ng trn (C) (A, B l 2 tip im). Bit AB i qua E(4;1)

Hng dn tm li gii Bi tp ny s cung cp cho cc bn 1 phng php lp phng trnh ng thng da theo tng qu tch.

+ Do M Oy M(0;m) + ng trn (C) ngoi tip t gic MAIB c tm

aF 2;2

l trung im MI,

bn knh 2MI 16 aR '

2 2+

= =

( )2

2

2

a(C ') : x 2 y2

16 a4

+

+=

+ Ta c ta A, B l giao ca (C) v (C) l nghim h phng trnh : ( )( )

2 2

2 22

x 4 y 44x ay 12 0a 16 a

x 2 y2 4

+ =

+ + = +

+ =

+ T y suy ra AB c phng trnh 4x ay 12 0 + + = , m E thuc AB m 4 M(0;4) = Bi 21: Cho hnh vung ABCD, trn tia i ca tia DA ly im P sao cho 0ABP 60= . Gi K, M, N ln lt l trung im BP, CP, KD. Tm ta D bit ta M(1;2), N(1;1)

Hng dn tm li gii + y l loi bi ton m hnh khng c phng trnh cc cnh nn ta s dng phng php tnh ra di cnh hnh vung. Nu gi cnh hnh vung l x, ta c: - on MN c di bng 1. - Gi E l trung im CK

01 1ME / /PB;ME PK PB;MEN PBA 602 4

= = = =

- PAB vung ti A, 0 xPBA 60 PB 2x ME2

= = = , m

DC xNE MEN2 2

= = u MN ME NE 1 x 2 = = = =

+ Nh vy ta tnh c cnh hnh vung bng 2, ta s i suy lun tm ta D - Gi D(a;b), m bi cho 2 im M, N bit ta ri, v

E(4;1)

F

B

I(4;0)

A

M

EN(1;1)

M(1;2)

K

P

D C

BA



vy hng suy ngh tip theo l i tnh DN v DM nh sau: - Ta c DKDN

2= , rng DPK c 0 PBDPK 30 ,PK 2

2= = = , vy cn tnh PD p

dng nh l hm s cos trong DPK th s tnh c DK. y 2 2PD AP AD PD AB AD 2 3 2= = = , quay tr li p dng nh l hm s cos trong DPK DK DN 2 3 = (1)

- Ta c 2 2PC PD DCDM 5 2 3

2 2+

= = = (2)

+ Cui cng, gii h phng trnh gm (1) v (2)

1 3D ;2 2

3 3D ;2 2

Bi 22: Cho hnh thang vung ABCD vung ti A v D c AB AD CD,B(1;2)= < , ng thng BD c phng trnh y 2= . Bit ng thng d : 7x y 25 0 = ct on thng AD, CD ln lt ti M v N sao cho BM BC v tia BN l tia phn gic ca MBC . Tm ta im D, bit D c honh dng.

Hng dn tm li gii + Ta c d(B;d) ... 2 2= = + Ta c BMN BNC = (do BN chung, MBN CBN;BM BC= = (do

BAM BHC) = BI BH 2 2 = = (2 ng cao tng ng ca 2 tam gic bng nhau)

BD BH. 2 4 = = (do BDH vung cn ti H) + Do D BD D(b;2) , gii phng

trnh d 3 0BD 4d 5 D(5;2)

=



+ M d M(1 2x;x) + Do ADMB l hnh ch nht t gic ADMB ni tip ng trn ng knh DB, m 0DHB 90= H thuc ng trn ng knh DB 5 im A, D, H, M, B nm trn ng trn ng knh DB t gic AHMB ni tip 0AHM 90 = (do 0ABM 90= ) n y ta gii phng trnh HA.HM 0 M(1;0)=

+ M AM // DC (do ADMC l hnh bnh hnh) ng thng DC i

qua H v song song vi AM DC : 5x 20y 39 0 + =

+ Ta c 1O 1;2

l trung im AM, gii tip h

9 12 1 7D ; B ;D DC 5 5 5 5OD OA 3 9 13 4D ; B ;

5 5 5 5

=

Bi 24: Cho hnh vung ABCD c A(3;4) . Gi M, N l cc trung im AD v DC. E l giao im BN v CM. Vit phng trnh ng trn ngoi tip BME , bit BN c phng trnh x 3y 1 0 + = v im B c ta nguyn. Hng dn tm li gii + Trc ht, quan st hnh v ta thy i vi bi tp dng ny, ta s chng minh c MC BN BEM vung ti E (bn t chng minh iu ny nh v chng ta lm vi ln ri) ng trn ngoi tip BEM c tm I l trung im MB, bn knh R = IB.

Nh vy im quyt nh l phi tm c ta B v I ( y bi cho B c ta nguyn nn chc chn s phi suy ngh n vic tm ta B ri) + B BN : x 3y 1 0 B(3b 1;b) + = , ta cn thit lp 1 phng trnh tm ra b = ?

By gi dng ti y v tip tc quan st hnh xem bn suy lun c g nh ! + Nu gi P l trung im BC, Q AP BN= s chng minh c AP l ng thng qua A v BN AP : 3x y 13 0 + =

+ Ta Q AP BN= , gii h c 19 8Q ;5 5

32AQ5

=

+ M AQ = BE (do AQB BEC = ) 32BE5

= , li c BE 2.BQ= , gii phng trnh

BE 2.BQ= 6b Z

B(5;2)5b 2

=

=

( tm c B ri nh - gn xong ri)

M

O

d:x+2y-1=0

D

CB

H( 35 ;95 )

A(-3;1)



+ By gi tm I nh: Gi I l trung im MB I l trung im AP (do ABPM l hnh ch nht) I AP I(x;13 3x) tm x, bn ch cn gii phng trnh IA = BI 7x

2 =

Nh vy bi ton ny c p s l 2 27 5 10

x y2 2 4

+ =

x - 3y + 1 = 0

E

Q

I

M

D NC

P

BA(3;4)

Bi 25: Cho hnh ch nht ABCD c AD AB 2= , AB c phng trnh 2x y 4 0+ + = , H(0;1) l trung im BC, M l trung im AD. I l giao im AC v BM. Vit phng trnh ng trn i qua 3 im B, I, C.

Hng dn tm li gii + Vi dng bi tp ny, theo kinh nghim ta s chng minh BIC vung ti I (y l quyt nh thnh cng). Tht vy:

Ta c 1

1 1AD AB. 2AM 22 2tan BAB AB AB 2

= = = = ,

1AB AB AB 2

tan CBC AD 2AB 2

= = = =

1 1B C = , m 0 0 01 2 1 2B B 90 C B 90 BIC 90 BIC+ = + = = vung ti I

+ Nh vy, ng trn i qua 3 im B, I, C c tm H(0;1), bn knh d(H;AB) 5R BH= = = .

Ta c p s cui cng ca bi: ( )22x y 1 5+ = Bi 26: Cho hnh vuong ABCD, A(-1;2). Cc im M, N ln lt l trung im AD, BC. E l giao im BN v CM. Vit phng trnh ng trn ngoi tip BME bit B c honh ln hn 2 v ng thng BN c phng trnh : 2x y 8 0+ =

121

I

2x+y+4=0

H(0;1)BC

DMA



Hng dn tm li gii + Nhn thy BME vung ti E (bn xem li cch chng minh nh - d thi) ng trn ngoi tip BME c tm F l trung im BM, bn knh R FB FM= = . Nh vy by gi ta phi i tm c ta B v M * Bc 1: Tm ta B

+ B BN B(b,8 2b) , m ( )2 1

2.( 1) 2 8 8d A;BN52 1

+ = =

+ (bn hy nh rng trong hnh

hc ta phng khi cho 1 im bit ta , 1 ng thng c phng trnh th ta lun c thi quen tnh khong cch t im n ng thng, c th y s l gi quan trng tm ra hng gii) + Nu gi I l trung im BC, H AI BN ABI= vung ti B, ng cao BH

22 2 2 28 8 ABAB AH.AI AB BI AB 4AB AB 4

25 5

= = + = + = =

Gii phng trnh AB = 4 7b 2

B(3;2)5b 3 2

=

* Bc 2: Tm ta M + Gi K BN AD D= l trung im AK (do KD DN 1

KA AB 2= = )

ng thng AK (i qua A, vung gc AB) : x + 1 = 0 K AK BN K( 1;10) D( 1;6) M( 1;4) =

Vy p s bi ton l : ( ) ( )2 2x 1 y 3 5 + =

K

FBN:2x+y-8=0

E

H

B IC

N

DMA(-1;2)



Bi 27: Cho ABC c A( 1;2),B(2;0),C( 3;1) . Gi M l im di ng trn BC. Gi 1 2R ;R ln lt l bn knh ng trn ngoi tip ABM v ACM . Hy xc nh ta

ca im M 1 2R R+ nh nht. Hng dn tm li gii

MC(-3;1)

A(-1;2)

B(2;0)

O2

O1

R2

R1

+ p dng nh l hm sin trong AMB c :

( ) ( ) ( )1 1AB AB 132R R

sin AMB 2.sin AMB 2.sin AMB= = =

+ p dng nh l hm sin trong AMC c :

( ) ( ) ( )2 2AC AC 52R R

sin AMC 2.sin AMC 2.sin AMC= = =

( ) ( )1 213 5R R

2.sin AMB 2.sin AMC + = +

+ Mt khc ta c : ( ) ( )sin AMB sin AMC= (do AMB,AMC l 2 gc b nhau) ( ) ( ) ( )1 2 1 2 MIN M

13 5R R R R sin AMB2.sin AMB AX

+ + = +

( ) 0sin AMB 1 AMB 90 = = AM BC M l hnh chiu vung gc ca A trn BC.

Nh vy, cc bn lp phng trnh BC v tm hnh chiu vung gc ca A trn BC

s c p s 33 17M ;26 26



Bi 28: Cho 2 2

x y(E) : 125 9

+ = c 2 tiu im 1 2F ;F . Gi s M l im thuc (E) sao cho

bn knh ng trn ni tip 1 2F MF bng 43

v M c tung dng. Vit phng trnh

ng thng (d) i qua M v to vi h trc ta mt tam gic c din tch bng 9. Hng dn tm li gii

NO F2F1

M

+ Ta thy ngay (E) c a 5;b 3;c 4= = = + Gi ( )

2 20 0

0 0x yM x ; y (E) 125 9

+ = (1) + Gi r l bn knh ng trn ni tip 1 2F MF , ta c:

( ) ( )1 2

1 2 1 2 1 2 1 2MF F 1 2

MF MF F F MF MF FF1S p.r .r .F F .d(M; ) .r2 2 2

Ox+ + + +

= = =

00

0 0

y 3 01 2a 2c 4.2c. y .

y 3 x 0 M(0;3)2 2 3=

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