1 THI V LI GII CHN I TUYN QUC GIAD THI OLYMPIC TON QUC TCA VIT NAM T NM 2005 N NM 20102 PHN I ***** BI3 THI CHN I TUYN QUC GIA D 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 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 FNng 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 ra k 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 33 3 33( ) ( ) ( ) 8a b ca b b c c a+ + + + + trong , , a b cl cc s thc dng. Bi 5. Cho s nguyn t( 3) p p > . Tnh: a) 12 22122pkk kSp p= ((= (( nu1 (mod 4) p . b) 1221pkkSp= (= ( nu1 (mod8) p . 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 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 qu2n sao cho vi 1 2, s s l hai s bt k thuc S ta lun c1 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 khi3 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 cho AL BM CNkAD 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 vi 12k = , 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 s thuc M bi ba mu xanh, vng, sao cho mi s c t bi mt mu v mi mu u c 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 rng1 22 S S > . (K hiu 3Ml tch - ccM M M ) . www.VNMATH.com7 THI CHN I TUYN QUC GIA D THI IMO 2009 *Ngy thi th nht. Bi 1. Cho tam gic nhn ABC ni tip ng trn (O). Gi 1 1 1, , A B Cv 2 2 2, , A B Cln lt l cc chn ng cao ca tam gic ABC h t cc nh A, B, C v cc im i xng vi 1 1 1, , A B C qua trung im ca cc cnh, , BC CA AB. Gi 3 3 3, , A B Cln lt l cc giao im ca ng trn ngoi tip cc tam gic 2 2 2 2 2 2, , AB C BC A CA Bvi (O). Chng minh rng: 1 3 1 3 1 3, , A A B B C Cng quy. Bi 2. Cho a thc3 2( ) 1 P x rx qx px = + + +trong , , p q rl cc s thc v 0 r > . Xt dy s( )naxc nh nh sau:21 2 33 2 11,, . . . , 0n n n na a p a p qa p a q a r a n+ + + = = = = Chng minh rng: nu a thc( ) P xc 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 bsao cho, a bvabu khng phil s chnh phng. Chng minh rng trong hai phng trnh sau: 2 22 211ax byax 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 2a b cr r r rb 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 vi ng 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 trn ng 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 qua mt im c nh khi M di ng pha trong ng trn. Bi 6. Mt hi ngh ton hc c tt c64 n +nh ton hc phi hp vi nhau ng 21 n + ln( ) 1 n . 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 1 n = . b/ Hi rng Ban t chc c th sp xp c ch ngi c hay khng vi mi 1 n > ?www.VNMATH.com8 THI CHN I TUYN QUC GIA D THI IMO 2010 * 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 Ol 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( ) Ov tip tuyn ti Q ca2( ) Ophi 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 hnh ch nht c kch thc 2 3 , b i 2 gc cho nhau (tc l c 4 vung nh) l hnh ch nht 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 cl cc s thc dng tha mn iu kin:1 1 116( ) a b ca b c+ + + + .Chng minh rng: 3 3 31 1 1 89 ( 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( ) 1 n k > > . Ngi ta chia. n k ngi ny thnh n nhm, mi nhm c k ngi sao cho khng c hai ngi 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 nSl tng bnh phng cc h s trong khai trin ca nh thc(1 )nx + , trong n l s nguyn dng; x l s thc bt k. Chng minh rng: 21nS +khng chia ht cho 3 vi mi n. www.VNMATH.com9 PHN II ***** LI GII10 LI GII THICHN 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 lt l 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 trnC ). 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 v t ca cc cp ng trn (O1) v (O2), (O2) v (O3), (O3) v (O1).Chng minh rng nu trong cc tm v t , c ba tm v t ngoi hoc hai tm v t trong, mt tm 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 Rny i mt khc nhau.Theo tnh cht v tm v t, ta c: 1 12 2( 1)aAO RR AO = . Tng t: 2 23 3( 1)bBO RR BO = , 3 31 1( 1)cCO RR CO = , trong , mi s , , a b cnhn 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 : 3 1 22 3 1. . 1CO AO BOAO BO CO =, theo nh l Menelaus o cho tam gic 1 2 3OO 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 Kthng 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. BCAO1O2O3www.VNMATH.com11 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 rng trc 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 ca on EF nn M thuc ng thng AI hay A, M, I thng hng. Tng t: B, N, I v C, P, I cng thng hng. Xt php nghch o tm I, phng tch 2rvi 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: : M A . Tng t:: , N B P C .Do :: MNP ABC . Gi E l tm ng 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 tam gic 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 trn ng thng OI.Ta c pcm. NMHKPFEDIOBCAHPNMFEDIOAB Cwww.VNMATH.com12 Bi 2.Trn mt vng trn c n chic gh c nh s t 1 n n. Ngi ta chn ra k 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. *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 dng1 2( , ,..., )ki i itrong 1 2, ,...,ki i il 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 tT 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 1tj 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 tj 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 b1 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= ' Tca chng cng phi khc nhau. Nhng iu ny l hin nhin do hai b ny l khc nhau nn tn ti ch sssao 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. www.VNMATH.com13 *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 xem nh c t mu ; giX 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 ' Xl cch t mu tha mn c mt im c t mu xanh thuc 1 2 3{ , , ,..., }pA A A Av '' \ ' X X X = , khi r rng, vi mi phn t thuc" Xth khng c im no c t mu xanh 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'iXtrong mi phn t ca'iXc ng mt im iAc t mu xanh, 1, i p = ; khi r rng' ' ,i jX X i j = v 1' 'piiX X==. Vi mi1, i p = , theo b trn, ta thy: 1 11 ( 1) 1k kn p k p n kpC C = , tc l cc tp'iXny c cng s phn t. Suy ra: 11'kn kpX pC = . Do : 11' ''k kn kp n kpX X X C pC = + = + . Thay3 p = , ta c s cch chn gh tng ng trong bi l 13 3 13k kn k n kC C + . Vy s cch chn gh tha mn tt c cc iu kin ca bi l: 13 3 13k kn k n kC C + . www.VNMATH.com14 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 10 n >trong tng trng hp n chn v n l.-Vi n l s l, t2 1 n k = + . Ta cn tm mt ng thc ng vi mi k m trong 2 1 k +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 38kxut hin trong( )32 1 k + , ta chn( )32 1 k ; 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 s nh 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 32 2 2 3 3 32 2 3 3 3(2 1) (2 1) ( ) ( )24 2 6 2(24 6 ) (2 2 ) 0k k ak b ak b c dk a bk b c dk a b b c d ((( + = + + + + = + + + + = Ta cn chn, , , a b c dsao cho 2 3 3 34,2 2 a b b c d = + + =trong 2 a . D thy2 a v nu2 a =th t 24 1 a b b = = , trng vi biu thc cn nh gi; do , 1, 4 a b = = , suy ra: 3 3126 c d + = , ta chn c5, 1 c 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, t2 2 n 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 , thay0 x y z = = = , ta c: 3 2(0) 3 (0) (0) 0 3 (0) 1 f f f f = = = . Do hm ny ch ly gi tr trn: f nn khng th c 23 (0) 1 f = , tc l(0) 0 f = . www.VNMATH.com15 -Thay0 y z = = , ta c: 3 3 3 3 3( ) ( ( )) ( (0)) ( (0)) ( ( )) f x f x f f f x = + + = . -Li thayy 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: -Vi1 k = , trong gi thit, thay1, 0 x y z = = = , ta c 3 3(1) (1) ( 1) (1) f f f f = = -Vi2 k = , trong gi thit, thay1, 0 x y z = = = , ta c 3 3(2) 2 (1) ( 2) 2 (1) f f f f = = -Vi3 k = , trong gi thit, thay1 x y z = = = , ta c 3 3(3) 3 (1) ( 3) 3 (1) f f f f = = -Thay2, 0 x y z = = = , ta c 3 3 3 3(8) (2) (2 (1)) 8 (1) ( 8) 8 (1) f f f f f f = = = = . -Thay2, 1, 0 x y z = = = , ta c 3 3 3 3(9) (2) (1) 9 (1) ( 9) 9 (1) f f f f f f = + = = . -Thay2, 1 x y z = = = , ta c: 3 3 3 3(10) (2) 2 (1) 10 (1) ( 10) 10 (1) f f f f f f = + = = . -Thay2, 1, 0 x y z = = = , ta c: 3 3 3 3(7) (2) (1) 7 (1) ( 7) 7 (1) f f f f f f = = = . -Thay2, 1 x 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 thay2 k = , 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 thay1 k = , 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 mi10 k . Vi10 k > , xt0 k > 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), 10 f k kf k = > . Vi10 k < 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, thay1, 0 x y z = = = , ta c: 3(1) (1) (1) 1 (1) 0 f f f f = = = . -Nu(1) 1 f = th( ) , f k k k = , th li thy tha. -Nu(1) 0 f =th( ) 0, f k k = , th li thy tha. -Nu(1) 1 f =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 = . www.VNMATH.com16 Bi 4. Chng minh rng: 3 3 33 3 33( ) ( ) ( ) 8a b ca b b c c a+ + + + + trong , , a b cl 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 21 1 1 11(1 ) (1 ) (1 ) (1 ) a b c d+ + + + + + +. Tht vy: Ta thy vi hai s thc dng ty th: 2 21 1 2(1 ) (1 ) 1 x y xy+ + + + (*) 2 22 2 222 2 22 2 3 3 2 22 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 ) 11x yx y xy xy x yxy x y xyx y x y xy xy x y xy x yx y x y x y xy x y xy xyx y x y x y xy xy xy x yx+ + + ( + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +2 2 2 2 2 2( ) 2 ( ) (1 ) 0 y x y xy x y xy x y xy + + + Do : 2 2 2 21 1 1 1 1 1 2 21(1 ) (1 ) (1 ) (1 ) 1 1 1 2ab cd ab cda b c d ab cd ab cd abcd ab cd+ + + ++ + + + = = =+ + + + + + + + + + +. Do b c chng minh. ng thc xy ra khi v ch khi1 a b c d = = = = . Trong b trn, thay, , , 1 a x b y c z d = = = = , ta c kt qu sau: Vi x, y, z l cc s thc dng v 1 xyz =th:2 2 21 1 1 3(1 ) (1 ) (1 ) 4 x y z+ + + + +. ng thc xy ra khi v ch khi1 x y z = = = . *Tr li bi ton cho: t, , , , 0; 1b c ax y z x y z xyza b c= = = > = .BT cho ban u tng ng vi: 3 3 33 3 31 1 1 3 1 1 1 38 (1 ) (1 ) (1 ) 8(1 ) (1 ) (1 )b c ax y za b c+ + + + + + ++ + +. Theo BT Cauchy cho cc s dng, ta c: www.VNMATH.com17 33 3 6 2 3 21 1 1 1 3 1 1 3 1 13 . .(1 ) (1 ) 8 8.(1 ) 2 (1 ) (1 ) 4 (1 ) 16 x x x x x x+ + = + + + + + +. Hon ton tng t:3 21 3 1 1.(1 ) 4 (1 ) 16 y y + +, 3 21 3 1 1.(1 ) 4 (1 ) 16 z z + +. Cng tng v cc BT ny li, ta c: 3 3 3 2 2 21 1 1 3 1 1 1 3.(1 ) (1 ) (1 ) 4 (1 ) (1 ) (1 ) 16 x y z x x x (+ + + + (+ + + + + + . Ta cn chng minh:2 2 23 1 1 1 3 3.4 (1 ) (1 ) (1 ) 16 8 x x x (+ + (+ + + 2 2 21 1 1 3(1 ) (1 ) (1 ) 4 x 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 khi1 x y z a b c = = == = . www.VNMATH.com18 Bi 5.Cho s nguyn t ( 3) p p > . Tnh: a. 12 22122pkk kSp p=| |((= | (( \ nu1 (mod 4) p . b. 1221pkkPp= (= ( nu1 (mod8) p . *Trc ht, ta s chng minh hai b sau:(1) B 1: Vi p l s nguyn t tha 1 (mod 4) p th mi s t nhin a vi: 112pa s tn ti duy nht st nhin b tha 112pb p+ v: 2 20(mod ) a b p + . *Chng minh: Theo nh l Wilson:( 1)! 1(mod ) p p . Vi mi 11, 2, 3,...,2pk= , ta thy:2(mod ) ( ) (mod ) p k k p k p k k p . Kt hp vi gi thit 11(mod 4) 22pp , ta c: 2 2121 11 ( 1)! ( 1) . ! ! (mod )2 2pp pp p| | | | | | | | ||||\ \ \ \ . t 21! 1(mod )2pp | |= |\ . Vi mi 112pa , ta chn 112pb 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 x bng 1 nu 1{ } 12x . ( y k hiuxl 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 2A B B C C Axy yz zx + + = + + = .Do : 222 2 2 2( )1 ( )( )( 1)( 1) ( )( ) ( )( )( )( )x xyx x y x zSy z y xy z xy y z y x z x z y++ + += = = =+ + + + + + + + 21( ) x y=+. p dng b trn, ta c: 9 94( ) 4Sxy yz zx =+ +. Vy gi tr nh nht ca S l 94 t c khi v ch khix y z = = hay ABC l tam gic u. www.VNMATH.com39 Bi 4.Tm tt c cc hm s lin tc: f tha mn: 21( ) ( )3 9xf x f x = + +vi mix. Ta c: 2 2 21 1 1 1 1( ) ( ) (( ) ),3 9 6 12 6 12xx x f x f x x + + = + + = + + .t 1 16 6y x x y = + = . Thay vo gi thit cho, ta c: 21 1( ) ( )6 12f y f y = + . (*) Xt hm s:( ) : g x tha mn: 1 1( ) ( ) ( ) ( ),6 6g x f x f x g x x + = = . (1) T (*), ta c: 21( ) ( ),4g x g x x = + , r rng ( ) g x cng lin tc. Ta s xc nh hm s( ) g xtha mn iu kin trn. Ta thy: 2 21 1, ( ) ( ) (( ) ) ( )4 4x g x g x g x g x = + = + = nn( ) g x l hm s chn.Ta ch cn xt0 x . Ta c hai trng hp: -Vi 012x : Xt dy s: 21 0 11, , 14n nu x u u n+= = + . Khi : 211( ) 02n n nu u u+ = nn dy( )nu tng. Mt khc, bng quy np, ta chng minhc 1,2nu n , tc l dy ny b chn trn. T suy ra n c gii hn. Gi t l gii hn th 21 14 2t t t = + = . Do : 01 1( ) ( ), 0;2 2g x g x (= ( . -Vi 012x > : Tng t nh trn, xt dy s 1 0 11, 0, 14n nv x v v n+= = . Khi : 211( )1204 14nn n n nn nvv v v vv v+ = = < +nn dy cho l dy gim. Cng bng quy np, ta chng minh c 12nv > , 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 2g x g x = + . t 1( )2g a = . T suy ra:( ) , 0 g 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. www.VNMATH.com40 Bi 5. Cho A l tp con cha 2007 phn t ca tp:{ } 1, 2, 3,..., 4013, 4014tha 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 aPcha tt c cc s nguyn dng c dng2 (2 1)na , trong n l mt s khng m. Khi , tp hp con A ca { } 1, 2, 3,..., 4013, 4014khng 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 aPni trn. Gi il 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 1 phi 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 nui j (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. www.VNMATH.com42 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 Anh hnh v. Trc ht, ta thy rng di cc cnh v cc ng cho ca hnh (H) ch thuc 4 gi tr khc nhau (nu gi R l bn knh ng trn ngoi tip ca (H) th ta d dng tnh c cc gi tr l 2 .sin9R,22 sin9R,32 sin9R,42 sin9R) ta t chng l 1 2 3 4, , , a a a atheo th t tng dn ca di. R rng cc tam gic c nh thuc cc nh ca (H) s c cnh c di thuc 1 trong 6 dng sau: 1 1 2 2 2 4 1 3 43 3 3 2 3 4 4 4 1( , , ), ( , , ), ( , , ),( , , ), ( , , ), ( , , )a a a a a a a a aa 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 cc cnh l 32 .sin9R; 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 l 32 .sin9R, 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 l 32 .sin9R, 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 s chia cho 3 cng d l 2, mu thun. Do , trong hai tam gic lp c, lun c mt cnh c di l 32 .sin9R. Suy ra trng hp ny lun c tam gic tha mn bi. a1a2a3a4987654321AAAAAAAAAwww.VNMATH.com43 -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 l1 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 l2 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 khng cn 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 trn nn phi c hai b trng nhau, tc l c t nht hai tam gic cn bng nhau v mt tam gic cn nhn mt trong ba gi tr thuc mt trong cc b trn lm cnh, khi lun c th chn t tam gic 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. www.VNMATH.com44 LI GII THI CHN I TUYN QUC GIA D THI IMO 2008 Bi 1. Trong mt phng cho gc xOy. Gi M, N ln lt l hai im ln lt nm trn cc 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 cho IM 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. 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 MON . Do d l phn gic ngoi ca

MON nn I chnh l trung im ca cung

MON , 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:

NIO NMO = . Mt khc: do IM = IN = IP = IQ nn t gic MNPQ ni tip trong ng trn tm I, ng knh PQ 2 PIN PMN = (gc ni tip v gc tm cng chn cung

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

OMN . xydJFEKPQIOMNwww.VNMATH.com45 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 caMON , suy ra K thuc phn gic trong ca

xOy , 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

MON ca tam gicMON , tc l K cng thuc phn gic trong 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 ng knh 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

MONi qua trung im ca cung

MJN . R rng M, N i xng nhau qua trung trc ca MN nn JM = JN, tc l J cng l trung im ca cung

MON .T suy ra: J thuc phn gic trong ca

MON hay O, K, J thng hng. Ta cn chng minh cc on JO, EN v MF trong JEF ng quy. Tht vy:. .sin sin.. .sin sinOEJOFJS OE JO JE OJE JE OJEOF S JO JF OJF JF OJF= = = . TrongJEF vMON , ta c : sin sin,sin sinJE JFE OM ONMJF JEF ON OMN= = . Mt khc :

sin sin,sin sinOJE ONMOJE OJN ONM OJF OJM OMNOJF OMN= = = = = . Kt hp li, ta c :sin sin sin. . .sin sin sinOE JFE OM OFN OM OFN OMOF JEF ON OEM ON OEM ON= = = sin .sin..sin sinOFN OM MOE MEON NOF OEM NF= = . Do :. 1 . . 1OE FN OE NF MJOF EM OF NJ ME= = . Theo nh l Ceva o, ta c OI, EN v MF ng quy. y chnh l pcm. www.VNMATH.com46 Bi 2. Hy xc nh tt c cc s nguyn dng m sao cho tn ti cc a thc vi h s thc( ), ( ), ( , ) P x Q x R x y tha mn iu kin:Vi mi s thc a, b m 20ma b = , ta lun c( ( , )) P R a b a =v( ( , )) Q R a b b = . Vi m l mt s nguyndng, ta xt cc trng hp : -Nu m l s chn, t2 , * m k k = . Suy ra : 20m ka b b a = = . Khi cn tm k sao cho cc a thc( ) ( ) ( ) , , , P x Q x R x ytha 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 ( ) 1 P x T x =haydeg ( ) deg ( ) 1 P x T x = = . Gi s( ) , , , 0 T x ux v u v u = + ,( ) '. ', ', ' , ' 0 P x u x v u v u = + th '( ) ' , . ' 1, ' ' 0 u ux v v x x u u u v v + + = = + =hay 1' , ' ( )v x vu v P xu u u= = = . Mt khc :( ( , )) ( ( )) ( ) ( ) ( ),kk 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 20ma 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 khng tha mn bi. -Nu m l s l. t 2( , ) ( )mP x x S x = . Suy ra : 2( ( )) , deg ( ).deg ( ) 2 P S x x x P x S x = = . Nu nh degS(x) = 2 thdeg ( ( )) Q S xl s chn, trong khi :deg ( ( )) degmQ 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 lmin(2, ) m m 2deg ( , ) 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 thc P(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.www.VNMATH.com47 Bi 3. Cho s nguyn 3 n > . 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 qu2n sao cho vi 1 2, s s l hai s bt k thuc S ta lun c1 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 2n, 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| |2ns 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 qu 2n khng t hn s cc s phn tln hn 2n. Xt hai tp hp sau: { | , }2nA x x S x = ,{ | , }2nB x x S x = > th, A B A B S = =v theo cch xc nh S nh trn thA B . Khi vi c l mt s nguyn dng no khng vt qu 2n, ta xt tp hp: { | } C x c x A = + . Ta c:| | | | A C = . DoA 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( ) | | | | | | | | A 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 3A B nA B++ .Hn na:, A B A B S = = v| | Sl s nguyn nn 2| | | | | |3nS A B (= + ( . Do s phn t ca tp khuyt S trong T lun khng vt qu 23n( ( . www.VNMATH.com48 *Ta s ch ra mt tp khuyt tha mn bi c ng 23n( ( phn t. Tht vy, xt hai tp hp A, B nh sau: 11, 2, 3,...,3nA + (= ` ( ),1, 2,...,3 3n nB n n n ( (= + + `( ( ) vS A B = . Chn 13 2n nc+(= ( . Ta thy: -Hiu hai phn t bt k trong A khng vt qu 1 113 3n n + +(( < (( . -Hiu hai phn t bt k trong B khng vt qu 1( 1) 13 3 3n n nn n+ ( (( + = ((( . Khi , r rngS A B = l mt tp khuyt trong T ng vi gi tr 13 2n nc+(= ( . Ta s chng minh rng 2| |3nS (=( . T cch xc nh A, B, ta c:1| | ,3 3n nA B+( (= = ( ( .Ta cn c: 1 2,3 3 3n n nn+( ((+ = ( (( (*). *Xt cc trng hp: -Nu n chia ht cho 3, tc l n c dng3 , m m . Suy ra: 1 3 1 3 6 223 3 3 3 3 3n n m m m nm m m+ +( ((((+ = + = + = = = ( (((( . -Nu n chia 3 d 1, tc l n c dng3 1, m m + . Suy ra: 1 3 2 3 1 6 6 2 223 3 3 3 3 3 3n n m m m m nm m m+ + + +( (((((+ = + = + = = = = ( ((((( . -Nu n chia 3 d 2, tc l n c dng3 2, m m + . Suy ra: 1 3 3 3 2 6 3 6 4 213 3 3 3 3 3 3n n m m m m nm m+ + + + +( (((((+ = + = + + = = = ( ((((( . T suy ra (*) c chng minh hay tp hp S cho l tp khuyt c 23n( ( . Vy gi tr ln nht ca s phn t ca tp khuyt S trong T l 23n( ( . www.VNMATH.com49 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: 11(2 3) (2 ) 3 .(2 ) .3 (2 ) 3 (mod6m)nn n n k n k k n nnkm m C m m=+ = + + +.Do ,6 | (2 3) 1nm m+ + 6 | (2 ) 3 1n nm m + + 2 | (3 1)nm +v3| (2 ) 1nm + . Cn chng minh rng: 2 | (3 1)nm +v3| (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 v3 1 2(mod8)n+ hoc3 1 4(mod8)n+ , trong khi4 8 m , tc l khng th c iu kin (2). - Xt iu kin (1): t m l s chn, suy ra3 1 4nn + l s l. Ta bit rng: s c dng 3n + 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 dng2(3 1), m k k = + . Suy ra:(2 ) 1 2 .2 .(3 1) 1 2(mod3) (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): t2 | (3 1)nm + suy ra m khng chia ht cho 3, t3| (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 th3 14n+ , kt hp vi( , 4) 1 m = , ta c4 | 3 1nm + , y chnh l iu kin (2). Do :(1) (2) . - Xt iu kin (2): t4 | 3 1 (4 | 3 1) ( | 3 1)n n nm m + + + suy ra n l s l v m c dng 3 1, k k + . Suy ra:(2 ) 1 2 . 1 1 1 0(mod3) 3| (2 ) 1n n n nm m m + = + + + ; t (2) ta cng trc tip c2 | (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 khi3 1n+chia ht cho 4m. y chnh l pcm. www.VNMATH.com50 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 cho AL BM CNkAD 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 12k = , 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. Cc ng 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 21 1 2 2A B A BkAC A C= = . Cc im A3, B3, C3 thuc a, b, c sao cho: 1 2 1 2 1 21 3 1 3 1 3A A B B C CA A B B C C= = .Khi , A3, B3, C3 thng hng v 3 33 3A BkA C = . (2)Cho ba ng thng phn bit a, b, c v ba ng thng phn bit khc a, b, c . Cc ng 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 nu 3 3 1 1 2 21 1 2 2 3 3A B A B A BAC A C A C= =th hoc 1 2 1 2 1 21 3 1 3 1 3A A B B C CA 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 31 2 1 2 1 2, 0A A B B CCk kA A B B CC= = = . Khi , A3, B3, C3 thng hng khi v ch khi1 k =hoc 12k = . (4)Cho tam gic ABC khng cn ngoi tip ng trn (I). ng trn (I) tip xc vi cc cnh BC, CA, AB ln lt ti D, E, F. ng thng EF ct BC ti M, ng thng AD ct (I) ti N (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). www.VNMATH.com51 *Chng minh b (3). + iu kin : -Vi k = 1, ta c A3, B3, C3 theo th t trng vi A2, B2, C2. V AA1, BB1, CC1 ng quy nn theo nh l Menelaus th 1 1 11 1 1. . 1A B BC C AAC B A C B = . V A2, B1, C1 thng hng nn 2 1 12 1 1. . 1A B BC C AA C B A C B = . Suy ra: 1 21 2A B A BAC A C= . Tng t: 1 2 1 21 2 1 2,BC B C C A C AB A B A C B C B= = . Nhn tng v cc ng thc trn, 2 2 2 1 1 12 2 2 1 1 1. . . . 1A B B C C A A B BC C AA C B A C B AC B A C B| | | | | |= = |||\ \ \ . Tc l A2, B2, C2 thng hng hay A3, B3, C3 thng hng. -Vi 12k = , A3, B3, C3 theo th t l trung im ca A1A2, B1B2, C1C2. Theo chng minh trn, ta c: 1 21 2A B A BAC A C= . Theo tnh cht t l thc th: 3 1 2 1 2 1 2 1 2 11 2 1 2 1 2 1 2 1 322A B A B A B AB A B AB A B AB A AAC A C AC A C AC A C AC A A A C+ = = = = = +Suy ra: 23 3 2 1 13 2 1 3 12.2A B A B A A ABA C A A A C AC| |= = |\ . Tng t: 2 23 3 1 13 1 3 1,B C C A BC C AB A B A C B C B| | | |= = ||\ \ . Nhn tng v cc ng thc trn, ta c: 2 2 23 3 3 1 1 13 3 3 1 1 1. . . . 1A B B C C A AB BC C AA C B A C B AC B A C B| | | | | |= = |||\ \ \ . Do , A3, B3, C3 th ng hng. + iu kin cn: Khi1 k , ta k hiu 3( ) 3( ) 3( ), ,k k kA B C thay cho A3, B3, C3. Gi s tn ti s k ng thi khc 1 v 12m 3( ) 3( ) 3( ), ,k k kA B C thng hng. Khi , cc im: 3( ) 3( ) 3( ), ,k k kA B Cv 3(1/ 2) 3(1/ 2) 3(1/ 2), , A B Ci mt khc nhau. D thy: 2 3(1/ 2) 2 3(1/ 2) 2 3(1/ 2)2 3( ) 2 3( ) 2 3( )1/ 2 11k k kA A B B C Ck A A B B C C= = =.B3A3C3C2B2A2A1C1B1AB CMC2B2A2A1C1B1AB CMwww.VNMATH.com52 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 kA B Ccng 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 hoc 3(1/ 2) 3(1/ 2)2 22 2 3(1/ 2) 3(1/ 2)A BA BA C A C= . + Nu 3(1/ 2) 3(1/ 2)2 22 2 3(1/ 2) 3(1/ 2)A BA BA C A C= th ch rng: 1 3(1/ 2) 1 3(1/ 2) 1 3(1/ 2)1 2 1 2 1 2A A B B C CA A B B C C= = , theo b (1) thA1, 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 21( )2A B A B A B = +

, 3(1/ 2) 3(1/ 2) 1 1 2 21( )2A C AC 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 thng hng, mu thun. Do ch c1 k =v 12k = tha mn. Vy b (3) c chng minh. *Chng minh b (4). Gi H l giao im ca EF v AI. Ta thy:IA EF . Tam gic AIF vung ti F c ng 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

090 IDM IHM = = .Do : 5 im, I, D, M, N, H cng thuc mt ng trn. Suy ra: IMNH ni tip hay

090 INM IHM MN IN = = . Vy MN l tip tuyn ca (I). B (4) c chng minh.A3(1/2)A3(k)C2B2A2A1C1B1ABCMC3(1/2)B3(1/2)B3(k)C3(k)IHMNEFDAB Cwww.VNMATH.com53 *Tr li bi ton. 1.Khi 12k =th L, M, N ln lt l trung im ca cc on AD, BE, CF.Gi H l trc tm caABC v l phng tch ca H i vi ng trn Euler i qua chn 3 ng cao caABC . Gi K l giao im ca ng thng AO1 vi ng thng BC.Ta s chng minh rng K nm trn (O1).Tht vy: DoABC l tam gic nhn nn O nm trong tam gic. Ta c: 02 90 AOB ACB OAB ACB = = . Khng mt tnh tng qut, gi s tia AD nm gia hai tia AO v AB. Khi :

0 090 902 2BAC BACOAD OAB DAB ACB KAD OAD ACB = = = = +. Mt khc: 12ADB 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 caABC . T suy ra phng tch ca H i vi ng trn (O1) chnh l . Hon ton tng t vi cc ng trn (O2), (O3).1HKOONMLDEFABCwww.VNMATH.com54 Do H c cng phng tch n cc ng trn (O1), (O2), (O3) nn H chnh l tm ng phng 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 l 2OA . Tng t nh vy, phng tch ca O i vi ng trn (O2) v (O3) ln lt l 2OB ,2OC , m O l tm ng trn ngoi tip caABC 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 ng trn (O1), (O2), (O3). Gi s ca 3 ng trn (O1), (O2), (O3) c 3 trc ng phng khc nhau th chng phi ng quy ti tm ng phng, m O v H cng l tm ng phng ca chng nn O phi trng vi H hayABC u, mu thun vi gi thitABC 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 3 ng trn, H th nm gia cc ng cao caABC 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 qua hai 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 khi0 k =hoc 12k = . Tht vy: *iu kin : - Khi 12k = , 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 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 cng c 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 ra O1 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, , OT O T OT(trong T1, T2, T3 l cc tip im).Ta c: 1 1 1 2 2 2 3 3 3, , OT O A O T O B OT O C = = = , tc l T1, T2, T3 cng tng ng thuc cc ng trn (O1), (O2), (O3).www.VNMATH.com55 Theo b (4) trn, (xt tam gic XYZ c (O) l ng trn ni tip) cc ng thng AT1,BT2, CT3 tng ng trng vi cc ng thng AX, BY, CZ.Hn na, XB = XC, YC = YA, ZA = ZA nn: . . 1AY BZ CXAZ 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 S nm trong (O). Do T1, T2, T3 nm trn (O) nn theo tnh cht phng tch: 1 2 321 3 / ( ) / ( ) / ( ). . .S O S O S OSAST SB ST SC ST P P P = = = =.Tng t cu 1/, ta c: 1 2 3/( ) / ( ) / ( ) O O O O O OP 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. ST3T2T1O2O3O1ZXYOCBAwww.VNMATH.com56 *iu kin cn: Vi mt gi tr0, 1 k k > , gi 1( ) 2( ) 3( ), ,k k kO O Oln 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 Othng 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 l giao 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 giao im 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 COv1( ) 2( )1 2,k kAO BOAL BMAD BE AO BO= = , 3( )3kCOCNCF CO= .Suy ra: 1( ) 2( ) 3( )1 2 3k k kAO BO COkAO BO CO= = = . (3) T (1), (2), (3), p dng b 3, ta c 1 k =hoc 12k = . Do , nu cc ng trn (O1), (O2), (O3) c ng hai im chung th 1 k =hoc 12k = . iu kin cn ca khng nh c chng minh. Vy tt c cc gi tr k cn tm l 1 k =v 12k = .Bi ton c gii quyt hon ton. O1(k)SO2O3O1ZXYOCBAO3(k)O2(k)www.VNMATH.com57 Bi 6.K hiu M l tp hp gm 2008 s nguyn dng u tin. T tt c cc s thuc M bi ba mu xanh, vng, sao cho mi s c t bi mt mu v mi mu u c 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 + + ; 31{( , , ) , S x y z M = trong x, y, z i mt khc mu v( ) 0 (mod 2008)} x y z + + . Chng minh rng1 22 S S > . (K hiu 3M l tch ccM 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: 31{( , , ) , S x y z M = trong x, y, z cng mu v( ) 0 (mod )} x y z n + + ; 32{( , , ) , 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) th 2 21| | S a ab b = + , 2| | 3 S 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 ba thuc tp S2 cha c x v y l:( , , );( , , );( , , );( , , );( , , );( , , ) x 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:( , , ); ( , , ); ( , , ) x 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 = y hon ton tng t, khng quan tm n mu ca chng na). Khi , ta cng c cc b 3 thuc tp 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 mu nhau cho ta ng 3 phn t trong tp T2 v mi phn t nh vy xut hin ng mt ln. Suy ra gi 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 l ab, t ta c: 2| | 3 S 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 = nn 2 2 2 21 2 1| | | | ( ) | | ( ) 3 S S a b S a b ab a ab b + = + = + = + . B c chng minh. www.VNMATH.com58 *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 v0 (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 v0 (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, v0 (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 ( ) ( ) ( ) 0 A B C ABC a b b c c a( + + = + + . Do : 1 22 | | | | 0 S S . ng thc khng xy ra v khng tn tia b c = =nguyn dng v2008 a 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. www.VNMATH.com59 LI GII THI CHN 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 Cv 2 2 2, , A B Cln lt l cc chn ng cao ca tam gic ABC h t cc nh, , A B Cv cc im i xng vi 1 1 1, , A B Cqua trung im ca cc cnh, , BC CA AB. Gi 3 3 3, , A B Cln lt l cc giao im ca ng trn ngoi tip cc tam gic 2 2 2 2 2 2, , AB C BC A CA Bvi (O). Chng minh rng: 1 3 1 3 1 3, , A A B B C Cng quy.

Ta s chng minh cc ng thng 1 3 1 3 1 3, , A A B B C Ccng i qua trng tm ca tam gic ABC. Tht vy: Gi M l trung trc ca BC, A l im i xng vi A qua trung trc ca BC. Ta s chng minh rng A trng vi A3 hay ng trn (AB2C2) ct (O) tiA. Ta c: A, A i xng nhau qua trung trc ca BC nn: , AB A C AC A B = = . Do A, B v C1, C2 cng i xng vi nhau qua trung im ca AB nn 2 1BC AC = . Tng t: 2 1CB AB = . Suy ra:

2 12 1''BC AC AC A BCB 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: 312AA AGGM AM= = 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 Ccng i qua G. Vy cc ng thng 1 3 1 3 1 3, , A A B B C Cng quy. Ta c iu phi chng minh. 333222111OBACCCBA ABABC122 11GA'BCCBM AABCwww.VNMATH.com60 Bi 2. Cho a thc3 2( ) 1 P x rx qx px = + + +trong , , p q rl cc s thc v 0 r > . Xt dy s sau:21 2 33 2 11,, . . . , 0n n n na a p a p qa 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 0 k k pk qk r + + + = (*) Theo gi thit, a thc 3 2( ) 1 P x rx qx px = + + +c ng mt nghim thc nn n cn c thm hai nghim phc lin hp na, ng thi 1k chnh l nghim ca P(x) do: 3 22 331 1 1 1( ) 1 0r qk pk kP r q pk k k k k+ + + | | | | | |= + + + = = |||\ \ \ . Xt dy s (un) xc nh bi cng thc:

1 3 2 1( )n n n nru 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 12 1 2( ) .( . )n n n n n n n n nn n nr kq ru pa qa ra p k a a ka a rak kkq r rk a a ak k+ + + + + + ++ ++= + + = += T (*)2 32( )kq rkq r pk k p kk+ + = + = + , do : 1 2 1( ( ) ) , 0,1, 2,... n n n n nru k a p k a a ku nk+ + += + + = =Trong (**), cho n = -1, ta c: 2 32 20 2 1 0( ) ( )r r pk qk r ku a p k a a p q p k p kk k k k+ += + + = + = = =Suy ra: 2 22 1( ) , 0,1, 2,... n nn n n nru k a p k a a k nk+ ++ += + + = = (***) www.VNMATH.com61 Gi s z l nghim phc ca phng trnh P(x) = 0 v, ln lt l modun v argument ca z trong : , , 0 > . Ta c:(cos sin )iz e i = = + v( ) [ ] P x x nn:( ) 0 ( ) 0 ( ) 0 P z P z P z = = =do : (cos sin )iz e i = = cng l nghim ca P(x). T (***), ta c: 2 22 1 2 1( ) ( ) , ( ) ( )n nn n n n n nr ra p z a a z a p z a a zz z+ ++ + + ++ + = + + = . Theo cng thc Moavre, ta c:(cos sin ) (cos sin )n nz i z n i n = + = +nn:( )| | | || |( )| | | |22 2 22222 1cos( 2) sin( 2) cos( 2) sin( 2)2 . sin ( 2) sin ( 2)2 . sin ( 2) .2 . sin sinnn n nnnnn nz z n i n n i nz zi n ni ni z z ++ + +++++ + = + + + + + =+ += + = = Tr tng v hai ng thc: 2 211 1( ) ( ) ( ) ( )n nn nz z a r a z zz z+ ++ = | |2 22 21 1 211 2( )( ) ( ) ( ). ( )sin ( 2).sinn nn nn n n nnn nz z r z zz z a r a z z a az z z znra a + ++ ++ +++ = + =+ + = Do0 > nn xt n0 l mt gi tr nguyn dng sao cho: | | | |00 00 0 11 2sin ( 2) sin ( 2)0 . 0 0sin sinnn nn nra a +++ +< < + < .V 20r> nn 0 01,n na a+ tri du vi nhau. Do , trong hai gi tr ny c mt s m. Ta thy khi n tin ti v cc, tn ti v s gi tr n0 sao cho | |0sin ( 2)0sinn +< , m ng vi mi gi tr n0 nh th ta li tm c mt s hng m ca dy cho, tc l dy (an) c v s s m. y chnh l iu phi chng minh. www.VNMATH.com62 Bi 3. Cho cc s nguyn dng a, b sao cho, a bv. a b u khng l s chnh phng.Chng minh rng trong hai phng trnh sau: 2 22 211ax byax by = = c t nht mt phng trnh khng c nghim nguyn dng. * Trc ht ta s chng minh b sau:Cho A, B l cc s nguyn dng v A, B, AB u khng l cc s chnh phng.Khi : nu gi (a, b) l nghim nguyn dng nh nht ca phng trnh Pell 2 21 x ABy =(do AB khng l s chnh phng nn phng trnh Pell ny lun c nghim nguyn dng, ngha l (a, b) tn ti) v (x0, y0) l nghim nguyn dng nh nht ca phng trnh 2 21 Ax By =th ta lun c h thc lin h sau: 2 20 00 02a Ax Byb x y = + = *Chng minh: Do (x0,y0) l nghim ca 2 21 Ax By =nn 2 20 01 Ax By = . t 2 20 0 0 0, 2 u Ax By v x y = + = ; khi , ta c: ( ) ( ) ( )2 222 2 2 2 2 20 0 0 0 0 0. 2 1 u ABv Ax By AB x y Ax By = + = = .Do , (u; v) l mt nghim ca 2 21 x ABy = .M (a, b) l nghim nguyn dng nh nht ca phng trnh Pell 2 21 x ABy = nn, u a v b .Ta s chng minh rng u = a, v = b.Tht vy, gi s ngc li, u > a, v > b. Ta c:( )( )( )( )( )2 20 0 0 00 0 0 0 0 01( )a b AB a b AB a b AB a ABba b AB Ax By Ax Byax Bby A ay Abx B Ax By < + = = + < + + < + Mt khc: ( )22 20 0 0 0 0 02 a b AB u v AB Ax By x y AB Ax By + < + = + + = +Do : ( )( ) ( )( ) ( ) ( )( ) ( )0 0 0 0 0 022 20 0 0 0 0 0 0 0 0 0( ) ax Bby A ay Abx B a b AB x A y Bx A y B x A y B Ax By x A y B x A y B = + < + = + = + t 0 0 0 0, s ax Bby t ay Abx = = , ta c: 0 00 0(1)(2)

s A t B x A y Bs A t B x A y B+ < + < + www.VNMATH.com63 Hn na:( ) ( ) ( )( )2 22 2 2 2 2 20 0 0 0 0 01.1 1 As Bt A ax Bby B ay Abx a ABb Ax By = = = =Ta thy s > 0 v:2 2 2 2 2 2 2 2 20 0 0 0 0 0 0 00 0 ( 1) s ax Bby ax Bby a x B b y a x Bb Ax > > > > > 2 2 2 2 2 20 0( ) a ABb x Bb x Bb > > , ng do B > 0. Ta cng thy0 t bi v nu t = 0 th:2 2 2 2 2 2 2 2 2 2 20 0 0 0 0 0 0 0 00 ( 1) ( 1) ay Abx ay Abx a y A b x y ABb Ab By y Ab = = = + = + =iu ny mu thun do A khng l s chnh phng. -Nu t > 0 th (s; t) l mt nghim dng ca phng trnh 2 21 Ax By = , t :, s a t b , suy ra: 0 0s A t B x A y B + > + , iu ny mu thun vi (1). -Tng t, nu t < 0 th (s; -t) l mt nghim dng ca phng trnh 2 21 Ax By = , t : , s a t b , suy ra: 0 0s A t B x A y B > + , iu ny mu thun vi (2). Do , iu gi s l sai, ngha l u = a, v = b. B c chng minh. * Tr li bi ton:Gi s ngc li, c hai phng trnh:2 22 21 (*)1 (**)

ax byax by = = u c nghim nguyn dng. Gi (m, n) l nghim nguyn dng nh nht ca phng trnh 2 21 x aby = ; (x1; y1) l nghim nguyn dng nh nht ca (*); (x2; y2) l nghim nguyn dng nh nht ca (**). Theo b trn, ta c cc h thc sau: 2 21 11 12m ax byn x y = + =v2 22 22 22m bx ayn x y = + = T (*)2 21 11 ax by = + , t (**)2 22 21 ay bx = , so snh cc ng thc trn, ta cng c:( )2 2 2 2 2 2 2 21 1 2 2 1 2 1 22 1 2 1 1 ax by bx ay by bx b y x + = + + = = . Nhng do b l s nguyn dng, khng phi l s chnh phng nn b > 1, ngha l ng thc trn khng th xy ra. Suy ra iu gi s trn l sai. Vy trong hai phng trnh (*) v (**) cho c t nht mt phng trnh khng c nghim nguyn dng. y chnh l iu phi chng minh. www.VNMATH.com64 Bi 4.Tm tt c cc s thc r sao cho bt ng thc sau ng vi mi s, , a b c dng: 31 2a b cr r r rb c c a a b| || || | | |+ + + + | | ||+ + +\ \ \ \ Ta s xt iu kin cn v tm cc gi tr r tha bi. * iu kin cn: Xt trng hp0 a b = > . t0cta= > . Ta c:322 3 31 21 1 1. .2 2 2 21a b cr r r rb c c a a ba c cr r r r r rca c a aa| || || | | |+ + + + | | ||+ + +\ \ \ \ | | || | | | | | | | | | + + + + + + | |||||+\ \ \ \ \ |+\ 2 3 2 32221 1 2 1 1. .1 2 2 1 1 2 22 3 1 2 3 1. . 01 2 2 ( 1) 1 4 1 8t r tr r r r r rt t tt t tr rt t t t (| | | | | | | | | | | | + + + + + + + ( ||||||+ + +\ \ \ \ \ \ ( | || | | | + + + + |||+ + + +\ \ \ Cho0 t , t bt ng thc trn, suy ra: 25 144 2 1 0(*)5 14rr rr

+

* iu kin :Ta s chng minh rng vi gi tr r tha mn (*) th bt ng thc cho ng vi mi s dng a, b, c. Tht vy:t: , , , , , 0a b cx y z x y zb c c a a b= = = >+ + +. Ta c:2 . . . . .( ) ( ) ( ) 2 ( )( )( )1( )( )( ) ( )( )( )a b b c c a a b cxy yz zx xyzb c c a c a a b a b b c b c c a a bab a b bc b c ca c a abc a b b c c aa b b c c a a b b c c a+ + + = + + + =+ + + + + + + + ++ + + + + + + + += =+ + + + + + Ta s chng minh cc bt ng thc sau:(1) 2( )2( )( ) ( )( ) ( )( )( )( ) ( )( ) ( )( ) 2 ( ) 2 ( ) 2 ( )( )( )( ) ( )( )( )x y z xy yz zxa b c ab bc cab c c a a b a c b c b a c a c b a ba a b a c b b c b a c c a c b ab a b bc b c ca c aa b b c c a a b b c c a+ + + + ( + + + + (+ + + + + + + + + + + + + + + + + + + + + + + + + + + + www.VNMATH.com65 3 3 33 3 32 2 2( ) ( ) ( ) 3 2 ( ) 2 ( ) 2 ( )3 ( ) ( ) ( )( ) ( ) ( ) 0( )( ) ( )( ) ( )( ) 0a b c ab a b bc b c ca c a abc ab a b bc b c ca c aa b c abc ab a b bc b c ca c aa a bc ab ac b b ca ba bc c c ab ca cba a b a c b b a b c c c a c b + + + + + + + + + + + + + + + + + + + + + + + + + + + + + Bt ng thc ny chnh l bt ng thc Schur vi cc s dng a, b, c. | | | |3(2)43( )( ) ( )( ) ( )( ) 43( ) ( ) ( ) ( )( )( )44 ( ) ( ) ( ) 3 ( ) ( ) ( ) 6( ) ( ) ( xy yz zxab bc caa c b c b a c a c b a bab a b bc b c ca c a a b b c c aab a b bc b c ca c a ab a b bc b c ca c a abcab a b bc b c ca c+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +2 2 2) 6( ) ( ) ( ) 0a abca b c b c a c a b + + Bt ng thc cui ng nn (2) ng vi mi s dng a, b, c. t 34t xy yz zx t = + + . Bt ng thc cho chnh l:33 2 3 221( )( )( ) ( )23 3 1( ) ( )2 4 83 3 1( ) ( ) 02 4 8r x r y r z rr x y z r xy yz zx r xyz r r rx y z r xy yz zx r xyz+ + + + + + + + + + + + + + + + + + + + Do2( ) x y z xy yz zx + + + + v 12 1 (1 )2xy yz zx xyz xyz xy yz zx + + + = = nn chng minh bt ng thc trn, ta chng minh bt ng thc sau: 222 22 223 3 1 12( ) ( ) (1 ) 02 4 2 83 3 1 1(2 ) ( ) (1 ) 02 4 2 81 3 3 3(2 )2 2 4 84 2 1 3(4 2 1) 3. (4 2 1) 02 8 4xy yz zx r xy yz zx r xy yz zxt r t r tt r r r rr r r rt r r t ( + + + + + + ( + + + + + + | | + |\ Bt ng thc cui ng do (*) v (2) nn bt ng thc cho l ng. Vy iu kin ca r cn tm l 5 14rhoc 5 14r . www.VNMATH.com66 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 vi ng 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 trn ng 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 qua mt im c nh khi M di ng pha trong ng trn. * Qua R k ng thng song song vi PQ ct NA ti C, qua S k ng thng song song vi PQ ct NB ti D. Gi I l trung im ca CD .Ta s chng minh rng CD // AB. Tht vy, do N nm trn ng trn ng knh AB nn:

090 ANB AN BN = , suy ra BN l tip tuyn ca ng trn ng knh PN.Do :( . ) BMN BNS g g V PQ l ng phn gic gc ngoi ca AMN nn:

SMP AMP QMR BMQ = = = .Mt khc:

SMP SNP =(gc ni tip cng chn cung PS ca ng trn ng knh PN),

QMR QNR = ( gc ni tip cng chn cung QR ca ng trn ng knh QN). Do :

SNP QNR SNP SNR QNR SNR CNR SNB = + = + = . Xt hai tam gic BNS v RNC c:

CNR SNB =v

RCN MPN NSM NSB = = = nn:( . ) BNS RNC g g . Suy ra cc tam gic ng dng:BMN BNS RNC .CDISRMPQNBOAwww.VNMATH.com67 Tng t, ta cng c:DSN RAN NAM . * Ta thy, t:. .NB NSBNS RNC NB NC NR NSNR NC = = . .NS NDDSN RAN NA ND NR NSNA NR = = . Suy ra:. .NA NCNA ND NB NCNB ND= = AB // CDTrung im ca AB, trung im ca CD v N l ba im thng hng.Tc l N, O, I thng hng. (1) * Hn na: . MN BN NB NCBMN RNC RCNC RC MN = = . . DN DS NA NDDSN NAM DSMN NA MN = = . Kt hp cc iu trn, ta c: RC = DS, m RC // DS (cng song song vi PQ) nn t gic RCSD l hnh bnh hnh. Do , hai ng cho CD v RS ca t gic ct nhau ti trung im ca mi ng. Suy ra I l trung im ca CD cng l trung im ca RS.Khi : NI chnh l ng trung tuyn ca tam gic NRS.(2) T (1) v (2), suy ra:trung tuyn NI ca tam gic NRS lun i qua O.Vy trung tuyn ng vi nh N ca tam gic NRS lun i qua I l im c nh khi M di ng khp pha trong ng trn (O).y chnh l iu phi chng minh. www.VNMATH.com68 Bi 6.Mt hi ngh ton hc c tt c64 n +nh ton hc phi hp vi nhau ng21 n +ln( ) 1 n . 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 ci 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 nu1 n = . b/ Hi rng Ban t chc c th sp xp c ch ngi c hay khng vi mi1 n > ? a. Vi1 n = , ta c bi ton nh sau: mt hi ngh ton hc c 10 nh ton hc, h phi hp vi nhau ng 3 ln v trong mi ln hp nh th, h phi ngi quanh mt ci bn 4 ch v mt ci bn 6 ch, cc v tr ngi chia u khp bn; ng thi, hai ngi ngi k nhau hoc i din nhau trong cuc hp ny th khng c ngi k nhau hoc i din nhau trong mt cuc hp khc. nh s th t cho 10 nh ton hc ang xt l (1), (2), (3), (10). Ta s ch ra mt cch sp xp tha mn bi trong trng hp ny. Ta c cc s sau: -Bui hp th 1:

-Bui hp th 2: -Bui hp th 3: 789610421353846109517273924106185www.VNMATH.com69 b. Ta s chng minh rng trong trng hp tng qut, Ban t chc lun c th xp ch ngi cho cc nh ton hc trong cc cuc hp. Ta chia64 n +nh ton hc thnh22 n + nhm. Mt nhm ch gm 1 ngi lun ngi mt v tr c nh ti bn 4 ch, t ngi ny l X0;21 n +nhm cn li chia ra t 63 n +nh ton hc, mi nhm c 3 nh ton hc.t cc nhm l 1 2 3 2 1, , ,...,nX X X X+. Cc nhm ny s ln lt ngi vo cc v tr cn li ca bn 4 ch cng vi X0, mi nhm ngi ng mt ln.* Vi cc bn 6 ch, ta c cch sp xp nh sau: bc th, 1 2 1 k k n + , vi hai nhm bt k,i jX Xtrong :(mod 2 1), i j k n + +1 , 2 1,i j n i j + th cc nh ton hc thuc hai nhm Xi, Xj s ngi vo hp cng nhau mt bn 6 ch no ; ng thi, nhng nh ton hc thuc cng mt nhm th ngi cc v tr to thnh mt tam gic u trn cc bn 6 ch, ngha l h s khng ri vo trng hp ngi i din nhau hoc ngi cnh nhau. * Vi bn 4 ch, ta c cch sp xp nh sau: - Nu k l s chn th bc ny, ch c mt nhm c ch s 2k l khng c ngi chung bn 6 ch vi nhm no, nh th nhm ny s ngi vo bn 4 ch cng vi X0. - Nu k l s l, bc ny; tng t trn, ch c mt nhm c ch s l 2 12k n + + khng c chung bn 6 ch vi nhm no, nhm ny s ngi vo bn 4 ch cng vi X0. D dng thy rng cch sp xp nh th tha mn mi yu cu ca bi ton. Vy Ban t chc c th sp xp c ch ngi mi1 n > . jiXXwww.VNMATH.com70 LI GII THI CHN I TUYN QUC GIAD THI IMO 2010 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 Ol 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 (O1), ng thng AC vi ng trn (O2). Chng minh rng: 1.Tip tuyn ti P ca 1( ) Ov tip tuyn ti Q ca2( ) Ophi 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. 1. V M l trung im ca BC nn 2 2MA MB = , suy ra M c cng phng tch n hai ng trn( ) ( )1 2, O Ohay M thuc trc ng phng ca .Hn na, hai ng trn ct nhau ti D nn D cng nm trn trc ng phng ca chng.T , suy ra DM chnh l trc ng phng ca( ) ( )1 2, O O , m A thuc ng thng DM nn A c cng phng tch n hai ng trn( ) ( )1 2, O O .Suy ra:. . AP AB AQ AC =hay t gic BPQC ni tip. T h thc trn ta cng thy rng nu P trng vi A th Q cng trng vi A, nu P thuc on AB th Q cng thuc on AC v ngc li.xSQPO2O1MABCDwww.VNMATH.com71 Khng mt tnh tng qut, gi s gc

ABCnhn, gi Px l tia tip tuyn ca ng trn (O1)sao cho gc

xPBnhn.Theo tnh cht tip tuyn, ta c:

xPB PBC = , m

PBC AQP = nn

xPB AQB = . Px cng l tip tuyn ca ng trn (APQ).Do : (O1) tip xc vi ng trn (APQ).Hon ton tng t: ta cng c (O2) cng tip xc vi ng trn (APQ). Suy ra: tip tuyn ti P ca 1( ) Ov tip tuyn ti Q ca2( ) O cng chnh l hai tip tuyn ca (APQ) ti cc im P, Q.Hn na, theo gi thit:

090 PAQ nn hai tip tuyn khng song song v do chng phi ct nhau (pcm). 2.Theo chng minh trn, ta c: S thuc tip tuyn ca (O1) v (O2),SP SQ =nn S c cng phng tch n hai ng trn( ) ( )1 2, O Onn S thuc trc ng phng ca hai ng trn ny, ngha l S nm trn AM.Vy khi D thay i trn AM th S cng di chuyn trn AM l ng thng c nh.Ta c pcm. www.VNMATH.com72 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. t( , , ) 11( ) 10( ), , ,k hf k h n k h n n k h n = + + + . Ta c:( , , ) ( , , ) f k h n f h k n =nn khng mt tnh tng qut, ta gi sh k . Nu (mod11) m n th : ( , , ) ( , , ) 11( ) 10( ) 11( ) 10( )10 ( ) ( ) 110 ( , , ) ( , , ) (mod110)k h k hk k h hf k h m f k h n k h m m k h n nm n m n f k h m f k h n (( = + + + + + + = ( = + T , ta ch cn xt cc gi tr n tha 1 11 n . Xt hiu: 5(6, 6, ) (1,1, ) 110 20 .( 1) f n f n n n = + . Nu 520 .( 1) n n chia ht cho 110 th gi tr n tng ng s khng tha. T , ta loi i cc gi tr 1, 3, 4, 5, 9,11 n = . Ta cng c 8 2 6 4 2 2 2 2(8, 2, ) (6, 4, ) 10( ) 10 ( 1) ( 1) f n f n n n n n n n n = + = + , vi10 n =th 2 2 2 210 ( 1) ( 1) 110 n n n + nn gi tr ny cng khng tha. Ta s chng minh rng cc gi tr2, 6, 7, 8 n =tha mn. Tht vy: Tnh ton trc tip, ta thy rng vi2, 6, 7, 8 n =th(mod11)k hn n vi (mod11) k h . Gi s ngc li, vi cc gi tr n nu trn, tn ti hai b( , ) ( ', ') k h k h (gi s' k k > ) sao cho: ( , , ) ( ', ', ) f k h n f k h n . Khi : ' '11( ' ') 10( ) 0 (mod110)k h k hk h k h n n n n + + + . Suy ra: ' ' ' ' '' '(mod10),(mod11) ( 1) ( 1) (mod11)k h k h k k k h h hk k h h n n n n n n n n + + . Do 11 l s nguyn t nn theo nh l Fermat nh th: ' ' (mod11)k k h hn n . D thy: ' '1 1 0 (mod11)k k h hn n nn t ng thc trn, suy ra: '(mod11) 'k hn n k h = . Do :' k h =hay( , ) ( ', ') k h k h , mu thun.Suy ra cc gi tr2, 6, 7, 8 n =u tha mn yu cu ca bi. Vy tt c cc gi tr n cn tm l:2, 6, 7, 8 (mod11) n . www.VNMATH.com73 Bi 3.Gi mt hnh ch nht c kch thc 1 2 l hnh ch nht n v mt hnh ch nht c kch thc 2 3 , b i 2 gc cho nhau (tc c c 4 vung con) l hnh ch nht 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 l2008 2010 . Tm s nh nht cc hnh ch nht n c th dng ghp. Ta s chng minh rng s hnh ch nht n nh nht tha mn bi l 1006. Tht vy: *iu kin cn: Ta xt mt cch ph hnh ch nht 2008 2010 tha mn bi (ch rng 2008 ch s hng v 2010 ch s ct). Gi, , , x y z tl s cc hnh ch nht 1 2,2 1,2 3,3 2 trong cch ph ( y thc ra cc hnh ch nht 1 2,2 1 u l cc hnh ch nht n ca bi, ch phn bit cch ph dc hay ngang, trong 1 2 c t ngang,2 1 c t dc; tng t vi cch phn bit2 3,3 2 ). T mu trng cho cc hng l, t mu en cho cc hng chn. tt c cc ca hng th ,1 2010 i i , ta nh s tng ng cc s t nhin i. Ta s chng minh cc nhn xt sau: -Nhn xt 1: ta lun c ng thc:2( ) 4( ) 2008.2010 x y z t + + + = . Mi hnh ch nht 1 2 hoc2 1 c cha hai vung, mi hnh ch nht2 3 hoc 3 2 c cha bn vung. Tng cc vung ny bng s vung ca c hnh ch nht ln nn 2( ) 4( ) 2008.2010 x y z t + + + = . -Nhn xt 2: Gi tr, x yl chn. Ta thy trn ton bng, cc hnh ch nht2 3,3 2 u c s trng bng s en; cc hnh ch nht2 1 c t dc nn cng c s trng bng s en. Suy ra, s hnh ch nht 1 2 cc hng c t mu trng bng s hnh ch nht cc hng c t en. Hn na, tng s cc hng l 2010 l chn nn gi trxphi l chn. T nhn xt 1, ta thy y cng phi chn. Tr li bi ton, ta xt tng ng i t tp hp cc hnh ch nht ang xt n cc s nguyn l hiu gia tng cc s vung c t en vi tng cc s vung c t trng ghi trn n.D dng thy rng:(3 2) 0; (2 3) 2; (2 1) 1 = = = .T suy ra: (3 2) 0; (2 3) 2 ; (2 1) z y = . (k hiu(3 2) l tng tnh trn tt c cc hnh ch nht3 2 c dng, nh ngha tng t vi cc hnh ch nht khc). Ta cng thy rng, tng cc s ghi trnxhnh ch nht 1 2 l mt s chn thuc| | 2;2.2008 , mxl s chn (nhn xt 2), ta c nh gi sau:( ) (1 2) 2.2008 22x .www.VNMATH.com74 Ta c: | |1004 10041 1(2008 2010) 2010. 2 (2 1) 2010. 2010.1004i ii i i= = = = = . Mt khc:(2008 2010) (2 1) (1 2) (3 2) (2 3) = + + + . T cc iu trn, suy ra:2010.1004 .(2.2008 2) 2 2010.1004 2007 22xy z x y z + + + + . Tip theo, ta xt hnh ch nht2010 2008 (tng t nh trn nhng c 2010 hng v 2008 ct), bt u li cc lp lun v s cc hnh ch nht 1 2,2 1,2 3,3 2 c dng.Ta xy dng c bt ng thc sau:2008.1005 2009 2 y x t + + . Cng hai bt ng thc ny li, ta c: 2008.1005 2010.1004 (2009 2 ) (2007 2 ) 2008 2010 2( ) y x t x y z x y z t + + + + + + = + + + .Hn na, theo nhn xt 1 th:2010.1004 ( ) 2( ) x y z t = + + + . T ta c:2008.1005 2007 2009 2009( ) x y x y + + . Suy ra: 20081005. 10042009x y + > , mx y +l s chn nn1006 x y + . Do , tng cc hnh ch nht n cn dng t nht l 1006. iu kin cn c chng minh. *iu kin : Ta s ch ra mt cch ghp hnh ch nht dng ng 1006 hnh ch nht n. Khi 4 Khi 3 Khi 2Khi 1

www.VNMATH.com75 Hnh trn m t cch ghp mt hnh ch nht 10 16 , trong : cc hnh ch nht khuyt c t bng 5 mu khc nhau (, hng, xanh lam, xanh l cy, xanh m) d dng phn bit; trn hnh cc khi c t mu xanh l m l cc hnh ch nht n chc chn phi dng, cc khi mu vng th ty trng hp, c th l hnh ch nht n m cng c th l hnh ch nht khuyt. * Hnh ch nht2010 2008 c th c to thnh t hnh trn bng quy tc sau: -Thm cc dng bng cch chn vo gia mi khi trn cc hnh c dng: Mi ln ghp nh th th ta c thm c hai hng mi, do 2010 chia ht cho 2 nn khi thc hin vic ny lin tip mt cch thch hp th khi ny s tng v chiu di, to thnh cc khi mi c kch thc2010 4 v mi khi nh vy, ta ch dng ng 2 hnh ch nht mu xanh l m. -Thm ct bng cch lp li cc khi 1, 2 , 3, 4 trn hnh (ch tnh tun hon gia cc khi: (1) tng ng vi (3), (2) tng ng vi (4)). Nh th th ta cn phi c tt c 502 khi dnh cho 2008 ct. ng thi, khi u tin v khi cui cng, ta cn dng thm mt hnh ch nht n mu vng, cc khi gia th dng cc hnh ch nht khuyt mu vng. Tc l: hai khi u tin v cui cng, ta cn dng 3 hnh ch nht n, cc khi gia ch cn dng 2 hnh ch nht n thi.Khi , tng s hnh ch nht n cn dng l:500.2 2.3 1006 + = . Xoay hnh ch nht2010 2008 li, ta c hnh ch nht2008 2010 cn phi ghp, hnh ch nht c ng 1006 hnh ch nht n tha mn bi. Do , iu kin c chng minh. Vy gi tr nh nht cc hnh ch nht n cn dng l 1006. Bi ton c gii quyt hon ton.www.VNMATH.com76 Bi 4.Cho, , a b cl cc s thc dng tha mn iu kin:1 1 116( ) a b ca b c+ + + + .Chng minh rng: 3 3 31 1 1 89 ( 2( )) ( 2( )) ( 2( )) a b a c b c b a c a c b+ + + + + ++ + + + +. Hi ng thc xy ra khi no? Theo bt ng thc Cauchy, ta c: 23 3( )( )2( ) ( ) 3 ( ). 32 2 2 2a c a c a c a b a ca b a c a b a b | |+ + + + ++ + + = + + + + = | |\ ( )31 227( )( )2( )a b a ca b a c + ++ + +. Tng t vi hai biu thc cn li. Do : ( )31 2 4( )27( )( ) 27( )( )( )2( )cyc syma b ca b a c a b b c c aa b a c+ + =+ + + + ++ + + . Hn na, ta thy vi mi a, b, c dng: 29( )( )( ) 8( )( ) ( ) 08( )( )( ) ( )( )9syma b b c c a a b c ab bc ca a b ca b b c c a a b c ab bc ca+ + + + + + + = + + + + + + + Do : ( )31 16( )2( )cycab bc caa b a c+ ++ + +.(1). Mt khc, ta cng c: 2( ) 3 ( ) ab ca ca abc a b c + + + + nn theo gi thit:1 1 1 3( ) 316( )16ab bc ca a b ca b c ab bc caa b c abc ab bc ca+ + + ++ + + + = + + + +. (2) T (1) v (2), suy ra: 3 3 31 1 1 89 ( 2( )) ( 2( )) ( 2( )) a b a c b c b a c a c b+ + + + + ++ + + + +. Ta c pcm. ng thc xy ra khi du bng tt c cc bt ng thc trn xy ra hay: , , 0141 1 116( )a b ca b c a b ca b ca b c>= = = = = + + = + +. www.VNMATH.com77 Bi 5.Trong mt hi ngh c n nc tham gia, mi nc c k i din ( 1 n k > > ). Ngi ta chia n.k ngi ny thnh n nhm, mi nhm c k ngi sao cho khng c hai ngi 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. Ta gi mt nc X v mt nhm Y no c lin h vi nhau nu trong nhm Y c ngi ca nc X. Khi , mt nc X bt k c k ngi i din nn c lin h vi k nc v mt nhm Y bt k c cha k ngi i din khc nhau ca cc nc khc nhau nn c lin h vi ng k nc. Do , mt tp hp bt k m nc no trong n nc cho s c lin h vi t nht . mkmk= nhm khc nhau.Gi , 1,ia i n =l n nhm cho v ,1,iX nl tp hp cc nhm c lin h vi nc th i.Theo iu va chng minh trn, ta thy vi mi:1 2 3, , ,..., {1, 2, 3,..., }ki i i i n , 1 k n th: 1jkijX k=.(*) Ta s chng minh rng tn ti 1 2 3 1 2 3( , , ,..., ) ...n na a a a X X X X v,i ja a i j . Tht vy, ta bt u b i phn t thuc mi tp Xi sao cho (*) vn c tha mn. Cui cng thu c cc tp hp mi 1 2 3' , ' , ' ,..., 'nX X X X(vi' )i iX X vn tha iu kin (*) c s phn t nh nht m nu b i thm bt c phn t thuc tp hp ' ,1,iX nth iu kin (*) s khng cn c tha. Ta chng minh rng ' 1, 1,iX i n = = .Tht vy, khng mt tnh tng qut gi s 1' Xc cha phn t khc nhau l , . Do nu b thm mt trong hai phn t hoc th iu kin (*) khng cn tha mn nn s c hai tp ch s P, Q sao cho:Vi 1( ' \ { }) 'ii PM X X = , 1( ' \ { }) 'ii QN X X = khng tha mn iu kin (*), tc l: 1, 1 , M P N Q M P N Q < + < + . Ta c: 1 1 1(( ' \ { }) ( ' \ { })) ( ' ' ) ' 'i i ii P i Q i P QM N X X X X X X = = 'ii P QX M N www.VNMATH.com78 T hai iu ny suy ra:1 M N P Q + ,M N P Q Theo nguyn l b tr, ta c: 1 1 P Q M N M N M N P Q P Q P Q + + = + + + = + +iu v l ny dn n khng nh' 1, 1,iX i n = = .R rng cc tp hp ny khng giao nhau v nu tn ti , ( ' ' ) ' ' 1 2i j i ji j X X X X = ). *Chng minh:Khng mt tnh tng qut, gi sm n >(num n =th b hin nhin ng). Trc ht, ta thy rng:( , ) 1, 1, 1 p i i p = = nn ! ( 1)!.!( )! !( )!kpp pC p pk p k k p k= = , tc l: 0 (mod ), 1, 1kpC p k p = . Ta c: 11( 1) 1 . 1(mod )pp p i p i ppix x C x x p=+ = + + +. (*) Ta s chng minh nhn xt: *( 1) 1(mod ),j jp px x p j + + bng quy np. Tht vy: - Vi1 j = , nhn xt ng theo (*). - Gi s nhn xt ny ng vi1 j h = . Ta s chng minh rng n cng ng vi1 j h = + . Ta c:( 1) 1 (mod )h hp px x p + + .Suy ra: ( ) ( )1 1( 1) 1 (mod ) ( 1) 1 (mod )h h h hh hp p p px x p x x p+ ++ + + + . Do nhn xt ng vi1 j h = + . Theo nguyn l quy np, nhn xt c chng minh. Ta xt khai trin sau: www.VNMATH.com80 0..0 0 0(1 ) (1 ) (1 ) (mod )kiiii ii iim k k m pm m p j j pmj i ix x x C x p== = =+ = + + . H s ca nx v(1 )mx +l nmC ; do biu din 1 21 2 1 0. . ... .k kk kn n p n p n p n p n= + + + + +l duy nht nn h s ca nx v .0 0iiim kj j pmj iC x= = l 0iikmniC=. T ta c: 0(mod )iikm mn niC C p=.B c chng minh. *Tr li bi ton: Ta c:22 221 0 0(1 ) (1 ) .( 1) . . . .nn n nn n n i n i i n i n i in n ni i ix x n C x C x C x = = =| | | |+ = + + = ||\ \ . ng nht h s ca 2nx hai v, ta c: 220 0. ( )n nn i n i in n n ni iC C C C= == = . Do , vi mi n t nhin th 2nn nS C = .Nh th ta cn chng minh rng: 241nnC + khng chia ht cho 3 vi mi n.Gi s: 02 .3 , , 1,kii iin a a i k== = . Xt hai trng hp: -Nu{0;1}, 1,ia i k =th2 {0; 2}, 1,ia i k =v tng{0;1}, 1,ia i k =l s chn, t 0202 , 2 2 4 1(mod3)kiik at tiia t t=== = = ; ta cng c: 04 2 .3 , , 1,kii iin a a i k== = . Theo b trn th 024 20 01 1 2 1 2 1 2(mod3)kii i iik k aa a nn ai iC C== =+ + + + . -Nu tn ti mt gi tr2ja = ; khng mt tnh tng qut, gi s y l s nh nht trong tp hp, 0,ia i k = . Khi : h s tng ng ti v trj khai trin theo ly tha 3 ca 4n l 1. M 210 C = nn 2 24 2 400 (mod3) 1 1 (mod3)iika n nn a niC C C= + . Vy trong mi trng hp, ta u c 21nS +khng chia ht cho 3. y chnh l pcm. www.VNMATH.com