[Vietmaths.com] Tonghop Chuyen de Boi Duong Hsg Toan Cua Khoi 11

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    Kon Tum, thng 05 nm 2013.

    http://www.vietmaths.com/http://www.vietmaths.com/http://www.vietmaths.com/http://www.vietmaths.com/
  • 8/10/2019 [Vietmaths.com] Tonghop Chuyen de Boi Duong Hsg Toan Cua Khoi 11

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    LI M U

    Ton hc l mn th thao ca tr tu

    Tht vy, vic hc ton gip rn luyn cho hc sinh mt t duy khe, l cng c

    cn thit gip cc em hc tp cc mn hc khc, gip x l tt cc tnh hung trong cuc

    sng v l hnh trang tt trn con ng hc tp v lm vic d l trong bt k lnh vc

    no.

    i vi hc sinh hc mn chuyn, vic t hc, t tm ti nghin cu l mt vic

    quan trng bc nht, quyt nh thnh cng cho cc em, to nim say m i vi v p

    ca ton hc. Tp san ny xin gii thiu nhng kt qu ca vic nghin cu tm ti, nhng

    kinh nghim nh rt ra c trong qu trnh hc tp ca cc em hc sinh chuyn ton lp

    11 t nm hc 2010-2011 n 2012-2013. y cng l ba nm hc m b mn ton ca

    Trng Trung hc Chuyn Nguyn Tt Thnh, tnh Kon Tum t c mt s thnh

    tch vt tri so vi cc nm hc trc; tiu biu c cc hc sinh nh: Phan Hng Hnh

    Trinh, Cao Thanh H, Trn Th T Trinh, Nguyn Ngc Khnh, L B Lc, T Quang

    Hi...

    Nhng ti ny ch l nhng bc i chp chng ca cc em vo lu i nguy ngaca Ton hc, nhng chnh l nn tng cho nhng bc i vng chi v sau v c bit

    l tin , l ti liu tt cho cc th h sau hc hi v tip ni. Chnh v th, ni dung cc

    chc chn cn rt nhiu va vp v thiu st, chng ti rt mong s gp ca qu thy c

    v cc bn a ch mail ca nhm ton lp 11: [email protected].

    Sau cng l li cm n v s quan tm, to iu kin ca ban gim hiu Trng

    Trung hc Chuyn Nguyn Tt Thnh ra i tp san s 1 ny.

    Kon Tum, ngy 30 thng 04 nm 2013.Ban bin tp: T ton

    Trng THPT Chuyn Nguyn Tt Thnh, Kon Tum.

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    HAI THC CC BT NG THC HNH HCTRONG TAM GICHc sinh thc hin:Trn Th T Trinh, Nguyn Th Tuyt Anh, NguynTh Hng, Nguyn Th Thanh Thy ( Nhm chuyn ton lp 11A1- Nmhc 2010-2011).

    Gio vin hng dn:V Th Ngc nh.

    Li dn:Ton hc, vi nhng v p huyn b v sc li cun mnh m ca n, vang lm cc nh ton hc trong nhiu th k qua. Ton hc ging nh mt kho bu v

    gi ca nhn loi m loi ngi lun kht khao c khm ph, chinh phc. V, chmtay vo chic cha kha m kho bu y, loi ngi ph tri qua mt qu trnh tch lylu di. Chng ti xin c v bt ng thc hnh hc nh mt vin kim cng trong kho qu . Bt ng thc hnh hc l mt mng rt th v, mi bt ng thc li c mtmi quan h mt thit vi mt s bt ng thc khc. Bi vy, chng ta c th thu cnhiu bi ton thng qua vic khai thc mt bt ng thc n gin.

    Vi mc ch hng ti vic s dng nhng bi ton tm c vo vic tm cc tr

    ca mt biu thc hnh hc, chng ti s ch tp trung vo nhng bt ng thc c dubng xy ra v xy ra khi tam gic u. Hy vng cc bn c th vn dng chng mt cchhiu qu.

    Ni dung chuyn

    I.Mt s kin thc lin quan:1. Mt s k hiu dng ch cc yu t trong tam gic:+ a, b, c tng ng l di ba cnh BC, AC, AB ca tam gic ABC.+ ma, mb, mctng ng l di cc trung tuyn k t cc nh A, B, C.

    + ha, hb, hctng ng l di cc ng cao k t cc nh A, B, C.+ la, lb, lctng ng l di cc phn gic dng t ba nh A, B, C.+ R, r tng ng l bn knh ng trn ngoi tip v ni tip ca tam gic ABC.+ S l din tch tam gic ABC.2. Cc kin thc c bn:

    ma2 = 1

    2(b2+c2) - 1

    4a2

    la=2 cos

    2A

    bc

    b c

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    1 1 1 1

    a b ch h h r

    S= 1 .2

    aa h =1

    .2

    bb h =1

    .2

    cc h

    = 1 sin2

    ab C= 12bcsinA= 1

    2acsinB

    = pr

    =4abc

    R

    = ( )( )( )p p a p b p c .

    sinA+sinB+sinC 3 32

    3 3cos cos cos2 2 2 2A B C

    cosA+ cosB+ cosC 32

    sin2A+sin2B+sin2C 94

    II. Bt ng thc hnh hc:T nhng ng thc phn I, cng thm mt cht bin ha, chng ta s thu c

    bt ng thc tng chng nh n gin nhng li v cng hu dng.1. Cc bt ng thc lin quan n ng trung tuyn:

    Ta c: ma2 = 14 (2b2+2c2-a2) (1)

    Ta c th vit li l: ma2=14

    (b2+c2+ (b2+c2-a2))

    mt cht ta thy: b2+c2a2 = 2bc.cosA

    Do : ma2=14

    (b2+c2+2bc.cosA) = 14

    ((b-c)2+4bc.cos2 2A

    ) bc.cos2 2A

    Hay ma cos2A

    bc

    Tng t i vi mb, mc ta c: mb cos2

    Bac , mc cos

    2

    Cab .

    Tip theo ch cn nhn v theo v cc bt ng thc va tm c vi nhau ta lic bi ton mi:

    Bi ton 1.1:Trong ABC ta c: ma.mb.mc abc.cos2A cos

    2B cos

    2C

    Chng minh:

    By gi hy vit li (1) theo mt cch khc: ma2=14

    (2(b2+c2) a2)

    Ta c: b2+c22( )

    2

    b c

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    Suy ra: ma214

    ((b+c)2 a2) = 14

    (b+c-a)(b+c+a) = p(p-a)

    Hay ma ( )p p a (*)

    p dng (*) va tm c ta c th gii quyt bi ton sau mt cch nhanh gn.Bi ton 1.2: Trong ABC, chng minh:a) ma+mb+mc (p p a p b p c )

    b) ma.mb.mc p.S

    Bi ton 1.3: Chng minh rng:

    ma+mb+mc 92R .

    Chng minh:

    Ta c: ma2

    + mb2

    + mc2

    =

    3

    4 (a2

    +b2

    +c2

    ) = 3R2

    (sin2

    A + sin2

    B + sin2

    C)

    Suy ra: ma2+ mb2+ mc2227

    4R

    Ta c: (ma2+ mb2+ mc2)2 3(ma2+ mb2+ mc2) 281

    4R ma+mb+mc

    92R .

    Chng ta c th s dng bt ng thc ny chng minh mt s bi ton sau:Bi ton 1.4:Chng minh rng:

    P= 21

    am+ 2 2

    1 1

    b cm m 2

    43R

    Chng minh:Ta c:

    P 3

    21 1 1

    3a b cm m m

    3 (2

    3

    a b cm m m

    P 3

    239

    2R

    = 3

    2

    2

    2 43 3R R

    (pcm).

    Bi ton 1.5: Chng minh rng:

    P= 2 2 2b c a

    a b c

    m m m

    m m m 2

    R

    Chng minh:p dng bt ng thc:

    2 2 2a b c

    b c a a+b+c

    Ta c: P 1 1 1

    a b cm m m 9

    a b cm m m 9 2

    92R R

    (pcm)

    Qua vic xy dng cc bt ng thc lin quan lin quan n ng trung tuyn trn, cc bn c t t cu hi rng: i vi ng phn gic v ng cao liu c thtng t ha hay khng? nu c th hy tip tc tm cu tr li cc phn tip theo.

    2. Cc bt ng thc lin quan n ng phn gic:

    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    Ta c: la=2 cos

    2A

    bc

    b c os

    2A

    bcc (v b+c 2 bc )

    Mt khc, ta li c:cosA =

    2 2 2

    2b c a

    bc

    cos2A =

    2 21 cos ( ) ( )2 4

    A b c a p p a

    bc bc

    Suy ra: la ( )p p a (**)Tng t cho lb, lc, ta thu c bi ton m sau:

    Bi ton 2.1: Trong ABC, ta c:

    a) la.lb.lc abc.cos cos cos2 2 2A B C (1)

    b) la.lb.lc pS (2)c) la+lb+lc 3p (3)Bt ng thc (3) c chng minh nh sau:T (**) ta c: la+lb+lc ( )p a p b p c

    3( ) 3p p a p b p c p (pcm)

    Ta thy: la ( )p p a m ma ( )p p a suy ra: la maKt hp vi nh l ng xin v ng vung gc, ta c c mi quan h gia

    ng trung tuyn, ng phn gic, ng cao trong mt tam gic nh sau: ha la ma(***)

    T (***) chng ta s pht trin c thnh cc bi ton mi m chng ti strnh by cc phn tip theo.

    3. Bt ng thc lin quan n ng cao:Ta c: ha laV th t phn 2 chng ta c th d dng suy ra c cc bt ng thc sau:

    Bi ton 3.1: Trong ABC, ta c:

    a) ha cos 2A

    bc

    b) ha ( )p p a

    Bi ton 3.2: Trong ABC, chng minh:

    a) ha.hb.hc abc cos cos cos2 2 2A B C

    b) ha+hb + hc 3p (4)c) ha.hb.hc pS (5)

    Ngoi ra, ta c: 1 1 1 1

    a b ch h h r

    1 1 1 1

    ( ) ( )( )a b c a b ca b c

    h h h h h hr h h h

    9.

    a b ch h h 9rVy ta thu c bi ton sau:

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    Bi ton 3.3:Trong ABC , chng minh:a b ch h h 9r (6)

    mt cht, ta thy (4) v (6) l 2 bt ng thc lin quan n tng a b ch h h

    nhng ngc chiu. Vy th hy th ht hp 2 bt ng thc ny li, thm mt cht bini na xem no!(4), (6) 3p 9r

    p 3 3 ry l 1 bt ng thc lin quan n p,rBt ng thc (5) cng em n cho chng ta s th v khng km

    Ta c: (5) 2a b ch h h S

    p p = r

    2 1 1 1. .a b c a b c

    p

    h h h h h h

    p2 hb.hc+ ha.hb + ha.hc

    Th y, ch cn m ba la mt ci chng ta thu thm c bi ton mi.Bi ton 3.4: Trong ABC , chng minh:

    p2 hb.hc+ ha.hb + ha.hc4. Mi quan h gia ng trung tuyn, ng phn gic, ng cao trong tam

    gic:Nh chng ti cp cui phn 2, gia ng trung tuyn, ng phn gic,

    ng cao trong tam gic c mi quan h sau:ha la ma (***)

    Liu t mi quan h ny, chng ta c th xy dng nn cc bi ton mi c haykhng?

    tr li cho cu hi ny, chng ta hy th i qua bi ton sau:

    Bi ton 4.1: Trong ABC , chng minh

    a) a b ch h h la+lb+lc ma + mb+ mc

    b) 1 1 1 1 1 1 1 1 1

    a b c a b c a b ch h h l l l m m m

    c) 12

    a b c a b c

    b b c c a a b c a

    m m m m m m

    h l h l h l m m m

    Bi ton 4.1 trn, bn c c th d dng chng minh da vo (***)Theo BT Nesbit ta c: 3

    2a b c

    b c a c a b

    m m m

    m m m m m m

    Li s dng BT (***) ta c bi ton:Bi ton 4.2: Trong ABC , chng minh:

    32

    a b c

    b c c a a b

    m m m

    h l h l h l

    Xt bi ton sau:Bi ton 4.3:Trong ABC, chng minh rng:

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    2 3a b c

    a b c

    m m m

    Chng minh:

    Ta c: 4ma2=2(b2+c2)-a2=2(a2+b2+c2)-3a2 (2ma)2+(a 3 )2=2(a2+b2+c2)Li c: (2ma)

    2+(a 3 )22(2maa 3 )

    Suy ra: 2(2maa 3 )2(a2+b2+c2) ama2 2 2a b c

    2 3

    2 2 21 2 3

    a b caam

    2 2 2

    2 3a b ca

    a a

    m

    (7)

    Tng t:

    2 2 2 2 2 2

    2 3 2 3;

    a b c a b cb c

    b b c c

    m m

    Do : 2 3a b c

    a b cm m m

    M halama, suy ra: 2 3a b c a b c a b c

    a b c a b c a b c

    h h h l l l m m m .

    Ch :

    T (7) ta cng suy ra:2

    2 2 2

    2 3a b c

    a am m

    a

    Tng t:2 2

    2 2 2 2 2 2

    2 3 2 3;

    a b c a b cb b c cm m m m

    b c

    Do : 2 2 22 3

    a b ca b cm m m

    a b c

    (ma2+mb2+mc2)=

    3 32

    Vy, liu c tng t ha bt ng thc trn i vi ng phn gic v ngcao c hay khng?

    Ta c:a b ch h h

    a b c = 2 2 2 2 2 2 2 2 2

    2 2 2 1 1 12 ( )

    4 sin sin sinS S S S

    a b c R A R B R C

    = 18

    (abcR 2 2 2 2 2 2

    1 1 1)

    sin sin sinR A R B R C

    = sinA.sinB.sinC( 2 2 21 1 1

    )sin sin sinA B C R rng, vi 90A cn 45B C th

    a b ch h h

    a b c = 5

    2

    3 32

    Do bt ng thc trn khng ng i vi ng cao. Vy cn i vi ngphn gic th sao? Cu hi ny xin c dnh cho bn c suy ngh!

    Bi ton 4.4:Trong ABC, chng minh rng:

    3 3a b b c c al l l l l l

    c a b

    (8)

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    Chng minh:Ta c:

    (8) 1 1 1 1 1 1( ) ( ) ( ) 3 3a b cl l l

    b c a c a b

    2 cos 2 cos 2 cos2 2 2. . . 3 3

    A B Cbc ac abb c a c a b

    b c bc a c ac a b ab

    3 3

    cos os os2 2 2 2A B C

    c c

    y l mt bt ng thc n gin trong tam gic.Vy, bt ng thc (8) c chng minh.M, trong tam gic ta lun c: hala, hblb, hclaV vy: t (8) ta c th suy ra c bt ng thc sau:

    Bi ton 4.5:3 3a b b c c a

    h h h h h h

    c a b

    T (8) chng ta li c th tip tc khai thc chuyn thnh mt bi ton mi

    Ta c: ( )( ) 9a b b c c aa b a c a b

    l l l l l l c b c

    l l l l l l c a b

    9

    3a b b c c aa b a c a b

    c b cl l l l l l l l l l l l

    c a b

    Hin nhin t y ta cng c th suy ra c cc bi ton sau:

    Bi ton 4.6:Cho ABC, chng minh:a) 3

    a b a c a b

    c b c

    l h l h l h

    b) 3a b a c a b

    c b c

    h h h h h h

    T tnga b a c a b

    c b c

    h h h h h h

    , ta li c th khai thc mt bi ton mi

    Bi ton 4.7:Tong ABC, chng minh rng:3

    2a b a c c b

    c b a R

    h h h h h h r

    Chng minh: Khng mt tnh tng qut, gi s: abc2 2 2

    a b c

    S S S

    h h h hahbhc

    hb+hc hc+ha ha+hb (1)

    c c a a b

    1 1 1h h h h h h

    (2)

    p dng bt ng thc Tr-b-sp vo hai dy (1), (2) ta c:

    b c c a a b

    1 1 1 1( )( )

    3 h h h h h ha b a c c b

    c b aa b c

    h h h h h h

    (3)

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    Ta c: a+b+c=2R(sinA+sinB+sinC) 3 3R (4)

    Mt khc:b c c a a b

    1 1 1 1 1 1 1 1( )

    h h h h h h 2 2a b ch h h r

    (5)

    T (3, (4), (5) suy ra:3

    2a b a c c b

    c b a R

    h h h h h h r

    T bi ton ny, ta li thu c bi ton sau:

    Bi ton 4.8: Trong ABC, chng minh:

    a) 32a b b c c a

    c a b R

    h l h l h l r

    b) 3

    2a b b c c a

    c a b R

    l l l l l l r

    c) 32

    a b b b c c c a a

    c a b R

    rh l h h l h h l h

    Cn rt nhiu cc bt ng thc khc c th khai thc, nhng chng ti xin c ktthc phn ny ti y i qua cc phn khc cng hp dn khng km.

    5. Bt ng thc lin quan n p, r, R:Bi ton 5.1:Trong ABC, chng minh:

    a2+b2+c2 9R2 (1)Chng minh:

    Ta c: (1)4R2(sin2A+sin2B+sin2C)9R2 sin2A+sin2B+sin2C 94 y l bt ng thc c bn trong tam gic nn ta c c ccm.Chng ta hy khai thc bi ton trn tm li gii cho mt bi ton mi.

    Ta c: (ab+bc+ac)( 1 1 1 ) 9ab bc ca

    2 2 2 2 2

    1 1 1 9 9 9 1ab bc ac a b c 9ab bc ca R R

    Vy ta c bi ton sau:Bi ton 5.2:Trong ABC, chng minh:

    21 1 1 1ab bc ca R Bi ton trn c rt nhiu cch gii, song chng ti ch gii thiu cho cc bn mt

    cch gii. Vi nim say m ton hc hi vng cc bn s tm c cch gii ring chomnh.

    Sau y, chng ti xin trnh by mt hng khai thc khc i vi bi ton 5.1:Ta c: ab+bc+ac 2 2 2a b c 29R M, ha+hb+hc9r

    Suy ra:2 2 2

    2 2

    ( ) (9 ) 99

    a b bh h h r r

    ab bc ac R R

    2 2 2 2 2

    2

    ( ) 9a b c a b bh h h h h h rbc ac ab ab bc ac R

    ng thc xy ra khi v ch khi ABC u.

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    Bi ton 5.3: Trong ABC, chng minh; R2rTht vy, theo cng thc Euler, ta c:

    OI2=R2-2Rr (vi O,I ln lt l tm ng trn ngoi tip, ni tip ABC)

    R

    2

    -2Rr 0

    R2rTrong qu rnh khia thc mi lin h ca ba ng trung tuyn, phn gic, ngcao phn 3, chng ti a ra cc bt ng thc m cc bn c th kt hp li an kt qu bi ton 5.3.

    Bi ton 5.4: Chng minh rng:2 2 2a b c r2+p2+4Rr

    Chng minh:Ta c: p2r2=S2=p(p-a)(p-b)(p-c)p2r2=p3-(a+b+c)p2+(ab+bc+ac)p-abc

    =-p3+(ab+bc+ac)p-4pRrr2=-p2+ab+bc+ac -4Rr ab+bc+ca=r2+p2+4Rr

    M

    2 2 2

    a b c

    ab+bc+car

    2

    +p

    2

    +4Rr (dccm)Thc cht trong li gii ca bi ton trn chng ta i thit lp cng thc :2 2 2a b c =r2+p2+4Rr

    2 2 2a b c =(a+b+c)2-2(ab+bc+ca) =2p2-8Rr-2r2T bi ton 5.4, suy ra:2p2-8Rr-2r2r2+p2+4Rrp23r2+12Rr3r(r+4R)p2

    3.4sin 2 2 2 2sin sin .4 (1 sin sin sin ) 16 os os os2 2 2 2 2 2 2 2 2A B C A B C A B C

    R R c c c

    3tanos os os

    2 2 2tan tan

    2 2 2 1 sin sin sin2 2 2

    A B Cc c cA B C

    A B C

    Vy, ta c bt ng thc sau: 3tanos os os

    2 2 2tan tan2 2 2 1 sin sin sin

    2 2 2

    A B Cc c cA B C

    A B C

    y l mt bt ng thc kh.Bi ton 5.5: Trong ABC, chng minh:

    p 3 32

    R.

    Chng minh:Ta c: 4( 2 2 2a b cm m m )=2 2 23( )a b c (a+b+c)2

    M 2 2 2a b cm m m 227

    4R

    Suy ra: (a+b+c)2227

    4R 4p2

    2274R pcm

    Trong nhng phn trn chng ta chng minh c bi ton sauBi ton 5.6:Trong ABC, chng minh:

    p3 3 rBi ton ny d dng dn n bi ton sau:

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    Bi ton 5.7:Trong ABC, chng minh:

    a) S2

    3 3p

    b) S2 2 2a b c

    4 3

    Sau y chng ti xin gii thiu mt bt ng thc c coi l mnh nht i vip, R, r trong tam gic (bt ng thc Bludon)Bt ng thc Bludon: Chng minh rng trong mi ABC ta c:

    2 2 2 2 22 10 2( 2 ) ( 2 ) 2 10 2( ) ( 2 )R Rr r R r R R r p R Rr r R r R R r Chng minh

    K hiu O l tm ng trn ngoi tip v I l tm ng trn ni tip ABC.Gi E l tip im ca ng trn ni tip vi cnh AB

    t AI = x th R d x R + d (d = OI)Ta c: AE = 2 2x r v AE = p aT tam gic vung AEI ta c:

    sin2A r

    , cos2 2

    2A x r

    x

    sinA=2 sin2A cos

    2 2

    2

    22A r x r

    x

    Theo nh l hm s sin ta c:

    A = 2RsinA =2 2

    2

    4Rr x r

    T : p = (p a) + a = 2 2 24

    1 Rrx r

    Xt hm s: p = p(x) = 2 2 24

    1Rr

    x rx

    (R d x R + d)

    Ta c: p(x) =2 2 3

    1 2 1 2

    3 2 2 3 2 2

    ( )( )( )( )4 8 x x x x x x xx Rrx Rr

    x x r x x r

    Trong : x1= 2 ( )r R d , x2= 2 ( )r R d D dng chng minh c 0 < R d x1 x2 R + dBng bin thin ca hm s p(x) nh sau:

    x R d x1 x2 R + d

    p(x) + 0 - 0 +

    p(x)

    rng d = 2 2R Rr ta c th chng minh c

    p(x1) = p(R + d) = 2 22 10 2( 2 ) ( 2 )R Rr r R r R R r v

    p(x2= p(R - d) = 2 22 10 2( 2 ) ( 2 )R Rr r R r R R r

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    Vy2 22 10 2( 2 ) ( 2 )R Rr r R r R R r p 2 22 10 2( 2 ) ( 2 )R Rr r R r R R r

    Ta c: 2 22 10 2( 2 ) ( 2 )R Rr r R r R R r

    = 4R2+ 4Rr + 3r2 [(R-2r) - ( 2 )R R r ]2 4R2+ 4Rr + 3r2v 2 22 10 2( 2 ) ( 2 )R Rr r R r R R r

    = [(R-2r) - ( 2 )R R r ]2+ 16Rr 5r2 16Rr 5r2Mi ng thc trn xy ra khi v ch khi R = 2r

    ABC l tam gic u

    Chng ti va trnh by mt s bt ng thc v h thc lng trong tam gic. Ccbn c th vn dng chng gii cc bi ton khc.

    III. Bi tp cng c:1.

    Cho ABC c bn knh ng trn ngoi tip R=1. Chng minh rng:sin sin sin

    3a b c

    A B C

    m m m

    Chng minh:

    Tac:2 2 2sin sin sin

    2 2 2 2 2 2a b c a b c a b c

    A B C a b c a b c

    m m m Rm Rm Rm am bm cm

    M,2 2 2

    2 2 22 2 2 2

    3 32 3 (2 2 )a

    a a a

    am a b ca b c a

    (theo C-si) (1)

    Tng t,2 2 2

    2 2 22 2 2 23 32 3 (2 2 )bb b bbm a b cb a c b

    (2)

    2 2 2

    2 2 22 2 2 2

    3 32 3 (2 2 )a

    c c c

    cm a b cc a b c

    (3)

    T (1), (2), (3) ta c: sin sin sin 3a b c

    A B C

    m m m

    Du = xy ra khi v ch khi a=b=c hay ABC u.

    2. Cho tam gic ABC c 3 gc nhn. Gi a,b,c l di ba cnh v ha, hb, hc l

    di cc ng cao tng ng. K hiu r, R ln lt l bn knh ng trn ni, ngoi tiptam gic . Chng minh:

    2 2 2

    9 1 1 1 12

    a b c b c a c a b

    R

    a b c r h h h h h h h h h

    Chng minh:Ta c:

    1 1 1 1 1 1 1 1 14 4 2a a b ca b c b ch h h hh h h h h

    Tng t:

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    1 1 1 1 1 14 2b c ab c a h h hh h h

    ; 1 1 1 1 1 14 2c a bc a b h h hh h h

    Do :1 1 1 1 1 1 1 1

    2 2a b ca b c b c a c a b h h h r h h h h h h h h h

    (1)

    Ta c:1 1 1 9

    a b c b c a c a b a b c b c c a a bh h h h h h h h h h h h h h h h h h

    9 9 9 91 1 12 4 ( )4a b c

    abc R

    h h h S bc ab ac ab bc caSa b c

    2 2 29R

    a b c

    (2)

    T (1),(2) suy ra pcm.

    3. Chng minh rng:1 1 1 6

    a b cRm m m

    Chng minh:Ta c:

    P= 1 1 1 3 3323

    a b c a b cm m m m m m R

    Suy ra : P 6R

    4.Chng minh rng:

    P=2 2 2

    3 3 3

    1b c a

    a b c

    h h h

    h h h r

    Chng minh:

    p dng bt ng thc:3 3 3

    2 2 2

    a b ca b c

    b c a

    Ta c: P 1 1 1 1

    a b ch h h r (pcm)

    5. Chng minh rng:

    a) ma.mb.mc327

    8R

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    b) (la+lb)(lb+lc)(lc+la)38

    3 3p

    Chng minh

    a) Ta c:ma.mb.mc

    2 3 33 273 2 8

    a b cm m m R R

    ma.mb.mc327

    8R

    b) Ta c:

    (la+lb)(lb+lc)(lc+la)32( )

    3a b cl l l

    3 32 83 3 3p p

    (pcm)

    6. Chng minh rng: ha.hb.hc 27r3

    Chng minhTa c:3

    3

    1 1 1 1 127.

    a b c a b cr h h h h h h

    Suy ra: ha.hb.hc 27r3 (pcm)

    7. Chng minh rng:

    P= 181 1 1 2 9

    a b c

    a b b c c a

    h h h r

    h m h m h m Rr

    Chng minhTa c:

    P= 1 1 1 91 1 1 1 1 1

    b c a a b ca b c a b c

    m m m m m mh h h h h h

    9 189 1 9 22

    rR Rr

    r

    (pcm)

    8.Chng minh rng:

    2 2 2 21 1 1 1 27

    93 2a b c a b c

    m m m RR m m m

    Chng minht a= 1

    3 am

    R, b= 1

    3 bm

    R, c= 1

    3 cm

    R. Ta c: a + b + c 3

    2

    BT tr thnh:

    a + b + c + 2 2 21 1 1 27

    2a b c

    Ta c: 3 13 2

    a b cabc

    VT 3 3 3 33 32 2 2 2 2 2

    3 13 3 8 8 15

    3

    abc abc abc abca b c a b c

    3 15 273 3 642 2

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  • 8/10/2019 [Vietmaths.com] Tonghop Chuyen de Boi Duong Hsg Toan Cua Khoi 11

    17/116

    - 16 -

    9. Trong ABC chng minh rng:a) 2 2 2 2 2 2a b c a b cr r r l l l

    b)rala+ rblb+ rclc p 2

    Chng minha)Ta c 2 2 2a b cr r r = p

    2 ( tan 22A

    tan 22

    B + tan 22C)

    V, tan 22A

    tan 22

    B tan 22C tan tan tan tan tan tan 1

    2 2 2 2 2 2A B B C C A

    Nn: 2 2 2 2a b cr r r p (1)Mt khc: 2 2 2 ( ) ( ) ( )a b cl l l p p a p p b p p c = p

    2 p(a+b+c-2p)=p2 (2)T (1),(2) suy ra 2 2 2 2 2 2a b c a b cr r r l l l

    b) Ta c:

    rala+ rblb+ rclc ( ) ( ) ( )S S Sp p a p p b p p cp a p b p c

    = ( )( ) ( )( ) ( )( )p b p c p a p c p a p b p ( p b + p c + p a ) = p2

    Vy rala+ rblb+ rclc p 2 (pcm)10. Trong ABC chng minh rng:

    9

    a b c

    a b c R

    r r r p

    Chng minh

    Ta c: 2 sin 2 sin 2 sin. tan . tan .tan

    2 2 2a b ca b c R A R B R C A B Cr r r p p p

    = 2 2 24 cos cos cos2 2 2

    R A B Cp

    = 2Rp

    (1 + cosA +1 + cosB + 1 + cosC)

    = 2Rp

    (3 + cosA + cosB + cosC)

    M: cosA + cosB + cosC

    3

    2 (bt ng thc c bn trong tam gic)Do : 9

    a b c

    a b c R

    r r r p (pcm)

    IV. Bi tp t gii:1. Cho ABC, chng minh rng:

    2 2 2

    9a b c

    a b cR

    r r r r r r

    2. Cho ABC l tam gic nhn. Gi R v r l bn knh ng trn ni tip v ngoitip tam gic. Gi H l trc tm tam gic, da, db, dctng ng l khong cch t H ti ba

    cnh BC, CA, AB. Chng minh rng:

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