chuyen de bat dang thuc luong giac (chuong 4)

8/14/2019 Chuyen de Bat Dang Thuc Luong Giac (Chuong 4)

1/22

Trung THPT chuyn L T Trng Cn Th Bt ng thc lng gicChng 4Mt schuyn bi vit hay,th v

lin quan n btng thc v lnggic

The Inequalities Trigonometry 77

Chng 4 :

Mt s chuyn bi vit hay,th v lin quan n bt ng thc v

lng gic

ng nh tn gi ca mnh, chng ny s bao gm ccbi vit chuyn v bt ngthc v lng gic. Tc gi ca chng u l cc gio vin, hc sinh gii ton m tc ginh gi rt cao. Ni dung ca ccbi vit chuyn u d hiu v mch lc. Bn cc th tham kho nhiu kin thc b ch t chng. V khun kh chuyn nn tc gich tp hp c mt s bi vit tht s l hay v th v :

Mc lc :

Xung quanhbi ton Ecds trong tam gic .78ng dng ca i s vo vic pht hin v chng minh bt ng thc trong tamgic82Th trv ci ngun ca mn Lng gic...............91Phng php gii mt dng bt ng thc lng gic trong tam gic.............94


2/22




Xung quanh bi ton Ecds trong tam gicNguyn Vn Hin

(Thi Bnh)

Bt ng thc trong tam gic lun l ti rt hay. Trong bi vit nh ny, chng tacng trao i v mt bt ng thc quen thuc : Bt ng thc Ecds.Bi ton 1 : Cho mt im M trong ABC . Gi cba RRR ,, l khong cch t M n

CBA ,, v cba ddd ,, l khong cch t M n ABCABC ,, th :

( ) ( )EdddRRR cbacba ++++ 2

Gii : Ta c :

a

bdcd

a

SS

a

SSdhR

bc

AMCAMB

BMCABC

aaa

+=

+=

=

22

22

Bng cch ly i xng M qua phn gic gc A

Tng t : ( )1

+

+

+

c

bdadR

b

cdadR

a

cdbdR

ab

c

ac

b

bc

a

( ) ++

++

++

+++ cbacbacba ddd

a

b

b

ad

a

c

c

ad

b

c

c

bdRRR 2 pcm.

Thc ra ( )E ch l trng hp ring ca tng qut sau :Bi ton 2 : Chng minh rng :

( ) ( )22 kck

b

k

a

kk

c

k

b

k

a dddRRR ++++

vi 01 > k Gii : Trc ht ta chng minh :B 1 : 0, > yx v 01 > k th :

( ) ( ) ( )Hyxyx kkkk ++ 12

Chng minh :( ) ( ) ( ) ( ) 0121121 11 ++=

+

+

kkk

k

kk

k

aaafy

x

y

xH vi 0>= a

y

x

V ( ) ( ) ( ) 021' 11 =+= kk aakaf 1= a hoc 1=k . Vi 1=k th ( )H l ng thcng.Do 0>a v 01 >> k th ta c :

( ) 00 > aaf v 01 >> k


3/22




( )H c chng minh.Tr libi ton 2 :T h ( )1 ta c :

+

+

k

b

k

ck

k

bck

a a

cd

a

bd

a

cd

a

bdR 12

( p dng b ( )H via

cdy

a

bdx bc == ; )

Tng t :

+

+

k

a

k

bkk

c

k

a

k

ckk

b

c

bd

c

adR

b

cd

b

adR

1

1

2

2

( )kck

b

k

a

k

kkk

c

kkk

b

kkk

a

kk

c

k

b

k

a

ddd

a

b

b

ad

a

c

c

ad

b

c

c

bdRRR

++

+

+

+

+

+

++

2

2 1

pcm.ng thc xy ra khi ABC u v M l tm tam gic. p dng ( )E ta chng minhcbi ton sau :Bi ton 3 : Chng minh rng :

( )3111

2111

++++

cbacba RRRddd

Gii : Thc hinphp nghch o tm M, phng tch n v ta c :

=

=

=

c

b

a

RMC

RMB

RMA

1*

1*

1*

v

=

=

=

c

b

a

dMC

dMB

dMA

1''

1''

1''

p dng ( )E trong '''''' CBA :

( )

++++

++++

cbacba RRRddd

MCMBMAMCMBMA111

2111

***2''''''

pcm.Mrng kt qu ny ta c bi ton sau :Bi ton 4 : Chng minh rng :

( ) ( )42 kck

b

k

a

k

c

k

b

k

a

kRRRddd ++++


4/22




vi 10 > k Hng dn cch gii : Ta thy ( )4 d dng c chng minh nh p dng ( )2 trongphp bin hnh nghch o tm M, phng tch n v. ng thc xy ra khi ABC uv M l tm tam gic.By givi 1>k th t h ( )1 ta thu c ngay :Bi ton 5 : Chng minh rng :

( ) ( )52 222222 cbacba dddRRR ++>++ Xutpht t bi ton ny, ta thu c nhng kt qu tng qut sau :Bi ton 6 : Chng minh rng :

( ) ( )62 kck

b

k

a

k

c

k

b

k

a dddRRR ++>++

vi 1>k Gii : Chng ta cng chng minh mt b :B 2 : 0, > yx v 1>k th :

( ) ( )Gyxyx kkk ++

Chng minh :

( ) ( ) ( ) 01111 >+=+>

+ k

k

k

kk

aaagy

x

y

xG (t 0>= a

y

x)

V ( ) ( ) 1;001' 11 >>>+= kaaakag kk

( ) 1;00 >>> kaag

( )G c chng minh xong.

S dng b ( )G vobi ton ( )6 :T h ( )1 :

k

b

k

c

k

bck

aa

cd

a

bd

a

cd

a

bd

R

+

>

+ (t a

cd

ya

bd

xbc

== ; )

Tng t :

k

a

k

bk

c

k

a

k

ck

b

c

bd

c

adR

b

cd

b

adR

+

>

+

>

( )k

c

k

b

k

a

kk

k

c

kk

k

b

kk

k

a

k

c

k

b

k

a

ddd

a

b

b

ad

a

c

c

ad

b

c

c

bdRRR

++

+

+

+

+

+

>++

2

pcm.Bi ton 7 : Chng minh rng :

( ) ( )72 kak

a

k

a

k

a

k

a

k

a RRRddd ++>++

vi 1


5/22




Hng dn cch gii : Ta thy ( )7 cng c chng minh d dng nh p dng ( )6 trongphp bin hnh nghch o tm M, phng tch n v. ng thc khng th xy ratrong ( )6 v ( )7 .Xt v quan h gia ( )cba RRR ,, vi ( )cba ddd ,, ngoi bt ng thc ( )E v nhng m

rng ca n, chng ta cn gp mt s bt ng thc rt hay sau y. Vic chng minhchng xin dnh chobn c :

( )( )( )

( )( )( )ccbbccaabbaacba

cbcabacba

c

ba

b

ca

a

cb

cbacba

dRdRdRdRdRdRRRR

ddddddRRR

R

dd

R

dd

R

dd

dddRRR

+++

+++

+

++

++

222)4

)3

3)2

8)1


6/22




ng dng ca i s vo vic pht hin v chngminh bt ng thc trong tam gic

L Ngc Anh

(HS chuyn ton kha 2005 2008Trng THPT chuyn L T Trng, Cn Th)

1/ Chng ta i tbi ton i ssau:Vi

x

0, 0, 0, 0,2222

ta lun c:

x x 2x< tg < < sinx < x

2 2 .

Chng minh: Ta chng minh 2 bt ng thc:2

sinx

x

> v2

2

x xtg

< .

t1

( ) sinf x xx

= l hm s xc nh v lin tc trong 0,2

.

Ta c:2

os x- sin x'( )

xcf x

x= . t ( ) os x- sin xg x xc= trong 0,

2

khi

( ) ( )' sin 0g x x x g x= nghch bin trong on 0,2

nn ( ) ( )0g x g< =0 vi

0,2

x

. Do ( )' 0f x < vi 0,

2x

suy ra ( )2

2f x f

> =

hay

2sin

xx

>

vi 0,2

x

.

t ( ) 1h x tgxx

= xc nh v lin tc trn 0,2

.

Ta c ( )2 2

sin' 0

2 os2

x xh x

xx c

= > 0,

2x

nn hm s ( )h x ng bin, do

( )2 2

xh x h

< =

hay

2

2

x xtg

< vi 0,

2x

.

Cn 2 bt ng thc2 2

x xtg > v sinx x< dnh cho bn c t chng minh.

By gimi l phn ng ch :Xt ABC: BC = a , BC = b , AC = b . GiA, B, Cl ln cc gc bng radian;

r, R, p, S ln lt l bn knh ng trn ni tip, bn knh ng trn ngoi tip, nachu vi v din tch tam gic; la, ha, ma, ra, tng ng l di ng phn gic, ngcao, ng trung tuyn v bn knh ng trn bng tip ng vi nhA...

Bi ton 1: Chng minh rng trong tam gic ABC nhn ta lun c:2 2 2os os os

4

p pAc x Bc B Cc C

R R

< + + <


7/22




Nhn xt:

Tnh l hm s sin quen thuc trong tam gic ta c: sin sin sinp

A B BR

+ + = v

bi ton i s ta d dng a ra bin i sau 2 2 24

os 2 os sin os2

AAc A tg c A A Ac A

< = < , t

a n li gii nh sau.Li gii:

Ta c: 2 2 24

os 2 os sin os2

AAc A tg c A A Ac A

< = < 2os sin

pAc A A

R< =

v 2 24

os sin os4

p pAc A A Ac A

R R

> = > . Ty suy ra pcm.

Trong mt tam gic ta c nhn xt sau: 12 2 2 2 2 2

A B B C C Atg tg tg tg tg tg+ + = kt hp

vi2

2

x xtg

< nn ta c

2 2 2 2 2 21

2 2 2 2 2 2

A B B C C A A B B C C Atg tg tg tg tg tg

+ + > + + =

2

. . .4

A B B C C A

+ + > (1). Mt khc2 2

x xtg > nn ta cng d dng c

12 2 2 2 2 2 2 2 2 2 2 2

A B B C C A A B B C C Atg tg tg tg tg tg+ + < + + = t y ta li c

. . . 4A B B C C A+ + < (2). T (1) v (2) ta c bi ton mi.Bi ton 2: Chng minh rng trong tam gic ABC nhn ta lun c:

2

. . . 44

A B B C C A

< + + <

Lu : Khi dng cch ny sng to bi ton mi th ton l ABC phi l nhn

v trong bi ton i sth 0, 2x

. Li gii bi ton tng t nh nhn xt trn.

Mt khc, p dng bt ng thc( )

2

3

a b cab bc ca

+ ++ + th ta c ngay

( )2 2

. . .3 3

A B C A B B C C A

+ ++ + = . Ty ta c bi ton cht hn v p hn:

2 2

. . .4 3

A B B C C A

+ +

By gita thi t cng thc la, ha, ma, ra tm ra cc cng thc mi.

Trong ABC ta lun c: 2 sin sin sin2 2a aA A

S bc A cl bl= = +


8/22




1 1 1 1

A 2 22 os2

a

b c b c

l bc b cbcc

+ + = > = +

1 1 1 1 1 1 1 1 1 1

2 sin sin sina b cl l l a b c R A B C

+ + > + + > + +

1 1 1 1 1 1 1

2a b c

l l l R A B C

+ + > + +

.

Nh vy chng ta c Bi ton 3.Bi ton 3: Chng minh rng trong tam gic ABC nhn ta lun c:

1 1 1 1 1 1 1

2a b c

l l l R A B C

+ + > + +

Mt khc, ta li c( )2 sin sin

A2 os 2sin

22 2

a

R B C bc b c

Alc

++= =

. p dng bi ton i s ta

c:

( )( )2

2 2a

B CRR B C bc

AA l

+

+> >

( ) ( )

( )

4

a

R B C R B C bc

B C l B C

+ +> >

+ +

4

a

bc RR

l

> > .

Hon ton tng t ta c:4

c

ab RR

l

> > v

4

b

ca RR

l

> > . Ty, cng 3 chui bt

ng thc ta c:Bi ton 4: Chng minh rngtrong tam gic ABC nhn ta lun c:

12 3c a b

R ab bc ca Rl l l

< + + <

Trong tam gic ta c kt qu sin b ch h

Ac b

= = , sin c ah h

Ba c

= = v sin a bh h

Cb a

= = ,

m t kt qu ca bi ton i s ta d dng c 2 sin sin sinA B C < + + < , m

( )1 1

2 sin sin sin aA B C hb c

+ + = +

1 1 1 1b ch h

c a a b

+ + + +

, t y ta c c Bi

ton 5.Bi ton 5: Chng minh rngtrong tam gic ABC nhn ta lun c:

1 1 1 1 1 14 2a b ch h h

b c c a a b

< + + + + + + + +

( )4 2R r aA bB cC > + +

Kt hp 2 iu trn ta c iu phi chng minh.Sau y l cc bi ton c hnh thnh t cc cng thc quen thuc cc bn luyn

tp:Bi ton: Chng minh rng trong tam gic ABC nhn ta lun c:a/ ( ) ( )2 8 2 2p R r aA bB cC p R r + < + + < + .

b/ ( ) ( ) ( )( ) ( ) ( ) 22

Sp a p b p b p c p c p a S

< + + < .

c/ ( ) ( ) ( )2 2 22

abc a p a b p b c p c abc

< + + < .

d/1 1 1 1 1 1

4 2a b cl l lb c c a a b

< + + + + + = ( )' 0f x > nn hm ( )f x ng bin .

Ch 3 bt ng thc i s:1.Bt ng thc AM-GM:

Cho n s thc dng 1 2, , ..., na a a , ta lun c:1 2

1 2

......n n

n

a a aa a a

n

+ + +

Du = xy ra 1 2 ... na a a = = = .

2.Bt ng thc Cauchy-Schwarz:

Cho 2 bn s ( )1 2, ,..., na a a v ( )1 2, ,..., nb b b trong 0, 1,ib i n> = . Ta lun c:


11/22




( )222 2

1 21 2

1 2 1 2

......

...nn

n n

a a aaa a

b b b b b b

+ + ++ + +

+ + +

Du = xy ra 1 2

1 2

... n

n

aa a

b b b = = = .

3.Bt ng thc Chebyshev:Cho 2 dy ( )1 2, ,..., na a a v ( )1 2, ,..., nb b b cng tng hoc cng gim, tc l:

1 2

1 2

...

...n

n

a a a

b b b

hoc 1 2

1 2

...

...n

n

a a a

b b b

, th ta c:

1 1 2 2 1 2 1 2... ... ....n n n na b a b a b a a a b b b

n n n

+ + + + + + + + +

Du = xy ra 1 2

1 2

...

...n

n

a a a

b b b

= = =

= = =.

Nu 2 dy n iu ngc chiu th i chiu du bt ng thc.Xt trong tam gic ABC c A B (A,B s o hai gc A,B ca tam gic theo

radian).

A B sin sin

A B

A B ( theo chng minh trn th hm ( )

xf x =

sinx)

2 2

A B

a b

R R

A a

B b , m A B a b . Nh vy ta suy ra nu a b th

A a

B b

(i).

Hon ton tng t : a b c

A B C

a b c v nh vy ta c

( )A

0B

a ba b

, ( ) 0

B Cb c

b c

v ( ) 0

C Ac a

c a

.Cng 3

bt ng thc ta c ( ) 0cyc

A Ba b

a b

( ) ( )2

cyc

AA B C b c

a+ + + (1).

-Cng A B C + + vo 2 v ca (1) ta thu c:( ) ( )3

A B C A B C a b c

a b c

+ + + + + +

(2)

-TrA B C + + vo 2 v ca (1) ta thu c: ( ) ( )2cyc

AA B C p a

a

+ + (3).

Ch rng A B C + + = v 2a b c p+ + = nn (2) 3 2cyc

Ap

a

3

2cyc

A

a p

(ii), v (3) ( )

2cyc

Ap a

a

(iii).


12/22




Mt khc ta c th p dng bt ng thc Chebyshev cho 2 b s

, ,A B C

a b c

v ( ), , .p a p b p c Ta c: a b c A B C

a b c

p a p b p c

( )( )

3 3 3cyc

A A B C p ap a p b p ca a b c

+ + + +

( )3

cyc

cyc

ApaA

p aa

. M

3

2cyc

A

a p

ta suy ra: ( )

32

3 3cyc

cyc

Ap p

aA pp a

a

hay ( )

3 2cyc

cyc

Ap

aAp a

a

(iv).

Ta ch n hai bt ng thc (ii) v (iii):

-p dng bt ng thc AM-GM cho 3 s , ,A B C

a b cta c:

13. .

3. .cyc

A A B C

a a b c

kt

hp vi bt ng thc (ii) ta suy ra13. . 3

3. . 2

ABC

a b c p

3. . 2

. .

a b c p

ABC

(v). Mt

khc, ta li c

1

3. .3

. .cyc

a a b c

A A B C

, m theo (v) ta d dng suy ra

1

3. . 2

. .

abc p

ABC

, t ta

c bt ng thc6

cyc

a p

A (vi).

-p dng bt ng thc Cauchy-Schwarz , ta c :

( )22 2

cyc cyc

A B C A A

a aA Aa Bb Cc Aa Bb Cc

+ += =

+ + + + (vii), m ta tm c

( ) ( )2 8 2 2p R r Aa Bb Cc p R r + < + + < + (bi tp a/ phn trc) nn

( )

2

2cyc

A

a p R r

>

(viii) (chng vi tam gic nhn).

-p dng bt ng thc AM-GM cho 3 s ( ) ( ) ( ), ,A B C

p a p b p ca b c

ta c:

( ) ( ) ( ) ( ) ( ) ( )2

3 3 3. . . . . . .

3 3 3. . 4 . 4 .

A B C ABC S ABC S ABC p a p b p c p a p b p c

a b c abc p S R p R + + = =

( )2

3. .

34 .cyc

A S A B C p a

a p S R (4)m ( )

3 2cyc

cyc

Ap aA

p aa

(theo iv) nn t (4)

32

43

. . 729 . . .3

4 . 3 2 4cyc

cyc

Ap

aS A B C S A B C Ap

p S R R a

3

4729 . . . 3

4 2

S A B C p

R p

354 . . . . .S A B C p R (ix).


13/22




Xt tng

2 22y yx z x z

Tb By a Ax a Ax c Cz c Cz b By

= + + + + +

.

Ta c: 0T

2 2 2

1 1 1 1 1 1. . . 2 0y z z x x y

x a A y b B z c C ab AB bc BC ca CA

+ + + + + + +

.

. . . 2 0y z bc z x ca x y ab c a b

x aA y bB z cC AB BC CA

+ + + + + + +

. . . 2y z bc z x ca x y ab a b c

x aA y bB z cC BC CA AB

+ + + + + + +

(5).

p dng bt ng thc AM-GM ta c:

13 6

3a b c abc p

ABCBC CA AB

+ +

(6).

T (5) v (6) ta c: 6. . .y z bc z x ca x y ab px aA y bB z cC + + ++ + (7).

Thay (x, y, z) trong (7) bng ( p-a, p-b, p-c) ta c:

( ) ( ) ( )

12bc ca ab p

A p a B p b C p c + +

(x)

Thay (x, y, z) trong (7) bng (bc, ca, ab) ta c:12b c c a a b p

A B C

+ + ++ + (xi).

3/ Chng ta xt bt ng thc sau:x

sinx

vi

x

0, 0, 0, 0,2222

(phn chng minh bt

ng thc ny dnh cho bn c).

Theo nh l hm s sin ta c sin2

aA

R= v kt hp vi bt ng thc trn ta c

2 4

2

a A a R

R A , t ta d dng suy ra

12

cyc

a R

A > .

4/ Bt ng thc:2 2

2 2

sin x - x

x + x vi ( ]x 0, 0, 0, 0, (bt ng thc ny xem nhbi

tp dnh cho bn c).

Bt ng thc trn tng ng2

2 2

sin 21

x x

x x

+

3

2 2

2sin

xx x

x

+(1).

Trong tam gic ta c: 3 3sin sin sin2

A B C + + (2) (bn c t chng minh).T (1)

v (2) ta thu c3 3 3

2 2 2 2 2 2

3 3sin 2

2 cyc

A B C A A B C

A B C

> + + + +

+ + +

3 3 3

2 2 2 2 2 2

3 32

2

A B C

A B C

> + +

+ + +

3 3 3

2 2 2 2 2 2

3 3

2 4

A B C

A B C

+ + >

+ + +.


14/22




Mt khc, p dng bt ng thc cho 3 gc A, B, Cta thu c2 2

2 2

sinA A

A A

>

+,

2 2

2 2

sinB B

B B

>

+v

2 2

2 2

sin C C

C C

>

+, cng cc bt ng thc ta c:

2 2 2 2 2 22 2 2 2 2 2

sin sin sinA B C A B C A B C A B C

+ + > + ++ + +

, t y p dng nh l hm s sin

sin2

aA

R= ta c

2 2 2 2 2 2

2 2 2 2 2 22 2 2

a b c

A B C R R R

A B C A B C

+ + > + +

+ + +hay

2 2

2 22

cyc

a AR

A A

>

+ .


15/22




Thtrv ci ngun ca mn lng gicL Quc Hn

i hc Sphm Vinh

Lng gic hc c ngun gc t Hnh hc. Tuy nhin phn ln hc sinh khi hcmn Lng gic hc (gii phng trnh lng gic, hm s lng gic ), li thy nnh l mt b phn ca mn i s hc, hoc nh mt cng c gii cc bi ton hnhhc (phn tam gic lng) m khng thy mi lin h hai chiu gia cc b mn y.

Trong bi vit ny, ti hy vng phn no c th cho cc bn mt cch nhn mi :dng hnh hc gii cc bi ton lng gic.

Trc ht, ta ly mt kt qu quen thuc trong hnh hc scp : Nu G l trng tmtam gic ABC v M l mt im ty trong mt phng cha tam gic th :

( ) ( )22222229

1

3

1cbaMCMBMAMG ++++= (nh l Lp-nt)

Nu OM l tm ng trn ngoi tip ABC th 2222 3RMBMBMA =++ nn p

dng nh l hm s sin, ta suy ra : ( )CBARROG 222222 sinsinsin9

4++=

( ) ( )1sinsinsin4

9

9

4 22222

++= CBAROG

Tng thc ( )1 , suy ra :

( )24

9sinsinsin 222 ++ CBA

Du ng thc xy ra khi v ch khi OG , tc l khi v ch khi ABC u.Nh vy, vi mt kin thc hnh hc lp 10 ta pht hin v chng minh c bt ng

thc ( )2 . Ngoi ra, h thc ( )1 cn cho ta mt ngun gc hnh hc ca bt ng thc( )2 , iu m t ngi nghn. Bng cch tng t, ta hy tnh khong cch gia O vtrc tm H ca ABC . Xt trng hp ABC c 3 gc nhn. Gi E l giao im caAH vi ng trn ngoi tip ABC . Th th :

( ) HAHEROHOH .22

/ ==

Do : ( )*.22 HEAHROH = vi :

ARC

ACR

C

AAB

C

AFAH cos2

sin

cossin2

sin

cos.

sin====

v CBABCBKHKHE cotcos2cot22 ===

CBRC

CBCR coscos4

sincoscossin2.2 ==

Thay vo ( )* ta c :

( )3coscoscos8

18 22

= CBAROH


16/22




Nu 090=BAC chng hn, th ( )3 l hin nhin. Gi s ABC c gc A t. Khi

( ) HEHAOHROH .22

/ == trong ARAH cos2= nn ta cng suy ra ( )3 .

T cng thc ( )3 , ta suy ra :

( )48

1

coscoscos CBA (Du ng thc xy ra khi v ch khi ABC u). Cngnh bt ng thc ( )2 , bt ng thc ( )4 c phthin v chng minh ch vi kin thc lp 10 v c mtngun gc hnh hc kh p. Cn nh rng, xanay cha ni n vic pht hin, ch ring vic chngminh cc bt ng thc , ngi ta thng phi dngcc cng thc lng gic (chng trnh lng gic lp11) v nh l v du tam thc bc hai.C c ( )1 v ( )3 , ta tip tc tin ti. Ta th s dng ng thng le.

Nu O, G, H l tm ng trn ngoi tip, trng tm v trc tm ABC th O, G, Hthng hng v : OHOG

3

1= . T 22

9

1OHOG = .

T ( )( )31 ta c :

( ) ( )CBACBA coscoscos814

1sinsinsin

4

9 222 =++

hay CBACBA coscoscos22sinsinsin 222 +=++ Thay 2sin bng 2cos1 vo ng thc cui cng, ta c kt qu quen thuc :

( )51coscoscos2coscoscos 222 =+++ CBACBA

Cha ni n vic pht hin ra ( )5 , ch ring vic chng minh lm nhc c khngbit bao nhiu bn tr mi lm quen vi lng gic. Qua mt vi v d trn y, hn ccbn thy vai tr ca hnh hc trong vic pht hin v chng minh cc h thc thunty lng gic. Mt khc, n cng nu ln cho chng ta mt cu hi : Phi chng cc hthc lng gic trong mt tam gic khi no cng c mt ngun gc hnh hc lm bnng ? Mi cc bn gii vi bi tp sau y cng c nim tin ca mnh.

1. Chng minh rng, trong mt tam gic ta c

=

2sin

2sin

2sin8122

CBARd trong

d l khong cch gia ng trn tm ngoi tip v ni tip tam gic .T hy suy ra bt ng thc quen thuc tng ng. 2. Cho ABC . Dng trong mt phng ABC cc im 1O v 2O sao cho cc tam

gic ABO1 v ACO2 l nhng tam gic cn nh 21,OO vi gc y bng 030 vsao cho 1O v C cng mt na mt phng bAB, 2O v B cng mt na mtphng bAC.

a) Chng minh :

( )ScbaOO 346

1 222221 ++=

b) Suy ra bt ng thc tng ng :


17/22




CBACBA sinsinsin32sinsinsin 222 ++

3. Chng minh rng nu ABC c 3 gc nhn, th :

2coscoscos

sinsinsinBA

v 02

3cos >

+

C

.

Ta c :

( )74

sin22

sin2

sin4

sin2

cos18

1

2cos2cos18

1

2

coscos

18

1

2

2sin

2sin

2

2sin

2sin

66663

3

32266

BABABABA

BABABA

BABA

++

+=

+

+

=

+

=

+

+

Tng t ta c : ( )84

3sin223sin

2sin 666

+

+

CC

Cng theo v ca ( )7 v ( )8 ta c :

( )964

3

6sin3

2sin

2sin

2sin

83sin4

43sin

4sin2

23sin

2sin

2sin

2sin

6666

6666666

=++

+++

++

++++

CBA

CBACBACBA

Trng hp tam gic ABC nhn, cc bt ng thc ( ) ( ) ( )9,8,7 lun ng.Th d 4. Chng minh rng vi mi tam gic ABC ta lun c :

( )( )( )

3

4

6

4

222sincossincossincos

++++ CCBBAA

Li gii. Ta c :

( )( )( )

=+++

4cos

4cos

4cos22sincossincossincos

CBACCBBAA

nn bt ng thc cho tng ng vi :


21/22




( )*4

6

4

2

4cos

4cos

4cos

3

+

CBA

- Nu { }4

3,,max

CBA th v tri ca ( )* khng dng nn bt ng thc cho

lun ng.

- Nu { }4

3,,max

>

>

CBA

nn ( )

+

+=

BABABA cos

2cos

2

1

4cos

4cos

( )1042

cos4

cos4

cos

42cos

2cos1

2

1

2

2

+

+

++

BABA

BABA

Tng t :

( )1142

3cos43

cos4

cos 2

+

C

C

Do nhn theo v ca ( )10 v ( )11 ta s c :

+

+

43cos

423cos

42cos

43cos

4cos

4cos

4cos 422

C

BACBA

3

3

4

6

4

2

43cos

4cos

4cos

4cos

+=

CBA

Do :

( )( )( )

3

4

6

4

222sincossincossincos

++++ CCBBAA

ng thc xy ra khi v ch khi tam gic ABC u.Mi cc bn tip tc gii cc bi ton sau y theo phng php trn.

Chng minh rng vi mi tam gic ABC, ta c :

( )NnCBA

CBA

n

nnn

++

++

2.3

2sin

1

2sin

1

2sin

1)2

3

1

2tan

2tan

2tan)1 333


22/22



( )314

2

4cos

4cos

4cos)3 +++

CC

BB

AA

( ) CBACBA coscoscos3122

1

4cos

4cos

4cos)4

3+

vi ABC nhn.

chuyen de bat dang thuc luong giac (chuong 4)

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