[TranPhuht.com] de Thi Va Loi Giai VN TST 2005 - 2010

7/28/2019 [TranPhuht.com] de Thi Va Loi Giai VN TST 2005 - 2010

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THI V LI GII

CHN I TUYN QUC GIAD THI OLYMPIC TON QUC T

CA VIT NAM

T NM 2005 N NM 2010


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PHN I*****

BI


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THI CHN I TUYN QUC GIAD THI IMO 2005

*Ngy thi th nht.Bi 1. Cho tam gic ABC c (I) v (O) ln lt l cc ng trn ni tip, ngoi tip.

Gi D, E, F ln lt l tip im ca (I) trn cc cnh BC, CA, AB. Gi , ,A B C ln lt l

cc ng trn tip xc vi hai ng trn (I) v (O) ln lt ti cc im D, K (vi ng

trn ); ti E, M (vi ng trn B ) v ti F, N (vi ng trn C ). Chng minh rng:

1. Cc ng thng , ,DK EM FN ng quy ti P.

2. Trc tm ca tam gic DEF nm trn on OP.

Bi 2. Trn mt vng trn c n chic gh c nh s t 1 n n. Ngi ta chn rak chic gh. Hai chic gh c chn gi l k nhau nu l hai chic gh c chn lin

tip. Hy tnh s cch chn ra k chic gh sao cho gia hai chic gh k nhau, khng c t

hn 3 chic gh khc.Bi 3. Tm tt c cc hm s :f tha mn iu kin:

3 3 3 3 3 3( ) ( ( )) ( ( )) ( ( ))f x y z f x f y f z+ + = + +

*Ngy thi th hai.Bi 4. Chng minh rng:

3 3 3

3 3 3

3

( ) ( ) ( ) 8

a b c

a b b c c a+ +

+ + +

trong , ,a b c l cc s thc dng.Bi 5. Cho s nguyn t ( 3)p p > . Tnh:

a)

12 22

1

22

p

k

k kS

p p

=

=

nu 1 (mod 4)p .

b)

122

1

p

k

kS

p

=

=

nu 1 (mod8)p .

Bi 6. Mt s nguyn dng c gi l s kim cng 2005nu trong biu din

thp phn ca n c 2005 s 9 ng cnh nhau lin tip. Dy ( ) , 1, 2,3,...n

a n = l dy tng

ngt cc s nguyn dng tha mn na nC< (C l hng s thc dng no ).

Chng minh rng dy s ( ) , 1, 2,3,...na n = cha v hn s kim cng 2005.


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* Ngy thi th nht.

Bi 1.Cho tam gic ABC c H l trc tm. ng phn gic ngoi ca gc BHC ct cccnh AB, AC ln lt ti D v E. ng phn gic trong ca gc BAC ct ng trn ngoi

tip tam gic ADE ti im K. Chng minh rng ng thng HK i qua trung im ca BC.

Bi 2. Hy tm tt c cc cp s t nhin ( );n k vi n l s nguyn khng m v k l

s nguyn ln hn 1 sao cho s : 2006 2 517 4.17 7.19n n nA = + + c th phn tch c thnh

tch ca k s nguyn dng lin tip.

Bi 3. Trong khng gian cho 2006 im m trong khng c 4 im no ngphng. Ngi ta ni tt c cc im li bi cc on thng. S t nhin m gi l s tt nu

ta c th gn cho mi on thng trong cc on thng ni bi mt s t nhin khng

vt qu m sao cho mi tam gic to bi ba im bt k trong s cc im u c hai

cnh c gn bi hai s bng nhau v cnh cn li gn bi s ln hn hai s .

Tm s tt c gi tr nh nht.

* Ngy thi th hai .

Bi 4. Chng minh rng vi mi s thc , , [1;2]x y z , ta lun c bt ng thc sau :

1 1 1( )( ) 6( )

y zx y z

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

+ + + .

Hi ng thc xy ra khi v ch khi no ?

Bi 5. Cho tam gic ABC l tam gic nhn, khng cn, ni tip trong ng trn tmO bn knh R. Mt ng thng d thay i sao cho d lun vung gc vi OA v lun ct cc

tia AB, AC. Gi M, N ln lt l giao im ca ng thng d v cc tia AB, AC. Gi s cc

ng thng BN v CN ct nhau ti K; gi s ng thng AK ct ng thng BC.

1. Gi P l giao ca ng thng AK v ng thng BC. Chng minh rng ng trn

ngoi tip ca tam gic MNP lun i qua mt im c nh khi d thay i.

2. Gi H l trc tm ca tam gic AMN. t BC = a v l l khong cch t im A n HK.

Chng minh rng ng thng HK lun i qua trc tm ca tam gic ABC.

T suy ra: 2 24l R a . ng thc xy ra khi v ch khi no?

Bi 6. Cho dy s thc ( )na c xc nh bi:

0 1

1 11, ( )

2 3n n na a a

a+= = + vi mi n = 1, 2, 3,

Chng minh rng vi mi s nguyn n, s2

3

3 1n nA

a=

l mt s chnh phng v n c t

nht n c nguyn t phn bit.


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*Ngy thi th nht.

Bi 1. Cho hai tp hp A,B l tp hp cc s nguyn dng tha mn A B n= =

(vi n l s nguyn dng) v c tng cc phn t bng nhau. Xt bng vung n n .Chng minh rng ta c th in vo mi vung ca bng mt s nguyn khng m

tha mn ng thi cc iu kin:

i/ Tng ca cc phn t mi hng l cc phn t ca tp A.

ii/ Tng ca cc phn t mi ct l cc phn t ca tp B.

iii/ C t nht 2( 1)n k + s 0 trong bng vi k l s cc phn t chung ca A v B.

Bi 2. Cho tam gic nhn ABC vi ng trn ni tip I. Gi ( )ak l ng trn c

tm nm trn ng cao ca gc A, i qua im A v tip xc trong vi ng trn (I) ti 1A .

Cc im 1 1,B C xc nh tng t .

1/ Chng minh 1 1 1, ,AA BB CC ng qui ti P.

2/ Gi ( ), ( ), ( )a b cJ J J ln lt l cc ng trn i xng vi ng trn bng tip

cc gc A, B, C ca tam gic ABC qua trung im BC, CA, AB.

Chng minh P l tm ng phng ca 3 ng trn ni trn.

Bi 3. Cho tam gic ABC. Tm gi tr nh nht ca biu thc sau:

2 2 2 2 2 2

2 2 2

cos cos cos cos cos cos

2 2 2 2 2 2cos cos cos

2 2 2

A B B C C A

S C A B= + + .

*Ngy thi th hai.

Bi 4. Tm tt c cc hm s lin tc :f tha mn:

2 1( ) ( )3 9

xf x f x= + + vi mi x .

Bi 5. Cho A l tp con cha 2007 phn t ca tp: {1, 2, 3,..., 4013, 4014} tha mn

vi mi ,a b A th a khng chia ht cho b. Gi mA l phn t nh nht ca A.

Tm gi tr nh nht ca mA vi A tha mn cc iu kin trn.

Bi 6. Cho a gic 9 cnh u (H). Xt ba tam gic vi cc nh l cc nh ca a gic(H) cho sao cho khng c hai tam gic no c chung nh.

Chng minh rng c th chn c t mi tam gic 1 cnh sao cho 3 cnh ny bng nhau.


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*Ngy thi th nht.

Bi 1. Trong mt phng cho gc xOy. Gi M, N ln lt l hai im ln lt nm trncc tia Ox, Oy. Gi d l ng phn gic gc ngoi ca gc xOy v I l giao im ca trung

trc MN vi ng thng d. Gi P, Q l hai im phn bit nm trn ng thng d sao

choIM IN IP IQ= = = , gi s K l giao im ca MQ v NP.

1. Chng minh rng K nm trn mt ng thng c nh.

2. Gi d1 l ng thng vung gc vi IM ti M v d2 l ng thng vung gc vi IN

ti N. Gi s cc ng thng d1, d2 ct ng thng d ti E, F. Chng minh rng cc ng

thng EN, FM v OK ng quy.

Bi 2. Hy xc nh tt c cc s nguyn dng m sao cho tn ti cc a thc vi hs thc ( ), ( ), ( , )P x Q x R x y tha mn iu kin:

Vi mi s thc a, b m 2 0ma b = , ta lun c ( ( , ))P R a b a= v ( ( , ))Q R a b b= .

Bi 3. Cho s nguyn n > 3. K hiu T l tp hp gm n s nguyn dng u tin.Mt tp con S ca T c gi l tp khuyt trong T nu S c tnh cht: Tn ti s nguyn

dng c khng vt qu2

sao cho vi 1 2,s s l hai s bt k thuc S ta lun c 1 2s s c .

Hi tp khuyt trong T c th c ti a bao nhiu phn t ?

*Ngy thi th hai.

Bi 4. Cho m, n l cc s nguyn dng. Chng minh rng (2 3) 1nm + + chia ht cho

6m khi v ch khi 3 1n + chia ht cho 4m.

Bi 5. Cho tam gic ABC nhn, khng cn c O l tm ng trn ngoi tip.Gi AD, BE, CF l cc ng phn gic trong ca tam gic. Trn cc ng thng AD, BE, CF

ln lt ly cc im L, M, N sao choAL BM CN

kAD BE CF

= = = (k l mt hng s dng).

Gi (O1), (O2), (O3) ln lt l cc ng trn i qua L, tip xc vi OA ti A ; i qua M, tip

xc vi OB ti B v i qua N, tip xc vi OC ti C.

1. Chng minh rng vi1

2k= , ba ng trn (O1), (O2), (O3) c ng hai im chung

v ng thng ni hai im chung i qua trng tm tam gic ABC.2. Tm tt c cc gi tr k sao cho 3 ng trn (O1), (O2), (O3) c ng hai im chung.

Bi 6. K hiu M l tp hp gm 2008 s nguyn dng u tin. T tt c cc sthuc M bi ba mu xanh, vng, sao cho mi s c t bi mt mu v mi mu uc dng t t nht mt s. Xt cc tp hp sau:

31 {( , , ) ,S x y z M = trong x, y, z c cng mu v ( ) 0 (mod 2008)}x y z+ + ;

32 {( , , ) ,S x y z M = trong x, y, z i mt khc mu v ( ) 0 (mod 2008)}x y z+ + .

Chng minh rng 1 22 S S> . (K hiu3 l tch - cc M M M ) .


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*Ngy thi th nht.Bi 1. Cho tam gic nhn ABC ni tip ng trn (O). Gi 1 1 1, ,A B C v 2 2 2, ,A B C

ln lt l cc chn ng cao ca tam gic ABC h t cc nh A, B, C v cc im i xngvi 1 1 1, ,A B C qua trung im ca cc cnh , ,BC CA AB . Gi 3 3 3, ,A B C ln lt l cc giao

im ca ng trn ngoi tip cc tam gic 2 2 2 2 2 2, ,AB C BC A CA B vi (O).

Chng minh rng: 1 3 1 3 1 3, ,A A B B C C ng quy.

Bi 2. Cho a thc 3 2( ) 1P x rx qx px= + + + trong , ,p q r l cc s thc v 0r > .

Xt dy s ( )na xc nh nh sau:

21 2 3

3 2 1

1, ,

. . . , 0n n n n

a a p a p q

a p a q a r a n+ + +

= = =

=

Chng minh rng: nu a thc ( )P x c mt nghim thc duy nht v khng c

nghim bi th dy s ( )na c v s s m.

Bi 3. Cho cc s nguyn dng ,a b sao cho ,a b v ab u khng phi l s chnh

phng. Chng minh rng trong hai phng trnh sau:2 2

2 2

1

1

ax by

ax by

=

=

c t nht mt phng trnh khng c nghim nguyn dng.

*Ngy thi th hai.

Bi 4. Tm tt c cc s thc r sao cho bt ng thc sau ng vi mi a, b, c dng: 31

2

a b cr r r r

b c c a a b

+ + + + + + +

Bi 5. Cho ng trn (O) c ng knh AB v M l mt im bt k nm trong (O),M khng nm trn AB. Gi N l giao im ca phn gic trong gc M ca tam gic AMB ving trn (O). ng phn gic ngoi gc AMB ct cc ng thng NA, NB ln lt ti P,Q. ng thng MA ct ng trn ng knh NQ ti R, ng thng MB ct ng trnng knh NP ti S v R, S khc M.

Chng minh rng: ng trung tuyn ng vi nh N ca tam gic NRS lun i quamt im c nh khi M di ng pha trong ng trn.

Bi 6. Mt hi ngh ton hc c tt c 6 4n + nh ton hc phi hp vi nhau ng2 1n + ln ( )1n . Mi ln hp, h ngi quanh mt ci bn 4 ch v n ci bn 6 ch, cc v

tr ngi chia u khp mi bn. Bit rng hai nh ton hc ngi cnh hoc i din nhau mt cuc hp ny th s khng c ngi cnh hoc i din nhau mt cuc hp khc.

a/ Chng minh rng Ban t chc c th xp c ch ngi nu 1n = .b/ Hi rng Ban t chc c th sp xp c ch ngi c hay khng vi mi 1n > ?


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* Ngy thi th nht.

Bi 1. Cho tam gic ABC khng vung ti A c ng trung tuyn AM. Gi D l mt

im di ng trn ng thng AM. Gi 1 2( ), ( )O O l cc ng trn i qua D, tip xc vi

BC ln lt ti B v C. Gi P, Q ln lt l giao im ca ng thng AB vi ng trn

1( )O , ng thng AC vi ng trn 2( )O . Chng minh rng:

1. Tip tuyn ti P ca 1( )O v tip tuyn ti Q ca 2( )O phi ct nhau ti mt im.

Gi giao im l S.

2. im S lun di chuyn trn mt ng thng c nh khi D di ng trn AM.

Bi 2. Vi mi s n nguyn dng, xt tp hp sau :

{ }11( ) 10( ) |1 , 10k hnT k h n n k h= + + + .Tm tt c gi tr ca n sao cho khng tn ti , ;na b T a b sao cho ( )a b chia ht cho 110.

Bi 3. Gi mt hnh ch nht c kch thc 1 2 l hnh ch nht n v mt hnhch nht c kch thc 2 3 , b i 2 gc cho nhau (tc l c 4 vung nh) l hnh chnht kp. Ngi ta ghp kht cc hnh ch nht n v hnh ch nht kp ny li vi nhau

c mt bng hnh ch nht c kch thc l 2008 2010 .Tm s b nht cc hnh ch nht n c th dng ghp.

* Ngy thi th hai.

Bi 4. Cho , ,a b c l cc s thc dng tha mn iu kin: 1 1 116( )a b ca b c

+ + + + .

Chng minh rng:

3 3 3

1 1 1 8

9( 2( )) ( 2( )) ( 2( ))a b a c b c b a c a c b+ +

+ + + + + + + + +.

Hi ng thc xy ra khi no?

Bi 5: Trong mt hi ngh c n nc tham gia, mi nc c k i din ( )1n k> > .

Ngi ta chia .n kngi ny thnh n nhm, mi nhm c k ngi sao cho khng c haingi no cng nhm n t cng mt nc.

Chng minh rng c th chn ra mt nhm gm n ngi sao cho h thuc cc nhm khc

nhau v n t cc nc khc nhau.

Bi 6: Gi nS l tng bnh phng cc h s trong khai trin ca nh thc (1 )n+ ,

trong n l s nguyn dng; x l s thc bt k.

Chng minh rng: 2 1nS + khng chia ht cho 3 vi mi n.


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PHN II*****

LI GII


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LI GII THI

CHN I TUYN QUC GIA D THI IMO 2005

Bi 1. Cho tam gic ABC c (I) v (O) ln lt l cc ng trn ni tip, ngoi tip.

Gi D, E, F ln lt l tip im ca (I) trn cc cnh BC, CA, AB. Gi , ,A B C ln ltl cc ng trn tip xc vi hai ng trn (I) v (O) ln lt ti cc im D, K (vi ng

trn A ); ti E, M (vi ng trn B ) v ti F, N (vi ng trn C ). Chng minh rng:

1. Cc ng thng DK, EM, FN ng quy ti P.2. Trc tm ca tam gic DEF nm trn on OP.

1. Trc ht, ta s chng minh b sau:Cho ba ng trn (O1), (O2), (O3) c bn

knh i mt khc nhau; A, B, C ln lt l tm vt ca cc cp ng trn (O1) v (O2), (O2) v(O3), (O3) v (O1).Chng minh rng nu trong cc tm v t , cba tm v t ngoi hoc hai tm v t trong, mttm v t ngoi th A, B, C thng hng.

*Chng minh:

Gi 1 2 3, ,R R R ln lt l bn knh ca cc ng

trn 1 2 3( ), ( ), ( )O O O , cc gi tr 1 2 3, ,R R R ny i

mt khc nhau.

Theo tnh cht v tm v t, ta c: 1 1

22

( 1)aAO R

RAO

= .

Tng t: 2 2

33

( 1)bBO R

RBO= , 3 3

11

( 1)cCO R

RCO= , trong

, mi s , ,a b c nhn gi tr l 0 (khi n l tm v t ngoi) hoc 1 (khi n l tm v t trong).

Theo gi thit trong a, b, c c ba gi tr l 0 hoc hai gi tr 0, mt gi tr 1. T :

31 2

2 3 1

. . 1COAO BO

AO BO CO= , theo nh l Menelaus o cho tam gic 1 2 3O O O , ta c: A, B, C thng hng.

B c chng minh.*Tr li bi ton:Gi P l tm v t trong ca hai ng trn (O) v (I). D thy: D l im tip xc ngoi

ca A v (I) nn cng chnh l tm v t trong ca hai ng trn ny; K l im tip xc trong

ca hai ng trn A v (O) nn l tm v t ngoi ca hai ng trn ny. Theo b trn th

', ,P D K thng hng hay ng thng DK i qua P. Tng t, cc ng thng EM v FN cng

i qua P; tc l ba ng thng DK, EM, FN ng quy v im P chnh l im P ca bi.

B

C

O1O2

O3


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2. Ta chng minh b sau:Cho tam gic ABC c (O), (I) ln lt l tm ng trn ngoi tip, ni tip ca tam gic

ABC. ng trn (I) tip xc vi cc cnh BC, CA, AB ln lt ti D, E, F. Chng minh rngtrc tm H ca tam gic DEF nm trn ng thng OI.

* Chng minh:Gi M, N, P ln lt l trung im ca cc

on EF, FD, DE. D thy AI l trung trc caon EF nn M thuc ng thng AI hay A, M, Ithng hng. Tng t: B, N, I v C, P, I cngthng hng. Xt php nghch o tm I, phng

tch 2r vi r l bn knh ng trn (I).D thy: tam gic IEA vung ti E c EM l

ng cao nn: 2 2.IM IA IE r= = , suy ra:

: A . Tng t: : ,N B P C .Do : : MNP ABC . Gi E l tmng trn ngoi tip ca tam gic MNP th

:E O , suy ra: E, I, O thng hng.

Hn na, I l tm ng trn ngoi tip tam gic DEF, E l tm ng trn ngoi tip tamgic MNP cng chnh l tm ng trn Euler ca tam gic DEF ny nn E, I, H thng hng.

T suy ra H, I, O thng hng. B c chng minh.

* Tr li bi ton:

Gi H l trc tm tam gic DEF th theo b trn:H, I, O thng hng.Theo cu 1/, im P nm trn on OI.Suy ra: 4 im H, I, P, O thng hng.T suy ra trc tm H ca tam gic DEF nm trnng thng OI.Ta c pcm.

K

PF

D

O

B C

FO

C


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Bi 2. Trn mt vng trn c n chic gh c nh s t 1 n n. Ngi ta chn ra kchic gh. Hai chic gh c chn gi l k nhau nu l hai chic gh c chn lin tip.Hy tnh s cch chn ra k chic gh sao cho gia hai chic gh k nhau, khng c t hn 3chic gh khc.

*Trc ht, ta chng minh b sau:Cho n im phn bit nm trn ng thng c t mt trong hai mu, xanh hoc

tha mn cc iu kin sau:- C ng k im c t mu xanh.- Gia hai im mu xanh lin tip c t nht p im c t mu (tnh t tri sang).- bn phi im mu xanh cui cng c t nht p im c t mu .

Khi , s cch t mu l: kn kpC .

*Chng minh: nh s cc im cho l 1,2,3,..., n . t tng ng mi cch t mu

vi mt b k cc s nguyn dng 1 2( , ,..., )ki i i trong 1 2, ,..., ki i i l cc im c t mu xanh.

D thy tng ng ni trn chnh l mt song nh t tp cc cch t mu n tp hp T sau:

1 2 1{( , ,..., ) | {1,2,..., }, 1, ; , 1, 1}k s s sT i i i i n p s k i i p i k += = > = .

Xt nh x sau:

1 2 1' {( , ,..., ) | {1,2,..., }, , 1, }k t t t T T j j j j n kp j j t k + = > = .

Ta s chng minh nh x ny l mt song nh.Tht vy:

*Xt mt b 1 2( , ,..., ) 'kj j j T . Khi , ta xt tip b: 1 2 3( , , 2 ,..., ( 1) )kj j p j p j k p+ + + .

Do 1 tj n kp nn phn t ln nht ca b ny l ( 1)kj k p+ c gi tr khng vt qu

( 1)n kp k p n p + = . T suy ra: ( 1) {1,2,..., }, 1tj t p n kp t+ .Hn na: [ ] [ ]1 1( 1) ( )t t t t j tp j t p j j p p+ ++ + = + > .

T suy ra b 1 2 3( , , 2 ,..., ( 1) )kj j p j p j k p T+ + + .

Do , tng ng ny l mt ton nh.

*Xt b 1 2( , ,..., )ki i i T . Khi , hon ton tng t trn, ta cng chng minh c b

1 2 2( , , 2 ,..., ( 1) ) 'ki i p i p i k p T .

Ta s chng minh rng nu c hai b khc nhau 1 2 1 2( , ,..., ), ( ' , ' ,..., ' )k ki i i i i i T th cc b tng

ng thuc T= 'T ca chng cng phi khc nhau. Nhng iu ny l hin nhin do hai b ny l

khc nhau nn tn ti ch s s sao cho 's si i , khi ( 1) ' ( 1)s si s p i s p .Suy ra, tng ng ny cng l mt n nh.Vy tng ng 'T T l mt song nh.Nhn xt trn c chng minh.

Do : | | | ' | kn kpT T C= = . B c chng minh.


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*Tr li bi ton:Ta xt tng qut gi tr 3 trong bi bi gi tr p tng ng vi b trn.

nh s cc gh trong bi theo chiu kim ng h l 1 2, ,..., nA A A (xem nh l cc im nm

trn mt vng trn) ; mi gh c chn xem nh c t mu xanh v khng c chn xemnh c t mu ; gi X l tp hp tt c cc cch t mu k im trong n im cho tha

mn bi.

Xt phn hoch: ' '' ' "X X X X X X= = + .

trong 'X l cch t mu tha mn c mt im c t mu xanh thuc 1 2 3{ , , ,..., }pA A A A v

'' \ 'X X X= , khi r rng, vi mi phn t thuc "X th khng c im no c t muxanh thuc 1 2 3{ , , ,..., }pA A A A , tc l mi im trong tp ny u c t mu . Ta ct ng

trn ngay ti im 1,p pA A + th r rng s to c mt ng thng tha mn tt c iu kin

ca b nu trn, suy ra: '' kn kpX C= . Ta ch cn cn tnh s phn t ca 'X .

Xt tp hp 'i

X trong mi phn t ca 'i

X c ng mt imi

A c t mu xanh,

1,i p= ; khi r rng ' ' ,i jX X i j = v1

' 'p

ii

X X=

= .

Vi mi 1,i p= , theo b trn, ta thy:1 11 ( 1) 1

k kn p k p n kpC C

= , tc l cc tp 'iX ny c cng s phn t. Suy ra:

11'

kn kpX pC

= .

Do : 1 1' ''k kn kp n kpX X X C pC

= + = + .

Thay 3p = , ta c s cch chn gh tng ng trong bi l 13 3 13k kn k n k C C

+ .

Vy s cch chn gh tha mn tt c cc iu kin ca bi l: 13 3 13k kn k n k C C

+ .


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Bi 3. Tm tt c cc hm s :f tha mn iu kin:3 3 3 3 3 3( ) ( ( )) ( ( )) ( ( ))f x y z f x f y f z+ + = + +

* Trc ht, ta s chng minh b sau:Vi mi s nguyn dng ln hn 10, lp phng ca n u c th biu din c di

dng tng ca 5 lp phng ca cc s nguyn khc c gi tr tuyt i nh hn n.

* Tht vy:Ta cn tm mi lin h vi s 10n > trong tng trng hp n chn v n l.

- Vi n l s l, t 2 1n k= + .Ta cn tm mt ng thc ng vi mi k m trong 2 1k+ l biu thc c gi tr tuyt

i ln nht, cc biu thc cn li phi l nh thc bc nht c h s ca k ln nht l 2. Khi

kh 38k xut hin trong ( )3

2 1k+ , ta chn ( )3

2 1k ; ta thy vn cn s hng cha k bc

hai trong , ta chn tip hai biu thc khc c cha k cng hai hng s bng cch dng tham snh sau:

Gi s hai biu thc cn tm c dng ( ), ( ); ,ak b ak b a b+ v hai s cn tm l

,c d , tc l:

( ) ( ) ( )

3 3 3 3 3 3

2 2 2 3 3 3

2 2 3 3 3

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

24 2 6 2

(24 6 ) (2 2 ) 0

k k ak b ak b c d

k a bk b c d

k a b b c d

+ = + + +

+ = + + +

+ =

Ta cn chn , , ,a b c d sao cho 2 3 3 34, 2 2a b b c d = + + = trong 2a .

D thy 2a

v nu 2a=

th t

2

4 1a b b= =

, trng vi biu thc cn nh gi; do ,1, 4a b= = , suy ra: 3 3 126c d+ = , ta chn c 5, 1c d= = .

Do :3 3 3 3 3 3(2 1) (2 1) ( 4) (4 ) ( 5) ( 1)k k k k + = + + + + + (1)

Th li, ta thy biu thc ny ng vi mi k.- Vi n l s chn, t 2 2n k= + .

Lp lun hon ton tng t, ta c c ng thc sau:3 3 3 3 3 3(2 2) (2 2) ( 8) (8 ) ( 10) ( 2)k k k k + = + + + + + (2)

B c chng minh.

*Tr li bi ton:Trong ng thc , thay 0x y z= = = , ta c:

3 2(0) 3 (0) (0) 0 3 (0) 1f f f f= = = .

Do hm ny ch ly gi tr trn :f nn khng th c 23 (0) 1f = , tc l (0) 0f = .


15/87

15

- Thay 0y z= = , ta c: 3 3 3 3 3( ) ( ( )) ( (0)) ( (0)) ( ( ))f x f x f f f x= + + = .

- Li thay z= , ta c:3 3 3 3 3 3( ) ( ( )) ( ( )) ( ( )) ( ( )) ( ( )) 0 ( ) ( ),f x f x f y f y f y f y f y f y y= + + + = = .

Ta s chng minh rng: 3: ( ) . (1)k f k k f = bng quy np. (*)

*Tht vy:- Vi 1k= , trong gi thit, thay 1, 0x y z= = = , ta c 3 3(1) (1) ( 1) (1)f f f f= =

- Vi 2k= , trong gi thit, thay 1, 0x y z= = = , ta c 3 3(2) 2 (1) ( 2) 2 (1)f f f f= =

- Vi 3k= , trong gi thit, thay 1y z= = = , ta c 3 3(3) 3 (1) ( 3) 3 (1)f f f f= =

- Thay 2, 0x y z= = = , ta c 3 3 3 3(8) (2) (2 (1)) 8 (1) ( 8) 8 (1)f f f f f f= = = = .

- Thay 2, 1, 0x y z= = = , ta c 3 3 3 3(9) (2) (1) 9 (1) ( 9) 9 (1)f f f f f f= + = = .

- Thay 2, 1x y z= = = , ta c: 3 3 3 3(10) (2) 2 (1) 10 (1) ( 10) 10 (1)f f f f f f= + = = .

- Thay 2, 1, 0x y z= = = , ta c: 3 3 3 3(7) (2) (1) 7 (1) ( 7) 7 (1)f f f f f f= = = .

- Thay 2, 1x y z= = = , ta c: 3 3 3 3(6) (2) 2 (1) 6 (1) ( 6) 6 (1)f f f f f f= = = .

- Trong ng thc (1) ca b trn, ta thay 2k= , suy ra:3 3 3 3 3 35 3 6 2 ( 5) ( 1)= + + + + hay 3 3 3 3 3 3(5 5 1 ) (3 6 2 )f f+ + = + + .

3 3 3 3 3 3 32 (5) (1 ) (3) (6) (2) (5) 5 (1) ( 5) 5 (1)f f f f f f f f f + = + + = = .

- Trong ng thc (2) ca b trn, ta thay 1k= , suy ra:3 3 3 3 3 34 0 9 7 ( 10) ( 2)= + + + + hay 3 3 3 3 3 3(4 10 2 ) (9 7 0 )f f+ + = + + .

3 3 3 3 3 3 3 3(4) (10) (2) (9) (7) (0) (4) 4 (1) ( 4) 4 (1)f f f f f f f f f f + + = + + = = .

Nh th, ta chng minh c (*) ng vi mi 10k .

Vi 10k > , xt 0k> th theo b trn, lp phng ca k u c th biu din c

di dng tng ca 5 lp phng khc c gi tr tuyt i nh hn n.

Hn na, d thy rng vi , , , , ,a b c d e f tha 3 3 3 3 3 3a b c m n p+ + = + + v ta c:3 3 3 3 3( ) (1), ( ) (1), ( ) (1), ( ) (1), ( ) (1)f b bf f c cf f m mf f n nf f p pf= = = = = th 3( ) (1)f a af= .

T , suy ra 3( ) (1), 10f k kf k= > .

Vi 10k< th 3 3( ) ( ) ( (1)) (1)f k f k kf kf= = = .

Do , theo nguyn l quy np (*) c chng minh.

Mt khc, trong gi thit cho, thay 1, 0x y z= = = , ta c: 3(1) (1) (1) 1 (1) 0f f f f= = = .

- Nu (1) 1f = th ( ) ,f k k k= , th li thy tha.

- Nu (1) 0f = th ( ) 0,f k k= , th li thy tha.

- Nu (1) 1f = th ( ) ,f k k k= , th li thy tha.

Vy tt c hm s cn tm l ( ) ,f k k k= ; ( ) ,f k k k= v ( ) 0,f k k= .


16/87

16

Bi 4. Chng minh rng:3 3 3

3 3 3

3

( ) ( ) ( ) 8

a b c

a b b c c a+ +

+ + +

trong , ,a b c l cc s thc dng.

*Trc ht, ta s chng minh b sau:Nu , , ,a b c d l cc s thc dng c tch bng 1 th:

2 2 2 2

1 1 1 11

(1 ) (1 ) (1 ) (1 )a b c d + + +

+ + + +.

Tht vy:

Ta thy vi hai s thc dng ty th:2 2

1 1 2

(1 ) (1 ) 1y xy+

+ + +(*)

2 22 2 2

2

2 2 2

2 2 3 3 2 2

2 2 2 2 2 2

( 1) ( 1) 1(*) ( 1) ( 1) (1 ) ( 1)

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

( 2 2 2) ( 2 2 2 )

( 2 2 2 ) (2 2 2 ) 1

1

x yx y xy xy x y

xy x y xyx y x y xy xy x y xy x y

x y x y x y xy x y xy xy

x y x y x y xy xy xy x y

x

+ + + + + + + + + +

+ + + + + + + + + + + + + + +

+ + + + + + + + +

+ + + + + + + + +

+ 2 2 2 2 2 2( ) 2 ( ) (1 ) 0y x y xy x y xy x y xy+ + +

Do :

2 2 2 2

1 1 1 1 1 1 2 21

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

ab cd ab cd

a b c d ab cd ab cd abcd ab cd

+ + + ++ + + + = = =

+ + + + + + + + + + +.

Do b c chng minh.

ng thc xy ra khi v ch khi 1a b c d = = = = .Trong b trn, thay , , , 1a x b y c z d = = = = , ta c kt qu sau:

Vi x, y, z l cc s thc dng v 1xyz = th:

2 2 2

1 1 1 3

(1 ) (1 ) (1 ) 4x y z+ +

+ + +. ng thc xy ra khi v ch khi 1y z= = = .

*Tr li bi ton cho:

t , , , , 0; 1b c a

y z x y z xyza b c

= = = > = .

BT cho ban u tng ng vi:

3 3 33 3 3

1 1 1 3 1 1 1 3

8 (1 ) (1 ) (1 ) 8(1 ) (1 ) (1 )b c a x y za b c

+ + + + + + ++ + +

.

Theo BT Cauchy cho cc s dng, ta c:


17/87

17

33 3 6 2 3 2

1 1 1 1 3 1 1 3 1 13 . .

(1 ) (1 ) 8 8.(1 ) 2 (1 ) (1 ) 4 (1 ) 16x x x x x x+ + =

+ + + + + +.

Hon ton tng t:

3 2

1 3 1 1.

(1 ) 4 (1 ) 16y y

+ +

,3 2

1 3 1 1.

(1 ) 4 (1 ) 16z z

+ +

.

Cng tng v cc BT ny li, ta c:

3 3 3 2 2 2

1 1 1 3 1 1 1 3.

(1 ) (1 ) (1 ) 4 (1 ) (1 ) (1 ) 16x y z x x x

+ + + + + + + + + +

.

Ta cn chng minh:

2 2 2

3 1 1 1 3 3.

4 (1 ) (1 ) (1 ) 16 8x x x

+ + + + +

2 2 2

1 1 1 3

(1 ) (1 ) (1 ) 4x x x + +

+ + +

vi x, y, z tha mn cc iu kin nu. (**)Theo b trn th (**) ng.

Vy ta c pcm.Du ng thc xy ra khi v ch khi 1x y z a b c= = = = = .


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18

Bi 5. Cho s nguyn t ( 3)p p > . Tnh:

a.1

2 22

1

22

p

k

k kS

p p

=

=

nu 1 (mod4)p .

b.1

22

1

p

k

kP

=

=

nu 1 (mod8)p .

*Trc ht, ta s chng minh hai b sau:

(1)B 1: Vi p l s nguyn t tha 1 (mod4)p th mi s t nhin a vi: 112

pa

s tn ti duy nht s t nhin b tha1

12

pb p

+ v: 2 2 0(mod )a b p+ .

*Chng minh: Theo nh l Wilson: ( 1)! 1(mod )p p .

Vi mi1

1,2,3,...,2

pk

= , ta thy: 2(mod ) ( ) (mod )p k k p k p k k p .

Kt hp vi gi thit1

1(mod 4) 22

pp

, ta c:

2 212 1 11 ( 1)! ( 1) . ! ! (mod )

2 2

p p pp

. t 2

1! 1(mod )

2

pp

=

.

Vi mi 112

pa , ta chn 1 12

p b p+ tha 2 2 2. (mod )b a p , d thy b tn ti v duy

nht. Khi : 2 2 2 2(1 ) 0(mod )a b a p+ + . B c chng minh.

(2)B 2:Vi x l s thc bt k th [ ] [ ]2 2 x bng 1 nu

1{ } 1

2x < v bng 0 nu

10 { }

2x < .

*Chng minh: Ta c: [ ] { }x x= + . Suy ra:

[ ] [ ] [ ] [ ] [ ] [ ] [ ]2 2 2[ ] 2{ } 2 [ ] { } 2{ } 2 { } 2{ }x x x x x x x x x = + + = = . Do :

-Nu 1 { } 12

x < th [ ] [ ] [ ]1 2{ } 2 2{ } 1 2 2 1x x x x < = = .

- Nu1

0 { }2

x < th [ ] [ ] [ ]0 2{ } 1 2{ } 0 2 2 0x x x x < = = .

B c chng minh. *Tr li bi ton:


19/87

19

1. Ta thy tng cho l:

12 22

1

22

p

k

k kS

p p

=

=

c ng

1

2

p s hng.

Theo b 2 th mi s hng trong tng nhn hai gi tr l 0 hoc 1. (1)

Theo b 1 th vi mi s t nhin a tha1

12

pa

th tn ti duy nht s t nhin b tha

11

2

pb p

+ sao cho 2 2 2 20(mod ) ( ) 0(mod )a b p a p b p+ + ; do , tn ti duy nht

s t nhin 'a tha1

1 '2

a

sao cho 2 2' 0(mod )a a p+ .

Gi ,y ln lt l s cc s d ca php chia 2k cho p (1

12

pk

) c gi tr ln hn

1

2

p

v nh hn1

2

p . Theo nhn xt trn th y= , hn na

1 1

2 4

p px y x y

+ = = = . (2)

T (1) v (2), ta c: 1.1 .04

pS x y = + = .

Do , tng cn tm l1

4

p .

2. Do 1 (mod8)p nn tn ti a sao cho 2 2(mod )a p .

(ta cng thy rng 1 (mod8) 1 (mod 4)p p ).

Ta c:

1 1 1 12 2 2 2 2 2 22 2 2 2

1 1 1 1

2 2 22

p p p p

k k k k

k k k k k k k P S

p p p p p p p

= = = =

= = =

.

Ta cn tnh:1 1 1 1 12 2 2 2 2 2 2 2 22 2 2 2 2

1 1 1 1 1

2 2 2 2p p p p p

k k k k k

k k k k k k k k k

p p p p p p p p p

= = = = =

= =

,

trong 1 (mod8)p .

Theo nhn xt trn th tp hp cc s d khi chia 21

,12

pk k

cho p trng vi tp hp cc s

d khi chia 21

2 ,12

pk k

cho p, tc l:

12 22

1

20

p

k

k k

p p

=

=

, suy ra:

1 12 2 2 22 2

1 1

2 1

24

p p

k k

k k k p

p p p

= =

= =

.

Vy2 1 1 ( 1)( 5)

24 4 24

p p p pP

= = .


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20

Bi 6. Mt s nguyn dng c gi l s kim cng 2005 nu trong biu din

thp phn ca n c 2005 s 9 ng cnh nhau lin tip. Dy ( ) , 1, 2,3,...na n = l dy tng

ngt cc s nguyn dng tha mn na nC< (C l hng s thc dng no ).

Chng minh rng dy s ( ) , 1,2,3,...na n = cha v hn s kim cng 2005.

Trc ht, ta s chng minh cc b sau:

(1)1

1lim

n

i n== + .

(2) Nu trong h c s m ( , 1)m m > : dy s ( )na tng v trong dy khng c s

hng no c cha ch s 1m th tng sau1

1n

i ia= hi t khi n tin ti v cc.

*Chng minh b (1):Ta cn chng minh BT: ln( 1), 0x x x> + > . Tht vy:

Xt hm s:1

( ) ln( 1), 0 ( ) 1 0, 01 1

xf x x x x f x x

x x= + > = = > >

+ +.

Do , hm s ( )f x ng bin trn (0; )+ . Suy ra: ( ) (0) 0 ln( 1), 0f x f x x x> = > + > .

Trong BT ny, thay x bi1

0> , ta cng c:

1 1 1 1 1ln( 1) ln( ) ln( 1) ln , 0

xx x x

x x x x x

+> + > > + > . p dng vo tng cn chng minh:

[ ]1 1

1ln( 1) ln( ) ln( 1) ln1 ln( 1)

n n

i i

n n n nn= =

> + = + = + , m [ ]lim ln( 1)n + = + nn:

1

1lim

n

i n== + . B c chng minh.

*Chng minh b (2):

t1

ks n= l tng cc s t nhin c cha k ch s vit trong h c s m v khng c

cha ch s 1m no.

Gi s mt s hng c k ch s no c dng: 1 2 1... k kb b b b , ch s th 1 phi khc 0 v

khc 1m nn c 2m cch chn, cc ch s cn li phi khc 1m nn c 1m cch chn.

Do , c ng1

( 2).( 1)k

m m

s c k ch s m trong biu din trong h c s m khng ccha ch s 1m , m mi s trong u ln hn 1km nn tng nghch o tng ng ca

chng s b hn1

1

( 2).( 1)k

k

m m

m

.


21/87

21

Hn na:1

ks n= l tng cc s hng c cha k ch s trong h s m v khng c cha

ch s 1m no nn n khng vt qu tng ca tt c cc s t nhin c cng dng m ta

va nh gi c, suy ra:1

1

( 2).( 1)kk k

m ms

m

< .

Do :1

11

1 1 1 1

1 ( 2).( 1) 1 2lim lim lim lim ( 2).( ) ( 2)

11

kn n n nk

k ki k k k i

m m m ms m m m

ma m mm

= = = =

= < = = =

.

Tc l tng ny hi t khi n tin ti v cc. B (2) c chng minh.

*Tr li bi ton:

t 200510 1m m= l s t nhin c cha ng 2005 s 9 lin tip khi vit trong hthp phn.

Ta cn chng minh trong dy cho, c v s s hng cha ch s 1m .Gi s trong dy ny khng c cha s hng no c ch s 1m . Khi , theo b (2)

trn:1

1lim

n

i ia= l hu hn.

Hn na, theo gi thit: ,na nC n< nn1 1 1

1 1 1 1lim lim .lim

n n n

i i iia nC C n= = => = . Theo b

(1), gii hn ny tin ti v cc. Hai iu ny mu thun vi nhau chng t iu gi s trn

l sai, tc l dy cho c t nht mt s hng cha ch s 1m , gi s l:0n

a .

Ta li xt dy con ca dy ban u:0 0 01 2 3

, , ,...n n n

a a a+ + +

Dy ny c y tnh cht ca dy cho nn cng cha t nht mt s hng c cha ch s

1m khc vi s0n

a trn (do y l dy tng).

Lp lun tng t nh th, dy con ny c thm mt s hng c cha ch s 1m .T suy ra dy cho c v s s hng cha ch s 1m .

Vy dy s ( ) , 1, 2,3,...na n = cha v hn s kim cng 2005. y chnh l pcm.


22/87

22

LI GII THI CHN I TUYN QUC GIA

D THI IMO 2006

Bi 1. Cho tam gic ABC c H l trc tm. ng phn gic ngoi ca gc BHC ctcc cnh AB, AC ln lt ti D v E. ng phn gic trong ca gc BAC ct ng trnngoi tip tam gic ADE ti im K.Chng minh rng ng thng HK i qua trung im ca on BC.

Trc ht ta s chng minh ADE cn ti A.

Tht vy: V HD l phn gic gc ngoi ca BHCnn:

01 1 1( ) (90 ) (90 )2 2 2

DHB HBC HCB ABC ACB BAC = + = + = .

Do : 0 01 1

90 902 2ADE DBH DHB BAC BAC BAC= + = + = .

Tng t, ta cng c: 01

902

AED BAC= , suy ra: ADE AED= , tc l tam gic ADE cn ti A.

Mt khc AK l phn gic DAEnn cng l trung trc ca on DE, do AK chnh l ng knh ca ng trnngoi tip ADE .T , ta c: KD AB , tng t ta cng

c: KE AC .

Gi P l giao im ca KD v HB, Q lgiao im ca KE v HC.Ta c: ,KP AB QH AB KP // QH.

Tng t, ta cng c: KQ // PH. Suy ra:KPHQ l hnh bnh hnh, tc l HK i quatrung im ca PQ.

Gi BB, CC l cc ng cao ca

tam gic ABC. Theo nh l Thals: DP // HC'

PB DB

PH DC = , QE // HB

'

QC EC

QH EB = .

Theo tnh cht ng phn gic: ,' ' ' '

DB HB EC HCDC HC EB HB

= = .

V B, C, B, C cng thuc ng trn ng knh BC nn theo tnh cht phng tch:

. ' . '' '

HB HCHB HB HC HC

HC HB= = . T cc iu ny, ta c:

PB QC

PH QH= PQ // BC.

V HK i qua trung im ca PQ nn cng i qua trung im ca BC. Ta c pcm.

C'

B'

P

Q

K

D

E

H

A

B C


23/87

23

Bi 2. Hy tm tt c cc cp s t nhin (n ; k) vi n l s nguyn khng m v k l s

nguyn ln hn 1 sao cho s : 2006 2 517 4.17 7.19n n nA = + + c th phn tch c thnh tchca k s nguyn dng lin tip.

Trc ht ta thy rng tch ca 4 s t nhin lin tip phi chia ht cho 8 v trong 4 s

c 1 s chia ht cho 4 v mt s chia 4 d 2.T 2006 2 517 4.17 7.19n n nA = + + , suy ra :

- Nu n l s chn, ta c :2006 2 5 2 1017 1 (mod8),4.17 4.1 (mod8),7.19 7.3 7.3 7 (mod8)n n n n

Suy ra : 12 4(mod 8)A , tc l A khng chia ht cho 8.

- Nu n l s l, cng tng t :2006 2 2 517 1 (mod8),4.17 4.1 (mod8),7.19 7.3 7.3 5(mod8)n n n

Suy ra : 10 2(mod 8)A , tc l A cng khng chia ht cho 8.

Tc l trong mi trng hp lun c A khng chia ht cho 8.Suy ra nu k tha mn bi th 4 {2,3}k k< .

Xt tng trng hp :- Nu 2k = : tn ti x t nhin sao cho ( 1)A x x= + .

+ Nu n = 0 th A = 12, x = 3, tha mn bi.

+ Nu n > 0 th r rng 1003 2 517 4.17 7.19n n n> + . Ta thy :2006 2 5 2006( 1) 17 4.17 7.19 17n n n nA x x= + = + + > , suy ra 100317 nx > nhng

2006 1003( 1) 17 17n nx A+ > + > , mu thun.

Do , trong trng hp ny khng c n tha mn bi.

- Nu 3k = : tn ti x t nhin sao cho ( 1)( 1), 1A x x x x= + ; d thy x phi l s chn (vnu ngc li th A chia ht cho 8, mu thun). Ta thy :

12.( 1) 2.( 1) (mod5)n nA trong khi 2( 1)( 1) ( 1) 0, 1(mod5)x x x x x + = , mu thun.

Do , trong trng hp ny khng c n tha mn bi.Vy tt c cc cp s tha mn bi l ( ; ) (0; 2)n k = .


24/87

24

Bi 3. Trong khng gian cho 2006 im m trong khng c 4 im no ng phng. Ngita ni tt c cc im li bi cc on thng. S t nhin m gi l s tt nu ta c th gncho mi on thng trong cc on thng ni bi mt s t nhin khng vt qu m saocho mi tam gic to bi ba im bt k trong s cc im u c hai cnh c gn bihai s bng nhau v cnh cn li gn bi s ln hn hai s .

Tm s tt c gi tr nh nht.

Do trong cc im cho khng c bn im no ng phng nn ba im bt k trongchung lun to thnh mt tam gic. Gi S(n) l gi tr nh nht cas ttng vi n im trongkhng gian (n l s t nhin), ta s xc nh gi tr ca S(2006). Ta ch xt cc gi tr 4n .

- Vi n = 4 th th trc tip, ta thy S(4) = 2. Bi v S(4) = 1 khngtha mn nn (4) 2S , ta s ch ra rng S(4) = 2 tha mn. C th

ta c th gn cc on thng nh sau : gn 4 on bt k bi s 1 v2 on cn li bi s 2, r rng cc tam gic to thnh u tha

mn bi.- Vi mt gi tr n > 4 bt k, ta s chng minh rng :

1( ) 1

2

nS n S

+ +

.

Gi a l s nh nht c gn cho cc on thng trong trng hp c n im. Trongtrng hp ti thiu, khng mt tnh tng qut, ta gi s rng a = 1, ta gi hai u mt ca onthng no c gn s 1 l X v Y.

Trong n 2 im cn li, nu c mt im c ni vi X v Y bi mt on thng gnbi s 1 th im cng vi X v Y s to thnh mt tam gic u khng tha mn bi.Do , nu gi A l tp hp tt c cc im ni vi X bi mt on thng gn s 1 (c tnh lun

im Y) v B l tp hp tt c cc im ni vi Y bi mt on thng gn s 1 (c tnh lun

im X) th gia A v B khng c phn t no chung hay A B n+ = .

*Ta c cc nhn xt sau :- Nu ly mt im bt k trong tp A v mt im bt k trong B th hai im cng phic ni bi on thng gn s 1 v nu khng th hai im s cng vi X s to thnh mttam gic khng tha mn bi (tam gic hoc khng c hai s c gn trn hai cnh bngnhau hoc c hai cnh bng nhau nhng cnh cn li gn s 1 nh hn).- Hai im bt k trong A c ni vi nhau bi mt on thng gn s ln hn 1 bi nukhng th khi chn thm mt im trong B, ta s c mt tam gic khng tha mn bi (tam

gic u). Tng t vi tp hp B. Tc l trong cc tp A v B u c cha cc s ln hn 1.Tip theo, ta li thy trong mi tp A, B nh vy u cn thm ( ), ( )S A S B s na

gn cho cc on thng. Gi s A B th1 1

12 2

n nA

+ + =

.

2

2 1

1

1

1


25/87

25

Ta hon ton c th gn cc s tp A trng vi cc s tp B nn cc s cn c thm

na l1

2

nS

+

, tnh thm s 1 nh nht c gn cho on XY ban u, ta c:

1( ) 1

2

nS n S

+ +

.

T , p dng lin tip kt qu ny, ta c: (ch rng S(4) = 2).(2006) 1 (1003) 2 (502) ... 9 (4) 11S S S A + + + = .

Tip theo, ta s chng minh rng gi tr 11 ny tha mn bi.* Tht vy :

Ta xy dng cch gn cc im t thp n cao bng cch ghp cc b im t hn li. C thnh sau :- u tin ta xy dng cho b 4 im. Cch gn tng t nh trn, nhng trong trng hp

ny gn 4 on bi s 11 v 2 on bi s 10.- Ghp 2 b ny li v tch ra t mt trong hai b ra 2 im, gn cho on thng ni 2 im

bi s 10, ta c tt c 8 im.- Tip tc ghp tng t nh vy theo th t nh sau :

4 8 16 32 63 126 251 502 1003 2006 (Cc trng hp t 32 n 63 hoc tng t ta phi b i 1 im no mt trong hai bra ngoi). Mi ln ghp hai b im li th s gn trn on c tch ra li gim i 1 n v,n khi ghp c 2006 im th s chnh l 1.

D thy cch gn s cho cc on thng ny tha mn bi.Vy gi tr nh nht cas ttcn tm l 11.


26/87

26

Bi 4. Chng minh rng vi mi s thc [ ], , 1;2x y z , ta lun c bt ng thc sau :

1 1 1( )( ) 6( )

x y zx y z

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

+ + +.

Hi ng thc xy ra khi v ch khi no ?

Trc ht ta thy rng :21 1 1 ( )

( )( ) 9x y

x y zy z xy

+ + + + = ,

2( )6( ) 9

( )( )

x y z x y

z z x x y y z z x

+ + =

+ + + + + .

Ta cn chng minh :2 21 1 1 ( ) 3( )

( )( ) 6( )( )( )

x y z x y x yx y z

y z y z z x x y xy y z z x

+ + + + + +

+ + + + +

vi mi s thc x, y, z thuc on [1 ; 2].

t1 3

( )( )xS

z x y x z

=

+ +

,1 3

( )( )yS

zx y x y z

=

+ +

,1 3

( )( )zS

xy z x z y

=

+ +

.

Bt ng thc cho vit di dng tng ng l:2 2 2( ) ( ) ( ) 0x y zS y z S z x S x y + + .

Khng mt tnh tng qut, ta gi s 2 1y z .

Ta s chng minh rng , 0x yS S . Tht vy:20 2 0xS x xy xz yz + + , ng.20 2 ( ) ( ) 0yS y yx yz zx x y z z z y x + + + + (do [ ], , 1;2x y z nn 0y z x+ ).

- Nu 0zS , ta c pcm.

- Nu 0zS < , ta chng minh c rng 2 0, 2 0x z y zS S S S + + .

Khi d dng thy rng v: 2 2 2( ) 2 ( ) ( )y y z z x + v 0zS < nn

2 2 2 2 2( ) ( ) ( ) ( 2 )( ) ( 2 )( ) 0x y z x z y zS y z S z x S x y S S y z S S z x + + + + +

Vy trong mi trng hp, ta lun c pcm.ng thc xy ra khi x = y = z hoc y = z = 1, x = 2 v cc hon v ca chng.


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27

Bi 5. Cho tam gic ABC l tam gic nhn, khng cn, ni tip trong ng trn tm O bnknh R. Mt ng thng d thay i sao cho d lun vung gc vi OA v lun ct cc tia AB,AC. Gi M, N ln lt l giao im ca ng thng d v cc tia AB, AC. Gi s cc ngthng BN v CN ct nhau ti K; gi s ng thng AK ct ng thng BC.1. Gi P l giao ca ng thng AK v ng thng BC. Chng minh rng ng trn

ngoi tip ca tam gic MNP lun i qua mt im c nh khi d thay i.2.Gi H l trc tm ca tam gic AMN. t BC = a v l l khong cch t im A n ngthng HK. Chng minh rng ng thng HK lun i qua trc tm ca tam gic ABC.

T suy ra: 2 24l R a . ng thc xy ra khi v ch khi no?

1. Khng mt tnh tng qut, gi s AB < AC (trng hp cn li hon ton tng t).

Do tam gic ABC khng cn nn AO khng vung gc vi BC v MN khng song song vi BC,do MN phi ct ng thng BC ti mt im, gi s l Q; gi I l trung im BC.

Theo nh l Menelaus cho ba im Q, M, N thng hng: . . 1NA MB QB

NC MA QC= .

Mt khc, theo nh l Cva cho cc on AP, BN, CM ng quy, ta c: . . 1NA MB PB

C MA PC = .

T , suy ra:PB QB

PC QC= hay Q, B, P, C l mt hng im iu ha, suy ra: 2 2.IP IQ IB IC= =

Do I l trung im BC nn 2 2 2 2OI BC QI BI OQ OB = , do :

2 2 2 2 2. . .QI QP QI QI PI QI IB OQ OB QB QC = = = =

(do theo tnh cht phng tch ca Q i vi (O) th 2 2 2 2 .OQ OB OQ R QB QC = = ).

M t gic BMNC cng ni tip v c NCB xAB AMN= = (vi Ax l tia tip tuyn ca (O)).

Suy ra . .QM QN QB QC = .

Q IP

K

N

M

O

A

BC


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28

T suy ra . .QM QN QP QI = , suy ra t gic MNIP ni tip hay ng trn ngoi tip tam gic

MNP lun i qua im I c nh. Ta c pcm.

2. Gi BD, CE l hai ng cao ca tam gic ABC, L l trc tm ca tam gic ABC; gi MF,

NG l hai ng cao ca tam gic AMN, H l trc tm ca tam gic AMN. Ta cn chng

minh rng H, K, L thng hng.

Xt ng trn (O1) ng knh BN v (O2) ng knh CM.

Ta thy: KM.KC = KB. KN nn K c cng phng tch n (O1), (O2), tc l K thuc trc ngphng ca hai ng trn ny.

ng thi, d thy rng cc im D, G thuc (O1) v M, F thuc (O2).

Do H, L l trc tm ca tam gic ABC v AMN nn LB. LD = LC. LE, HN. HG = HE. HM; tc

l H, L cng thuc trc ng phng ca hai ng trn (O1), (O2).

T suy ra H, K, L cng thuc trc ng phng ca (O1), (O2) nn chng thng hng.

T suy ra l AL .

Mt khc do tam gic ABC nhn nn2

2 2 22 44

BCAL OI R R a= = = .

Do 2 24AL l R a= . y chnh l pcm.

n y, ta s tm v tr ca d sao cho ng thc xy ra.

Gi s d ct AB, AC ti M v N tha mn .AN AM

k MN k BC AB AC

= = = .

FG

H

L

E

D

Q IP

K

N

MO

A

BC


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29

Gi R, S ln lt l trung im ca BN v CM; suy ra R, S cng chnh l tm ca hai

ng trn (O1), (O2).

Ta thy khi ng thc xy ra th AL vung gc vi trc ng phng ca (O1), (O2), tc

l AL song song vi ng ni tm RS ca hai ng trn ny, m AL vung gc vi BC nn

RS phi vung gc vi BC.

Ta c: 2RS BC NM= +

, m . 0 ( ). 0RS BC BC NM BC= + =

. Do gc to bi MN v BC

chnh l MQB ANM ACB ABC ACB= = nn t ng thc trn suy ra:

2 1. . .cos( )cos( )

BC BC MN BC kBC B C kB C

= = =

.

Vy ng thc xy ra khi v ch khi1

cos( )k

B C=

, tc l ng thng d ct AB ti M,

AC ti N sao cho1

cos( )

AN AM

AB AC B C

= =

.


30/87


31/87

31

+ Nu k c t nht k + 1 c nguyn t khc nhau th r rng (***) ng.

+ Nu k c ng k c nguyn t i mt khc nhau, gi s l 1 2 3, , ,.., kp p p p .

Khi : ( , ), 1,k iA p i k= l 1 hoc 3.

Gi s 1kA + ch c ng k c nguyn t i mt khc nhau l cc gi tr trn th cn phi c

*3 3 , , 2mkA m m+ = .Nhng khi th 3(mod 9)kA khng phi l s chnh phng, mu thun.

T dn n 1kA + phi c mt c nguyn t no khc k c c, tc l c t nht k + 1 c

nguyn t i mt khc nhau hay (***) ng vi 1n k= + .Do (***) c chng minh.

Vy vi mi n nguyn dng, s2

3

3 1n nA

a=

l mt s chnh phng v n c t nht n

c nguyn t phn bit, bi ton c gii quyt hon ton.


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32

LI GII THI CHN I TUYN QUC GIA

D THI IMO 2007

Bi 1.

Cho hai tp hp A,B l tp hp cc s nguyn dng tha mn A B n= = (vi n l snguyn dng) v c tng cc phn t bng nhau. Xt bng vungn n .

Chng minh rng ta c th in vo mi vung ca bng mt s nguyn khng mtha mn ng thi cc iu kin:

i/ Tp hp tng cc s mi hng l tp A.ii/ Tp hp tng cc s mi ct l tp B.

iii/ C t nht 2( 1)n k + s 0 trong bng vi k l s cc phn t chung ca A v B.

Trc ht, ta thy rng nu mt gi tr k sao

cho tn ti 2 phn t bng nhau mi tp lk ka b t= = th ta in s t vo vung nm hng th

k v ct th k, cc cn li ca hng th k v ct th ku in vo s 0; nh th th tng cc s hng v ctny tha mn bi v khng nh hng n cc hngv ct khc. Do , khng mt tnh tng qut, ta xttrng hp A B = (trng hp c cc phn tchung th in thm vo cc hng v ct theo cchtng t nh trn), tc l s phn t chung ca hai tp

l 0k= .Ta s chng minh bi ton ny bng quy np. Gi l tp hp cc iu kin i/, ii/, iii/

nh trn (iu kin iii/ tng ng vi trng hp xt s nguyn dng n).

Vi n = 1, bi ton hin nhin ng.

Gi s bi ton ng vi mi s tp hp c n 1 phn t. Ta s chng minh rng vi haitp A, B c n phn t, ta cng c th xy dng mt bng n n tha mn iu kin .

Tht vy, xt hai tp hp 1 2 3 1 2 3{ , , , ..., }, { , , , ..., }n nA a a a a B b b b b= = trong :

1 2 3 ... na a a a< < < < , 1 2 3 ... nb b b b< < < < (hai tp ny khng c phn t no chung).

Gi s 1 1a b< . Do tng cc phn t hai tp bng nhau nn tn ti mt ch s i tha mn

1 1 1 1 1( ) 0i ia b b a a b a> > > . Ta xt hai tp hp A*, B* nh sau:

2 3 1 1 1* { , ,..., , ,..., }i i nA a a a a b a a= + , 2 3* { , ,..., }nB b b b= .

0 00

0

0

000

0

0

0

0

n

i

i

2

1

n21


33/87

33

Hai tp hp ny c cng s phn t l n 1 nntheo gi thit quy np, tn ti mt bng c kch thc( 1) ( 1)n n tha mn iu kin (trong bng ny c

t nht 2( 2)n s 0).

Ta thm vo bn tri bng mt ct v bn trn bng mthng na nh hnh v. gc bn tri v pha trn, ta

in s 1a , hng th i ca bng ban u (hng c tng

cc phn t bng 1 1ia b a + ), ta in s 1 1b a ; cn tt

c cc cn li ca hng v ct va thm vo, ta invo cc s 0. Khi , bng ny c tng cc phn t mi

hng l tp A v tng cc phn t mi ct l tp B, s cc s 0 bng va lp c khng nh

hn 2 2( 2) 2( 1) 1 ( 1)n n n + = v do n tha mn iu kin .

Do , bi ton cng ng vi mi tp hp c n phn t.

Theo nguyn l quy tp, bi ton ny ng vi mi s nguyn dng n.

Vy ta c pcm.

n - 1

i

2

1

n - 1321


34/87

34

Bi 2.

Cho tam gic nhn ABC vi ng trn ni tip I. Gi ( )ak l ng trn c tm nm

trn ng cao ca gc A v tip xc trong vi ng trn (I) ti 1A. Cc im 1 1,B C xc

nh tng t .

1/ Chng minh 1 1 1, ,AA BB CC ng qui ti P.2/ Gi ( ), ( ), ( )a b cJ J J ln lt l cc ng trn i xng vi ng trn bng tip cc

gc A, B, C ca tam gic ABC qua trung im BC, CA, AB.Chng minh P l tm ng phng ca 3 ng trn ni trn.

1/ Trc ht, ta s chng minh b sau:Cho tam gic ABC ngoi tip ng trn (I) c D l tip im ca ng trn bng tip

gc A ln BC. Gi M, N l giao im ca AD vi (I) (N nm gia A v M). Gi s IM ct ngcao AH ti K. Chng minh rng: KA = KM.

* Tht vy:Gi E l tip im ca (I) ln

BC. Gi s IE ct (I) ti im th hai lN khc E. Qua N v ng thngsong song vi BC ct AB v AC lnlt ti B v C. D thy tn ti mtphp v t bin tam gic ABC thnhtam gic ABC. Php v t cng bintip im N ca ng trn bng tip(I) ca ABC ln BC thnh tip

im D ca ng trn bng tip (J)ca ABC ln BC. Suy ra A, N, Dthng hng hay N trng vi N. Khi ,tam gic IMN ng dng vi KMA(do IN // AK), m IMN cn ti I nn KAM cn ti K hay KA = KM.

Ta c pcm.T y suy ra: ng trn c tmthuc ng cao gc A, i qua A vtip xc vi (I) ti M th M thucAD.

D thy ng trn l duy nht.*Tr li bi ton:Gi D, E, F ln lt l tip im ca ng trn bng tip cc gc A, B, C ca tam gic

ABC ln cc cnh BC, CA, AB. Theo b trn, ta thy: 1 1 1, ,A AD B CF C BE .

Suy ra: 1 1 1, ,AA BB CC ng quy khi v ch khi AD, BE, CF ng quy. (1)

C'B'

K

D

J

BC


35/87


36/87


37/87

37

Bi 3.Cho tam gic ABC. Tm gi tr nh nht ca biu thc sau:

2 2 2 2 2 2

2 2 2

cos cos cos cos cos cos2 2 2 2 2 2

cos cos cos2 2 2

A B B C C A

SC A B

= + +

* Trc ht, ta s chng minh b :

Vi mi a, b, c khng m v khng ng thi bng 0, ta c:

2 2 2

1 1 1 9

( ) ( ) ( ) 4( )a b b c c a ab bc ca+ +

+ + + + +.

Du ng thc xy ra khi a = b = c hoc a = b, c = 0 v cc hon v.

* Chng minh:

Quy ng v khai trin BT trn, ta cn chng minh rng:

5 4 2 3 3 4 3 2 2 2 2(4 3 ) ( 2 ) 0sym sym

a b a b a b a bc a b c a b c + + (*)

Theo BT Schur cho cc s khng m a, b, c, ta c:

( )( ) 0sym

a a b a c , nhn vo hai v cho s abc khng m v khai trin, ta c:

4 3 2 2 2 2( 2 ) 0

sym

a bc a b c a b c + . (1)

Hn na, theo BT Cauchy, ta c:

5 4 2 3 3 5 4 2 5 3 3(4 3 ) ( ) 3( ) 0sym sym sym

a b a b a b a b a b a b a b = + . (2)

Cng tng v cc BT (1) v (2) li, ta c BT (*).

ng thc xy ra trong (*) khi v ch khi ng thc xy ra trong (1) v (2), tc l khi

a = b = c hoc a = b, c = 0 v cc hon v. B c chng minh.

* Tr li bi ton:

Ta c: 22

1cos

2 tan 12

AA

=+

. Tng t vi 2 2cos ,cos2 2

B C.


38/87

38

Thay cc bin i ny vo biu thc cho, ta c:

2

2 2

1

( 1)( 1)

xS

y z

+=

+ + , trong tan , tan , tan ; , , 0

2 2 2

A B Cx y z x y z= = = > .

( y k hiu l tng i xng ly theo cc bin x, y, z).Mt khc, trong tam gic ABC, ta lun c:

tan tan tan tan tan tan 1 12 2 2 2 2 2

A B B C C Ay yz zx+ + = + + = .

Do :

22

2 2 2 2

( )1 ( )( )

( 1)( 1) ( )( ) ( )( )( )( )

x xyx x y x zS

z y xy z xy y z y x z x z y

++ + += = = =

+ + + + + + + +

21( )x y=

+ .

p dng b trn, ta c:9 9

4( ) 4S

xy yz zx =

+ +.

Vy gi tr nh nht ca S l9

4t c khi v ch khi x y z= = hay ABC l tam gic u.


39/87

39

Bi 4. Tm tt c cc hm s lin tc :f tha mn:

2 1( ) ( )3 9

xf x f x= + + vi mix .

Ta c: 2 2 21 1 1 1 1

( ) ( ) (( ) ),3 9 6 12 6 12

xx x f x f x x+ + = + + = + + .

t1 1

6 6y x x y= + = . Thay vo gi thit cho, ta c: 2

1 1( ) ( )

6 12f y f y = + . (*)

Xt hm s: ( ) :g x tha mn:1 1

( ) ( ) ( ) ( ),6 6

g x f x f x g x x+ = = . (1)

T (*), ta c: 21

( ) ( ),4

g x g x x= + , r rng ( )x cng lin tc.

Ta s xc nh hm s ( )x tha mn iu kin trn.

Ta thy: 2 21 1

, ( ) ( ) (( ) ) ( )

4 4

x g x g x g x g x = + = + = nn ( )x l hm s chn.

Ta ch cn xt 0x . Ta c hai trng hp:

- Vi 01

2x : Xt dy s: 21 0 1

1, , 1

4n nu x u u n+= = + . Khi :

21

1( ) 0

2n n nu u u+ = nn

dy ( )nu tng. Mt khc, bng quy np, ta chng minh c1

,2n

u n , tc l dy ny b chn

trn. T suy ra n c gii hn. Gi t l gii hn th 21 1

4 2t t t= + = .

Do : 01 1

( ) ( ), 0;2 2

g x g x = .

- Vi 01

2x > : Tng t nh trn, xt dy s 1 0 1

1, 0, 1

4n nv x v v n+= = .

Khi :

2

1

1( )1 2 0

4 14

n

n n n n

n n

vv v v v

v v+

= = , tc l dy cho b chn di, suy ra n c gii hn.

Gi k l gii hn th 1 14 2k k k= = .

Do : 01 1

( ) ( ), ( ; )2 2

g x g x= + . t1

( )2

g a= .

T suy ra: ( ) , 0g x a x= hay ( ) ,g x a x= . (2)

T (1) v (2), ta c: ( ) ,f x a x= . y l tt c hm s cn tm.


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40

Bi 5.

Cho A l tp con cha 2007 phn t ca tp:{ }1,2,3,...,4013,4014 tha mn vi mi

,a b A th a khng chia ht cho b. Gi mA l phn t nh nht ca A.

Tm gi tr nh nht ca mA vi A tha mn cc iu kin trn.

Chia tp hp {1,2,3,...,4013,4014} thnh 2007 phn 1 2 2007...P P P (mi tp hp

cha t nht mt phn t ca { }1,2,3,...,4013,4014 ) tha mn tp hp aP cha tt c cc s

nguyn dng c dng 2 (2 1)n a , trong n l mt s khng m. Khi , tp hp con A ca

{ }1,2,3,...,4013,4014 khng th cha hai phn t cng thuc mt trong 2007 tp hp v nu

khng th r rng c mt s s chia ht cho s cn li, mu thun. Mt khc, A li c ng 2007

phn t nn A cha ng 1 phn t ca mi tp aP ni trn.

Gi i l mt phn t ca A vi , 1,2007i iP i = . Xt cc phn t 1 2 5 1094, , ,... ln

lt c cc dng 2 72 , 3.2 , 3 .2 ,..., ,3 .2n n n n ; r rng mi phn t ch c hai c nguyn t l 2 v

3. Ta cng thy rng ly tha ln nht ca 2 trong 1phi ln hn ly tha ln nht ca 2 trong

2 v nu ngc li th 2 1 , mu thun.

Hon ton tng t vi cc phn t khc trong dy 1 2 5 1094, , ,... , tc l nu i j< th

ly tha ca 2 trong i phi ln hn ly tha ca 2 trong . Do , ly tha ca 2 trong dy

1 2 5 1094, , ,... l mt dy gim thc s. Do c 8 phn t trong dy trn (tng ng vi ly

tha ca 3 thay i t 0 n 7) nn gi tr ca 1 t nht l72 128= . Hn na, cc phn t ca

dy c gi tr ca ly tha 3 tng dn (nu ngc li, gi tr ca ly tha 3 gim m ly tha 2cng gim nn c mt s chia ht cho s khc, mu thun).

Suy ra: 1 chnh l gi tr nh nht trong dy trn.

Hon ton tng t vi cc s nguyn t khc, ta cng xt s c dng 2 .3 ...i i kn n ni kp = ,

trong 2, 3, , pk l cc c nguyn t khng vt qu 4014. T suy ra 1 cng chnh l

phn t nh nht trong cc phn t ca tp A. Do phn t nh nht mA ca tp A chnh l 1 ,

nh chng minh trn 1 128 hay 128Am .

Ta s chng minh rng 128 chnh l gi tr nh nht ca mA bng cch ch ra mt tp hp

con A ca { }1,2,3,...,4013,4014 tha mn iu kin bi.

Xt tp hp: { }( ) ( )2 (2 1) |1338 3 .(2 1) 4014f i f iiA i i= = .


41/87

41

R rng tp hp ny c ng 2007 phn t thuc { }1,2,3,...,4013,4014 .

Ta s chng minh rng x khng chia ht cho y , vi x y> (tc l tp hp ny tha mn bi).

Tht vy, gi s ngc li tn ti x, y tha mn x y , khi :( ) ( )2 (2 1) 2 (2 1)f x f yx y ,

tc l ( ) ( )f x f y v (2 1) (2 1)x y .

T cch xc nh cc gi tr u, v; ta c: (2 1) (2 1) 2 1 3(2 1)x y x y , ng thi( ) 1 ( ) ( ) 13 (2 1) 4014 3 (2 1) 3 (2 1) ( ) ( )f y f x f xy x y f y f x+ + > > . Mu thun ny chng t tt

c cc phn t ca A u tha mn vi mi ,a b A th a khng chia ht cho b.

Vy 128 chnh l gi tr nh nht ca mA cn tm.


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42

Bi 6.Cho a gic 9 cnh u (H). Xt ba tam gic vi cc nh l cc nh ca a gic (H)

cho sao cho khng c hai tam gic no c chung nh.Chng minh rng c th chn c t mi tam gic 1 cnh sao cho 3 cnh ny bng nhau.

K hiu hnh (H) cho l a gic

1 2 3 8 9...A A A A A nh hnh v. Trc ht, ta thy

rng di cc cnh v cc ng cho cahnh (H) ch thuc 4 gi tr khc nhau (nugi R l bn knh ng trn ngoi tip ca(H) th ta d dng tnh c cc gi tr l

2 .sin9

R

,2

2 sin9

R

,3

2 sin9

R

,4

2 sin9

R

)

ta t chng l1 2 3 4

, , ,a a a a theo th t tng

dn ca di. R rng cc tam gic c nhthuc cc nh ca (H) s c cnh c dithuc 1 trong 6 dng sau:

1 1 2 2 2 4 1 3 4

3 3 3 2 3 4 4 4 1

( , , ), ( , , ), ( , , ),

( , , ), ( , , ), ( , , )

a a a a a a a a a

a a a a a a a a a.

Gi s 3 tam gic c ly ra l 1 2 3, , .

Xt cc trng hp sau:

- Nu trong cc tam gic c mt tam gic u, r rng, tam gic ny phi c di cccnh l

32 .sin

9R

; khng mt tnh tng qut, gi s l tam gic 1 4 7A A A . Do cc tam gic

1 2 3, , khng c hai nh no trng nhau nn ta s lp mt tam gic c cc nh l mt trong

hai nh ca cc tp hp 2 3 4 5 7 8{ , },{ , },{ , }A A A A A A . Ta s chng minh rng tam gic phi c t

nht 1 cnh c di l3

2 .sin9

R

, tc l hai nh c ch s c cng s d khi chia cho 3. Gi s

ngc li, trong hai tam gic cn lp, khng c tam gic no c cnh l3

2 .sin9

R

, khi nh

A2 phi ni vi A4 v A4 phi ni vi A8, nhng khi A8 c ni vi A2 l hai nh c ch schia cho 3 cng d l 2, mu thun. Do , trong hai tam gic lp c, lun c mt cnh c

di l3

2 .sin9

R

. Suy ra trng hp ny lun c tam gic tha mn bi.

a1

a2

a3

a4

9

8

7

6

5

4

3

21


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- Nu trong cc tam gic , khng c tam gic no u . Khi cc tam gic c xt

khng c ba nh cng thuc mt trong ba tp hp sau: 1 1 4 7{ , , }A A A = , 2 2 5 8{ , , }A A A = ,

3 3 6 9{ , , }A A A = . Ta thy mt on thng ni hai im bt k thuc hai tp khc nhau s nhn 1

trong 3 gi tr l 1 2 4, ,a a a . Hn na, khng c tam gic no c di 3 cnh l 1 2 4( , , )a a a nn ta

c hai nhn xt:

(1) Mt tam gic c cc nh thuc c ba tp 1 2 3, , ni trn th s c hai cnh no c

di bng nhau (cc cnh ca n c th l 1 1 2( , , )a a a , 2 2 4( , , )a a a , 4 4 1( , , )a a a ) tc l n phi cn.

(2) Mt tam gic c hai trong ba nh thuc cng mt tp th tam gic cc cnh c di

l 2 3 4( , , )a a a hoc l 1 3 4( , , )a a a , tc l tam gic khng cn.

* Ta xt tip cc trng hp (cc tam gic xt di y l cn nhng khng u):

+ C hai tam gic cn v mt tam gic khng cn: khi theo nhn xt (1), hai tam gic cn

phi c nh thuc cc tp hp khc nhau trong ba tp 1 2 3, , ; khi , r rng tam gic cn li

cng phi c nh thuc cc tp hp khc nhau, tc l n phi cn, mu thun.

Vy trng hp ny khng tn ti.

+ C mt tam gic cn v hai tam gic khng cn: khi theo nhn xt (2), hai tam gic khngcn phi c hai nh thuc cng mt tp hp v nh cn li thuc tp hp khc, gi s mt

tam gic c hai nh thuc 1 v mt nh thuc 2 ; r rng tam gic khng cn cn li phi c

hai nh thuc 2 , mt nh thuc 3 , suy ra tam gic cn li c hai nh thuc 3 , mt nh

thuc 1 nn n l tam gic cn, mu thun.

Vy tng t nh trn, trng hp ny khng tn ti.

+ C ba tam gic u khng cn: khi theo nhn xt (2), tam gic thuc mt trong hai dng

2 3 4( , , )a a a hoc l 1 3 4( , , )a a a , tc l cc tam gic ny lun cha 1 cnh c di l a3.

Trong trng hp ny, bi ton c gii quyt.

+ C ba tam gic u cn: khi , cc tam gic c di l 1 1 2( , , )a a a , 2 2 4( , , )a a a , 4 4 1( , , )a a a .

R rng khng tn ti trng hp c di cc cnh ln lt nhn c ba gi tr nh ba b trnnn phi c hai b trng nhau, tc l c t nht hai tam gic cn bng nhau v mt tam gic cnnhn mt trong ba gi tr thuc mt trong cc b trn lm cnh, khi lun c th chn t tamgic ny mt cnh bng vi cnh y hoc cnh bn ca hai tam gic cn bng nhau kia.Trong trng hp ny, bi ton cng c gii quyt.Vy trong mi trng hp, ta lun c pcm.


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LI GII THI CHN I TUYN QUC GIAD THI IMO 2008

Bi 1. Trong mt phng cho gc xOy. Gi M, N ln lt l hai im ln lt nm trncc tia Ox, Oy. Gi d l ng phn gic gc ngoi ca gc xOy v I l giao im ca trungtrc MN vi ng thng d. Gi P, Q l hai im phn bit nm trn ng thng d saochoIM IN IP IQ= = = , gi s K l giao im ca MQ v NP.

1. Chng minh rng K nm trn mt ng thng c nh.2. Gi d1 l ng thng vung gc vi IM ti M v d2 l ng thng vung gc vi IN ti

N. Gi s cc ng thng d1, d2 ct ng thng d ti E, F. Chng minh rng cc ngthng EN, FM v OK ng quy.

1. Xt trng hp cc im M, Q v N, P nm cng pha vi nhau so vi trung trc ca MN.Khi giao im K ca MP v NQ thuc cc on ny.:

Gi I l giao im ca d vi ng trn ngoi tip ON . Do d l phn gic ngoi caMONnn I chnh l trung im ca cung ON, do : IM = IN hay I chnh l giao im

ca trung trc MN vi d. T , suy ra: 'I I hay t gic MION ni tip.Ta c: IO NMO= .Mt khc: do IM = IN = IP = IQ nn t gic MNPQ ni tip trong ng trn tm I, ng knh

PQ 2PIN PMN = (gc ni tip v gc tm cng chn cung PN).

T cc iu trn, ta c: 2NMO PMN= MP l phn gic trong ca OMN.

x

y

d

J

F

E

K

P

Q

IO

M

N


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Tng t, ta cng c: NQ l phn gic trong ca ONM .Do K l giao im ca MP v NQ nn K chnh l tm ng trn ni tip ca ON , suy ra K

thuc phn gic trong ca Oy , tc l K thuc mt ng thng c nh (pcm).

- Nu giao im K nm ngoi cc on MP v NQ: ta cng c lp lun tng t v c

c K l tm ng trn bng tip

MONca tam gic ON , tc l K cng thuc phn gictrong ca xOy , l mt ng thng c nh.

Trong mi trng hp, ta lun c pcm.

2. Gi J l giao im ca d1 v d2. Ta thy t gic MINJ ni tip trong ng trn ngknh IJ. Hn na: MION cng l t gic ni tip nn 5 im M, N, I, J, O cng thuc mt ng

trn. Do : phn gic trong gc ON i qua trung im ca cung JN.R rng M, N i xng nhau qua trung trc ca MN nn JM = JN, tc l J cng l trung

im ca cung ON.

T suy ra: J thuc phn gic trong ca ONhay O, K, J thng hng.

Ta cn chng minh cc on JO, EN v MF trong JEF ng quy.

Tht vy:. .sin sin

.. .sin sin

OEJ

OFJ

SOE JO JE OJE JE OJE

OF S JO JF OJF JF OJF = = = .

Trong JEF v ON , ta c :sin sin

,sin sin

JE JFE OM ONM

JF JEF ON OMN= = .

Mt khc : sin sin

,sin sin

OJE ONM OJE OJN ONM OJF OJM OMN

OJF OMN = = = = = .

Kt hp li, ta c :sin sin sin

. . .sin sin sin

OE JFE OM OFN OM OFN OM

OF JEF ON OEM ON OEM ON = = =

sin .sin.

.sin sin

OFN OM MOE ME

ON NOF OEM NF = = .

Do : . 1 . . 1OE FN OE NF MJ

OF EM OF NJ ME = = .

Theo nh l Ceva o, ta c OI, EN v MF ng quy. y chnh l pcm.


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Bi 2. Hy xc nh tt c cc s nguyn dng m sao cho tn ti cc a thc vi h sthc ( ), ( ), ( , )P x Q x R x y tha mn iu kin:

Vi mi s thc a, b m 2 0ma b = , ta lun c ( ( , ))P R a b a= v ( ( , ))Q R a b b= .

Vi m l mt s nguyn dng, ta xt cc trng hp :

- Nu m l s chn, t 2 , *m k k= . Suy ra : 2 0m ka b b a = = . Khi cn tm k

sao cho cc a thc ( ) ( ) ( ), , ,P x Q x R x y tha mn c hai iu kin :

(1) ( ( , )) , ( ( , )) ,k k kP R x x x Q R x x x x= = .

(2) ( ( , )) , ( ( , )) ,k k kP R x x x Q R x x x x = = .

Xt a thc mt bin T(x) tha mn : ( , ) ( ),kR x x T x x= . Theo gi thit th

( ( )) ( ( , )) ,kP T x P R x x x x= = . T suy ra : deg ( ).deg ( ) 1P x T x = hay deg ( ) deg ( ) 1P x T x= = .

Gi s ( ) , , , 0T x ux v u v u= + , ( ) '. ', ', ' , ' 0P x u x v u v u= + th

'( ) ' , . ' 1, ' ' 0u ux v v x x u u u v v+ + = = + = hay 1' , ' ( )v x vu v P xu u u= = = .

Mt khc : ( ( , )) ( ( )) ( ) ( ) ( ),k

k k kx uQ R x x Q T x Q ux v x Q x P x xu

= = + = = =

.

Suy ra : ( ( , )) ( ( , )) , ,k kQ R a b P R a b b a a b= = tha 2 0ma b = . Nhng theo iu kin ban

u th b cng c th l ka . Mu thun ny cho thy cc gi tr m trong trng hp ny khngtha mn bi.

- Nu m l s l. t 2( , ) ( )mP x x S x= . Suy ra :2( ( )) , deg ( ).deg ( ) 2P S x x x P x S x= = .

Nu nh degS(x) = 2 th deg ( ( ))Q S x l s chn, trong khi : deg ( ( )) deg mQ S x x m= =

vi m l s l, mu thun. Suy ra : degS(x) = 1, degP(x) = 2.

Mt khc, trong a thc 2( , )mR x x , bc ca n c th t gi tr nh nht l min(2, )m m2deg ( , ) deg ( ) 1mR x x S x= = nn m = 1.

Ta s chng minh rng gi tr m = 1 ny tha mn bi bng cch ch ra cc a thcP(x), Q(x), R(x, y) tha mn bi.

Tht vy :

Xt cc a thc 2( ) , ( ) , ( , )P x x Q x x R x y y= = = .

Khi vi m = 1 th ta c quan h ca a vi b chnh l : 2a b= . Suy ra :( , )R a b b= , 2( ( , )) ( )P R a b P b b a= = = , ( ( , )) ( )Q R a b P b b= = , tha mn bi.

Vy tt c cc gi tr m tha mn bi l m = 1.


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Bi 3. Cho s nguyn 3n > . K hiu T l tp hp gm n s nguyn dng u tin.Mt tp con S ca T c gi l tp khuyt trong T nu S c tnh cht: Tn ti s nguyn

dng c khng vt qu2

nsao cho vi 1 2,s s l hai s bt k thuc S ta lun c 1 2s s c .

Hi tp khuyt trong T c th c ti a bao nhiu phn t ?

Trc ht ta thy rng: Nu S l tp khuyt trong T th tp ' { | }S n x x S = cng l mt

tp khuyt trong T.Tht vy: Gi s ngc li S khng phi l tp khuyt, khi tn ti hai s nguyn

dng 1 2' , ' 's s S sao cho 1 2| ' ' |s s c = vi c l mt s nguyn dng no khng vt qu 2

n,

Khi xt tng ng hai phn t 1 1 2 2' , 's n s s n s= = th r rng 1 2,s s S v

1 2 1 2 1 2| | | ( ' ) ( ' ) | | ' ' |s s n s n s s s c = = = , tc l tn ti hai phn t 1 2,s s S v

1 2| | 2

n

s s c = trong khi S l tp khuyt. Mu thun ny suy ra nhn xt trn c chng minh.Hn na, do | | | ' |S S= nn khi S c s cc phn t l ln nht th tng ng cng c tp

S c s phn t ln nht bng vi S.

T , ta thy c th xt cc tp khuyt S c s cc s phn t khng vt qu2

nkhng

t hn s cc s phn t ln hn2

n. Xt hai tp hp sau:

{ | , }2

nA x x S x= , { | , }

2

nB x x S x= > th ,A B A B S = = v theo cch xc nh S

nh trn th A B .

Khi vi c l mt s nguyn dng no khng vt qu2

n, ta xt tp hp:

{ | }C x c x A= + . Ta c: | | | |A C= . Do A S nn A cng l mt tp khuyt v khi r rng

,A C B C = = (v nu ngc li th tn ti hai phn t thuc S m hiu ca chng l c,

mu thun).Suy ra tt c cc phn t thuc tp A hoc B hoc C u l mt s nguyn dng khng

vt qu n, tc l ( ) | | | | | | | |B C T A B C T n + + = .

Kt hp cc iu ny li, ta c: 2 | | | |A B n+ . Do :4 | | 2 | | 2

| | | | 3 3

B nA B

+

+ .

Hn na: ,A B A B S = = v | |S l s nguyn nn2

| | | | | |3

nS A B = +

.

Do s phn t ca tp khuyt S trong T lun khng vt qu2

3

n

.


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Bi 4. Cho m, n l cc s nguyn dng.

Chng minh rng (2 3) 1nm + + chia ht cho 6m khi v ch khi 3 1n + chia ht cho 4m.

Theo khai trin nh thc Newton th:1

1(2 3) (2 ) 3 .(2 ) .3 (2 ) 3 (mod 6m)

nn n n k n k k n n

nkm m C m m

=+ = + + + .Do , 6 | (2 3) 1nm m + + 6 | (2 ) 3 1n nm m + + 2 | (3 1)nm + v 3 | (2 ) 1nm + .

Cn chng minh rng: 2 | (3 1)nm + v 3 | (2 ) 1nm + (1) 4 | 3 1 (2)nm + .

Xt cc trng hp:* Nu m l s chn:- Xt iu kin (2):

3 1n + khng chia ht cho 4m v 3 1 2(mod8)n + hoc 3 1 4(mod8)n + , trong khi 4 8m , tc l

khng th c iu kin (2).

- Xt iu kin (1): t m l s chn, suy ra 3 1 4n n+ l s l. Ta bit rng: s c dng3n + 1 ch c c nguyn t l ng d vi 1 modun 4. T , suy ra m tha mn: 2 | (3 1)nm +

phi c dng 2(3 1),m k k= + . Suy ra: (2 ) 1 2 .2 .(3 1) 1 2(mod 3) (2 ) 1n n nm k m+ = + + +

khng chia ht cho 3, tc l iu kin (1) cng khng tn ti.* Nu m l s l:

- Xt iu kin (1): t 2 | (3 1)nm + suy ra m khng chia ht cho 3, t 3 | (2 ) 1nm + suy ra n

phi l s l v nu ngc li th (2 ) 1 2(mod3)nm + , mu thun. M n l s l th 3 1 4n + ,

kt hp vi ( , 4) 1m = , ta c 4 | 3 1nm + , y chnh l iu kin (2). Do : (1) (2) .

- Xt iu kin (2): t 4 | 3 1 (4 | 3 1) ( | 3 1)n n n

m m+ + + suy ra n l s l v m c dng3 1,k k+ . Suy ra: (2 ) 1 2 . 1 1 1 0(mod 3) 3 | (2 ) 1n n n nm m m+ = + + + ; t (2) ta cng trc

tip c 2 | (3 1)nm + . Do : (2) (1) .

Kt hp cc iu trn li, ta c: (1) (2) .

Vy (2 3) 1nm + + chia ht cho 6m khi v ch khi 3 1n + chia ht cho 4m.

y chnh l pcm.


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Bi 5. Cho tam gic ABC nhn, khng cn c O l tm ng trn ngoi tip.Gi AD, BE, CF ln lt l cc ng phn gic trong ca tam gic. Trn cc ng thng

AD, BE, CF ln lt ly cc im L, M, N sao choAL BM CN

kD BE CF

= = = (k l hng s dng).

Gi (O1), (O2), (O3) ln lt l cc ng trn i qua L, tip xc vi OA ti A ; i qua M tip

xc vi OB ti B v i qua N tip xc vi OC ti C.

1. Chng minh rng vi 12

k= , ba ng trn (O1), (O2), (O3) c ng hai im chung v

ng thng ni hai im i qua trng tm tam gic ABC.2. Tm tt c cc gi tr k sao cho 3 ng trn (O1), (O2), (O3) c ng hai im chung.

Trc ht, ta nu 4 b sau:(1) Cho ba ng thng i mt phn bit a, b, c v hai ng thng phn bit d, d. Ccng thng d, d theo th t ct a, b, c ti 1 1 1 2 2 2, , ; , ,A B C A B C tha mn iu kin:

1 1 2 2

1 1 2 2

A B A Bk

AC A C= = . Cc im A3, B3, C3 thuc a, b, c sao cho: 1 2 1 2 1 2

1 3 1 3 1 3

A A B B C C

A A B B C C= = .

Khi , A3, B3, C3 thng hng v 3 3

3 3

A Bk

A C= .

(2) Cho ba ng thng phn bit a, b, c v ba ng thng phn bit khc a, b, c . Ccng thng a, b, c theo th t ct a, b, c ti 1 1 1 2 2 2 3 3 3, , ; , , ; , ,A B C A B C A B C (cc im ny i

mt phn bit). Khi nu3 31 1 2 2

1 1 2 2 3 3

A BA B A B

AC A C A C= = th hoc1 2 1 2 1 2

1 3 1 3 1 3

A A B B C C

A A B B C C= = hoc a, b, c i

mt song song.

(3) Cho tam gic ABC v M bt k. Cc tia AM, BM, CM ln lt ct BC, CA, AB A1, B1,C1. Cc ng thng A1B1, B1C1, C1A1 ct cc ng thng AB, BC, CA ln lt A2, B2, C2.Cc im A3, B3, C3 theo th t nm trn cc ng thng BC, CA, AB sao cho

1 3 1 3 1 3

1 2 1 2 1 2

, 0A A B B C C

k kA A B B C C

= = = . Khi , A3, B3, C3 thng hng khi v ch khi 1k= hoc1

2k= .

(4) Cho tam gic ABC khng cn ngoi tip ng trn (I). ng trn (I) tip xc vi cccnh BC, CA, AB ln lt ti D, E, F. ng thng EF ct BC ti M, ng thng AD ct (I) tiN (khc D). Chng minh rng: MN tip xc vi (I).

Cc b (1), (2) c th chng minh d dng bng cc biu din theo vect.Di y trnh by cc chng minh cho b (3), (4).


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Theo chng minh iu kin th hai b im A2, B2, C2 v 3(1/2) 3(1/2) 3(1/2), ,A B C thng

hng, m theo iu gi s trn th 3( ) 3( ) 3( ), ,k k kB C cng thng hng nn theo b (2), hoc

ng thng A2B2C2 v 3(1/ 2) 3(1/ 2) 3(1/ 2)A B C song song hoc3(1/2 ) 3(1/2 )2 2

2 2 3(1/2) 3(1/2)

A BA B

A C A C= .

+ Nu 3(1/2 ) 3(1/2 )2 2

2 2 3(1/2) 3(1/2)

A BA B

A C A C= th ch rng: 1 3(1/2) 1 3(1/2) 1 3(1/2)

1 2 1 2 1 2

A A B B C C

A A B B C C= = , theo b (1) th

A1, B1, C1 thng hng, mu thun.

+ Nu A2B2C2 v 3(1/ 2) 3(1/ 2) 3(1/ 2)A B C song

song vi nhau th ch rng

3(1/2) 3(1/2) 3(1/2), ,A B C theo th t l trung

im ca 1 2 1 2 1 2, ,A A B B C C . Ta c:

3(1/2) 3(1/2) 1 1 2 2

1

( )2A B A B A B= +

,

3(1/2) 3(1/2) 1 1 2 2

1( )

2A C A C A C= +

. Suy ra:

A1B1 song song vi A2B2 v 3 (1/ 2 ) 3 (1/ 2)A B ,

A1C1 song song vi A2C2 v 3 (1/ 2 ) 3 (1/ 2 )A C .

T suy ra, A1, B1, C1 cng thnghng, mu thun.

Do ch c 1k= v1

2k= tha mn.

Vy b (3) c chng minh.*Chng minh b (4).

Gi H l giao im caEF v AI. Ta thy: IA EF .Tam gic AIF vung ti F cng cao FH nn :

2 2. .IF IH IA ID IH IA= = .Suy ra: ( . . )IDH IAD c g c .

Do : IHD IDA= .Mt khc: tam gic IDN cn ti

I nn IND IDN IDA= = . T

, ta c: IND IHD= .

T gic IDNH ni tip. Hn na, t gic IDMH cng ni tip v c 090IDM IHM= = .Do : 5 im, I, D, M, N, H cng thuc mt ng trn. Suy ra: IMNH ni tip hay

090INM IHM MN IN= = . Vy MN l tip tuyn ca (I). B (4) c chng minh.

A3(1/2)

A3(k)

C2

B2

A2A1

C1

B1

A

B C

M

C3(1/2)

B3(1/2)

B3(k)

C3(k)

I

H

M

N

E

F

D

A

B C


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*Tr li bi ton.

1. Khi1

2k= th L, M, N ln lt l trung im ca cc on AD, BE, CF.

Gi H l trc tm ca ABC v l phng tch ca H i vi ng trn Euler i qua chn 3ng cao ca ABC . Gi K l giao im ca ng thng AO1 vi ng thng BC.

Ta s chng minh rng K nm trn (O1).Tht vy:Do ABC l tam gic nhn nn O nm trong tam gic. Ta c:

02 90AOB ACB OAB ACB= = .

Khng mt tnh tng qut, gi s tia AD nm gia hai tia AO v AB. Khi :

0 090 902 2

BAC BACOAD OAB DAB ACB KAD OAD ACB= = = = +

.

Mt khc: 1

2DB DAC DCA BAC ACB= + = + nn KAD KDA= .

Ta cng c 1 1 1O A O L AO L= cn ti O1 nn

1 1O AL O LA= .

T suy ra: 1O LA KDA= hay O1L // KD, m L l trung im ca AD nn O1 l trung im ca

AK hay K thuc ng trn (O1). Do (O1) ct BC ti chn ng cao ca ABC . T suyra phng tch ca H i vi ng trn (O1) chnh l .Hon ton tng t vi cc ng trn (O2), (O3).

1

H

K

O

O

NM

L

D

E

F

A

B

C


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Do H c cng phng tch n cc ng trn (O1), (O2), (O3)nn H chnh l tm ngphng ca 3 ng trn (O1), (O2), (O3).

Hn na: do OA l tip tuyn ca (O1) ti A nn phng tch ca O i vi (O1) chnh l2OA . Tng t nh vy, phng tch ca O i vi ng trn (O2) v (O3) ln lt l2OB , 2OC , m O l tm ng trn ngoi tip ca ABC nn OA = OB = OC hay O c cng

phng tch n cc ng trn (O1), (O2), (O3), suy ra: O cng l tm ng phng ca 3 ngtrn (O1), (O2), (O3).

Gi s ca 3 ng trn (O1), (O2), (O3)c 3 trc ng phng khc nhau th chng phing quy ti tm ng phng, m O v H cng l tm ng phng ca chng nn O phi trngvi H hay ABC u, mu thun vi gi thit ABC khng cn.

Do , iu gi s trn l sai v 3 ng trn cho phi c 1 trc ng phng chung,trc ng phng chnh l ng thng i qua O v H. Ta cng thy rng O nm ngoi c 3ng trn, H th nm gia cc ng cao ca ABC nn n nm trong c 3 ng trn.Suy ra ng thng OH ct c 3 ng trn ti 2 im no .

Vy 3 ng trn (O1), (O2), (O3) c ng 2 im chung, hn na, ng thng i quahai im chung chnh l ng thng OH v do , n cng s i qua trng tm ca tam gic(ng thng Euler). Ta c pcm.

2. Ta s chng minh rng ba ng trn (O1), (O2), (O3) c ng hai im chung khi v ch

khi 0k= hoc1

2k= . Tht vy:

*iu kin :

- Khi1

2k= , khng nh chng minh cu 1/.

- Ta s tip tc chng minh rng vi k = 1, ba ng trn (O1), (O2), (O3) ln lt i quaL, tip xc vi OA ti A ; i qua M tip xc vi OB ti B v i qua N tip xc vi OC ti C cngc ng hai im chung. Tht vy:

- Khi k = 1, cc im L, M, N tng ng trng vi cc im D, E, F.Theo chng minh cu 1/, ng trn (K, KA) i qua D v tip xc vi OA ti A nn

chnh l ng trn (O1) ang c xt. Gi d1, d2, d3 l tip tuyn ca ng trn (O) ln lt

ti A, B, C. Gi X, Y, Z theo th t l giao im ca 2 3 3 1 1 2, ; , ; ,d d d d d d .

V O1 thuc ng thng BC v OA tip xc vi (O1) ti A nn O1 thuc d1, t suy raO1 chnh l giao im ca BC v d1.

Tng t: O2, O3 ln lt chnh l giao im ca CA v d2, AB v d3.Qua cc im O1, O2, O3 v cc tip tuyn ti ng trn (O) ln lt l 1 1 2 2 3 3, ,O T O T O T

(trong T1, T2, T3 l cc tip im).

Ta c: 1 1 1 2 2 2 3 3 3, ,O T O A O T O B O T O C = = = , tc l T1, T2, T3 cng tng ng thuc cc ng

trn (O1), (O2), (O3).


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Theo b (4) trn, (xt tam gic XYZ c (O) l ng trn ni tip) cc ng thngAT1, BT2, CT3 tng ng trng vi cc ng thng AX, BY, CZ.Hn na, XB = XC, YC = YA, ZA = ZA nn:

. . 1AY BZ CX

AZ BX CY= AX, BY, CZ ng quy (theo nh l Ceva o trong tam gic XYZ).

Do : AT1, BT2, CT3 ng quy. t im chung ca ba ng thng l S, r rng Snm trong (O). Do T1, T2, T3 nm trn (O) nn theo tnh cht phng tch:

1 2 321 3 /( ) /( ) /( ). . . S O S O S OSA ST SB ST SC ST P P P = = = = .

Tng t cu 1/, ta c:1 2 3/( ) / ( ) / ( )O O O O O O

P P P= = , tc l OS l trc ng phng chung ca ba

ng trn O1), (O2), (O3).Mt khc, S nm trong c ba ng trn, O nm ngoi c ba ng trn nn ng thng

OS ct c ba ng trn ti hai im, tc l (O1), (O2), (O3) c ng hai im chung.Vy trong trng hp k = 1, ba ng trn (O1), (O2), (O3) cng c ng hai im chung.

iu kin ca khng nh trn c chng minh.

S

T3

T2

T1

O2

O3

O1

Z

X

Y

O

CB

A


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*iu kin cn:Vi mt gi tr 0, 1k k> , gi 1( ) 2( ) 3( ), ,k k kO O O ln lt l tm ca cc ng trn i qua

L, tip xc vi (O) ti A; i qua M, tip xc vi (O) ti B, i qua N, tip xc vi (O) ti N.Gi s cc ng trn

( ) ( ) ( )1( ) 2( ) 3( ), ,

k k kO O O ni trn c

ng hai im chung, tc l ba

tm ca chng l 1( ) 2( ) 3( ), ,k k kO O O

thng hng. (1)Gi d1, d2, d3 l tip tuyn

ca ng trn (O) ln lt ti A,B, C. Gi X, Y, Z theo th t lgiao im ca d2, d3; d3, d1; d1, d2.Chng minh tng t nh trn,AX, BY, CZ ng quy. (2)

t O1, O2, O3 l giaoim ca BC vi YZ, CA vi ZX,AB vi XY. D thy rng:

1( ) 2( ) 3( ), ,k k kO O O ln lt thuc cc

on thng 1 2 3, ,AO BO CO

v 1( ) 2( )

1 2

,k kAO BOAL BM

AD BEAO BO= = ,

3( )

3

kCO CN

CFCO = .

Suy ra:

1( ) 2( ) 3( )

1 2 3

k k kAO BO CO

AO BO CO= = = . (3)

T (1), (2), (3), p dng b 3, ta c 1k = hoc1

2k= .

Do , nu cc ng trn (O1), (O2), (O3) c ng hai im chung th 1k = hoc1

2k= .

iu kin cn ca khng nh c chng minh.Vy tt c cc gi tr k cn tm l 1k = v

1

2k= .

Bi ton c gii quyt hon ton.

O1(k)

S

O2

O3

O1

Z

X

Y

O

CB

A

O3(k)

O2(k)


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Bi 6. K hiu M l tp hp gm 2008 s nguyn dng u tin. T tt c cc sthuc M bi ba mu xanh, vng, sao cho mi s c t bi mt mu v mi mu uc dng t t nht mt s. Xt cc tp hp sau:

31 {( , , ) ,S x y z M = trong x, y, z c cng mu v ( ) 0 (mod 2008)}x y z+ + ;

3

1

{( , , ) ,S x y z M = trong x, y, z i mt khc mu v ( ) 0 (mod 2008)}x y z+ + .

Chng minh rng 1 22 S S> . (K hiu3 l tch cc M M ) .

*Trc ht ta s chng minh b sau:Vi n l s nguyn dng, xt tp hp M = {1, 2, 3, , n}. T mu cc phn t ca S bi

mu xanh hoc . Xt cc tp hp sau:3

1 {( , , ) ,S x y z M = trong x, y, z cng mu v ( ) 0 (mod )}x y z n+ + ;3

2 {( , , ) ,S x y z M = trong x, y, z khc mu v ( ) 0 (mod )}x y z n+ + .

Gi s rng trong M c a s c t mu v b s c t mu xanh (vi a + b = n) th2 2

1| |S a ab b= + , 2| | 3S ab= .

Tht vy:Ta chn mt s x c t mu , mt s y c t mu xanh,x y . Gi s z l mt s

thuc S m | ( )n x y z+ + , r rng z tn ti v duy nht (z c th trng vi x ho y).Khng mt tnh tng qut, gi s z c t mu , cng mu vi x.

Xt hai trng hp:- Nu z khc c x v y. Khi c ba s x, y, z l phn bit. Khi , ta c tt c 6 b bathuc tp S2 cha c x v y l: ( , , ); ( , , ); ( , , ); ( , , ); ( , , ); ( , , )y z x z y y x z y z x z x y z y x .Nu khng tnh n th t ca x v y th ch c 3 b trong cc b trn thuc S2. Tht vy: khi xt

x ng trc z trong cc b trn, ta c 3 b ba l: ( , , ); ( , , ); ( , , )y z x z y y x z .Tng t, li xt cc b khng tnh n th t ca z v y (ch rng trm, ta gi s x

v x cng mu), xt x ng sau z trong cc b ny, ta cng c 3 b ba na l:( , , ); ( , , ); ( , , )z x y z y x y z x . Do , trong trng hp ny, ta c tt c 3 b thuc S2.

- Nu z bng x hoc z bng y. Khng mt tnh tng qut, gi s z = x (trng hp z = yhon ton tng t, khng quan tm n mu ca chng na). Khi , ta cng c cc b 3 thuctp S2 cha c x v y l: ( , , ); ( , , );( , , )x x y x y x y x x (ch xt cc b khng tnh n th t ca x, y).

Do , trong c hai trng hp, mi b khng tnh n th t (x, y) vi x, y khc munhau cho ta ng 3 phn t trong tp T2 v mi phn t nh vy xut hin ng mt ln. Suy ragi tr ca |S2| bng 3 ln s b khng tnh n th t (x, y) nu.

Mt khc: c a s c t mu , b s c t mu xanh nn s b (x, y) ni trn chnh lab, t ta c: 2| | 3S ab= . Vi x, y cho trc th s z tha mn | ( )n x y z+ + l duy nht.

C x v y c chn ng n ln nn 2 21 2| | ( )S S n a b = = + , hn na: 1 2S S = nn2 2 2 2

1 2 1| | | | ( ) | | ( ) 3S S a b S a b ab a ab b+ = + = + = + .

B c chng minh.


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*Tr li bi ton:Gi s s cc s c t mu xanh, vng, ln lt l a, b, c th : 2008; , , *a b c a b c+ + = .

-Xt tp hp:3{( , , )A x y z M= | x, y, z c t cng mu xanh v 0 (mod 2008)}x y z+ + .

Cc tp B, C nh ngha tng t, ng vi cc mu vng v .

-Xt tp hp: 3{( , , )AB x y z M= | x, y, z c t bi hai mu xanh, vng v 0 (mod 2008)}x y z+ + Cc tp BC, CA c nh ngha tng t, ng vi cc cp mu vng, v , xanh.

- Xt tp hp :3{( , , )ABC x y z M= |x, y, z c t bi c ba mu xanh, vng, v 0 (mod 2008)}x y z+ +

Tip theo, ta s dng b trn nh gi s phn t ca cc tp hp trn:Gi c l mu i din cho hai mu xanh v vng, khi : s b ba c t cng mu chnh l:A B C AB v s b ba t khc mu chnh l: ABC BC CA .Ta c: 2 2| | | | | | | | ( ) ( ) , | | | | | | 3 ( )A B C AB a b c a b c ABC BC CA c a b+ + + = + + + + + = + .Hon ton tng t, ta c:

2 2

| | | | | | | | ( ) ( ) ,| | | | | | 3( ( )A B C BC b c a b c a ABC CA AB a b c+ + + = + + + + + = + .2 2| | | | | | | | ( ) ( ) ,| | | | | | 3 ( )A B C CA c a b c a b ABC AB BC b c a+ + + = + + + + + = + .Theo cch xc nh nh trn th:

1 1| | | | | | | |S A B C S A B C = = + + , 2 2| | | |S ABC S ABC = =

Cng tng v tng ng ca nhm th nht ri nhn vi hai, ta c:2 2 26(| | | | | |) 2(| | | | | |) 3( )A B C AB BC CA a b c+ + + + + = + + .

Cng tng v tng ng ca nhm th hai, ta c:3 | | 2(| | | | | |) 6( )ABC AB BC CA ab bc ca+ + + = + + .

Suy ra: 2 2 26(| | | | | |) 3 | | 3 ( ) ( ) ( ) 0A B C ABC a b b c c a + + = + + .

Do : 1 22 | | | | 0S S .ng thc khng xy ra v khng tn ti a b c= = nguyn dng v 2008a b c+ + = .Vy bt ng thc trn l thc s, tc l 1 2 1 22 | | | | 0 2 | | | |S S S S > > .

y chnh l pcm.


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LI GII THICHN I TUYN QUC GIA

D THI IMO 2009

Bi 1. Cho tam gic nhn ABC ni tip ng trn (O). Gi 1 1 1, ,A B C v 2 2 2, ,A B C

ln lt l cc chn ng cao ca tam gic ABC h t cc nh , ,A B C v cc im ixng vi 1 1 1, ,A B C qua trung im ca cc cnh , ,BC CA AB. Gi 3 3 3, ,A B C ln lt l cc

giao im ca ng trn ngoi tip cc tam gic 2 2 2 2 2 2, ,AB C BC A CA B vi (O).

Chng minh rng: 1 3 1 3 1 3, ,A A B B C C ng quy.

Ta s chng minh cc ng thng

1 3 1 3 1 3, ,A A B B C C cng i qua trng tm ca tam gic

ABC. Tht vy:

Gi M l trung trc ca BC, A l im i xng vi Aqua trung trc ca BC. Ta s chng minh rng A trngvi A3 hay ng trn (AB2C2) ct (O) ti A.

Ta c: A, A i xng nhau qua trung trc ca BC

nn: , AB A C AC A B= = . Do A, B v C1, C2cng i xng vi nhau qua trung im ca AB nn

2 1BC AC= . Tng t: 2 1CB AB= . Suy ra:

2 1

2 1

'

'

BC AC AC A B

CB AB AB A C = = = .

Kt hp vi 3 3' 'C BA B CA= ( cng chn cung AA)

ta c: 2 2 2 2 2 2' ' ( . . ) ' ' ' 'C BA B CA c g c BC A CB A AC A AB A = = .

Do , t gic AC2B2A l t gic ni tip hay A trng vi A3. Gi G l giao im ca

trung tuyn AM vi A1A3. Do AA3 // A1M nn: 3

1

2AAAGGM A M

= = G l trng tm ca tam gic

ABC hay ng thng A1A3 i qua trng tm G ca tam gic ABC.

Tng t: 1 3 1 3,B B C C cng i qua G.

Vy cc ng thng 1 3 1 3 1 3, ,A A B B C C ng quy. Ta c iu phi chng minh.

3

3

3

2

2

21

1

1

O

B

A

C

C

C

B

AA

B

A

B C

1

2

2 1

1

G

A'

B

C

C

B

MA

A

B C


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Bi 2.

Cho a thc 3 2( ) 1P x rx qx px= + + + trong , ,p q r l cc s thc v 0r > .

Xt dy s sau:2

1 2 3

3 2 1

1, ,

. . . , 0n n n n

a a p a p q

a p a q a r a n+ + +

= = =

=

Chng minh rng: nu a thc ( )P x c mt nghim thc duy nht v khng c

nghim bi th dy s ( )na c v s s m.

* Gi s k l mt nghim (thc hoc phc) ca a thc:3 2( )Q x x px qx r = + + + , do r > 0 nn 3 20 0k k pk qk r + + + = (*)

Theo gi thit, a thc 3 2( ) 1P x rx qx px= + + + c ng mt nghim thc nn n cn c thm

hai nghim phc lin hp na, ng thi1

k chnh l nghim ca P(x) do:3 2 2 3

3

1 1 1 1( ) 1 0

r qk pk k P r q p

k k k k k

+ + + = + + + = =

.

Xt dy s (un) xc nh bi cng thc:

1 3 2 1( )n n n nr

u a p k a ak+ + + +

= + + (**)

Mt khc, theo gi thit: 3 2 1 , 0,1, 2,...n n n na pa qa ra n+ + += =

Ta c:1 2 1 2 1 2 1

2 12

( ) .

( . )

n n n n n n n n n

n n n

r kq r u pa qa ra p k a a ka a ra

k kkq r r

k a a ak k

+ + + + + + +

+ +

+= + + =

+=

T (*) 2 32

( )kq r

kq r pk k p k k

+ + = + = + , do :

1 2 1( ( ) ) , 0,1, 2,...n n n n nr

u k a p k a a ku nk+ + +

= + + = =

Trong (**), cho n = -1, ta c:2 3

2 20 2 1 0( ) ( )

r r pk qk r k u a p k a a p q p k p k

k k k k

+ += + + = + = = =

Suy ra: 2 22 1( ) , 0,1, 2,...n n

n n n n

ru k a p k a a k n

k+ +

+ += + + = = (***)


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G

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