# Nguy n Qu c Qu n - Tr ng THPT chuyên Nguy n Quang thpt-c ?· Nguy n Qu c Qu n - Tr ng THPT chuyên…

• Published on
04-Sep-2018

• View
212

0

Transcript

• Nguyn Quc Qun - Trng THPT chuyn Nguyn Quang Diu

1

H tr hc Ton Hnh hc 10

(hc k 1)

Vectn Cc cng thc c bn cn nh

1/ Qui tc 3 im.

a/ Qui tc cng: ACBCAB =+ (A, B, C bt k)

b/ Qui tc tr: ABOAOB = (O, A, B bt k)

2/ Qui tc hnh bnh hnh:ABCD l hnh bnh hnh th: ACBDAB =+

3/ Nu a = k b th a v b cng phng. C th:

k > 0 th a vb cng hng ; k < 0 th a vb ngc hng v akak = ,

=

==

0a

0k0a.k

4/ Cng thc lin quan trung im: O l trung im on AB th:

a/ 0=+ OBOA ; b/ ( )MBMAMO +=2

1 ( M bt k)

5/ Cng thc lin quan trng tm tam gic. G l trng tm tam gic ABC th:

a/ 0=++ GCGBGA b/ ( )MCMBMAMG ++=3

1 ( M bt k)

Bi tp Bi 1/ Cho bn im A , B , C , D . Tnh :

a/ CABDDCABu +++= ; b/ DABCCDABv +++=

Gii

a/ CABDDCABu +++= = ( ) ( ) 0==+=+++ AADAADCADCBDAB b/ DABCCDABv +++= = ( ) ( ) 0==+=+++ AACAACDACDBCAB Bi 2/ Cho 6 im A , B , C , D , E , F. Chng minh: CDBFAECFBEAD ++=++ Gii chng minh T = P ta c th chng minh T P = 0

( ) ( )CDBFAECFBEAD ++++ = ( ) ( ) ( )CDCFBFBEAEAD ++ = DFFEED ++ = ( ) FEDFED ++ = DEED + = 0 Suy ra iu phi chng minh

Bi 3/ Cho tam gic ABC . Gi M, N, P ln lt l trung im ca BC, CA, AB.

Chng minh: 0=++ CPBNAM Gii S dng cng thc trung im

( )CACBBCBAACABCPBNAM +++++=++2

1 = ( ) ( ) ( )[ ]CBBCCAACBAAB +++++

2

1 = 0

S dng cng thc trng tm

( ) 0GCGCBGA2

3CPBNAM =++=++

Bi 4/ Cho tam gic ABC . Gi M l trung im ca AB v N l im trn cnh AC sao cho NC = 2NA.

Gi K l trung im ca MN.

a/ Chng minh: ACABAK61

41

+=

• Nguyn Quc Qun - Trng THPT chuyn Nguyn Quang Diu

2

K

N

M

DCB

A

b/ Gi D l trung im ca BC . Chng minh: ACABKD31

41

+=

Gii

a/ M l trung im AB nn: ABAM2

1= , NC = 2NA nn ACAN

3

1=

K l trung im MN nn: ( )ANAMAK +=2

1 =

+= ACABAK

3

1

2

1

2

1

= ACAB6

1

4

1+

b/ Gi D l trung im ca BC . Chng minh: ACABKD31

41

+=

Gii

D l trung im AC nn ( )ACABAD +=2

Vy: ACABACABACABKD3

1

4

1

6

1

4

1

2

1

2

1+=+=

Bi 5/ Cho tam gic ABC.

a/ Tm I sao cho : 02 =+ IBIA b/ Tm K sao cho : CBKBKA =+ 2 ;

c/ Tm M sao cho : 02 =++ MCMBMA Gii

a/ Tm I sao cho : 02 =+ IBIA 02 =+ IBIA BIIA 2= hay IBAI 2= AI v IB cng hng v AI = 2IB

Vy I thuc on AB v chia AB thnh thnh 3 on bng nhau th c hai im chn im I v pha B

b/ CBKBKA =+ 2 KCKBKBKA =+ 2 thay KBKCCB =

0=++ KCKBKA Vy K G l trong tm tam gic ABC

c/ Tm M sao cho : 02 =++ MCMBMA Gii. lu cng thc trung im c s 2, bi ton v tm t c n gin nht

Gi D l trung im on AB, ta c: MDMBMA 2=+

02 =++ MCMBMA 022 =+ MCMD 0=+ MCMD Vy M l trung im CD

Bi 6/ Cho hnh bnh hnh ABCD v im M ty . Chng minh : MDMBMCMA +=+ Gii

Gi O = AC BD O l trung im ca AC v BD

Ta c:

=+

=+

MOMDMB

MOMCMA

2

2 MDMBMCMA +=+

Bi 7/ Cho tam gic ABC v im M ty .

a/ Chng minh rng vect MCMBMAv 32 += khng ph thuc v tr im M b/ Dng im D sao cho vCD = . CD ct AB ti K . Chng minh: 02 =+ KBKA v

CKCD 3= Gii

a/ Chng minh rng vect MCMBMAv 32 += khng ph thuc v tr im M

• Nguyn Quc Qun - Trng THPT chuyn Nguyn Quang Diu

3

Tm im c nh lin quan ABC v bin i v mt M l xong

( ) ( )MCMBMCMAv += 2 = CBCA 2+ iu phi chng minh b/ Dng im D sao cho vCD = . CD ct AB ti K . Chng minh: 02 =+ KBKA v CKCD 3= T ( ) ( )MCMBMCMAv += 2 = CBCA 2+ lm mt s 2 i Dng im E sao cho E l trung im CE CBCE 2= . Khi :

CECAv += . Dng hnh bnh hnh CADE vCD = Gi O = CD EA O l trung im ca CD v EA

K = CO AB K l trng tm tam gic ACE KBKA 2=

02 =+ KBKA .

Bi 8/ Cho tam gic ABC ni tip ng trn tm O, H l trc tm tam gic, D l im i xng vi A qua O. a/ Chng minh rng HBDC l hnh bnh hnh .

b/ Chng minh rng OHOCOBOAHOHCHBHA =++=++ ,2

c/ Gi G l trng tm tam gic ABC . Chng minh : OGOH 3= ng thng qua 3 im O, H, G gi l ng thng le

Gii a/ Chng minh rng HBDC l hnh bnh hnh . AD l ng knh nn DC AC v BD HC V: DC AC v BH AC nn DC // BH (1)

V : DB AB v CH AB nn DB // CH (2)

(1) v (2) BHCD l hnh bnh hnh

b/ Chng minh rng OHOCOBOAHOHCHBHA =++=++ ,2

BHCD l hnh bnh hnh nn: HDHCHB =+

HOHDHAHCHBHA 2=+=++ ( V O l trung im AD)

C: HAOHOA += , HBOHOB += v HCOHOC +=

Nn: HCHBHAOHOCOBOA +++=++ 3

Bi 9/Cho tam gic ABC.Gi D, E, F ln lt l trung im ca AB, BC,CA v im M ty . Chng minh:

a/ MFMEMDMCMBMA ++=++

b/ 0=++ CDBFAE

Gii

a/ MFMEMDMCMBMA ++=++

( ) ( )MFMEMDMCMBMA ++++ = ( ) ( ) ( )MFMCMEMBMDMA ++ = FCEBDA ++ = AFFDDA ++ = DDFDAFDA =++

b/ 0=++ CDBFAE ( cng thc trung im)

CDBFAE ++ = ( )CACBBCBAACAB +++++2

1 = 0

D

K

O

E

A

B

C

H O

D

C

B

A

F

E

D

CB

A

• Nguyn Quc Qun - Trng THPT chuyn Nguyn Quang Diu

4

NM

D

CB

A

NH

G

M CB

A

( ) 02

1

2

Bi 10/ Cho t gic ABCD v M, N ln lt l trung im ca on AB, CD. Chng minh:

Gii

)()( BCBDACAD +++ = BNAN 22 + = ( )NBNA + 2 = NM2.2

Bi 11/ Cho tam gic ABC vi G l trng tm, H l im i xng ca B qua G, M l trung im BC. Chng minh:

a/ ABACAH3

1

3

2= b/ ABACCH

3

1

3

1= c/ ABACMH

6

5

6

1=

Gii

a/ ABACAH3

1

3

2=

Gi N l trung im AC, ta c: AGNC l hnh bnh hnh

GKGCAH3

2== ( K l trung im AB)

( ) CBACCBCAAH3

1

3

1

2

1.

3

2=+= = ( )ACABAC

3

1

3

1 = ACAC

3

1

3

2

b/ ABACCH3

1

3

1=

AMGACH3

2== = ( )ACAB +

2

1.

3

2 = ACAB

3

1

3

1

c/ ABACMH6

5

6

1=

( )HCHBMH +=2

1 = ( )AGGB + 2

2

1 = ( )GBAB +

2

1 = BGAB

2

1

2

1+ = BNAB

3

2.

2

1

2

1+

( )BCBAABMH ++=2

1.

3

1

2

1 = ( )ACBABAAB +++ .

6

1

2

1 = ABABAC

3

1

2

1

6

1

ABACMH6

5

6

1=

Bi 12/ Cho tam gic ABC u cnh a, trng tm O. Tnh:

a/ ACAB + b/ ACAB c/ OBOA + d/ ACAO +

Gii

Nhc li di ng cao tam gic u bng: dicnh.2

3

a/ ACAB + = AM2 = 2AM = 3a

b/ ACAB = CA = CA = a

c/ OBOA + . Gi M l trung im BC H

K

M

O

CB

A

• Nguyn Quc Qun - Trng THPT chuyn Nguyn Quang Diu

5

OMOMOBOA 22 ==+ = AM3

1.2 =

3

3a

d/ ACAO + . Gi K l trung im OC. AKACAO 2=+ = 2AK

Gi H l hnh chiu ca K ln AM, trong tam gic AHK c:

44

1

2

1 aBCMCHK === ; OMAMOHAOAH

2

1

3

2+=+= = AMAM

6

1

3

2+ =

AM6

5=

2

3.

6

5 a

12

35aAH = nn : AK2 = AH2 + HK2 =

144

84

16144

75 222 aaa=+ AK =

6

21a

Vy: 3

21aACAO =+

Bi 13/ Cho hnh vung ABCD cnh a, tm O. Tnh:

Gii

a/ ADAB + = ACAO =2 = AC = 2a b/ ADAB = DA = DA = a

c/ ACAB + . Gi M l trung im BC

AMACAB 2=+ = 2AM = 2

5

4

2222 aa

aBMAB =+=+

d/ ABAO + . Gi H l trung im OB. AHABAO 2=+ = 2AH

Trong tam gic AOH c: AH2 = AO

2 + OH

2 =

16

10

16

2.5

16

5

4

5

4

222222 aaACAOOB

AO ====+

Hay 4

10aAH = . Vy:

2

10aABAO =+

Bi 14/Cho lc gic u ABCDEF v im M ty .Chng minh rng:

MFMDMBMEMCMA ++=++ . Gi O l tm ng trn ngoi tip lc gic cho

Gii. ( ) ( )MFMDMBMEMCMA ++++ = ( ) ( ) ( )MFMEMDMCMBMA ++ = FEDCBA ++

= AOOBBA ++ = BBOBAOBA =++

O

H

M

DC

BA

• Nguyn Quc Qun - Trng THPT chuyn Nguyn Quang Diu

6

Gi tr lng gic ca mt gc ( t 00 n 1800) Trong h trc (Oxy) cho ng trn tm O qua cc im A(1 ; 0 ), A/(1 ; 0) v B(0

; 1). V cung AM c s o l ( tng ng gc c hai tia OA, OM ). Tm ta im M . Nu: M( xM ; yM )

cos = xM sin = yM

cos

sintan = ( 90

0)

sin

coscot = ( 0

0 v 180

0 )

Nu a + b = 1800 th: sina = sinb v cosa = cosb ; tana = tanb ; coaa = cotb Cc h thc lng gic cn nh

1/ sin2x + cos

2x = 1 2/ tanx =

x

x

cos

sin 3/ cotx =

x

x

sin

cos

4/ tanx.cotx = 1 5/ x

2cos

1 = 1 + tan

2x 6/

x2sin

1= 1 + cot

2x

Bi tp.

Bi 1/ Cho sinx = 13

5 ( 90

0 < x < 180

0). Tnh cc gi tr lng gic cn li

Gii

cos2x = 1 sin

2x =

169

144

169

251 =

13

12cos =x v ( 90

0 < x < 180

0) nn: cosx < 0

12

5

12

13.

13

5tan ==x ,

5

12cot =x

Bi 2/ Bit cot150 = 2 + 3 . Tnh cc gi tr lng gic cn li ca gc 150 Gii

3232

115tan 0 =

+= ; ( )32415tan1

15cos

1 0202

=+=

( ) 432

324

115cos 02

+=

=

2

3215cos 0

+= ;

sin150 = tan15

0.cos15

0 = ( ) ( )( )( )

2

32

2

323232

2

3232

=

+=

+

Bi 3/ Cho tan = 3 . Tnh:

cos11sin4

cos3sin2/

+a b/

33 cos17sin

cos2sin3

Gii

Cch 1/

cos11sin4

cos3sin2/

+a =

11tan4

3tan2

+

( chia 2 v cho cos ) = 11

Cch khc: tan = 3 sin = 3cos . Thay vo biu thc

Cch 2/ b/

33 cos17sin

cos2sin3

=

23 cos

1.

17tan

2tan3

= ( )

23

tan117tan

2tan3+

= 7

Cch khc: tan = 3 sin = 3cos . Thay vo biu thc Bi 4/ Cho tana + cota = m , hy tnh theo m.

a/ tan2a + cot

2a , b/ tan

3a + cot

3a , c/ | tana cota|

Gii a/ tan

2a + cot

2a = (tana + cota)

2 2tana.cota = m

2 2

b/ tan3a + cot

3a = (tana + cota)

3 3tana.cota(tana + cota) = m

3 3m

c/ | tana cota| = ( ) 2cottan2cottan 22 =+ maaaa

• Nguyn Quc Qun - Trng THPT chuyn Nguyn Quang Diu

7

Bi 5/ Cho sina + cosa = m , hy tnh theo m. a/ sina cosa b/ | sina cosa| c/ sin

3a + cos

3a

d/ sin4a + cos

4a e/ sin

6a + cos

6a

Gii

a/ sina cosa = ( )

2

1

2

1cossin 22

=

+ maa

b/ | sina cosa| = ( )2cossin aa = ( ) aaaa cossin4cossin 2 + =

2

14

22

m

m

| sina cosa| = 22 m

c/ sin3a + cos

3a = (sina + cosa)

3 3sinacosa)(sina + cosa) =

2

13

23

m

mm

sin3a + cos

3a =

2

23 3mm

d/ sin4a + cos

4a = (sin

2a + cos

2a)

2 2sin

2acos

2a = 1 2(sinacosa)

2 =

22

2

121

m

sin4a + cos

4a =

2

21 42 mm +

Bi 6/ Chng minh rng:

a/ aa

aa22

22

coscot

sintan

= tan

6a b/ aaa

a

aa 323

tantantan1cos

cossin+++=

+

c/ sin2atan

2a + 4sin

2a tan

2a + 3cos

2a = 3

Gii

a/ aa

aa22

22

coscot

sintan

= tan

6a

aa

aa22

22

coscot

sintan

=

( )( )aa

aa22

22

sin1cot

cos1tan

=

aa

aa22

22

coscot

sintan

b/ aaaa

aa 323

tantantan1cos

cossin+++=

+

aa

aa

a

aa23 cos

1.

cos

cossin

cos

cossin +=

+= ( )( )aa 2tan11tan ++

c/ sin2atan

2a + 4sin

2a tan

2a + 3cos

2a = 3

VT = sin2atan

2a + sin

2a tan

2a + 3sin

2a + 3cos

2a = sin

2a(1 + tan

2a) tan

2a + 3(sin

2a +

cos2a)

= 3

Bi 7/ Chngminh cc ng thc sau: a/ cos

4x sin

4x = 2cos

2x 1 b/ cot

2x cos

2x = cos

2x.cot

2x

c/ tan2x sin

2x = tan

2x.sin

2x d/ (sinx + cosx)

2 + (sinx cosx)

2 = 2

Gii a/ cos

4x sin

4x = 2cos

2x 1

cos4x sin

4x = (cos

2x sin

2x)(cos

2x + sin

2x) = cos

2x (1 cos

2x) = 2cos

2x 1

b/ cot2x cos

2x = cos

2x.cot

2x.

VP = (1 sin2x)cot

2x = cot

2x sin

2xcot

2x = cot

2x cos

2x

c/ tan2x sin

2x = tan

2x.sin

2x.

VP = tan2x( 1 cos

2x) = tan

2x tan

2x.cos

2x = tan

2x sin

2x

Bi 8/ Rt gn cc biu thc sau:

• Nguyn Quc Qun - Trng THPT chuyn Nguyn Quang Diu

8

D/C/

D

C

BAB/

A/B

AO

a/ 2(sin6x + cos

6x) 3(sin

4x + cos

4x) b/ 2cos

4x sin

4x + sin

2xcos

2x +3sin

2x

c/ (sin4x + cos

4x 1)(tan

2x + cot

2x + 2)

Gii. sin

4x + cos

4x = (sin

2x + cos

2x)

2 2sin

2xcos

2x = 1 2sin

2xcos

2x

sin6x + cos

6x = (sin

2x + cos

2x)

3 3sin

2xcos

2x(sin

2x + cos

2x) = 1 3sin

2xcos

2x

a/ 2(sin6x + cos

6x) 3(sin

4x + cos

4x) = 2(1 3sin

2xcos

2x) 3(1 2sin

2xcos

2x) = 1

b/ B = 2cos4x sin

4x + sin

2xcos

2x +3sin

2x = cos

4x sin

4x + cos

4x+ sin

2xcos

2x

+3sin2x

= (cos2x sin

2x)(cos

2x + sin

2x) + cos

2x(cos

2x + sin

2x) + 3sin

2x

= cos2x sin

2x + cos

2x + 2sin

2x = 2(cos

2x + sin

2x) = 2

c/ C = (sin4x + cos

4x 1)(tan

2x + cot

2x + 2) = 2sin

2xcos

2x(tan

2x + cot

2x + 2)

= 2sin4x 2cos

4x 4sin

2xcos

2x = 2(sin

2x + cos

2x)

2 = 2

d/ (sinx + cosx)2 + (sinx cosx)

2 = 2

D = sin2x + 2sinxcosx + cos

2x + sin

2x 2sinxcosx + cos

2x = 2(sin

2x + cos

2x) = 2

TCH V HNG Gc gia hai vect: Cho hai vect a v b . T im O tu , dng aOA = v

bOB = . Gc AOB gi l gc gia hai vect a v b . ( )ba ; = AOB Tch v hng ca hai vect a v b . K hiu: ba. v ba. = ( )baba ;cos.. Tnh cht.

1/ ba. = ab. 2/ ( )cba + = caba + 3/ ( )22

aa = 4/ ( ) 222 2 bbaaba += 5/ Cho hai vect OA v OB . B/ l hnh chiu ( vung gc) ca B ln ng thng qua

hai im O, A. Ta c: 'OBOAOBOA = ( Hnh tri: A/ l hnh chiu ca A ln OB, B

/ l hnh chiu ca B ln OA th:

OBOAOBOAOBOA ... // ==

Hnh phi: C/, D/ ln lt l hnh chiu ca C,D ln AB th :

//.. DCABCDAB = )

6/ Cho ng trn (O) tm O bn knh R v mt im M. Qua M k ng thng ct

(O) ti hai im AB, ta lun c: 22. RMOMBMA = Chng minh K ng knh BC,ta c: CA AB A l hnh chiu ca C ln AB ( hay )

( )( )OMOBOMOCMBMCMBMA == = ( )( )OMOCOMOC = ( OC2 OM2) Hay : 22. RMOMBMA =

22. RMOMBMA = .Gi l phng tch ca im M i vi ng trn (O) v K

hiu: PM/(O) Nu M ngoi (O) v MT l tip tuyn ca (O) ( T l tip tuyn ) , ta c:

2. MTMBMA =

Hn nhin. ng thng qua M ct (O) ti A, B; ng thng qua (O) ct (O) ti

C,D th MDMCMBMA .. =

Bi tp Bi 1. Cho tam gic ABC c AC = 9, BC = 5 , ACB = 900. Tnh ACAB. Gii.

• Nguyn Quc Qun - Trng THPT chuyn Nguyn Quang Diu

9

C

B

A

H

O

M CB

A

O

D C

BA

Cch 1/ B l hnh chiu ca C ln AC,

nn: 810cos... 02 === ACACACACAB Cch 2/

819

.9.cos...22

22=

++==

BCACBCACAACABACAB

Cch 3/ ( ) 81... 22 ==+=+= ACACCBACACCBACACAB ( v 0. =ACCB ) Bi 2. Cho tam gic u ABC cnh a, tm O v M l trung im BC. Tnh:

Gii. Lu : Tnh gc gia hai vect ta a hai vect v hai vect chung gc v gc ca chng l gc kp gia hai mi tn

1/ OMOA. v BCOA. 2/ OBOA. v ABOA.

Gii

1/ OMOA. v BCOA.

Gc gia OA v OM l 1800. Nn )1.(.. = OMOAOMOA = 22OM 2

3

12.

= AMOMOA =

2

2

3

3

12

a=

6

2a

OA BC nn 0. =BCOA

2/ OBOA. v ABOA.

Gc gia OA v OB l 1200 nn: OBOA. =

2

32OA =

2

1

2

3

3

22

a =

6

2a

Mt pht hin th v l M l hnh chiu ca B ln OA nn OBOA. = OMOA.

Gc gia OA v AB l 1500 nn: ABOA. = OA.AB.cos1500 =

ti sao khng p dng iu pht hin trn: M l hnh chiu ca B ln OA nn ABOA. =

ABOA.

Bi 3. Cho hnh vung ABCD tm O, cnh a. Tnh:

1/ ABOA. v ACAB + 2/ DCAB. v ADOB.

Gii

1/ ABOA. . O l hnh chiu ca B ln AO nn:

Gi ABOA. = AOOA = OA2 = 22

2 22

aa=

Gi M l trung im BC, ta c: ACAB + = AM2 = 54

22

2a

aa =+

AB v DC cng hng nn DCAB. = AB2 = a2

O l hnh chiu ca A ln OB nn ADOB. = DOOB = OB2 = 2

2a

Bi 4. Cho tam gic ABC c AB = 5 , BC = 7, CA = 8.Tnh ACAB. , suy ra gi tr ca gc A

Gii

ABACBC = nn ( ) ( )22 ABACBC = hay BC2 = AC2 ACAB.2 + AB2

• Nguyn Quc Qun - Trng THPT chuyn Nguyn Quang Diu

10

H

M CB

A

2

496425

2.

222+

=+

=BCACAB

ACAB = 20

2

1

.

.cos ==

ACAB

ACABA A = 600

Bi 5. Cho hai im A, B v O l trung im ca AB. M l im tu . Chng minh :

22. OAOMMBMA =

Gii

( )( )OMOBOAMOMBMA +=. = ( )( )OAMOOAMO + = OM2 OA2

Bi 6. Cho 4 im M, A, B, C. Chng minh: 0... =++ ABMCCAMBBCMA Gii

VT = ( ) AB.MCCA.MBACBAMA +++ = AB.MCCA.MBAC.MABA.MA +++ = CA.MBAC.MAAB.MCBA.MA +++ = ( ) ( )MBAMCAMCAMAB +++ = AB.CAACAB + = ( )CAACAB + = 0 Bi 7.Cho tam gic ABC vi H l trc tm v M l trung im BC.

Chng minh: 24

1. BCMAMH =

Gii. S dng t nht 2 trong 3 sau: AH BC, BH AC v CH AB

MAMH . = ( )( )ACABHCHBAMHM ++=4

1. = ( )ACHCABHB +

4

1

= ( )( )[ ]BCABBCHBABHB +++4

1

=

+++

2

4

1BCABBCACHBABHB = 2

2BC

4

1BCHCAB

4

1=

+

Bi 8. Cho tam gic ABC . Tm tp hp cc im M sao cho: ABACABAM .. = Gii

ABACABAM .. = ( ) 0= ACAMAB 0. =CMAB . Vy M thuc ng thng qua C v

vung gc AB

Bi 9. Cho tam gic ABC .

a/ Tm tp hp cc im M sao cho: ( )( ) 0=++ MCMAMBMA b/ Tm tp hp cc im M sao cho: ( )( ) 0MCMBMCMBMA =+++ Gii

a/ Tm tp hp cc im M sao cho: ( )( ) 0=++ MCMAMBMA

Gi H, K ln lt l trung im ca AB v AC

( )( ) 0=++ MCMAMBMA 0. =MKMH Vy M thuc ng trn ng knh HK b/ Tm tp hp cc im M sao cho: ( )( ) 0MCMBMCMBMA =+++ Gi G l trng tm tam gic ABC v I l trung im BC

( )( ) 0MCMBMCMBMA =+++ 0MI2.MG3 = 0MI.MG = . Vy M thuc ng trn ng

knh GI

Bi 10. Cho hai im M, N nm trn ng trn ng knh AB = 2R. I l giao im ca hai ng thng AM v BN

1/ Chng minh: AIABAIAM .. = v BIBABIBN .. =

• Nguyn Quc Qun - Trng THPT chuyn Nguyn Quang Diu

11

I

N

M

BA

M

D

HCB

A

M

T/

T

B

A

O/

O

H

I

F

E

O

M

BA

2/ Tnh AIAM . + BIBN. theo R

Gii 1/ AB l ng knh nn AM BM v AN BN Nn M l hnh chiu ca B ln AI v N l hnh chiu ca A ln BI

Vy AIABAIAM .. = v BIBABIBN .. = 2/ Da vo kt qu trn ta c:

AIAM . + BIBN. = BIBAAIAB + = ( ) 2ABIBAIAB =+ = AB2 = 4R2 Bi 11. Cho tam gic ABC cn nh A, ng cao AH. Gi D l hnh chiu ca H ln AC, M l trung im HD. Chng minh: AM BD Gii

= ( ) HCADCDHCAH .. ++ = HCADCDAH + = DCADCDAD + ( H l hnh chiu ca H ln AC ) = 0 . Vy. AM BD Bi 12. Cho hnh ch nht ABCD. Chng minh:

1/ MDMBMCMA .. = 2/ MA2 + MC2 = MB2 + MD2 Gii

1/ MCMA. = ( )( )DCMDBAMB ++ = MCBADCMBMDMB ++ = ( )MCMBDCMDMB +

= CBDCMDMB + = MDMB

2/ Gi O = AC BD MOMCMA 2=+ v MOMDMB 2=+ MDMBMCMA +=+

2222

2.2 MDMDMBMBMCMCMAMA ++=++ 2222

MDMBMCMA +=+ ( do 1/

)

Bi 13. Cho tam gic ABC c G l trng tm v M tu . Chng minh: MA

2 + MB

2 + MC

2 = 3MG

2 + GA

2 + GB

2 + GC

2

Gii

( ) 22222 2 GAGAMGMGGAMGMAMA ++=+== ( ) 22222 2 GBGBMGMGGBMGMBMB ++=+== ( ) 22222 2 GCGCMGMGGCMGMCMC ++=+== Suy ra: MA

2 + MB

2 + MC

2 = 3MG

2 + ( )GCGBGAMG ++2 + GA2 + GB2 + GC2

Hay: MA2 + MB

2 + MC

2 = 3MG

2 + GA

2 + GB

2 + GC

2

Bi 14 . Cho hai ng trn (O) v (O/) ct nhau ti A, B. TT/ l on tip tuyn chung ngoi. ng thng qua A, B ct TT/ ti M. Chng minh M l trung im TT/ Gii i vi ng trn (O) ta c: MA.MB = MT2. i vi ng trn (O/) ta c: MA.MB = MT/2

Vy: MT2 = MT/2 hay MT = MT/ Bi 15/ Cho im M thuc ng trn (O) ng knh AB. ng trn tm M tip xc AB ti H v ct (O) ti E v F. EF ct MH ti I. Chng minh I l trung im MH Gii

i vi ng trn (M), ta c: IHIMIFIE .. = v 22. MHIMIFIE =

• Nguyn Quc Qun - Trng THPT chuyn Nguyn Quang Diu

12

K

M

D

C

BA

N

M

H CB

A

D

CB

A

N

M

H

DC

BA

i vi ng trn (O), ta c: 22. OMOIIFIE = Ta c: IM2 MH2 = OI2 OB2 ( pytago trong 2 tam gic OMH v OIH c) IM2 MH

2 = OH

2 + IH

2 OH

2 MH

2 IM

2 = IH

2

Hay IM = IH

Bi 16/ Cho ng trn tm O,bn knh R. T im M bn trong ng trn (O), v hai dy AMB v BMD vung gc nhau. Gi K l trung im BD. Chng minh MK vung gc CD Gii

Ta c MD.MCMB.MA = ( cng bng OM2 R2 )

AC.MK = ( )( )MCAMMDMB2

1++

= ( )MD.MCMD.AMMC.MBMB.AM2

1+++

= ( )MD.MCMD.AMMC.MBMB.MA2

1+++ = 0

( V : MBMC, MAMA v MD.MCMB.MA = )

Bi 17/ Cho tam gic ABC vung ti A,ng cao AH. Gi M, N l lt l trung im ca AH v HC. Chng minh BM vung gc AN Gii

( )( )ACAHBHBAAN.BM4 ++= = AC.BHAH.BHAC.BAAH.BA +++ = HC.BHAH.HA + ( AB AC, v AH BC,cng thc chiu) = AH

2 + BH.AC = 0. Vy BM AC

Bi 18/ Cho hnh ch nht ABCD, v BHAC (HAC). Gi Gi M, N ln lt l trung im ca AH v DC. Chng minh tam gic BMN vung ti M Gii

( )( )MCMDBHBAMN.BM4 ++= = MC.BHMD.BHMC.BAMD.BA +++ = MD.BHMC.BAMD.BA ++

= ( ) ( )CDMC.BHMC.BAADMA.BA ++++ = BA.BHMC.BAMA.BA ++ ( BH MC, BA AD v BACD = )

= ( )BHHC.BA + ( HMMA = ) = BC.BA = 0. Vy BM MN Bi 19/ Cho hnh bnh hnh ABCD c AB = 13, AD = 19 v AC = 24.Tnh BD Gii

Ta c: ADABAC += v BCBABD +=

Nn: AC2 + BD

2 = 2(AB

2) + 2 ( )BC.BAAD.AB +

2) + 2 ( )CBADAB +

2)

Nn: BD2 = 2(AB

2) AC

2 = 2(169 + 361) 576 = 484 BD = 22

H thc lng trong tam gic

Cho tam gic ABC , ta nhc li cc k hiu thng dng: * a = BC ; b = CA ; c = AB

* S : din tch tam gic ABC * R : bn knh ng trn ngoi tip tam gic ABC * r : bn knh ng trn ni tip tam gic ABC * ma , mb , mc : di cc trung tuyn k t A, B , C * ha , hb , hc : di cc ng cao k t A, B , C .

• Nguyn Quc Qun - Trng THPT chuyn Nguyn Quang Diu

13

* p : na chu vi : 2

cbap

++=

I.nh l cosin v nh l sin trong tam gic. 1/ nh l cosin trong tam gic. Vi mi tam gic ABC , ta c: a

2 = b

2 + c

2 2bc.cosA ; b

2 = a

2 + c

2 2ac.cosB ; c

2 = a

2 + b

2 2ab.cosC

H qu.

cosA =

222

2bc

acb +; cosB =

222

2ac

bac +; cosC =

222

2ab

cba +

2/ nh l sin trong tam gic. Vi mi tam gic ABC , ta c:

R2Csinc

Bsinb

Asina

===

3/ di trung tuyn

4c

2ba

m,4b

2ca

m,4a

2cb

m222

2c

2222b

2222a

+=

+=

+=

4.Din tch tam gic

ahaS .2

1= = chb.

2

1= chc.

2

1

CbaS sin..2

1= = Acb sin..

2

1Bac sin..

2

1=

S = pr

R

abcS

4=

))()(( cpbpappS = cng thc H - rng

II.Trong tam gic vung. Cho tam gic ABC vung ti A, ng cao AH. BC

2 = AB

2 + AC

2 ; AB

2 = BH.BC ; AC

2 = CH.CB

AH.BC = AB.AC ; AH2 = HB.HC ;

222

111

ACABAH+=

III. H thc lng gic c bn

sin2x + cos

2x = 1 ; x

x

2

2tan1

cos

1+= , (cosx 0) ; x

x

2

2cot1

sin

1+= ,(sinx 0)

• Nguyn Quc Qun - Trng THPT chuyn Nguyn Quang Diu

14

H

M

CB

A

D

C

BA

N

G

M

C

B A

Bi tp Bi 1.Cho tam gic ABC vung ti C, ng cao CD, DA = 9m, DB = 16m.Tnh :CD, AC,BC

Gii CD

2 = DA.DB = 9.16 = 144 CD = 12

2 + CD

2 = 81 + 144 = 225 AC = 15

BC2 = BD

2 + CD

2 = 256 + 144 = 400 BC = 20

Bi 2/ Cho tam gic ABC vung ti A, M l trung im AB v H l hnh chiu ca M ln BC. Chng minh: HC2 HB2 = AC2

Gii Trong tam gic MHC c HC

2 = MC

2 MH

2

Trong tam gic MBH c HB2 = BM

2 MH

2

Trong tam gic AMC c MC2 = AM

2 + AC

2 = BM

2 + AC

2

HC2 HB

2 = (MC

2 MH

2) ( BM

2 MH

2) = MC

2 BM

2

= BM2 + AC

2 BM

2 = AC

2

Bi 2. Cho tam gic ABC vung ti A. Cc trung tuyn l AD, BE, CF. Chng minh: BE

2 + CF

2

Gii Gi a = BC , b = AC v c = AB, Tam gic ABC vung ti A nn BC = a = 2 AD

42

2222 bca

BE +

= , 42

2222 cba

CF +

= v 44242

2222222 aaaacb

=

BE2 + CF

2 =

42

222bca

+

+ 42

222cba

+

= =+++

4

2222 222222 cbabca

4

5

4

4 2222 acba=

2

Bi 3. Cho tam gic ABC. Chng minh: cotA + cotB + cotC = S

cba

4

222++

Gii

abc

acbR

a

R

bc

acb

A

AA

222222 2.

2sin

coscot

+=

+==

abc

bcaR

b

R

ac

bca

B

BB

222222 2.

2sin

coscot

+=

+==

abc

cbaR

c

R

ab

cba

C

CC

222222 2.

2sin

coscot

+=

+==

cotA + cotB + cotC = abc

cbaR

222++

= S

cba

4

222++

Bi 4. Cho tam gic ABC c hai trung tuyn BM v CN vung gc nhau. Chng minh rng: AB2 + AC2 = 5BC2 Gii

Gi G l trng tm tam gic ABC. Ta c: BMGB3

2= v CNGC

3

2=

v GC2 + GB

2 = BC

2

GC2 + GB

2 = BC

2 ( ) 222

9

4BCCNBM =+

2222222

4

22

4

22

9

4a

cbabca=

++

+ 2

222

9

4a

cba=

++ b

2 + c

2 = 5a

2

Bi 6. Chng minh trong mi tam gic ta u c: a = b.cosC + c.cosB

• Nguyn Quc Qun - Trng THPT chuyn Nguyn Quang Diu

15

C/B/

A/

O

CB

A

b.cosC + c.cosB = ac

bcac

ab

cbab

22

222222+

++

= a

bca

a

cba

22

222222+

++

= a

Bi 5. Cho tam gic ABC c trung tuyn AM = 2c

. Chng minh:

a/ 2b2 = a

2 c

2

b/ sin2A = 2sin

2B + sin

2C

Bi 7. Cho tam gic ABC c B = 600 , R = 2. I l tm ng trn ni tip tam gic ABC. Tnh bn knh ng trn ngoi tip tam gic ACI Gii

B = 600 A + C = 1200 060

2=

+ CA gc AIC = 1200 = B .

Gi RB l bn knh ng trn ngoi tip tam gic AIC ,

Trong tam gic AIC, c: BRAC

2120sin 0

= m 42sin

== RB

AC RB = 2

Bi 8. Cho tam gic ABC c BC = 13 cm, AB = 12 cm v AC = 5 cm a/ Tnh din tch tam gic ABC b/ Tnh di cc ng cao, cc ng trung tuyn

c/ Tnh R v r

Bi 9. ng trn ni tip tam gic ABC ln lt tip xc BC, CA, AB ti A, B, C. Chng

minh: ''' CBAS = R

pr

2

2

Gii Ta c cc cp gc C/AB/ v C/OB/ , C/BA/ v C/OA/, A/CB/ v A/OB/ b nhau nn sin ca chng bng nhau Bn knh r ca ng trn ni tip tam gic ABC l bn knh ca ng trn ngoi tip tam gic A/B/C/

/////////BOACOACOBCBA

SSSS ++= = C

r

B

r

A

r

sin2

1

sin2

1

sin2

1 222++

=

++

R

c

R

b

R

ar

2222

1 2 = R

pr

2

2

Bi 15 . Tnh A , B , ha , R v r ca tam gic ABC bit:

a/ a = 6 , b = 2 v c = 3 + 1

b/ a = 32 , b = 2 2 v c = 26

Phng php to trong mt phng 1/ H trc to Oxy.

jyixa += a = (x ; y)

jyixOM MM += M(xM ; yM) (xM: honh ; yM: tung im M )

2/ Cc php ton trn vect. Cho 2 vect: a = ( x ; y ) v b = (x/ ; y/ )

=

==

/

/

yy

xxba

( )// ; yyxxba = ; ( )kykxak ;= Tch v hng ca 2 vect: ba. = x.x/ + y.y/

3/ Cc biu thc lin quan im: Cho A(xA ; yA), B(xB ; yB), C(xC ; yC)

• Nguyn Quc Qun - Trng THPT chuyn Nguyn Quang Diu

16 P N

M CB

A

AB = (xB xA ; yB yA ) ; AB = ( ) ( )22

ABAB yyxxAB +=

I l trung im on AB th: I

++

2;

2

BABA yyxx

G l trng tm tam gic ABC th: G

++++

3;

3

CBACBA yyyxxx

4/ Cho 2 vect: a = ( x ; y ) v b = (x/ ; y/ )

2222 ''

'.'.);cos(

yxyx

yyxxba

++

+= ( )

( ) ( )2/2/22

//

.

..;cos

yxyx

yyxxba

++

+=

a cng phng b '' y

y

x

x= ; a b x.x

/ + y.y

/ = 0

Bi tp I.PHNG PHP TA 1/ Cho ba im A(1 ; 1) , B(1 ;3) , C(2 ; 0).Chng minh A, B, C thng hng

Gii. Nu AB v AC cng phng th ng thng qua A,B v ng thng qua A,C song song hoc trng nhau. Khng song song c ri

( )2;2=AB v ( )3;3 =AC . V ACAC3

2= ( ta c th dng t s ta tng ng

bng nhau) nn AB v AC cng phng. Vy A, B, C thng hng 2/ Cho ba im A(3 ;4) , B(2 ;5) , C(7 ; x).Tm x A, B, C thng hng

( )1;1=AB v ( )4;10 = xAC . A, B, C thng hng khi: 1

4

1

10 =

x x = 14

3/ Cho A(1 ; 3), B(2 ; 4), C(0 ; 1) a/ Tm ta trung im M ca AC b/ Tm ta im D sao cho ABCD l hnh bnh hnh Gii

a/

=+

=

=+

=

22

2

1

2

CA

M

CA

M

yyy

xxx

2;

2

1M

b/ ABCD l hnh bnh hnh nn M cng l trung im BD.

==

==

02

32

BMD

BMD

yyy

xxx. Vy D(3

; 0)

4/ Cho ( ) ( )4;7,2;3 == ba v ( )22;19=c . Biu din c theo a v b Gii

Gi s byaxc += , m byax + = ( 3x + 7y ; 2x + 4y )

byaxc +=

=+

=+

2242

1973

yx

yx

=

=

4

3

y

x. Vy: bac 43 +=

5/ Cho tam gic ABC vi A(1 ;1 ) , B(3 ;1) , C(6 ; 0) . Tnh gc B

( )2;4 =BA v ( )1;3 =BC

cosB = ( ) ( )( )2

1

210

10

19416

1.23.4;cos =

=

++

+=BCBA B = 1200

6/ M(3 ; 1), N(0 , 4), P(2 , 2) ln lt l trung im BC ; CA ; AB .tm ta A, B, C Gii

• Nguyn Quc Qun - Trng THPT chuyn Nguyn Quang Diu

17

MBNC l hnh bnh hnh nn PNMC =

M : ( )1;3 += CC yxMC v ( )6;2=PN

Nn:

=+

=

61

23

C

C

y

x

=

=

5

1

C

C

y

x C(1 ; 5)

M l trung im BC nn:

==

==

32

12

CMB

CMB

yyy

xxx B(1 ; 3)

N l trung im CC nn:

==

==

32

12

CNA

CNA

yyy

xxx A(1 ; 3)

7/ Hnh tnh t gic ABCD v tnh din tch ca n a/ A( 2 ; 1) , B(0 ; 3 ) ,C(6 ; 6 ) , D(8 ; 2 )

b/ A(0 ; 2) , B(1 ; 1 ) , C(4 ; 0) , D(6 ; 4)

c/ A(2 ;1) , B(6 ; 4) , C ( 3 ; 8) , D(1 ; 5)

d/ A(2 ; 4) , B(3 ; 1) , C(6 , 0) , D(5 ; 3)

Gii a/ A(2 ; 1) , B(0 ; 3 ) ,C(6 ; 6) , D(8 ; 2 )

( )4;2 =AB v ( )4;2 =DC . DCAB = ABCD l hnh bnh hnh

Vy ABCD l hnh ch nht, c din tch: S = AB.AD = 936164 ++ = 30 b/ A(0 ; 2) , B(1 ; 1 ) , C(4 ; 0) , D(6 ; 4)

( )3;1 =AB , ( )1;3=BC , ( )2;6=AD V: BCAD 2= v 0. =ADAB nn ABCD l hnh thang vung ti A v B

1= ( ) 9119436

2

1++++ = 15

c/ A(2 ;1) , B(6 ; 4) , C ( 3 ; 8) , D(1 ; 5)

( )3;4=AB , ( )3;4=DC v ( )4;3=AD

V: AB = DC , AB . AD = 0 v AB = AD nn ABCD l hnh vung. Din tch S = AB2 = 25

d/ A(2 ; 4) , B(3 ; 1) , C(6 , 0) , D(5 ; 3)

( )3;1 =AB , ( )3;1 =DC v ( )1;3=AD

V: AB = DC v AB = AD nn ABCD l hnh thoi.

Din tch S = 2

1AC.BD =

2

144.1616 ++ = 8