Chuyen de Tong Quat Phan 1

8/14/2019 Chuyen de Tong Quat Phan 1

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Chuyn bt ng thc hnh hc Nhm 5

3

PHN I: BT NG THC HNH HC TRONG MTPHNG.

BI 1: PHNG PHP KO THEO

I . Slc v phng php ko theo:Xut pht t cc bt ng thc bit, vn dng cc tnh cht ca bt ng thc suyra bt ng thc cn chng minh. Sau y l cc v d:

Vd 1: Cho tam gic ABC, Mthuc AC. Chng minh rng:1 1

. ; .2 2

ABC ABC S AB AC S BM AC

Gii:

GiBHl ng cao ca tam gic ABC BH AB 1 1

. .

2 21 1

. .2 2

ABC

ABC

S BH AC AB AC

M BC BH BM S BH AC BM AC

=

=

Bt ng thc c chng minh xong.

Vd2: Cho tam gicABC, AMl trung tuyn. Chng minh:2

BCAM th 90oBAC v

ngc li.

Gii:

MA C

B

H

MA C

B

D

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4

a) Gi s 90oBAC< .Gi D l im i xng caA quaM. Suy raAD=2AM. Ml trung im hai on thngBCvAD.

& / / 180

O

AB DC AB DC BAC ACD

=

+ = m

90

o

BAC< 90O ACD BAC ACD > <

Xt tam gicABCv tam gic CDB c:AB=DC, BCl cnh chung, BAC ACD<

Do : BC (V l).

90oBAC

Vd 3: Cho t gic liABCD sao choAB ct CD ti E, AD ctBCti F, v E,F,Ccng

thuc na mt phng c bBD. t , AED AFB = = ; v ABCDS S= . Chng minhrng: . .sin . .sin 2 . . AB CD AD BC S AB CD AD BC + + .

Gii:

D thy:

ABF

ACE

>

>

* Trong ABD ta ly im K sao cho/ /

/ /

BK DE

DK BF

T ta c1 1

. .sin . .sin2 2 ACK ADK

S S S AB BK AD DK S + +

. .sin . .sin 2 (1) AB BK AD DK S + D thyDKBCl hnh bnh hnh.

(2) BK CD

BC DK

=

=

Thay (2) vo (1) ta c:. .sin . .sin 2 (1) AB CD AD BC S +

K

E

F

B A

D

C

P


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5

* Trong na mt phng c blBD ta ly im P sao cho DP BC

BP CD

=

=.

D thy

1 1. sin . .sin2 2

1 1. .2 2

ABCD ABPD ADP ABPS S S S AD DP ADP BA BP ABP

AD DP BA BP

= = + = +

+

2 . .S ABCD AD BC + Vy . sin . .sin 2 . . AB CD AD BC S AB CD AD BC + +

*Mt s kin thc thng dng gii tan cc tr trong mt phng:- S dng quan h gia ng vung gc v ng xin, hnh chiu.- Trong cc tam gic vung (c th suy bin thnh on thng) c cnh gc vungAHvcnh huyn AB th AH AB . Xy ra du bng khi H B .- Trong cc on thng ni tim n ng thng, on no vung gc vi ngthng l on thng c di nh nht.- Trong cc on thng ni 2 im thuc hai ng thng song song, on thng vunggc vi hai ng thng song song c di ngn nht.- Trong hai ng xin k t 1 im n cng mt ng thng, ng xin ln hn khiv ch khi hnh chiu ca n ln hn.- Mt t gic li b cha trong mt t gic khc (khng nht thit l li) th chu vi ca tgic b cha s nh hn chu vi ca t gic cha n bn trong.- di on thng nm trong mt a gic li khng ln hn di ng cho lnnht..- Trong tt c cc dy cung qua mt im cho trc trong mt ng trn th dy cung c di nh nht l dy cung vung gc vi on thng ni tm ng trn vi im .- Trong cc tam gic c cng chu vi th tam gic u c din tch ln nht.- Mt ng thng c th ct nhiu nht hai cnh ca mt tam gic.(nguyn tcDirichlet).

* Mt s v d:

Vd1: Cho on thng AB c di 2a. V v mt pha ca AB cc tia Ax v By vunggc vi AB. Qua trung im Mca AB c 2 ng thng thay i lun vung gc vinhau v ctAx,By ln lt ti C,D. Xc nh v tr ca cc im C,D sao cho MCD cdin tch nh nht. Tnh din tch .

Gii:

a

21H

D

MA B

C


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6

Gi K l giao im ca CMvDB,

( ) MAC MBK gcg MC MK = =

DCK cn

1 2D D = K MH CD Do M thuc phn gic gcD nnMH=MB=a.1

.2MCD

S CD MH = .

Do 2 &CD AB a MH a = = nn:21 2 .

2MCDS a a a CD Ax = . Cc im C,D c xc nh trn Ax, By sao cho

AC=BD=a

* Trong li gii trn, SMCDc biu th bi1

.

2

CP MH. Sau khi chng minhMHkhng

i, ta thy SMCD nh nht khi v ch khi CD nh nht.- Nu bi ton trn khng cho Ml trung imAB th ta phi gii quyt ra sao?

1 . ,2MCD

S MC MD MAC MDB = = = (cng ph BMD )

,cos sin

a b MC MD

= = nn

1

2 sin cosMCDab

S

=

Do a,b,c l hng s nn SMCD nh nht khi v ch khi 2sin osc ln nht.2 22sin cos sin cos 1 MCDS ab + =

min sin cos tan 1 45MCD

oS ab = = = =

Cc im C,D trnAx, Byc xc nh sao choAC=AM, BD=BMy c xem l bi ton tng qut.

Vd 2: Cho ABC c B l gc t,D di ng trnBC. Xc nh v tr caD sao cho tngcc khang cch tB v tCn ng thng AD c gi tr ln nht.

ba

D

A BM

C


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7

Gii:

Ta c :1 1 1

. . .2 2 2ABC

S AH BC BE AD CF AD= = +

2 ABCS BE CFAD

+ = . Do

( ) max min BE CF AD+

AD nh nht khi v ch khi hnh chiu HD nh nht. HD HB vHD=HB khi D B Suy ra pcm.

Vd3: Cho tam gicABCvung c di cnh gc vungAB=6cm, AC=8cm.Ml imdi ng trn cnh huynBC. GiD v El chn cc ng vung gc k tMnABvAC. Tnh din tch ln nht ca t gicADME.

Gii:t AD x= th ME x= . Theo Thalet:

4 48

6 8 3 3

EM CE x CE CE x AE x

AB CA= = = =

Ta c:

( ) ( ) ( )

2

22 2

4 4. 8 8

3 3

4 4 46 6 9 12 3 12 12

3 3 3

ADMES AD AE x x x x

x x x x x

= = =

= = + + = +

212 3ADMES cm x D= = l trung im caAB, Ml trung imBC, El trung im

AC.

Vd4: Cho tam gic ABC, im Mdi chuyn trn cnh BC. Qua Mk cc ng thngsong song vi ACvAB, chng ctAB vACtheo th t D v E. Xc nh v trMsaochoADMEc Smax.

Gii:

Gi SABC=S, SBDM=S1, SEMC=S2.

Ta nhn thy ( ) 1 2ADME 1 2S max min minS S

S SS

+ +

Cc & DBM EMC ng dnh vi ABC nn:

E

F

HC

A

B D

yx

21

K

EH

D

B C

A


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8

( )

2 2

1 2

2 2 2 21 2

22

;

1

2

S S BM MC

S BC S BC

S S BM MC x y

S BC x y

= =

+ + + = =

+

Nh vy1

max 2ADMES S= . Du = xy ra khi v ch khi x y= . Khi Ml trung im

caBC.

Vd 5: Gi s 1 1 1, ,C B A l cc im ty trn cc cnhAB,CA,BCca tam gicABC.K

hiu 1 2 3, , ,S S S S l din tch cc tam gic 1 1 1 1 1 1, , ,ABC AB C BC A CAB CMR:

1 2 3

3

2S S S S+ +

Gii:

BT cho tng ng vi 31 2 32

SS SS S S

+ +

1 1 1 1 1 1 1 1 1 1 1 1. . . 1 3

. . . 2 2

AB AC BA BC CB CA AB AC BC BA CA CBVT

AB AC AB BC AC BC AC AB AB BC BC AC = + + + + + + + =

MT S BI TON CHN LC:

1.Cho tam gicABCnhn. Dng mt tam gic c chu vi nh nht ni tip tam gicABC,

tc l c 3 nh nm trn ba cnh ca tam gicABC.

Gii:

Xt MNP ni tip ABC mt cch ty . ( ), ,M AB N BC P AC . VE,Fsao cho

AB l trung trc caNEvACl ng trung trc caNF.Chu vi MNP MN MP PN EM MP PF FE = + + = + +

1 22 2 2EAF A A BAC = + =

FAE l tam gic cn c gc nh khng i nn cnh y nh nht khi v ch khicnh bn nh nht.

21

B C

A

N

M

PE

F


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9

Ta c: EFnh nht min min AE AN AN BC Ta c nhn xt rng khi Nl chn ng vung gc k tA th Mv P cng l chn 2ng cao cn li ca tam gic.Chng minh nhn xt trn nh sau:

HMP :AB l ng phn gic ca EMH,ACl ng phn gic ca FPH.Ta c: AB,ACgp nhau ti A nn AHl tia phn gic ca gc trong ca tam gic ti H

hayHA l tia phn gic MHP . V AH HC nnHCl ng phn gic gc ngai caA ti nhH.

Theo trn,ACl ng phn gic ngai ti nh P,HCgpACti CnnMCl tia phngic gc trong tiM.MB vMCl cc tia phn gic ca hai gc k b nn , MB MC PC PB . Chu vi MNP min khiM,N,P l chn 3 ng cao ca tam gicABC.Do ABC nhn nnM,N,P thuc bin ca tam gic.

2. Hai anh em chia ti sn l mt ming t hnh tam gic ABC. H mun chia i dintch ming t bng mt bro ngn nht. Tnh di m ca bro ny theo din tch Sv gc nh nht ca tam gic.

Gii:Bro phi ct hai cnh ca tam gic. Gi s gc ti nh l , di ca bro l

IK=m v khang cch tnh ca gc A ti hai u bro lx vy.

( )2 2 2 2 cos 1 IK x y xy A = +

t , , , AIK ABC S S S S AI x AK y= = = = , ta c: ' 2S

S const = = .

Do1

ysin2

S x A = m S v A khng i nnxy khng i.

( )2 2min min IK x y + . p dng bt ng thc AM-GM, ta c

( )2 2 min x y x y+ = Nh vy xt bro chn gcA th di bro ngn nht khi v ch khi AIK cn tiA.

2 ' tan . '2 2

A S IK S Do S= = nn 2tan

2

AIK= .

P

M

B C

A

H

F

E


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10

Vy di bro ngn nht { }( )2 tan min , ,2m A B C

= = =

3. Cho ABC ni tip ng trn tm O bn knh R c a,b,c l di 3 cnh vma,mb,mc l trung tuyn ln lt tng ng vi 3 nh A,B,C. Cc trung tuyn ca tam

gic (theo th t trn) ct ng trn tiA,B,C. Tm GTLN ca:2 2 2 2 2 2

c a b

a b b c c a

m m m

+ + ++ +

Gii:

Trc ht ta c2

2 2 222aa

b c m+ = +

Ta s chng minh:2 2

4a

b cR

m

+

Theo h thc lng ng trn:2 2

1 1 1. . 4 4a a

a a MA MA MB MC m MA MA

m= = =

Ta li c: 1 1 2 MA MA AA R+ = 2

2 2

2

2 4 84

2 4 .2

a a aa

a a

am R m a Rm

m

am R m

+ +

+

2 22 2 4 4a

a

b cb c Rm R

m

++ .

Mt cch tng t, v cng cc bt ng thc li v theo v ta c pcm.Du = xy ra khi v ch khi:

1

1

1

2

2

2

AA R

BB R ABC

CC R

=

= =

u. Khi d=2R.

4. Gi H l trc tm ABC nhn v r l bn knh ng trn ni tip tam gic. Cm:6 HA HB HC r + + . Du bng xy ra khi no?

Gii:

Ta thy: 2 3. 2

BC HA S S= +

( )2 3 2 322

a x S S ax S S = + = +

Tng t:


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11

( )

( )

1 3

1 2

2ax 2.2

2ABC

by S Sby cz S

cz S S

= + + + =

= +

Ta cn chng minh: x y z 6r+ + Gi s: a b c . Theo quan nim vng xin v hnh chiu x y z

Ty ta s chng minh ( ) ( ) ( ) ( )3 ax 2a b c x y z by cz

+ + + + + + Tht vy:( ) ( ) ( ) ( )

( ) ( )

( )( ) ( )

2 3ax 3 3 0

0

0 3

a x y z b x y z by c x y z cz

a y x x z

y z a b

+ + + + + + + +

V a b c , x y z nn (3) ng.T (1) v (2)

( )( ) 3.4 12

26

a b ca b c x y z S r

x y z r

+ ++ + + + =

+ +

Du bng xy ra khi v ch khi:a b c

a b c ABC x y z

= = = =

= =u.

5. ABC c cnh BC a= khng i v A = khng i. Hy xc nh v tr caAABC c chu vi nh nht.

Gii:

Xt A nm trn mt na mt phng bBC. Ta c A di chuyn trn cung cha gc dng trnBC.Trn tia i ca tiaAB lyD sao choAD=AC.Chu vi ABC bng AB BC CA AB BC a+ + = + + Chu vi ABC ln nht ( )max max AB AC BD + .

a

cb

x

yz

3 2

1

B C

A

H

a

2

D'

K

A'

B C

A

D


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12

M 2

BDC D

= di ng trn cung cha2

dng trn on BC(c gii hn bi tip

tuyn tiB) l cung KC.Do vyBD max khi v ch khiBD l ng knh ca cung cha gc trn (tm ca cung

cha gc2

chnh l im A , im chnh gia cung cha gc )

ABC c chu vi ln nht khi n l tam gic cn tiA c ,BC a A = =

6. Tam gic ,ABC M l im trong tam gic. bn ngai tam gic k cc ng thngsong song vi cc cnh, cch chng mt khang bng khang cch t M n cnh .Mi ng thng to vi mt cnh ca tam gic v cc ng thng cha hai cnh kiamt hnh thang. Chng t rng tng din tch ca ba hnh thang khng nh hn7

3 ABCS .

Gii:

Gi din tch cc tam gic , , ,ABC MBC MAC MAB v cc hnh thang ln lt l' ' '

1 2 3 1 2 3, , , , , ,S S S S S S S .

Ta c: ( ). . ADE ABC g g g 2

1 2 2

2 2

2

1 1 2 1

2

21

1 1

1

2

ADE

ABC

S AA AA MM

S AA AAS S S MM S

DoS S AA S

SS S

S

+ = =

+

= + =

= +

Tng t ta c:2232

2 2 3 32 ; 2SS

S S S SS S

= + = +

B C

A

M

A2 M2

HG

D

E

F K

A1


11/121


13

M: ( ) ( ) ( ) ( ) ( ) ( )2 2 2 2 22 2 2

1 2 3 1 2 3 1 2 2 3 3 1 1 2 33 S S S S S S S S S S S S S S S+ + = + + + + + + +

( )2

2 2 21 2 3 1 2 33

SS S S Do S S S S + + = + +

Do : 1 2 3 1 2 37

23 3

SS S S S S S S S + + + + + .

Ta c pcm.

7. Cho ABC , trn 2 cnhAB vACly 2 im E,Fsao cho c im a trong AEF tha2

3 ABE AGF

ACE AGE

S S

S S

=

=. Chng minh rng:

4

9 BEGF CFGE ABC S S S+ .

Gii:

Ta t: 1 2; ; ABC AEG AFGS S S S S S= = = Ta c:

2 1 21 2

1 1 2

3(1)

3

ABF BEGF BEGF CFGE

ACE CFGE

S S S S SS S S S

S S S S S

= = + + + = +

= = + +

Ta c:

( )1 21 1 1

3 3 3ACE ABF

ACE ABF

S S AE AF S S S S S S

S S AB AC + = + = + = +

1 2

1 21 2

1 2

1 2.2

3 3

2

34

(2)9

AEFSAE AFS S S S AB AC S

S SS S S

S

S S S

+ =

+ +

+

T (1) v (2) suy ra4

9 BEGF CFGE S S S+

Du = xy ra 1 2S S

G EF

=

F

E

BC

A

G


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14

8.Cho ng trn tm O, ng knh AB v bn knh R. im M nm trn AO v

1AM

kMO

= > . TMk dy CD bt k. Tm maxABCDS

.

Gii:

( )2

2 1 ons ABCD ACD BCD

OCD OCD

S S S AM BM AB AOk c t

S S MO MO MO MO

+= = + = = = + =

ABCDS ln nht OCDS ln nht

M ta li c:1

. .sin2OCD

S OC OD = nn OCDS khi sin ln nht nh nht.

M ta d thy: 90 180< < . sin ln nht nh nht.Ta d thy CD OA .Khi ta c: ' 'max OCDS OC DS=

Ta c:

( ) ( )( )

2 22 2

2 2

1 11 1

' 21 1

OM OM ROM MA k R k k

R R MC R k k

k k

= = =+ +

= = + + +

( )

( )

( )

( )

( )

2

' ' 2

2

' 21

2' ' 2

1

1. ' ' 2

2 12

21ABCD

OC D

S

R MC k k

kR

C D k k k

RS MO C D k k

kR

Max k k k

= ++

= ++

= = +

+

= ++

9. Gi s c mt tam gic nhn din tch S1, v c mt tam gic vung cha tam gicnhn ni trn, c din tch S2 . Hy tm S2 nh nht v so snh S1 v S2.

Gii:

Gi tam gic nhn lABCc A l gc ln nht. GiMl trung imBC. TrnBClyE,Fsao choME=MF=MA.

BC a = . Tam gicAEFvung tiA.Khi AEF c din tch S2 nh nht.Ta c: v ln nht:

A B BC AC

Ta c: 180o AMF AME+ = ( )max ; 90o AMF AME

Gi s 90oAMF

D'

C'

D

BOA

C

M


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15

Suy ra tam gicAMFt.2 2 2 2 2 AM ME AC BC a +

Ta c:2

2 2 3

2 2

a a AM a AM +

Ta c: 32 2 32

2

AEF

ABC

aS EF MA

aS BC BC = = =

3 AEF ABC S S

* Bi tp tluyn:

1. Cho ABC c cc cnh khng bng nhau, gi cc im G,I,Hln lt l trng tm,tm ng trn ni tip, trc tm ca tam gic. Chng minh rng: 90oGIH> .

2. Phn gic ca gcA,B,Ctrong ABC ct ng trn ngoi tip ti 1 1 1, , A B C. Gi s

, ,o o o A B C ln lt l tm ng trn bng tip gcA,B,C. Chng minh:

1 1 1) 2 ;

) 4 .o o o

o o o

A B C AC BA CB

A B C ABC

a S S

b S S

=

3. Cho ABC cn tiA v c 060BAC . Hy tm imMtrnmt phng sao cho tng

MB MC MA+ nh nht c th.4. Tm im O trong ABC cho trc sao cho tng khang cch tO ti ba nh ca tamgicABCnh nht c th (im Toricenli).

5..ABCD l mt t gic ni tip. Tm ng trn ngoi tip nm bn trongABCD. Cnh

ngn nht c di bng 24 t v cnh di nht c di bng t, vi 2 2t< < . Cctip tuyn ti A v B ct nhau ti A , cc tip tuyn ti B v Cct nhau ti D v A ctnhau ti D .Tm gi tr ln nht v gi tr nh nht ca t s

MF

A

C B


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16

BI 2: PHNG PHP SDNG H THC LNG

I. Slc v phng php:

Khng ch trong bt ng thc tam gic ta mi s dng h thc lng h trcho victnh ton v chng minh, m trong cc bi ton v tam gic ni chung, h thc lngtrong tam gic cng trthnh mt cng c mnh tnh ton, n gin ha vn ,.Qua vic s dng h thc lng trong tam gic ta c th s dng cc cng c tnh tanmnh hn na nh p dng cc bt ng thc cin, hay cc bt ng thc trong tamgic (c nhiu bt ng thc trong tam gic rt c o m nu khng chc hn ccbn khng th thy c vp huyn b ca n). Hoc t vic p dng h thc lng tac th bin mt bi ton hnh hc n thun trnn phc tp, kh nhn hn v n bngiu sau mt lat cc cng thc m nu khng nm vng kin thc chc hn khng phiai cng lm c.

* Mt s h thc lng v bt ng thc cbn trong tam gic:Cho tam gic ABC, vi , , ; , , , , ,a b c a b cAB c AC b BC a m m m d d d = = = ln lt l cctrung tuyn v cc phn gic ng vi cc cnh 1 2, , ; ,a b c d d ln lt l khang cch t

tm ng trn ngoi tip ti trng tm tam gic v tm ng trn ni tip ti trng tmtam gic, p l na chu vi tam gic, S l din tch tam gic; ,r R l cc bn knh ng

trn ni tip v ngoi tip tam gic; , ,a b cr r r l ng trn bng tip cc cnh , ,a b c . Ta

c cc h thc sau:2 2 2

2 2 2

4ab c a

m+

= ;2 2 2

2 2

9

a b cd R

+ +=

( )2 2 2

2

1

5 169d p r Rr = + ; , ,a b cS S S

r r rp a p b p c= = =

4

abcR

S= ;

Sr

p= ; ( ) ( )( )S p p a p b p c=

2 3

3 3

2

a b c

a b c

a b c

m m m

m m m

a b c

+ +

+ +

; ( ) ( )( )a b c a c b b c a abc+ + +

( ) ( )9 5

20 4a b b c c aab bc ca m m m m m m ab bc ca+ + < + + < + +

By gichng ta n vi mt bi ton.Cho tam gicABCvi , ,AB c BC a CA b= = = . Gi S l din tch tam gic ny, v rvR

l bn knh ng trn ni tip v ngoi tip tam gic; , ,a b cm m m ln lt l di cc

trung tuyn xut pht tA,B,C. Ml mt im bt k, , , l cc s thc. Ta c btng thc sau l hin nhin:

( )2

0 MA MB MC + +


15/121


17

Bnh phng hai v ca MA MB BA =

ta c 2 2 22 . MA MB MA MB AB= +

v ty qua cc php bin i tng ng ta s c c bt sau:

( ) ( ) ( )2 2 2 2 2 2 1 MA MB MC a b c + + + + + +

Du bng xy ra khi v ch khi 0 MA MB MC + + =

Nhng khi choMl trng tm tam gic th ta s c (1) trthnh:

( ) ( ) ( ) ( )2 2 2 2 2 29 1.14a b c

m m m a b c + + + + + +

- Chn , ,a b c = = = thay vo (1.1) ta c:

( )2 2 29

1.24a b c

am bm cm abc+ +

Bin i tng ng bt ng thc (1.2) ta s c cc bt ng thc sau:2 2 2 9

22 2

a b ca b c

a b c

m m m abc R S ah bh ch

h h h R + + = = = =

( ) ( ) ( )3 3 3 2 2 23 9 2a b c abc a b c a b c+ + + + + + +

- Chn , ,2

a b ca p a b p b c p c p + + = = = =

thay vo (1.1).

Ta c:

( ) ( ) ( ) ( )( )2

22

1 1

1

a p b p c ap p a p b p c p a p

p a p r S S pr ra

p a p p a

=

= =

Tng t vi ( )( ) ( ) ( )2 2&b p a p c c p a p b ri p dng h thc:

1 1 1 4R r

p a p b p c pr

++ + =

Ta thu c bt ng thc sau:

( ) ( ) ( ) ( )2 2 2 9a b cp a m p b m p c m S R r + +

- Chn , ,a cb

a b c

m mm = = = thay vo 1.2 ta sc:

( )9

1.34b c a c b a

am m bm m cm m abc+ +

- Chn , ,bc ca ab = = = thay vo 1.2 ta c:

( )

2 2 2 3 3 39

44a b cm m m a b c

a b c ab bc ca

+ +

+ + + +

Ch cn da vo vic chn b s , , thch hp chng ta c c nhng bt ngthc p, cc bt ng thc ny cng c ng dng rt nhiu trong vic gii cc bi btng thc hnh hc, h thc lng.

II. Mt s v d:


16/121


18

Vd 1: Cho ABC , tim FtrnACv cc on thng EG//AB, v EF//AC (F,G thuconAB vAC). Cm 2 16 ABC BEF CEGS S S .

Gii:

Qua bi ta d nhn thy rng chng minh bt ng thc trn ta s dng bt ngthc AM-GM v cng th Herong l ph hp. D thyAFEG l hnh bnh hnh.

AF GE AG EF=

=

Ta t :

2 1

1 2

1 2

; ;

; ;

; ;

BC a BE a CE a

AB c AF c FB c

CA b AG b GC b

= = =

= = =

= = =

Ta c:

1 1 1 1 1 1 1 2 2 2 2 2 2 216 16 ( )( )( ) ( )( )( ) BEF CEGS S p p a p b p c p p a p b p c=

1 2 1 1 2 2 1 1 2 2 1 1 2 22 2 ( )( )2 ( )( )2 ( )( )p p p a p a p b p c p c p b=

( ) ( ) ( ) ( ){ }1 2 1 2 1 2 1 2 1 2 1 2 1 2p p p p a a p p b b p p c c + + + + + + + ( ) ( )( ) 2ABCp p a p b p c S= =

Vy ta c: 2 16 ABC BEF CEGS S S Du = xy ra khi v ch khi El trung imBC.

Vd 2: Cho tam gicABCvung ti C. Kng cao CHv phn gic CEca gcACH,

CEca gcBCH. Chng minh rng 2 1ABC

CEF

S

S + .

Gii:t ; ;AB c BC a CA b= = =

Ta c: ABC

CEF

S AB

S EF=

Ta c: CBF ACH= ACF FCB FBC CFA = + = ACF cn ti A. AC AF b = =

c2

c1 b2

b1

a2 a1

B

A

C

F

E

G

F

E

H

BC

A


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19

Tng t: tam gicBCEcn tiB suy raBC=BE=aEF BE AF AB a b c = + = +

chng minh:

( )

( )

2 1 2 1

2 1 2 (1)

ABc EF

EF

c a b c c a b

+

+ +

Ta c:2 2 2 2 2

2

2 2 2

2 2 (2)

c a b ab c ab c

c ab c

= + +

+

M ta c:

( )22 2 2

2

2 2

2 (3)

ab c a b ab a b

ab c a b

+ = + + = +

+ = +

T (1),(2),(3) ta c:

2 1 2 1ABC

CEF

Sc

EF S + + (pcm)

Vd 3: Cho ABC . LyMthucAB, NthucAC, tha 1 AM AN

k AB AC

= = < ; DngAMON

lm thnh hnh bnh hnh. K 1 ng thng bt k dctAB,ACti E,Fsao cho G khngnm ngai AEF . Chng minh: 24 AGE AGF ABC S S k S+

Gii:Ta c:

. cos . .cos ANB AMC S AN AB AM AC S = = =

1 2,h h ln lt l khang cch tMvB tiAC.

Khi ta c:

1 1

2 2

AGFAMF

ABF ABF

Sh S hAMk k

h AC S S h= = = = =

Tng t ta c: AGE

ACE

Sk

S=

Ta c:

( ) ACE ABE AGE AGF ACE ABF ABC ABC

S SS S k S S k

S S

+ = + = +

.2

.

. .cos2 2 2

. .cos

ABC ABC

AGE AGFAEF ABC ABC ABC

ABC ABC

AE AF AE AF kS kS

AB AC AB AC

S SSAE AFkS kS kS

AB AC S S

= +

= =

22 AGE AGF ABC S S k S + (pcm)

F

E

BC

A

G

M

N


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20

Du = xy ra

/ /

AGE AGF

EF BC

G EF

S S

=

Vd 4: Trong tt c cc t gic c 3 cnh l a. Gi S l din tch cc t gic . CM:23 34

aS

Gii:

Ta c: ( ) ( ) 360oA D B C + + + = ( ) ( )min ; 180oA D B C + +

Qua Ckd//AB.V(A;2a) ct d ti D . Khi ta c t gic ABCD cng tha iu kin ca t gic nura.

Ta c: ' 90o DAC DA C> Suy ra khang cch tD nACnh hn khang cch tD nAC

' ' ' ' ' ADC AD C ABC ADC AD C ABC ABCD AB C DS S S S S S S S < + < + < Nn t gic trn c din tch ln nht th c iu kin 2 cnh ny song song. Ta xtdin tch hnh thang .KBE//AD.Khi ABED l hnh thoi. BEC l tam gic cn tiB.

t ADE = ta c: 2EBC = Ta c:

( )

( )

2 2

2

1sin sin 2

2

1sin sin 2 12

ABCD ABED BCE S S S a a

a

= + = +

= +

a

a

a

d

A B

A' C

D


19/121


21

[ ] ( ) ( ) ( ) ( ) ( 4

2 2 32 2

1sin sin 2 sin sin cos sin (1 cos ) 0

2

1 1 6sin (1 cos ) 1 cos 1 cos 3 3cos 1 cos

3 3 4

S

S AM G

= + = + = + >

= + = + = +

( )

3 3

24S

T (1) v (2) 23 3

4ABCDS a . Bt ng thc c chng minh xong.

Du = xy ra

13cos 1 cos cos 602

oB = + = =

Tc l t gic l na lc gic u.Khi qut ha bi ton ta c bi ton sau:

Vd 5: Cho t gic c mt cnh c di ln hn 1. Chng minh:3 3

4S

Gii:Bng cch tng t bi trn ta chng minh rng t gic phi c 2 cnh song songnhau.Gi s BC//AD. K CE//AB ct AD ti E. Khng mt tnh tng qut gi s:CD AB .Ta c:

( )( )

1. sin . sin

2

1sin sin2

ABCD ABCE ECDS S S BC CE CE CD

= + = +

+ +

( )1

sin sin2ABCD

S T + + =

Qua Cta kon thng to viAD gc .Khi ta c:* Nu 180 90o o > +

( )90 sin sin 2o < +

1 3 3sin sin 2

2 4

T +

Du = xy ra

1

3

AB CE BC

= = =

= =

Tc lABCD l na lc gic u.

E

A

D

B C


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22

Vd 6: Cho lc gic ABCDEF ni tip trong ( );O R v ; ; AB BC CD DE EF FA= = = .

Chng minh:

ACE BDF S S

Gii:Ta t:

CAE

CEA

ACE

=

=

=

Khi :

( )

( )

( )

1

21

21

2

DFB

DBF

BDF

= +

= +

= +

Khi :

21 1. .sin 2 sin .2 sin sin 2 sin sin sin2 2ACE

S AC CE R R R = = =

Tng t:

22 sin sin sin2 2 2BDF

S R + + + =

Suy ra ta c: ACE BDF S S sin sin sin sin sin sin cos cos cos

2 2 2 2 2 2

1sin sin sin

2 2 2 8

+ + +

Bt ng thc cui cng ng. Du bng xy ra khi v ch khi lc gic cho u.

III. Mt s bi ton chn lc:

1. Cho tam gic ABCni tip trong mt ng trn. t aa a

ml

M= tng t vi ,

b c

l l .

Vi , ,a b cm m m l di cc phn gic k tA,B,Ctng ng v , ,a b cM M M l

di cc phn gic ko di, tnh t cc nh tng ng A,B,Cn cc giao im cachng vi ng trn ngoi tip tam gicABC. Chng minh rng:

2 2 23

sin sin sina b cl l l

A B C + +

V du ng thc xy ra khi v ch khi tam gic ABC u.

F

O

A

B

C

D

E


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23

Gii:Gi sng phn gic gcA ctBCti P v ct vng trn ngoi tip tam gicABCQ.p dng nh l hm s sin cho tam gicABP ta c:

( )0

sin sin1

sin 180 sin2 2

AP B BA AAB

B B= =

+

Vi ch rng C BQA= v

2

A ABQ ABC CBQ ABC CAQ B= + = + = +

p dng nh l hm s sin cho tam gicABQ ta c:

( )sin

2sin

2

AB C

AAQB

= +

Nhn (1) v (2) v theo v ta c:

2

sin sinsin sin

sin2

a

a a

m AP B C l B C

A M AQ B= = =

+

Hon ton tng t ta c:

2

2

sin sinsin sin

sin2

sin sinsin sin

sin2

bb

b

cc

c

m A Cl A C

BMC

m B Al B A

CMA

= = +

= = +

T suy ra2 2 2 2 2 2

32 2 2

sin sin sin sin sin sin


sin sin sin sin sin sin3 3

sin sin sin

a b cl l l B C C A A B

A B C A B C

B C C A A B

A B C

+ + + +

+ + =

Du bng xy ra khi v ch khi: sin sin sin 12 2 2

A B C B C A

+ = + = + =

tc l khi

tam gicABCu.2. Cho m,n,p l cc s thc tha mn , , &m n n p p m mn np mp+ + + + + l cc s m.

t a,b,c l di 3 cnh, S l din tch tam gicABC. Khi :2 2 2 4ma nb pc mn np mpS+ + + +

Gii:Theo nh l hm s cosin, ta c:


22/121


24

( )

( ) ( ) ( )

2 2 2

2 2 2 2

4

2 cos 2 sin

2 cos

ma nb pc mn np mpS

ma nb p a b ab C ab C mn np mp

a bm p n p mn np mp p C

b a

+ + + +

+ + + + +

+ + + + + +

p dng bt ng thc BCS ta c:

( ) ( )( ) ( )( )2

2 2 2cos sin cosmn np mp p C p mn np pm C C m p n p+ + + + + + + = + +

Mt khc: ( ) ( ) ( ) ( )2

4a b

m p n p m p n pb a

+ + + + +

Do ta c iu phi chng minh.Cu hi t ra cho chng ta l ng thc xy ra khi no? ng thc xy ra khi ng thcbt ng thc BCS m chng ta s dng chng minh xy ra. Tc l:

( ) ( )

( ) ( )

2 2

2

cos sin

cos sin 1

a bm p n p

b aC C

p mn np mp

a b

n p m p

C C

p mn np mp m p n p

+ = +

= + +

= + +

= = + + + +

Thay cos ,C b tng ng vo biu thc: 2 2 2 2 cosc a b ab C = + ta thu c:

( )( )

2 2 2 2

2

2

1 2

m p m p pc a a a

n p n p m p n p

c m p p

a n p n p

c m n a c

a n p n p m n

+ += +

+ + + +

+ = + + +

+ = =

+ + +

Tng t ta c ng thc xy ra khi v ch khia b c

n p m p m n= =

+ + +

Bi ton ny l mt nh l quen thuc v c rt nhiu ng dng trong vic gii cc biton ln hn. V d ta c th s dng nh l ny chng minh cc bi tan sau:

t a,b,c l di 3 cnh tam gicABCv S l din tch ca tam gic .3. (Bt ng thc Hadwiger-Finsler)Chng minh:

( ) ( ) ( )2 2 22 2 2 4 3a b c S a b b c c a+ + + + +

Gii:Thng thng khi nhn vo bi ny chng ta s nghngay n cng thc Herong v mts bt ng thc cin quen thuc v a n v mt bi ton i s khng hn khng


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25

km. Nhng nu tinh nhn ra s lin h gia bi ton ny v bi ton 2 th bi ton trnn v cng d dng.Bt ng thc cho tng ng vi:

2 2 2 4 3b c a c a b a b c

a b c Sa b c

+ + + + +

Theo nh l ta chng minh trn th ta ch cn chng minh:

( ) ( )

( )( )

3

0

b c a c a bab

a a b a c

+ +

y li chnh l bt ng thc Schur.

4. Chng minh:3

2 2 2 4 28 27a b b c c a S+ +

Gii: bi ton ny ta d dng thy rng n c lin h g vi bi ton 3. V t bi ton 3 ta

d dng suy ra c rng:4 3ab bc ca S+ +

By gita c th trv vic p dng nh lbi ton 2. Ta c:3

2 2 2 4 24 8 27a b b c c a ab bc caS S+ + + + Bt ng thc c chng minh xong.

5. Chng minh: 2 2 23 4abc a b c S + +

Gii:

Cng theo tng l s dng nh lbi ton 2, nhng bi ton ny chng ti mungii thiu ti bn c 2 cch gii:Cch 1:Cch gii ny ging nh cch gii bi ton 3:

Bt ng thc cho tng ng: 2 2 2 2 2 24bc ca ab

a b c a b c Sa b c

+ + + +

Theo nh l trn th ta c:2 2 23 4abc a b c S + +


Cch 2:Cch ny li hon ton chng lin h g ti nh l trn nhng th v mt iu l li giica n v cng ngn gn v n gin.Bt ng thc cho c th vit di dng 2 2 2 29R a b c + + , p dng nh l hm s sinta s c c bt ng thc tng ng vi:

2 2 2 9sin sin sin4

A B C + +

y l mt bt lng gic cbn, bt c ngi hc ton no cng bit nn vic chngminh bt ny chng ti s b qua, cc bn c th t tm chng minh cho mnh.


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26

6. Chng minh: 2 2 2 2 3 x y z

a b c Sy z x z x y

+ + + + +

vix,y,z l cc s thc dng.

Gii:

chng minh bi ton ny ta c mt b nh:

( ) ( ) ( )( )

( ) ( ) ( )2 2 2

3

( )( ) 4

0

xy yz xz

y z x z y x z x x y z y

x y z y z x z x y

+ + + + + + + +

+ +

B c chng minh xong.Gita mi chng minh tip tc bt ng thc ny. p dng nh l trn ta c:

( )( ) ( ) ( ) ( )( )2 2 2 4

x y z xy yz xza b c S

y z z x x y y z x z y x z x x y z y+ + + +

+ + + + + + + + +

Do ta c pcm.

7. (Bt ng thc Pedoe) t a,b,c l di cc cnh cu tam gic ABC vi din tch S.t 1 1 1, ,a b c l di cc cnh ca tam gic 1 1 1A B C vi din tch 1S . Chng minh

rng:

( ) ( ) ( )2 2 2 2 2 2 2 2 2 2 2 21 1 1 1 1 1 1 1 1 116a a b c b b c a c b c a SS+ + + + + +

Gii: bi ny chng ti cng a ra hai cch chng minh, mt cch s dng nh l trn vmt cch khc s dng h thc lng trong tam gic ri a vnh l trn hon tt

vic chng minh.Cch 1:p dng trc tip nh l ta c:

( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( )

( )( )( ) ( )

2 2 2 2 2 2 2 2 2 2 2 21 1 1 1 1 1 1 1 1

2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 21 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1 1

4 .

4 16

a a b c b b c a c b c a

S a b c b c a b c a b c a b c a a b c

S a b c a b c a b c a b c SS

+ + + + + +

+ + + + + + + + + +

= + + + + + + =


Cch 2:Cch 2 ny l cch m t ngi nghti.

Ta c:2 2 22 cos

4cot1

sin2

bc A b c aA

Sbc A

+ = =

Vy th bt ng thc cho tng ng:2 2 2

1 1 1cot cot cot 4a A b B c C S+ +

M 1 1 1 1 1 1cot cot cot cot cot cot 1 A B C A B C + + = nn theo nh l ta c pcm.


25/121


27

(Da vo vic tm trng hp ng thc xy ra khi no ca nh l cc bn hy tmtrng hp xy ra ng thc ca cc bt ng thc trn, vic ny hon ton d dng.)

IV. Bi tp tluyn:

1. Chng minh bt ng thc sau vi mi tam gic ABC khng cn tha mn0 a b c< < < :

( )

( )( )

( )

( ) ( )

( )

( ) ( )

2 2 2

4ab a b bc b c ca c a

Sb c c a c a a b a b b c

+ + >

2. Chng minh bt ng thc sau vi mi tam gic ABC khng cn tha mn0 a b c< < < :

( )( ) ( )( ) ( ) ( )4 2

a b b c c aabc S

b c c a c a a b a b b c

+ +

3. Cho tam gicABCvi , ,AB c BC a CA b= = = . , ,a b cm m m ln lt l di cc trungtuyn xut pht tA,B,C. Chng minh:

a) ( )( )2 2 2 4 4 43

42a b c a b c

m m m bcm acm abm a b c a b c + + + + + +

b)( )( ) ( )

( )

3 3 3 3 3 32 2 2 9

4 4a b c

a b c a b c a b cm m m

abc a b c ab ca bc

+ + + + + + + +

+ +

c)3 3 3

2a b ca b c a b c

m m mabc

+ + + ++ +

d) 2 2 2 2 2 232a b c

am bm cm a b b c c a+ + + +

e)

( ) ( ) ( ) ( ) ( ) ( )22 2 2 2 2 2 2 2 24 a b cbcm acm abm a b c a b c + + + + + + + +

f) ( ) ( ) ( ) ( )2 23 3 3 2 2 2 2 2 24 a b ca b c a b c bcm acm abm a b c+ + + + + + + +

Gi : s dng bi ton xt phn l thuyt, cc php bin i, cc h thc lng,im t caMv cc bt ng thc cin.


26/121


27/121


29

( )1 2 1 2

2 3

1 3

. cos , cos

. cos

. cos

v v v v C

v v A

v v B

= =

=

=

Nn

( )0 3 2 cos cos cos3

cos cos cos2

A B C

A B C

+ +

+ +


Vd 2: Chng minh vi mi tam gicABCv ba s thcx,y,z bt k ta lun c:

( )2 2 2 2 cos 2 cos 2 cos 1 x y z xy C yz A zx B+ + + +

Gii:

Li chn 1 2 3, ,v v v

nh vd1 trn, p dng tch v hng cho cc vect 1 2 3, , xv yv zv

ta

c:

( ) ( )2

2 2 21 2 30 2 cos cos cos xv yv zv x y z xy C xz B yz A + + = + + + +

Ty ta c ngay iu phi chng minh.

Vd 3: Chng minh rng vi mi tamABCv ba s dng m,n,p ty , lun c:

2 2 2

. . 1 1 1sin sin sin

2 2 2 2

A B C m n pm n p

m n p

+ + + +

Gii:

Ta c:

( )2 2 2 2 2 22 2 2. . 1 1 1 1 1

2 2 2

m n p np mp mnVP n p m p m n

m n p m n p mnp

= + + = + = + +

Bt ng thc tng ng:

( )2 2 2 2 2 2 2 sin sin sin 12 2 2

A B C n p m p m n mnp m n p

+ + + +

t ; ;mn x mp y np z= = = , bt ng thc (1) trthnh:

2 2 2 2 cos cos cos 2( cos cos cos )2 2 2

B C A C A B x y z xy zx yz xy xz yz

+ + + + + + + = + +

Vi ; ;2 2 2

B C A C A B + + += = = to thnh 3 gc mt tam gic. Ta thy bi ton ny

trv vd 2 m ta va xt ban ny.

Vd 4: Chng minh rng vi mi tam gicABCv mi s thcx ta lun c:

( )21

1 cos (cos cos ) 12

x A x B C + + +


28/121


30

Gii:

t ( )21

( ) 1 cos (cos cos ) 12

f x x A x B C = + +

Ta cn chng minh ( ) 0, f x x .Tuy nhin vi phng php ny ta thy vic tnh ton rt kh khn v khng hiu qu. Tas chn phng php tch v hng cho bi ny:

Li chn 1 2 3, ,v v v nh trn ta c:

( ) ( )( )

22 2 221 2 3 1 2 1 3 3 2 2 3

2

0 2 2

0 2 cos cos 2cos 2

v xv v v x x v v v v v v v v

x x C B A

= + + = + + + + +

+ +

T ta suy ra c pcm.Tip tc vn dng tng trn vo bi ton hnh hc khng gian c sc sau, phn chngminh chng ti s dnh cho bn c t tm hiu.

Vd 5: Chng minh tam gicABCc cc trung tuyn ng vi cc cnhAB vBCvung

gc th ta c4

cos 5B .

Gii:

t ,BC a BA c= =

ta c:1 1

,2 2a c

m a c m c a= =

do a cm m

nn

( )2 21 1 2

02 2 5

a c c a ac a c = = +

Vy ta c pcm.

Vd 6: Cho tam gic ABC. Chng minh rng khang cch gia trc tm v tm ngtrn ngoi tip b hn ba ln bn knh ng trn ngoi tip.

Gii: bi ton ny, that u nhn vo ta s ngh ngay n dng h thc lng trong tamgic, hoc dng phng php hnh hc thng thng, t ai ngh rng vi phng phpvecto th bi ton trnn v cng n gin.Ta bit rng tm O ca ng trn ngoi tip, trng tm G v trc tmHnm trn mtng thng (ng thng Euler) v ta c: 3OH OG= . T ta c:

3OH OG OA OB OC = = + +

Suy ra: 3OH OA OB OC R + + =

Vi R l bn knh ng trn ngoi tip. ng thc xy ra khi v ch khi A,B,C thnghng. Nhng iu ny khng xy ra c theo gi thit. T ta c pcm. Vy l ta gii quyt xong mt bi APMO na mt cch kh d dng.

Mt s bi ton chn lc:


29/121


31

1. Cho t gic ABCD. Cc im M&N thuc cc on AD,BCchia chng theo cng ts. Chng minh rng { }max , MN AB CD .

Gii:M,NchiaAD,BCtheo cng t s ( ); 0k M AD N BC k

Vi mi im O ta c:

1

1

OA kODOM

k

OB kOCON

k

=

=

( ) ( )

( )

1 1

0

1 1

OB OA k OC OD AB kDC MN ON OM MN

k k

MN AB k DC AB kDCDo k

k k

= = =

+ =

2. Cho nim 1 2, ,..., n X X X v s dng 1 ,.., na a .Mthuc onAB. Chng minhrng:

1 1 1

max ,n n n

i i i i i ii i i

a MX a AX a BX = = =

Gii:

Mthuc onAB nn chia n theo t s:MA

MB

vi mi im , 1,i X i n= ta c:1

i i

i

MAX A X BMBX MMA

MB

+=

+

( ).

1,i ii i i

i i i i i i

MB X A MAX B MB MA X M X A X B i n

MA MB AB MB MB MA

a MX a AX a BX AB AB

+ = = + =

+

+

{ } { }max , max ,i i i i i i i i i i MB MA

a MX a AX a BX a AX a BX AB AB

+ =



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32

BI 4: PHNG PHP I S HA.

I. Slc v phng php:

Vi vic t di cc on thng hay so cc gc trong tam gic l cc ch hay cc khiu ton hc ta s lm cho vic tnh ton trnn d nhn hn. Bc th hai l ta s pdng phng php i s ha. l vic chng minh da trn nhng tnh ton, nhngbt ng thc cin, nhng con s, nhng phn s cbn, nhng tnh cht cbn trongi s,.

* Mt s v d:

Vd 1: Cho t gic liABCD c khong cch tAnBCnh hn khong cc tBnDC. CA vBD c giao im K. TrnBClyN, trnAD lyM, trnACly G, sao cho KN

// DC; GN // DC; KG // AB. Chng minh:8

27GMNK ABCDS S<

Gii:Ta c: KN // DC; GN // DC; KG // AB

/ / AM GM AG BK BN

MN AB AC DC AD BD BC

= = = = GMNKl hnh bnh hnh.

Theo bi ta c khang cch t An CD nh hn t BnDC

ACD BCD

AKD AKD BKC

BKC

S S

S ayS S

S bx

>+ +

Ta c:22 2

AGM

ACD

S AG BK x

S AD BD x y

= = = +

Tng t:.

.AGK

ACD

S AG AK a x

S AD AC a b x y= =

+ +

( )1GKM

ACD

S x x a

S x y x y a b

= + + +

Ta c:

( )2 ACD ACD

ABC ABCD

S Sy y

S x S x y= =

+

T (1) v (2) ta c:

MG

NK

D

A

B

C


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33

( ) ( )( )2 2

2 23

GKM

GMNK ABCD ABCD

x x a yS

x y x y a b x y

xy x a xy xS S S

x y a b x y x y x y

= + + + +

=


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34

( )

( )( )

2

2

2

2

1

1 1

1

1

31

APM MNB CPM

ABC

ABC MNP

ABC

MNP

ABC

S S S k

S k k

S S k k

S k

S kS k

+ + = + + +

+ + =

+

=+

M ta c:

( )

( )( )

2

2

1 4

14

41

k k

k

k

+

+

T (3) v (4) suy ra iu phi chng minh.

Vd 3: Cho ABC . Qua imMthuc cnhAB, dng hai tia ln lt song song vi cc

trung tuyn AD v BE v chng ct BC v CA ti PQ. Dng hnh bnh hnh MPSQ.Chng minh:3

8 MPSQ ABC S S

Gii:GiNv Tnh hnh v. Ta c:

( )

2

3

2

3 4. .sin . .sin

94

19

MNGE

MPSQ

MN BG

MQ BE

MT AG

MP ADS MN ME NQ MN

S

= =

= =

= =

=

Ta t: MA kMB= Ta c:

( )

1

1

1

12

1 MNP APM MNP APM MNPAPM

ACN MCB ACN MCB ABC

MN AM

GB AB k MT BM k

GA BA k NMT NGT

S S S S SS

k S S S S S

= =+

= =

+=

+ + = = = =

+ +

T (1) v (2) suy ra:

N

G

S

T

P

Q

E

D

A B

C

M


33/121


35

( ) ( )

2

2

2

1

1 1

1

1

31

APM MNB CPM

ABC

ABC MNP

ABC

MNP

ABC

S S S k

S k k

S S k k

S k

S k

S k

+ + = + + +

+ + =

+

= +

M ta c:

( )

( )( )

2

2

1 4

14

41

k k

k

k

+

+

T (3) v (4) suy ra:

( )4 MNP ABC S S dpcm

Vd 4: Cho ABC c cc phn gicAM,BN,CP. Chng minh rng: 14 MNP ABC

S S

Gii:Trc tin ta nhn thy y chnh l phn mrng ca bi 3.

Ta t:

AB c

BC a

CA b

=

= =

V ta c:

( ) ( )

.

.BPM

ABC

BP a BP a

PA b BA a b BN c BM c

MC b BC b cS BP BM ac

S AB BC a b b c

= = +

= =+

= =+ +

Tng t vi cc & MNC APN .M

( )

( )( ) ( )

2

1

MNP ABC PN BPM MNC

MNP PN BPM MNC

ABC ABC

S S S S S

S S S S abc

S S a b b c c a

= + +

+ +

= = + + +

Ta c:

( ) ( ) ( )

2 1

4

abc

a b b c c a

+ + +pcm.

Vi trng hp tam gicABCcn th ta a v bi 3.

M

NP

B C

A


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36

BI 5: SDNG CC NH L, NH NGHA V CCNG THNG, NG TRN

I. Slc v phng php:

C th ni, t cp hai chng ta c tham kho nhiu sch vit v cc nh l ln, ccng trn, ng thng, cc h thc, c s dng rng ri trong vic chng minhcc bi ton hnh hc.

1. nh l Menelaus:Cho tam gicABC, giM,N,Kln lt l cc im thuc cc ng thngAB,BC,CA(c th nm trn phn ko di ch khng nht thit phi nm trn on thng) chiacc cnh tam gic theo t s l m,n,k (u khc 1). Th ta c M,N,K thng hngkhi v ch khi m.n.p = 1.nh l ny theo ti nh l chng trnh sch gio khoa cp II khng c cp n,nhng nu cc bn chu kh tham kho cc sch tham kho ngai th trng, th cth bit c rng nh l ny c gii thiu vi hc sinh t nhng nm cp II.

Nm u cp III, cc bn c hc v vecto th nh l ny li c mrng thm,v pht biu chnh xc hn, ty ta bit c chnh xc vnh l Ceva.

2. nh l Ceva:Gi D,E, F l ba im tng ng cc dng thng BC, CA, AB ca tam gic ABC.Chia cc cnh tam gic theo t s l m,n,k(u khc 1). Lc ba ng thngAD, BE, CF ct nhau ti mt im O hoc song song khi v ch khi 1mnp = .* Ch : cc ng AD, BE, CF gi l cc cevian.nh l ny cn c pht biu di dng lng gic nh sau:Gi D,E, F l ba im tng ng trn cc cnh BC, CA, AB ca tam gic ABC.Lc ba ng thng AD, BE, CF ct nhau ti mt im O khi v ch khi

sin sin sin 1sin sin sin

ABE BCF CADCBE ACF BAD

=

Tnh l Menelaus v Ceva ny ta c th suy ra cch chng minh nh l Desargues.

3. nh l Desargues:Trong mt mt phng cho hai tam gic ABC v A'B'C'.Nu cc ng thngAA',BB',CC' ng qui ti mt im v cc cp ng thng BC,B'C' ; AC,A'C' ;AB,A'B' u ct nhau th cc giao im ca chng thng hng.y l mt bi ton c gii bng kin thc THCS, bn c c th t chng minh.

4. ng thng Simson:ng thng Simson l mt bi ton kh ni ting trong chng trnh ton hc phthng. nh l c pht biu nh sau:T mt im D bt k nm trn vng trn ngoi tip tam gic ABC k 3 ng vunggc xung 3 cnh ca tam gic ny. Khi khi D chuyn ng trn ng trn chn3 ng vung gc ny lun thng hng.

5. ng thng Steine:


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37

ng thng steine l ng thng i xng ca ng thng Simson qua cc cnhca tam gic. N lun lun i qua trc tm ca tam gic vi mi M thuc (ABC).Ngai ra cng c mt im khc cng lin quan n ng thng Simson l imMiquel:

6. im Miquel:

Cho 4 ng thng ct nhau ti 6 im to thnh 4 tam gic. Cc ng trn ngoitip 4 tam gic ny c mt im chung (gi l im Miquel).

7. ng thng Euler, ng trn Euler, h thc Euler:- ng trn Euler:Chn ba ng cao ca mt tam gic bt k, ba trung im ca ba cnh, ba trung imca ba on thng ni ba nh vi trc tm, tt c chn im ny cng nm trn mtng trn. ng trn ny thng c gi l ng trn Euler hay cn gi lng trn Feuerbach hay ng trn chn im.Gi R l bn knh ng trn ngoi tip ca tam gic th ng trn Euler c bn

knh l R/2 v tm ca n l trung im on ni trc tm v tm ng trn ngoitip ca tam gic .- nh l Feuerbach:ng trn Feuerbach ca mt tam gic tip xc vi ng trn ni tip v ba ngtrn bng tip ca tam gic .- ng thng Euler:ng thng Euler l ng thng ni cc im l Trc tm, Trng tm v Tm vngtrn ngoi tip tam gic. ng thng Euler l mt nh l rt ni ting ca hnh hcscp m mi hc sinh u bit.- H thc Euler:Cho ( ) ( ); , ;O R O r ln lt l ng trn ngoi tip v ni tip tam gicABCth ta

c: 2 2 2OO R Rr = . y chnh l h thc Euler.- Cng thc Euler:Cng thc Euler, hay cn gi l ng nht thc Euler, l mt cng thc ton hctrong ngnh gii tch phc, c xy dng bi nh ton hc ngi Thy SLeonhardEuler. Cng thc ch ra mi lin h gia hm s lng gic v hm s m phc.C th, vi mi s thc x, ta c:

cos sinixe x i x= + y e l cs logarit t nhin, i l n v ca s phc.Khai trin t cng thc trn, cc hm s sin & cosx x c thc vit di dng sau:

( )

( )

1cos

21

sin2

ix ix

ix ix

x e e

x i e e

= +

=

Trng hp c bit: khi x = , ta c: cos sin 1ie i = + = t dn n cngthc rt gn ni ting: 1 0ie + = (Cch chng minh cc h thc ny cc bn c th tham kho trang wikipedia.orghoc t tm ti sng to ra cch gii ring cho mnh nhng cng thc Euler th bn


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38

c cn c kin thc v s dng s phc, php tnh vi tch phn, vi phn, chuiTaylor mi c th hiu c phn ny).

8. nh l Stewart:- nh l Stewart 1 :(Tnh cht ng phn gic)Cho tam gic ABC, D l im trn BCsao cho AD sao cho AD l ng phn gic

ca gcA. Khi : AB AC BD CD

=

- nh l Stewart:Cho tam gic ABC, D l im trn BC. Khi ta c:

2 2 2. . . .AB CD AC BD AD BC BC BD DC+ = - nh l Apolonius :(cho ng trung tuyn):Cho tam gicABCc cnh a, b,c v di ng trung tuyn am . Khi ta c:

2 2 22

2 4ab c a

m+

=

+ H qu 1:(Tng bnh phng ba ng trung tuyn)

( )2 2 2 2 2 234a b c

m m m a b c+ + = + +

+ H qu 2 :(cng thc ng phn gic)

( )2a

bcp p ad

b c

=

+

9. ng trn Apollonius:Cho hai im phn bitA,B v s thc 1k . Chng minh rng tp hp nhng im

Msao choMA

kMB

= l mt ng trn, chnh l ng trn Apollonius ng vi 2

imA,B v t sk.

10. Bt ng thc v nh l Ptolemy:nh l Ptolemy v tnh cht ca t gic ni tip l mt trong nhng kt qu kinhin v p ca hnh hc scp.C th ni, bt ng thc Ptolemy v nh l Ptolemy p t cc cch chng minh adng n nhng ng dng phong ph trong cc bi ton chng minh, trong tnh tonhnh hc v trong cc bi ton bt ng thc hnh hc.Bt ng thc Ptolemy l hqu ca bt ng thc tam gic? Ai cng bit bt ng thc tam gic: ViA, B, Clba im bt k trn mt phng, ta c ( )1 . AB BC AC+ Du bng xy ra khi v ch

khi A, B, Cthng hng v B nm gia A v C. Ni cch khc BCkAB = vi k lmt s thc dng.Trong khi , bt ng thc Ptolemy khng nh: Vi 4 imA, B, C, D bt k trnmt phng, ta c ( ). . . 2 . AB CD AD BC AC BD+

R rng, theo mt quan im no th bt ng thc Ptolemy chnh l mrng cabt ng thc tam gic. V sao vy? Xin gii thch l do:Chia hai v ca (2) choBD, ta c


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39

ACBD

ADBC

BD

CDAB +

Nu chnD xa th ty ta s suy ra AB BC AC + .iu ny nghe cng ngc nhin, tuy nhin li ch em li ca sc bit ho nykhng nhiu, v chng l li dng bt ng thc Ptolemy cao siu chng minh btng thc tam gic vn c coi nh tin ?

Tuy nhin, mt logic rt t nhin dn chng ta n mt tng hu ch hn: Nh vybt ng thc Ptolemy c lin quan n bt ng thc tam gic. Vy c th l btng thc Ptolemy c thc chng minh nh vo bt ng thc tam gic? iuny qu l nh vy. chng minh cho lun im ny ta c th dng ba php chngminh tiu biu: S dng tnh cht tam gic ng dng v bt ng thc tam gic, Sdng php nghch o v bt ng thc tam gic, S phc. Ta cng c th chngminh nh l Ptolemy bng cch s dng ng thng Simson.- Nhng kt qu kinh in:Trc ht ta xem xt ng dng ca bt ng thc Ptolemy v trng hp c bit can nh l Ptolemy trong vic chng minh cc kt qu kinh in ca hnh hcphng

+ im Toricelli:Xt bi ton Cho tam gic ABCbt k. Hy tm imMtrong mt phng tamgic sao cho MA MB MC+ + t gi tr nh nht.imMtm c c gi lim Toricelli ca tam gicABC.C th gii ngn gn bi ton ny bng cch s dng bt ng thc Ptolemy nhsau:Trn cnh BC, dng ra pha ngoi tam gic uBCA . p dng bt ng thcPtolemy cho t gic MBA C ta c . . .BM CA CM BA BC MA + T, do CA BA BC = = nn ta c BM CM MA+ Nh th AM BM CM MA MA AA + + + Tc l ( ) AM BM CM AA const + +

Du bng xy ra khi v ch khi1. T gic BMCA ni tip2. Mnm giaA v A

D thy ta c th tm c imMtho mn c hai iu kin ny khi v ch khitt c cc gc ca tam gicABCu khng ln hn 1200.Nu chng hn, gc 0120A > th imMcn tm s chnh l imA (bn c tchng minh!).R rng phng php ni trn c th p dng cho bi ton tng qut hn: Chotam gicABCv cc s thc dng m, n, p. Hy tm imMtrong mt phngtam gic sao cho . . .m MA n MB p MC+ + t gi tr nh nht.

Tt nhin, chng ta cng s gp phi tnh hung tng t nh tnh hung tamgicABCc 1 gc ln hn 1200 nhtrn.Nu ch n xut pht im ca bt ng thc Ptolemy, chng ta c th ddng xy dng li gii trc tip cho bi ton im Toricelli m khng qua btng thc ny bng cch s dng vic v thm cc tam gic ng dng.Chng hn vi bi ton im Toricelli. Xt php quay tm Cgc 600 bin Mthnh ,M B thnh B th CMM l tam gic u vMB M B = , do

AM BM CM AM MM M B AB + + = + +


38/121


39/121


41

Cho t gicABCD ni tip trong mt ng trn, khi

( )2 2 2. . . 8 ACD ABD BCD BD S CD S AD S= +

+ nh l Carnot:Trong tam gic nhnABCni tip trong ng trn tm O bn knhR. Gix, y,z l cc khong cch tOnBC, CA, AB tng ng. Khi

x y z R r + + = + trong rl bn knh ng trn ni tip tam gic.Vit di dng lng gic, nh l Carnot chnh l h

thc cos cos cos 1r

A B C R

+ + = + . Ch h thc ny ng vi mi tam gic.

Vi h thc hnh hc, nh l Carnot vn ng trong trng hp tam gic t,nhng nu chng hn A t th ta c .x y z R r + + = +

- Mrng nh l Ptolemy v bt ng thc Ptolemy:nh l Ptolemy v bt ng thc Ptolemy c nhiu hng mrng khc nhau. Thm

ch t bt ng thc Ptolemy, pht sinh ra hn mt khi nim gi l khng gian metricPtolemy, th Ptolemy Di y, chng ta xem xt mt s mrng ca nh lPtolemy (v cng l ca bt ng thc Ptolemy)

+nh l BretschneiderCho t gicABCD c di cc cnhAB, BC, CD, DA ln lt l a, b, c, dv di hai ng cho AC, BD l m, n. Khi ta c

( )2 2 2 2 2 2 2 .cosm n a c b d abcd A C = + +

R rng nh l Ptolemy v c bt ng thc Ptolemy u l h qu ca nh lBretschneider.

+nh l Casey (nh l Ptolemy mrng)Cho t gicABCD ni tip ng trn (C). Bn ng trn , , , tip xcvi (C) ln lt tiA, B, C, D. Gi t l di on tip tuyn chung, trong

t l di on tip tuyn chung ngoi nu , cng tip xc ngoi hoc

cng tip xc trong vi (C) v t l di on tip tuyn chung trong trong

trng hp ngc li. Cc i lng ,t t c nh ngha tng t. Khi

ta c. . . .t t t t t t + = (9)

nh l Ptolemy chnh l trng hp c bit ca nh l Casey, khi0.x y z t = = = =

nh l Casey c th pht biu mt cch khc, nh sau: Cc ng trnA, B, C,D tip xc vi ng trn (O); a,b,c,d,x,y l di cc tip tuyn chung ca cccp ng trn A v B,B v C,Cv D,D v A,A v C, B v D tng ng. Khi . . .x y a c b d = + . Ch ta ly di tip tuyn chung trong hay tip tuynchung ngoi theo nguyn tc cp trn. Cui cng, im c th coi nhng trn bn knh 0 v tip tuyn ca hai ng trn im chnh l ng


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42

thng i qua chng. iu ny sc dng n trong phn ng dng ca nhl Casey.

+nh l 1. Cho hai ng trn (O1), (O2) tip xc ngoi nhau ti I v cngtip xc trong vi ng trn (O). Mt tip tuyn chung ngoi ca (O1) v (O2)

ct O ti B v C, trong khi tip tuyn chung trong ca chng ct (O) ti imA cng pha vi I. Khi I l tm ng trn ni tip tam gic ABC.

+nh l 2. Cho tam gic ABC ni tip trong ng trn (O). ng trn (C)tip xc vi dy cung BC ti D v cc cnh AB, AC tng ng ti P v Q. Khi trung im ca PQ l tm ng trn ni tip tam gic ABC.

11. Trng tm ca mt him:

Cho n b ( ),i iA m vi 1,i n= , trong iA l cc im cn im l cc s thc dng.

Ta ni trng tm ca hn b ( ),i iA m l mt im T sao cho:

1

0n

i ii

m TA=

=

(C th hiu im l cc trng lng t vo v tr iA . Khi 3n = v 1 2 3 1m m m= = = , tali gp khi nim trng tm ca mt tam gic).nh ngha ny hp l v ta c tnh cht sau: Vi mi n b nh ni trn, trng tmlun lun tn ti v duy nht.

12. Bao li ca hnim:Trong mt phng cho nim. Ta ni bao li ca hnim ny l a gic li nh nhtcha tt c cc

im

, ngha l

a gic ny khng cha bt c mt

a gic li no

khc cng vi tnh cht . (Ta ni mt a gic l li nu ko di mt cnh bt k thn s khng ct bt c cnh no khc). C th chng minh c rng bao li ca mth hu hn im lun lun tn ti v duy nht.

13. nh l Pick:Cho a gic (P) khng t ct nhau trong mt phng ta vi cc nh c ta nguyn. K hiuB l s tt c cc im c ta nguyn nm trn bin ca (P),Ils tt c cc im c ta nguyn nm bn trong ca (P). Khi , din tch ca (P)l:

( )1

1

2

P I B= +

14. nh l Poncelet:

Gi s c mt ng trn c t bn trong mt ng trn khc. Gi , 1,i A i n= l

cc im trn ng trn ln sao cho mi on trong ng gp khc 1i iA A+ u tip

xc vi ng trn nh. Khi , nu , 1,i B i n= , l cc im trn ng trn ln sao


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43

cho mi on trong ng gp khc 1i iB B + u tip xc vi ng trn nh th 1nB B

l tip tuyn ca ng trn nh.

II. Mt s v d:Php chng minh bt ng thc Ptolemy cng nh cch t bt ng thc Ptolemy suy ra btng thc tam gic cho thy bt ng thc ny c th p dng nh gi di cc on

thng. Vic dng tam gic u BCA ra pha ngoi trong li gii bi ton Toricelli chnh lmt cch lm mu mc p dng c bt ng thc Ptolemy.

tng chung l: nh gi tng . .p MA q MB+ , ta c th dng im N saocho . . p NA q NB= . Sau p dng bt ng thc Ptolemy th c

. . .NA MB NB MA AB MN+ T

. . . . .

. . . . .

. . .

p NA NB p NP MA AB MN

q NB MB p NB MA AB MN

MN

p MA q MB AB NB

+

+

+

Ch l im N l cnh, nh th p.MA + q.MB c nh gi thng qua MN. tng ny l cha kho gii hng lot cc bi ton cc tr hnh hc. Ta xem xt mt sv d:

Vd 1: Cho imMnm trong gc nhnxOy. Hai imA,B ln lt thay i trn Ox, Oy sao

cho 2 3OA OB= . Tm v tr caA, B sao cho 2 3 MA MB+ t gi tr nh nht.

Gii:p dng bt ng thc Ptolemy cho t gic OAMB, ta c

. . .OA MB OB MA OM AB+ T

2 .. 2. . 2. .

3 . 2. . 2. .

2 3 2. .

OA MB OB MA OM AB

OB MB OB MA OM AB

AB MA MB OM

OB

+

+

+

V tam gic OAB lun ng dng vi chnh n nnAB

OBl mt i lng khng i. T

suy ra 2 3 MA MB+ t gi tr nh nht bng 2. . ABOMOB

. Du bng xy ra khi v ch khi t

gic OAMB ni tip.

Vd 2 : Mt lc gic c di 6 cnh u bng 1. Chng minh rng lc gic c t nht mtng cho nh hn hay bng 2.

Gii :


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44

Khng nggi cho li gii bi ton ny li l mt ng thc lp mt: 1 vi 1 l 2 . V thc hin php cng hai cnh thnh ra ng cho , ta s p dng bt ng thcPtolemy.Xt lc gic ABCDEF. Xt tam gic ACE. Khng mt tnh tng qut, c th gi sCE lcnh ln nht trong tam gic. p dng bt ng thc Ptlemy cho t gicACDE, ta c

. . .AC DE AE CD AD CE+

T, do 1CD DE = = v CE AC,CE AE nn ta suy ra 2AD (pcm).

Vd3. (IMO SL 1997) Cho lc gic liABCDEFc , , AB BC CD DE EF FA= = = . Chng

minh rng3

2

BC DE FA

BE DA FC + + . Du bng xy ra khi no?

Gii:p dng bt ng thc Ptolemy cho t gicACDEta c . . . DE AC DC AE DACE+ . S

dng DE DC= , ta c ( ) . DE AC AE DACE+ hayDE CE

AC AEDA

+Tng t, ta

c FA EA BC EA&CE CA EA ECFC BE

+ +

. Cng cc bt ng thc ny li v s dng bt ng

thc Nesbitt ta thu c iu phi chng minh. c du bng ta phi c du bng ba bt ng thc Ptolemy v bt ng thc Nesbitt.

Du bng bt ng thc Nesbitt xy ra khi tam gic ACE u, nh th 60oCAE= . VACDEl t gic ni tip nn gcD phi bng 120o. By gicc tam gic ABC, CDE, EFAphi bng nhau (Tam gicABCcn, v vy cc gc ca n bng 30o, 120o, 30o v cnhAClcnh ca tam gic u). Nh th lc gic c tt c cc cnh u bng nhau v tt c cc gcbng 1200, vy n l lc gic u. Ngc li, hin nhin l vi lc gic u, ta c du bngxy ra.

Vd 4: (IMO 2001) Cho tam gicABCvi trng tm G v di cc cnh, ,a BC b CA c AB= = = . Tm im P trn mt phng tam gic sao cho i lng

. . .AP AG BP BG CP CG+ + t gi tr nh nht v tm gi tr nh nht theo a, b, c.

Gii:

Vng trn ngoi tip tam gic BGC. Ni ditrung tuynAL ct ng trn ny ti K. GiM, Nl trung im cc cnhAC, AB tng ng.p dng nh l hm s sin cho tam gicBGL, ta

csin sin

BG BL BLG BGK

= .

Tng t, p dng nh l hm s sin cho CGL, ta

csin sin

CG CL

CLG CGK = . Nhng L l trung im

ca BC vsin sin BLG CLG= ,

nnsin

sin

BG CGK

CG BGK = . Ta c


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45

( )2 .sin BK R BGK = , trong R l bn knh ng trn ngoi tip. Tng

t 2 sinCK R CGK = , do CK BG

BK CG= , v ty

BG CG

CK BK =

Tng t,sin sin

sin sin

AG BGN BGN

BG AGN CGK = = (gc i nhau). Hn na

( ) ( )2. .sin 2. .sin BC R BKC R BGN = = (VBGCKl t gic ni tip nn BKC BGN= . T

sin

sin

BC BGN AG

CK CGK BG= = , t

BG AG

CK BC = v : : : : BC CK BK AG BG CG= .

By gip dng bt ng thc Ptolemy cho t gic PBKC: . . .PK BC BP CK CP BK + . T . . .PK AG BP BG CP CG + . Suy ra ( ) . . . AP PK AG AP AG BP BG CP CG+ + + v cui

cng . . . .AK AG AP AG BP BG CP CG + + vi du bng xy ra khi v ch khi (1) P nm trnng trn gia CvB ( c ng thc BDT Ptolemy) v (2) P nm trnAK( c ngthc trong bt ng thc tam gic). Do gi tr ny t c khi P G= .

D dng tnh c rng2 2 2

2 2 2

3

a b c AG BG CG

+ ++ + = .

C th thy y l trng hp c bit ca bi ton Toricelli tng qut m chng ta xemxt phn u. Ch rng t ba onAG, BG, CG c th dng c 1 tam gic . Ta chcn dng tam gic BCKng dng vi tam gic l c. Cch gii nu trn ch ra cchdng tng minh cho im K.

Vd 5: Cho tam gicABC. GiD l trung im caBC, E, Fln lt l haiim trnAB vAC. Chng minh rng nuAD, BF, CEng quy th EFsong song viBC.

Gii :

p dng nh l Ceva ta c:1

EA DB FC

EB DC FA=

V DB DC= nn 1EA FC EA FA

EB FA EB FC = =

Vy EFsong songBC.

Vd 6:Trn cc cnh AB, BC, CA ca tam gic ABC, ta ly cc im 1 1 1, ,C A B sao cho cc

ng thng 1 1 1, , AA BB CCng quy ti im O. ng thng v qua O song song viACct

cc ng thng 1 1A B v 1 1B C tng ng ti K, M. Chng minh:OK OM = .

Gii :V quaBng thng song songAC(nh hnh v)Mnh cn cm 1 1 BM BK =

T cc tam gic ng dng 1 1ABCv 1 1BM C suy ra

O

DB C

A

FE


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46

11 1

1

BC BM AB

AC=

Tng t ta c: 11 11

BA BK CB

CA=

Chia cc ng thc v theo nh l Ceva ta c:

1 1 1 1

1 1 1 1

1 BM AB CA BC BK CB BA AC

= = (pcm)

Vd 7: Cho tam gic ABC. V pha ngoi cc tam gic ABX, BCYv CAZcn tiX, Y, Zvng dng vi nhau. Chng minhAY, BZv CXng quy.

Gii:Gi x l so gc y ca tam gic cn v cc giao im L, M, N (nh hnh v).

Ta c:( )

( )

( )

( )

1. .sin .sin2

1. .sin

2

ABY

ACY

AB BY B x BA B xSBL

LC S CA C xCA CY C x

+ += = =

++

Tng t .

T ta c: 1BL CM AN

LC MA NB=

VyAL, BM, CNng quy (pcm).

Vd 8: Cho na ng trn (C) nm v mt pha ca ng thng (d). CvD l cc im trn(C) . Cc tip tuyn ca (C) ti CvD ct (d) tiB vA, v tm ng nm gia hai imny. Gi El giao im caACvBD, Fl im nm trn (d) sao cho EFvung gc vi (d).Chng minh EFl phn gic gc CFD.

Gii:Gi P l giao im caAD vBC. Qua P dng PHvung gc (d).

Ta c tam gic PAHv ODAng dng nn AH HP

AD DO=

Tng t: HB HP HP

BC CO DO= = , t ta c:

AH BH

AD BC =

K1

M 1 C1 B1

A1

K

M

O

C B

A

B C

A

X

Y

Z

L

MN


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47

Suy ra: . . 1 AH BC PD

HB CP DA=

Theo nh l Ceva AC, BD, PH ngquy, suy ra PHtrng FED thy 5 im P, D, H, O, C cng nmtrn mt ng trn. T :

DHP DOP COP CHP= = =

Vd 9: Cho tam gic ABC . Ly im Mtrong tam gic.AMctBCti E, CMctAB ti F.

Gi N l im i xng ca B qua trung im ca EF. Chng minh rng ng thng MNlun i qua mt im cnh khi im Mdi chuyn bn trong tam gicABC.

Gii :

Dng hnh bnh hnh ABCD. Gi ccgiao im nh hnh v.Ta cBENFl hnh bnh hnh.

Suyra:

( ); ; 1 IF FA KN FN HE PE

IN EN KE EC HF FN

= = =

Mt khc:

( ). .

2.

PE NL EC NE EC PE NE EC PE hay

NE DL FA FA FN FN FA= = = =

T (1) v (2) ta suy ra: . . 1 IF KN HE

IN KE HF=

Theo nh l Ceva :NH, FK, EIng quy, suy raD, N, Mthng hng. VyMNlun iqua im cnhD khiMdi ng trong tam gicABC.

Vd 10: Cho tam gicABCvi AB AC > . Gi P l giao im ca ng trungtrcBCv ng phn gic trong ca gcA. Dng cc imXtrnAB v YtrnACsao cho PXvung gcAB v PYvung gcAC.GiZl giao im caXYvBC. Xc

nh gi tr t sBZ

ZC.

Gii:Ta c PAX PAY = , suy ra AX AY = v PX PY = Suy ra PYC PXB = , suy ra CY BX = V , ,X Y Zthng hng, p dng nh l Menelaus ta c

d

E

C

OA

P

B

D

HF

K

I

M

A D

B C

NL

P E

F


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48

Vd 11: Cho ( ),C O R l ng trn ni tip tam gicABC.Cho imMnm trong tam gic.

Gi 1 1 1, , A B Cl hnh chiu caMln ba cnhBC,AC,AB.Chng minh rng:

1 1 1

2 2

2

| |

4A B C

ABC

S R OM

S R

=

Gii:

Xt v trimMnm trong BAC v nm ngoi ( )O (cc trng hp khc tng t). Ta c

t gic 1 1MACB ni tip ng trn ng knhMCnn 1 1 sin A B MC C = , tng t th

1 1 sin B C MA A= .GiD l giao im caMCvi ng trn ngoi tip tam gicABC. Ta c:

1 1 1 MB A MCA BAD= = v

1 1 1C B M C AM= .

Mt khc:

1 1 1 1 1 1 1 1 A B C MB A MB C DAB MAC MAD= = =

Xt tam gicADM, theo nh l sin ta c:

1 1 1

AM DM DM

sinADM sinMAD sinA B C= =

Suy ra: 1 1 1 AMsinA B C DMsinADM= . T ta c:

1 1 1 1 1 1 1 1 1 1 1 1 1

1 1 1. . . . . . .

2 2 2A B CS B A C B sinA BC MCsinC MAsinAsinA BC MC MD sinAsinBsinC= = =

Mt khc, ta li c: 22ABCS R sinAsinBsinC= v2 2

/( ). | |M O MC MD P OM R= = nn:

1 1 1

2 2

2| |

4A B C

ABC

S R OM

S R

=

Vd 12: Cho tam gic v ba im 1 1 1, , A B Ctng ng nm trn ba cnhBC, CA, AB sao cho

cc ng thng 1 1 1, , AA BB CCct nhau ti O. Gi s ba cpAB v 1 1 1 1, ,A B BC v B C CAv

1 1C A ln lt ct nhau ti 2 2 2, ,C A B . Chng minh 2 2 2, ,C A B thng hng.


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49

Gii :

p dng nh l Menelaus cho tam gicBCD vi cc im1 1 2

1 1 21 1 2

1 1 21 1 2

1 1 2

1 1 21 1 2

1 1 2

& , , : 1 (1)

& , , : 1 (2)

& , , : 1 (3)

AA OB BCOAB A B C

OA BB AC

OC BB CAOBC B C A

CC OB BA

OA CC ABOAC A C B

AA OC CB

=

=

=

Nhn v theo v ta c: 2 2 2

2 2 2

1 BC CA AB

AC BA CB=

p dng nh l Menelaus (phn o) ta suy ra pcm.

Vd 13: ChoABCl mt tam gic khng cn. Cc ng trung tuyn k tA, B,Cln lt ctng trn ngoi tip tam gic ti cc im th haiL,M,N. Gi s LM LN= , chng minh:

2 2 22 BC AB AC = + .

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

LN LG NLG AGL AC GC =

.

Tng t: LM GL

AB BG= ,

T suy ra: AB BG

AC CG=

B2

A2

C2

B1C1

A1

O

A

B C

G

I

K J

A

B C

L

N

M


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50

p dng nh l Stewart ta c:

( )( )

2 2 2 2

2 2 2 2

2 2 2 2 2

2 2

2 2

2 0

AB AB BC AC

AC AC BC AB

AC AB BC AB AC

+ =

+

=

Do tam gic ABC khng cn nn 2 2 22 BC AB AC = +

III. Mt s bi ton chn lc:

1. Cho tam gic ABC tho ( )1

2b a c= + chng minh rng: ( )cos 4cos 3 A C B + =

Gii:Nu a c= , kim tra ta thy ng.Nu a c , lyD trnACsao choDB bng a. t AD x=

Theo nh l Stewart ta c:( ) ( )

( ) ( )

2 2 2

2 2 2 2 2 2 0

c b x b x bc a b a x

bx c b a x b a c

+ + + = +

+ + =

Thay ( )1

2b a c= + ta c:

( ) ( ) ( )2 2 21

5 3 0 22

x c a a c x a c+ = =

p dng nh l cosin cho tam gicABD ta c:

( ) ( ) ( )2 2 2 2 21 1

cos 8 3 3

2 2

A C a c x ac a c

ac ac

= + =

p dng nh l cosin cho tam gicABCta c

( ) ( )2 2 2 2 21 1

4cos 4 3 3 22 2

B a c b a c acac ac

= + = +

T suy ra: ( )cos 4cos 3 A C B + =

2. Tam gicABCc 2 2 2sin ,sin ,sin A B C lp thnh mt cp s cng v c tng bng3

2.

ng cao k tA v ng phn gic trong ca gcB ca tam gic ABCct nhauti I, Ithuc min trong tam gic ABC. Chng minh rng: ( , IAC IBC IAC IBC S S S S= l

din tch tam gicIAC,IBC)

Gii:Vng caoAH, phn gic trongBD, CIctAB tiM.t : , ,BC a CA b AB c= = = VIthuc min trong tam gicABC, nn cc gcB v Cu nhn.

T gi thit, ta suy ra : 2 2 21

sin &sin sin 12

B A C = + =


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51

sin cos &sin sin B B A C = = Theo nh l Cva :

1

cos

cossin sin1

cos sin

MA HB DC MA HC DA

MB HC DA MB HB DB MA b C c

MB c B a MA B C

BM B A

= =

=

= =

Do Ml trung im caAB.HAEvBFvung gc vi CMth AE BF= T suy ra : IAC IBC S S=

3. Cho tam gicABC. Mt ng trn bt k ct cc cnhAB, BC, CA ln lt ti Q, M,S, P, N, R nh hnh v { } { } { }, ,PQ RS X RS MN Y PQ MN Z = = = .Cmr AX,

BY, CZng quy.

Gii :p dng nh l Ceva cho tam gicAQR.

( )1

2

sin sin sin1 1

sin sin sin

A XQR XRA

A XQA XRQ=

tng t cho tam gicBMS, tam gic CNP ta c:

( )

( )

1

2

1

2

sin sin sin1 2

sin sin sin

sin sin sin1 3

sin sin sin

B YSM YMB

B YSB YMS

C ZNP ZPC

C ZNC ZPN

=

=

Nhn (1), (2), (3) ng thi, lu rng sin sin XQR YSB= , ta c:

1 1 1

2 2 2

sin sin sin1

sin sin sin

A B C

A B C = pcm.

4. Cho t gic liABCD tha iu kin 090BAD > .Chng minh rng nuMNvBD ctnhau tiIthIA vung gcAC.

Gii :Nu M C ( hay N C ) th I D (hay I B ). Lc bi ton ng.Nu I B (hay I D ), p dng nh l Menelaus cho tam gicBCD vi 3 imM, N,Ita c:

5 3 1

4 2

sin sin sin1 1 1

sin sin sinANC AMB AID

AMC AND AIB

S AB A AC AS S AD AMB NC ID

MC ND IB S S S AC A AD A AB IAB= = =


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52

( )

( ) ( )

( ) ( ) ( ) ( )

( )

1 3 4 2 5

1 3 2 3 1 2

1 3 1 3 1 2 3 1 3

1 2 3


sin sin cos cos

1 1cos cos cos 2 cos

2 2cos 0

A A A IAB do A A

A A A A A A

A A A A A A A A A

A A A

= =

= + +

+ = + +

+ + =

Vy IA AC .

5. Cho lc gic li ABCDEF tha iu kin , & AB BC CD DE EF FA= = = . Chngminh rng:

3

2

BC DE FA

BE DA FC + +

ng thc xy ra khi no?

Gii:t , , .AC a CE b AE c= = = p dng nh l Ptolemy cho t gic ACEF ta c:

. . .AC EF CE AF AE CF+

V EF FA= nn suy ra:FA c

FC a b

+

Tng t ta cng c: & DE b CB a

DA c a BE b c

+ +. T suy ra:

( )3

12

BC DE FA a b c

BE DA FC b c c a a b+ + + +

+ + +

ng thc xy ra khi cc bt ng thc (1) ng thi xy ra, tc l khi cc t gic, , ACEF ABCE ACDE ni tip c, ngha l khi ABCDEF l mt lc gic ni tip.

Ngai ra cho ng thc xy ra, bt ng thc cui cng cng phi trthnh ng thc,tc l ta phi c a b c= = Vy ng thc xy ra khi v ch khi ABCDEF l mt lc gicu.(Bt ng thc cui cng ta c th chng minh bng nhiu cch, chng ti sa ra haicch chng mnh c coi l cbn v d hiu:Cch 1:

ta b c

Sb c c a a b

= + ++ + +

. Ta c: ( )1 1 1

3S a b cb c c a a b

+ = + + + + + + +

M theo BCS ta c:

( )

1 1 1 9

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

+ + + + +

Suy ra pcm.

Cch 2:t , ,x a b y a c z b c= + = + = + . Khi :

1 33

2 2

a b c x y x z y z

b c c a a b y x z x z y

+ + + + + + + + + +


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53

6. Gi s ( )n r l s tt c cc im c ta nguyn trn mt ng trn c bn knh

1r> . Chng minh rng: 3 2( ) 6n r r< .

Gii:Xt ng trn bn knh r c cha n im c ta nguyn trn . Ta cn chng minh

3 26n r< .V 1r> v 36 8 > nn ta c th gi s 8n > Gi n im c ta nguyn ni trn l

1 2, ,..., nP P P , nm theo th t ngc chiu kim ng h. Do tng cc cung

1 3 2 4 2, ,..., nPP P P P P bng 4 nn mt trong cc cung 2i iPP+ s c so ln nht l4

n

, v

khng mt tnh tng qut, ta gi s l cung 1 3PP .

Xt tam gicABCni tip trong cung cha gc4

n

. Din tch tam gic ny ln nht khi

hai im A,C trng vi hai u cung, v im B l giao im ca trung trc AC vi

ng trn (khi y, khang cch tBn ng thngACln nht). Ti y, ta c: 0 2& 180CAB BCA ABC

n n

= = =

Do : [ ]

22 sin 2 sin 2 sin

4 4

r r rabc n n n

ABCr r

= =

2 3

3

22 2 2

4

4

r r rrn n n

r n

=

V tam gic1 2 3

PP P ni tip trong cung cha gc4

n

nn theo kt qu trn ta c:

[ ]2 3

1 2 3 3

4rPP P

n

M 1 2 3, ,P P P l cc im c ta nguyn nn gi tr b nht ca [ ]1 2 3PP P l1

2(c th

chng minh iu ny bng nh l Pick). V vy ta c:

[ ]2 3

3 33 2 3 2 21 2 3 3

1 48 2 6

2

rPP P n r n r r

n

<

IV. Bi tp tluyn:1. Cho tam gicABCni tip trong ng trn (O) v 2 AC AB= . Cc ng thng tip xcvi ng trn (O) tiA, Cct nhau ti P. Chng minh rngBPi qua im chnh gia cacungBAC.

2. Cho tam gicABCcIl tm ng trn ni tip, O l tm ng trn ngoi tip v trng

tm G. Gi s rng 090OIA = . Chng minh rngIG song song viBC.


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54

3. (IMO Shortlist) Gi s M,N l cc im nm trong tam gic ABC sao

cho MAB NAC= , MBA NBC= . Chng minh rng:

1.

.

.

.

.

.=++

CBCA

CNCM

BCBA

BNBM

ACAB

ANAM

4. (VMO 1997) Trong mt phng, cho ng trn tm O bn knh R v im P nm trongc trn ( ).OP d R= < Trong tt c cc t gic liABCD ni tip trong ng trn (O) v

c hai ng choACvBD vung gc v ct nhau ti P, hy tm t gic c chu vi ln nhtv t gic c chu vi nh nht. Tnh cc gi tr ln nht v nh nht ny theoR v d.

5. (Bulgaria 2007) Cho tam gic ABCc BC AB AC > > v11

cos cos cos8

A B C + + = . Xt

cc imXthuc BCv YthucACko di v pha Csao cho BX AY AB= = .

a) Chng minh rng2

ABXY= .

b) GiZl im nm trn cung AB ca ng trn ngoi tip tam gic khng cha C

sao cho ZC ZA ZB= + . Hy tnh t sZC

XC YC+

6. Cho tam gic ABCvi BE, CFl cc ng phn gic trong. Cc tia EF, FEct ngtrn ngoi tip tam gic theo th t tiMvN. Chng minh rng:

CMBNANAMCNBM

111111+++=+

7. Cho tam gic ABCni tip ng trn (O). ng trn ( )O nm trong (O) tip xc vi

(O) ti T thuc cungAC(khng cha B). K cc tip tuyn , , AA BB CC ti ( )O . Chng

minh rng: . . . BB AC AA BC CC AB = + .

8. (nh l Thebault) Cho tam gic ABCni tip trong ng trn (O).D l trung im caBC. Gi ( ) ( )1 2,O O l cc ng trn nm trong (O), tip xc vi (O), BCv AD. Khi

ng thng ni tm ca ( ) ( )1 2,O O i quaI. Hy chng minh.

9. (CMO 1988, Trung Quc) ChoABCD l mt t gic ni tip vi ng trn ngoi tip ctm O v bn knh R. Cc tia AB, BC, CD, DA ct ng trn tm O bn knh 2R ln ltti , , , A B C D . Chng minh rng chu vi t gic A B C D khng nh hn hai ln chu vi t

gicABCD.

10. Cho ng trn (O) v dy cungBCkhc ng knh. Tm imA thuc cung lnBCca ng trn 2 AB AC + t gi tr ln nht.11. Lc gic liABCDEFcABFl tam gic vung cn tiA, BCEFl hnh bnh

hnh. 3AD = , 1, 2 2 BC CD DE = + = . Tnh din tch lc gic.


53/121


55

12. Cho tam gicABCni tip ng trn tm O, P v Q l hai im thuc (O) sao cho PQsong song viAB.AD l ng cao ca tam gicABC. Gi a l ng thng simson caim Png vi tam gicABC, b l ng thng simson ca im Qng vi tam gicABC.CMR: (a), (b) vADng quy.

13. Cho n gic u v mt im M chuyn ng trn ng trn ngoi tip a gic y.

Chng minh rng:2

1

n

ii

MA const=

= .


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56

PHN II: BT NG THC HNH HCTRONG KHNG GIANCHNG I: TDIN.

BI 1: C LNG HNH HC.

1. Vi c lng cc yu t ca mt tam gic:

Ta c nhn xt: Mt tam gic u vi cnh bng 1, c ba ng cao bng3

2, din tch

bng3

4, v bn knh (hnh trn ni tip) bng

3

2. Nhn xt ny l xut pht im ca

bi ton sau y vc lng.

Bi ton 1: Cho mt tam gic, di mi cnh ca n khng vt qu 1. Chng minhrng tam gic y c:

a) t nht hai ng cao vi di3

2 .

b) Din tch3

4 .

c) Bn knh ng trn ni tip3

2 .

Gii:a) Trong mt tam gic, ng cao c di khng vt qu di trung tuyn xutpht cng mt nh, nn CM a) ta CM mt kt qu mnh hn: mt tam gic tha mn

iu kin bi ton th c t nht hai trung tuyn vi di3

2

Gi di cc cnh tam gic cho l a, b, c. Lun c th coi rng: 0 b a 1c< .Gi , , ,a b cm m m l cc di ca cc trung tuyn ng vi cnh y. Theo cng thc tnh

trung tuyn, ta c:2 2

2 2 22

2 22 2 2

2

1 32

2 21 2 2

1 322 21 2 2

a

b

a cm b c

b

b am c ac

= + + + =

= + + + =

T suy ra3

2am v

3

2bm .


55/121


57

Nhn xt: Tam gic cn vi di ba cnh bng 1,1,2c c mt ng cao (ng thi l

ng trung tuyn) c di 23

12

c > , khi c kh nh. iu chng t khng

th ci tin hn kt qu pht biu a).

b). 3

2 4

aa hS =

c) Ta c nhn xt: mt tam gic c cha trong mt tam gic (ln hn) ng dngvi n, vi t sng dng 1k ; hn na, nu mt tam gic c cha trong mt tamgic khc (khng cn ng dng), th bn kinh ni tip ca tam gic th nht khng vtqua bn knh ni tip ca tam gic th 2, bi v hnh trn ni tip ca mt tam gic lhnh trn ln nht c cha trong tam gic y.Nu tam gic ABCc cnh ln nht a = BC ln hn 1, th ta xt tam gic ng dng

11k

a= > . Nn ta c th coi rng 1, 1, 1a BC b AC c AB= = = =

Khi , tam gic ABCc cha trong tam gic cong BCD chn bi BD, CD ca haing trn tmB, Cv bn knh 1BC= .(hnh)GiIl trung im ca on ,BC A l hnh chiu vung gc caA ln ,DI r l bn knhni tipca tam gic A BC . Tam gic A BC c cha trong tam gic uBCD ( c bn

knh ni tip bng3

2) nn

3

2r . V vy, chng minh c) ta ch vic chng minh

rng r r . rng, ABC l tam gic cn c cng yBCv cng di ng caoAI = AHnhtam gicABC. Bt r r l h qu ca bi ton 2 sau y:

Bi ton 2: Gi s tam gicABCc din tch

3

4S . CMR tam gic c:a) t nht mt ng cao

b) Bn knh ni tip3

2r .

Bi ton 3: CMR nu c ba ng cao ca mt tam gic l3

2 th tam gic c bn

knh ni tip3

2r .

Bi ton 4:CMR nu tam gic c bn knh ni tip

3

2r th n c t nht 1 ng cao3

2h .

2. Vi c lng cc yu t ca mt tdin:


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58

Trong khng gian, t din u vi di cnh bng 1 c 4 ng cao bng6

3, th tch

bng2

12v bn knh hnh cu ni tip bng

6

12.

T nhn xt trn ta c bi ton:

Bi ton 1: Cho t din vi di mi cnh khng vt qu 1. CMR t din y c t

nht 2 ng cao6

3 .

a) Th tch 212

.

b) Bn knh mt cu ni tip 612

.

Gii:a) Ta c nhn xt: di trung tuyn ca mt t din khng nh hn di ng caoxut pht t cng nh. Vy chng minh ta chng minh mt kt qu mnh hn: Mt t

din tho mn iu kin bi ton phi c t nht hai trung tuyn vi di6

3 .

Trc tin ta biu din di trung tuynAG ca t dinABCD theo cc cnh vi G ltrung im ca tam gicBCD.GiIl trung im CD, ta cBG = 2GI. GiHl hnh chiu caA lnBI. V Hc thnm trong hoc ngoi onBInn ta s dng di i s.

Ta c :

( )

( )

( )

2 2 2 2 2 2 2

2 2 2 2 2

2 2 2 2 2

2 2 2

2 2 2 2

( ) 24 2

2 19 3

( )

1 12 . 2

9 3

2(1) & (2) 3 2 3

3

AH AB BH BG GH AB BG GH BGGH

AH GH AG AB BI BGGH do BI BG

AH AI HI AI HG GI

AG AI BI HG GI do BI GI

AG AI AB BI

= = + = + = = =

= = +

= =

= +

V AI l trung tuyn ca tam gic ACD, BIl trung tuyn ca tam gic BCD, nn theo

cng thc trung tuyn ca mt tam gic ta c:

( )

( )

2 2 2 2

2 2 2 2

12 4

21

2 52

AI AC AD CD

BI BC BD CD

= +

= +

Vy cui cng, t (3), (4), (5) ta suy ra

( ) ( )2 2 2 2 2 2 21

33

AG AB AC AD BC CD DB= + + + +


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59

l cng thc tnh trung tuyn ca mt t din theo cc cnh ca t din y. Trongcng thc ta t:

( ) 2 2 2d A AB AC AD= + + (l tng bnh phng cc cnh xut pht tnhA)

( ) 2 2 2s A BC CD DB= + + ( l tng bnh phng cc cnh ca mt i nhA).

Cng nh trn ta k hiu d(B), d(C), d(D) ln lt l tng bnh phng cc cnh xut

pht tnhB, C, D v k hiu s(B), s(C), s(D) l tng bnh phng cc cnh ca mt idin cc nhB, C, D. K hiu 2kl tng bnh phng 6 cnh ca t din. Hin nhin tac

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2d A s A d B s B d C s C d D s D k + = + = + = + =

v ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 4d A d B d C d D s A s B s C s D k + + + = + + + =

t ( ) ( ) ( ) ( ); ; ;d A k d B k d C k d D k = + = + = + = + th ta c:

( ) ( ) ( ) ( ); ; ;s A k s B k s C k s D k = = = =

0 + + + =

Khng mt tnh tng qut, ta c th coi rng ( ) ( ) ( ) ( )d A d B d C d D

Tc l . Th th ( )4 0 6 + + + =

V 3 0 + + + + = do ( )4 7

Gi &A Bm m l di cc trung tuyn ca t din, xut pht t cc nhA vB. Theo

(5) v (6), ta c

( ) ( ) ( ) ( ) ( )21 1 2 4 2

3 2 83 3 3 3 3A

m d A s A k k k = = + = +

Bi v theo gi thit, di mi cnh ca t din 1 , vy 2 6k . T (8) ta suy

ra6

3Am theo (6) v (7) , ta li c

( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

2 1 1 2 4 2 133 3 3 3 3 3

1 1 1 12

3 3 3 3

Bm d B s B k k k k

k k d B d A

= = + = + +

= + + +

Bi do gi thit ca bi ton, ta c ( ) ( )3& 3d B s A . Thnh th ta c

6

3Em (pcm).

Ta c th kim nghim rng t din c 5 cnh bng 1, cnh th 6 bng a, vi 0a >

kh nh, c 2 ng cao6

3> , v vy khng th ci thin hn kt qu nu phn a)

ca bi ton.

b) Theo bi ton 1, mi mt ca t din c din tch3

4 , do nu ly mt ng vi

mt ng cao6

3 , th c lng c th tch ca t din cho


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60

( )1 3 6 2

. . 1.23 4 3 12

V =

c) Gi S l din tch ton phn ca t din cho, th th 3V rS= vi rl bnh knhmt cu ni tip ca t din y. Ta c bt ng thc b

Trong mt tdin tu vi thtch V v din tch ton phn S, ta c2 3

216 3V S

(9)(Mrng ca bt ng thc 23 3S p lin h din tch S v na chu vi p ca mt tamgic)

Theo b) ta c2

12V , v vy

32 3

3216 3 27

VV S

r = ,

do

3

3 1 6648 28 3

Vr

=

hay6

12r . (pcm)

i n bt ng thc (9), trc tin ta phi chng minh

Bi ton 2: Trong tt c cc t din c cng th tch Vv c yABCcho trc, hy xcnh t din c din tch ton phn nh nht.

Gii: GiD l nh th t ca t din,Hl chn ng cao h tD xung yABC. VHc th nm trong hay ngoi tam gic ABCnn hy xc nh khong cch i s tHn cc ng thngBC, CA, AB nh sau

, , x HA y HB z HC = = =

Vi , , A B C l cc hnh chiu vung gc caHlnBC, CA, AB. i vi du cax, ta

coi rng 0x > nuHvAcng v mt pha i vi ng thngBC,x = 0 nuHnmtrn ng thng BC, 0x < nu Hv Av 2 pha i vi ng thng BC. Tng tvi cch xt du cay vz.Vi a = BC, b = CA, c = AB, c th thy rng trong mi trng hp, ta lun c

ax + by + cz = 2S (10)vi S l din tch tam gic ABC.Theo gi thit ca bi ton, h = DHl khng i, hn na

( ) 2 2 2 2 2 22 .dt DBC BC OA a h x a h a x= = + = +

( ) 2 2 2 2 2 22 .dt DCA CAOB b h y b h b y= = + = +

( )

2 2 2 2 2 22 .dt DAB AB OC c h z c h c z= = + = +

Nh vy, ta phi xc nhx, y, z thoiu kin (10), sao cho biu thc2 2 2 2 2 2 2 2 2 2 2 2T a h a x b h b y c h c z= + + + + +

t gi tr nh nht.S dng bt ng thc Minkowski, ta c

2 2 2 2 2 2 2 2( ) ( ) 4 4 2T ah bh ch ax by cz p h s p h s + + + + + = + = +

ng thc xy ra khi v ch khi


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ax by cz

ah bh ch= =

Tc lx = y = z; khi Tt gi tr nh nht

( ) ( )2 22 2 2 2

min 2T a b c h ax by cz p h s= + + + + + = +

Vip l na chu vi tam gicABC.

Tm li ta c kt lun:Trong tt c cc t din OABCc cng th tch V, v c cng yABCcho trc t dinc din tch ton phn nh nht l t din c chnHl ng cao OHtrng vi tm ni

tip ca tam gicABC. Din tch ton phn nh nht y bng 2 2 20S s p h s= + + trong

s,p l din tch v na chu vi ca tam gicABCv3

.V

h OHs

= =

T bi ton 2, ta chuyn sang

Bi ton 3: Trong cc t din c cng t

Chuyen de Tong Quat Phan 1

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