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…

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  • 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

    1 m AKADKD =

    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

    c/ 0=++ CFBEAD

    Gii

    a/ MFMEMDMCMBMA ++=++

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

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

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

    1 = 0

    c/ 0=++ CFBEAD

    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

    1=++=++ CABCABCFBEAD

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

    MNBCACBDAD 4=+++

    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:

    a/ ADAB + b/ ADAB c/ ACAB + d/ ABAO +

    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 =+

    2/ DCAB. v ADOB.

    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

    ( )( )HDBHADAHBDAM ++=2 = HDADBHADHDAHBHAH ... +++ = HCADHDAH .. + ( v AH BH v AD HD v H l trung im BC)

    = ( ) 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 + AD

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

    = 2(AB2 + AD

    2) + 2 ( )CBADAB +

    = 2(AB2 + AD

    2)

    Nn: BD2 = 2(AB

    2 + AD

    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

    AC2 = AD

    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 = 5AD

    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

    AD ==+

    =

    BE2 + CF

    2 =

    42

    222bca

    +

    + 42

    222cba

    +

    = =+++

    4

    2222 222222 cbabca

    4

    5

    4

    4 2222 acba=

    ++ = 5AD

    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

    ( )3;6 =AD . V ADAB. = (2).6 + ( 4)(3) = 0 AB AD

    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

    Din tch: ( )ABBCADS +=2

    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

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