Chuyende-Toanhoc9 Ptnk

8/10/2019 Chuyende-Toanhoc9 Ptnk

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TRN NAM DNG (ch bin)V QUC B CN - L PHC L - PHM HY HIU

T NGUYN THI SN - L VIT HI

Chuyn

TON HCS 9

d

TP H CH MINH, THNG 10 NM 2010


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d


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LI NI U

Chuyn Ton hc s 9 ca trng Ph thng Nng khiu m cc bn ang cm trntay l mt n phm c nhiu iu c bit.

Th nht, n c thai nghn trong mt khong thi gian di k lc: t nht l 1 nm.Th hai, n c ra i cch chuyn trc 5 nm. Th ba, tham gia ng gpcho Chuyn ln ny khng ch l cc hc sinh v thy c ca trng Ph thng Nngkhiu m cn ca nhiu bn hc sinh, sinh vin, cc thy c cc trng khc. Internet to ra mt th gii khc hn, v Ton hc cng khng nm ngoi s thay i .

Lm Chuyn Ton hc s 9 ny, chng ti bng nh v cc th h hc sinh ca trng

Ph thng Nng khiu, nhng tc gi ca Chuyn Ton hc s 1, 2, 3, 4, 5, 6, 7, 8. l nhng L Long Triu, Trn Quang nh, V Tm Vn, L Quang Nm, Lu Minh c,Nguyn L Lc, Trnh L Tun, L Minh Tun, Phm Quc Vit, Hong Thanh Lm,Trn nh Nguyn, Nguyn Cm Thch, Phm Tun Anh, Lng Th Nhn, Trn VnhHng, Nguyn Tin Khi, Trn Quang, Trn Anh Hong, Nguyn ng Khoa, NguynAnh Cng, Trn Chiu Minh, Kha Tun Minh, . . . Chng ti ang c mong mun thchin mt cun tuyn chn cc bi vit t cc Chuyn ny.

Hy vng rng Chuyn ton hc s 9 vi ni dung kh phong ph v hnh thc p sl mt mn qu tng ngha i vi cc bn tr yu Ton. Trong bin c mnh mng

ca kin thc, nhng ngi thc hin chuyn ch mong n phm ca mnh s l mtgit nc trong c ch.

Chuyn Ton hc s 9 c hon thnh vi s h tr ca Cng ty c phn gio dcTitan. Ban bin tp xin chn thnh cm n s h tr ny.

BAN BIN TP

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4 Chuyn Ton hc s 9

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V NH TI TR

Cng ty c phn gio dc Ti tan (Titan Education) hot ng trong lnh vc cung cpdch v gio dc ph thng, trong c bit l cc sn phm c ni dung Ton hc.Sn phm chnh ca cng ty l trang web o to Ton hc trc tuyn cho hc sinh lp1 12, kho d liu Ton hc dnh cho hc sinh, sinh vin, gio vin v cc nh nghincu Ton hc, cc lp hc trc tip rn luyn t duy Ton hc, bi dng v pht trinnng khiu Ton hc, cc lp hc nghip v dnh cho gio vin Ton.

Ngoi ra, cng ty cn t vn, cung cp ti liu, sch tham kho v Ton cho hc sinhv gio vin cc cp mi trnh .

Titan Education mong mun bng nhit huyt v nng lc ca mnh gp phn nh bvo s nghip gio dc v o to th h tr cho t nc Vit Nam.

Mi chi tit xin lin h:

Cng ty c phn gio dc Ti tan

a ch: 18A Nam Quc Cang, P. Phm Ng Lo, Q. 1, TP. H Ch Minh.

in thoi: 0854 542305 0839 260137.

Fax: 0839 260140 Hotline: 012345 39976.

Website: www.titan.edu.vn.Email: [email protected].

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6 Chuyn Ton hc s 9

d


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MC LC

Li ni u . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

V nh ti tr . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

Duyn s ca Ton hc Vit Nam vi gii Fields

Ng Vit Trung . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

Thng tin Ton hc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .15

Khm ph mt s tnh cht ca dy s truy hi tuyn tnh cp hai

o Hong Nh . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

nh l thng d Trung Hoa

Phm Hy Hiu . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

Rn luyn k nng gii cc bi ton Hnh hc phng

L Phc L . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

Nhn Hnh hc bng con mt i s

T Nguyn Thi Sn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69

Ngun gc bi ton Hnh hc s 5 trong thi Vit Nam TST 2009

L B Khnh Trnh . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85

Nh m khng nh

V Quc B Cn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97

Bt ng thc Bernoulli

Trng Tn Sang . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .113

Enlightening Trigonometrical Substitutions

Vardan Verdiyan, Daniel Campos Salas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .141

V mt bi ton Bt ng thc

Nguyn Vn Huyn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .161

T hp v cng thc C2k

ng Hong Linh . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167

M rng t mt bi ton

T Nguyn Thi Sn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .171

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8 Chuyn Ton hc s 9

ng dng ca Ton hc trong vic hin th hnh nh bng my vi tnh

Phm Mng Bo . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 177

Cp s cng v Phng trnh hm trn NNguyn Trng Tun . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .187

Sng to Phng trnh hm t cc hng ng thc

L Vit Hi, o Thi Hip . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .193

Li gii thi vo lp 10 chuyn Ton nm hc 2010 - 2011 . . . . . . . . . . . . . . . . . . . . . . 213

p n thi chn i tuyn Ton nm hc 2008 - 2009 . . . . . . . . . . . . . . . . . . . . . . . . . .219

p n thi chn i tuyn Ton nm hc 2009 - 2010 . . . . . . . . . . . . . . . . . . . . . . . . . .229

Mt s thi Olympic Ton . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .239


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DUYN S CA TON HC VIT NAMVI GII FIELDS

Ng Vit Trung

Vin Ton hc Vit Nam

Ni l c duyn l v rt nhiu nc giu hn, c truyn thng Ton hc hn Vit Namnhng khng c gii Fields. Nc c, mt cng quc v Ton hc mi ch c mi mtgii Fields. n v Trung Quc l nhng ci ni Ton hc trong lch s v l nhngnc c nhiu nh Ton hc ni ting, nhng cng cha mon men c n gii Fields.C chu cho n nay mi c ba gii Fields, u l ca Nht. Chu M Latin v chuPhi khng c gii Fields.

Theo mt ngha no , ot gii Fields cn kh hn gii Nobel v gii Nobel c traohng nm, trong lc gii Fields c trao 4 nm mt ln. Ni mt cch nm na, ngic gii Fields phi c kt qu xut sc nht trong 4 nm ch khng phi trong mtnm nh gii Nobel. Ngoi ra gii Fields ch c trao cho ngi khng qu 40 tui.Andrew Wiles, ngi gii quyt bi ton Fermat khng c gii Fields khi cn tuiv li gii ban u c l hng. n khi khc phc c l hng trong li gii th ng qu 40 tui. Ni nh vy thy gii Fields l mt c duyn thc s. Trong lch s hn70 nm ca gii Fields mi c 48 ngi c gii. Th m Vit Nam li c mi lin h

cht ch vi nhiu ngi trong s h.Chng ta hy bt u cu chuyn vi GS L Vn Thim. ng Thim l ngi Vit chc v tin s Ton hc u tin v c coi nh l cha ca nn Ton hc Vit Nam.Sau khi bo v tin s ng Thim lm tr l cho GS Rolf Nevanlinna trng i hcZurich hai ln vo nhng nm 1946 v 1948. Sinh thi ng Thim thng coi mnh lhc tr ca Nevanlinna, cng trnh ni ting nht ca ng Thim l v bi ton ngcca Nevanlinna. Trong Ton hc c mt tp ch tn l Tp ch trung tm v Ton hc,chuyn ng bi gii thiu cc cng trnh mi cng b. Ngi gii thiu bi bo ca ngThim l ng Lars Ahlfors, mt trong hai nh Ton hc c trao gii Fields ln u

nm 1936. Ahlfors l hc tr ca ng Nevanlinna (theo ngha bo v lun n tin s).Do ng Nevanlinna khng phi l gio s hng dn lun n ca ng Thim nn ta cth coi ng Thim l con nui ca ng Nevanlinna v mt Ton hc. V vy ta c thcoi Ahlfors l anh ca ng Thim.

Gii Fields c trao ln th hai nm 1950. Mt trong hai ngi c gii l nh Tonhc Php Laurent Schwartz. ng Schwartz l mt trong nhng ngi sng lp ra yban quc gia v Vit Nam ca Php nm 1966 v Ta n quc t Russel x ti ditchng ca M Vit Nam nm 1967. ng Schwartz sang thm Vit Nam nhiu ln, lnu tin vo nm 1968 trong chin tranh chng M. Nm 1990 ng c B i hc mi

sang tham quan v nh gi nn gio dc Vit Nam. ng vit mt bn bo co 40

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Duyn s ca Ton hc Vit Nam vi gii Fields 11

ng Smale c coi l mt nh bc hc trong Ton hc v ng quan tm ngin cu nhiuchuyn ngnh Ton hc khc nhau v chuyn ngnh no ng u t c nhng ktqu xut sc. Nhng nm 60 ng l lnh t phong tro tr thc chng chin tranh Vit

Nam M. Nm 1965 ng t chc cho sinh vin bi kha i hc California v chntu ch lnh M Berkeley. Nm 1966 ng t chc hp bo chng chin tranh Vit Nambn thm i hi Ton hc th gii khi nhn gii Fields. V nhng hot ng chngchin tranh m ng b Qu khoa hc quc gia M ct tin ti tr nghin cu. Nm 2004Vin Ton hc mi GS Smale sang Vit Nam ging bi vi s ti tr ca Qu gio dcVit Nam (VEF). Trong bui ni chuyn vi sinh vin ti i hc bch khoa H Ning khc v xin li v chin tranh Vit Nam. ng Smale c mt hc tr ngi Vitl H Quang Minh, hin ang lm vic i hc Humboldt Berlin.

i hi Ton hc th gii tip theo nm 1970 c hai gii Fields lin quan n Vit Nam.

Ngi th nht l nh Ton hc Nht Heisuke Hironaka. Nm 1968 ng Hironaka dyv L thuyt k d cho cc nh Ton hc tr chu u. Trong lp hc c mt sinhvin Vit Nam tn l L Dng Trng mi tui i mi. Sau ny L Dng Trng trthnh mt trong nhng chuyn gia hng u th gii v L thuyt k d. GS L DngTrng l ngi a Hi Ton hc Vit Nam gia nhp Lin on Ton hc th gii l tchc xt v trao gii Fields. GS Hironaka rt quan tm n vic gip Ton hc VitNam. ng l ngi vn ng Hi Ton hc Nht thnh lp Chng trnh trao iTon hc gia Nht v Vit Nam. ng sang thm Vit Nam mt vi ln vi t cchc nhn. Nm 1977 ng cng b mt cng trnh Ton hc ni ting ca mnh trong Tpch Acta Mathematica Vietnamica ca Vin Ton, c trch dn rt nhiu. Ngi th

hai l nh Ton hc Nga Sergey Novikov. ng Novikov l thy ca L T Quc Thng,huy chng vng Olympic Ton quc t nm 1982. Hin nay L T Quc Thng l mtchuyn gia hng u th gii trong lnh vc T p chiu thp.

Cn hai gii Fields na sang lm vic Vit Nam. Ngi th nht l nh Ton hcM David Mumford c gii Fields nm 1974. ng ny lm bo co mi tai Hingh Ton quc t do Vin Ton phi hp vi ai hc Quy Nhn t chc nm 2005.Ngi th hai l nh Ton hc New Zealand Vaughan Jones, ngi c gii Fields nm1990. ng ny lm bo co mi ti Hi ngh quc t v Tp lng t do Vin Tont chc nm 2007 v cng b mt cng trnh ca mnh trong tp ch Acta Mathematica

Vietnamica ca Vin Ton.Nm 1978 c nh Ton hc Php Pierre Deligne c gii Fields. ng Deligne l hctr ca ng Grothendieck v l thy ca GS L Dng Trng (ng hng dn). ngtng l thnh vin Hi ng bo v ca GS Hong Xun Snh. Do GS Hong Xun Snhcng l hc tr ca ng Grothendieck nn c th coi GS Hong Xun Snh l em v NgBo Chu l chu h ca GS Deligne v mt Ton hc.

c bit hn, bn cng thy ca Ng Bo Chu l Laurent Lafforgue cng c giiFields nm 2002. Hc tr u tin ca Lafforgue l Ng c Tun, ngi tng ot

huy chng vng hai ln thi Olympic Ton quc t nm 1995 v 1996. Hin nay Ng


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12 Chuyn Ton hc s 9

c Tun ang lm vic ti i hc Paris 13. Gn y nht c Terence Tao l nh Tonhc c c gii Fields nm 2006 cng c lin quan n VN. Tao c mi quan h cngtc thn thit vi V H Vn, hin l mt chuyn gia hng u th gii trong lnh vc

T hp. H vit chung 15 cng trnh v mt cun sch chuyn kho. Ngoi ra, Taoc cng thy vi Dng Hng Phong, cng l mt nh Ton hc VN hng u M.Hin nay, Tao c mt nghin cu sinh ngi Vit l L Thi Hong, huy chng vngOlympic Ton quc t nm 1999. Vi nhng ngi tr tui nh Ng c Tun v LThi Hong theo ui nghip Ton, bit u VN li c c may c gii Fields ln na.

Grothendieck v GS Ng Thc Lanh (pha sau) v GS Hong Ty (bn phi nh)ti ni s tn ca i hc Tng hp i T, Thi Nguyn.

Bn chp trch on bo co nh my ca Grothendieck: C mt nn Ton hc

tht s ng ngha nc Vit Nam Dn ch Cng ha (gch di).


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Duyn s ca Ton hc Vit Nam vi gii Fields 13

Bn chp on Hi k ca Schwartz v bn bo co v gio dc i hc, trong c cu Vit Nam thng trong chin tranh v thua trong ha bnh.

V chng ng Schwartz v GS T Quang Bu (bn tri nh) i thm Vit Bc.


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THNG TIN TON HC

Mt ma h vi nhiu hot ng si ni dnh cho gio vin v hc sinh chuynTon kt thc. Nm nay, ln u tin Kha bi dng chuyn mn nghip vh dnh cho gio vin Ton (gi tt l Trng h bi dng gio vin Ton) c t chc ti HQG TP H Ch Minh t 5 11/8 vi s tham gia ca trn150 gio vin n t cc trng THPT chuyn trn ton quc. GS TSKH NguynVn Mu, thy Nguyn Khc Minh, TS Nguyn Chu Gia Vng, TS L B KhnhTrnh, TS Nguyn Vn Minh Mn, TS Trn Nam Dng, ThS Nguyn Trng Tun,thy Nguyn c Tn n ging bi v chia s nhiu kinh nghim qu bu choi ng cc thy c gio dy Ton.

Tip ni thnh cng ca Trng h dnh cho gio vin, t ngy 16 22/8, chngtrnh Gp g Ton hc ln 2 c i hc Quc gia thnh ph H Ch Minh

phi hp vi trng THCS THPT u Lc (qun G Vp) t chc vi s thamgia ca 50 hc sinh n t cc a phng: thnh ph H Ch Minh, Cn Th,ng Nai, Ninh Thun, Qung Nam v Nng. Bn cnh bi ging ca ccthy c gio c nhiu kinh nghim, cc bn hc sinh cn c s hng dn nhittnh ca cc anh ch sinh vin trng thnh t phong tro chuyn Ton. Chngtrnh hot ng vn ngh th thao cng ht sc si ni vi cc bui chiu luyntp v thi u th thao, mt ngy i d ngoi ti thc Giang in, mt m linhoan vn ngh ht mnh. c bit cc hc sinh tham gia chng trnh c GS

Egorov v hc tr ca ng l TS Nguyn Thnh Nam, tng gim c Tp onFPT gh thm v trao i thn mt v Ton hc.

Ngy 19/9, kha 3 ca cu lc b Ton hc dnh cho cc hc sinh lp 10 ckhai ging ti trng THPT chuyn L Hng Phong. Cc hc sinh s sinh hottrong vng 5 thng vi nhiu hnh thc hot ng phong ph: Nghe ging chuyn, tham gia seminar gii ton, lm bi kim tra, tham gia cc cuc thi ng i,nghe ni chuyn v Ton hc v Khoa hc, i d ngoi, . . . C trn 60 hc sinhn t cc trng THPT trong thnh ph tham gia cu lc b, trong ngnht l cc trng L Hng Phong, Ph thng Nng khiu, Trn i Ngha.

Theo thng tin t ban t chc ca cuc thi Gii thng Ton hc Shing Tung Yaudnh cho hc sinh ph thng th c hai n (project) ca hai nhm hc sinh Phthng Nng khiu (gm 4 bn Phm Hy Hiu, Nguyn Mnh Tin, T NguynThi Sn, Phm Anh Tun) lt vo vng bn kt ca cuc thi v s c trnhby trc tip trc ban gim kho ti Singapore vo chiu 16/10/2010. c bitmi project lt vo bn kt s c thng 1000 USD. Cc project lt vo chung ktc thng 2000 USD v cc tc gi s c i bo v ti M. Thng tin v cucthi c th xem chi tit ti: http://www.yau-awards.org/overseas/index.html.

Theo thng bo t Hi Ton hc Vit Nam, ngy 22/4/2011, GS Frank Morgan,

chuyn gia hng u th gii nhn chuyn i thm v lm vic ti Vit Nam s gh

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16 Chuyn Ton hc s 9

thnh ph H Ch Minh v c bi ging public vBong bng x phng v Tonhc. c bit Frank Morgan khng ch ni ting vi nhng cng trnh khoa hctm c ca mnh m c rt ni ting nh mt din gi xut sc, c th lm cho

hc sinh ph thng hiu c nhng iu cao siu. Thng tin chi tit v FrankMorgan c th c ti y: http://math.williams.edu/morgan/.

Phong tro t chc seminar Ton hc v cu lc b Ton hc ang c lan rng.Hin nay ngoi seminar Gii tch v Ton s cp t chc ti trng i hc Khoahc T Nhin i hc Quc gia H Ni (do GS Nguyn Vn Mu ch tr),seminar cc phng php Ton s cp ti trng Ph thng Nng khiu cthm seminar ton s cp ti Vnh Long, An Giang, Long An. c bit, k tthng 10/2010, Vin Ton hc Vit Nam s t chc cu lc b Ton hc dnh chohc sinh THPT ti Vin Ton (1 thng 1 bui). c bit 3 bi ging u tin(vo cc thng 10, 11, 12/2010) s do GS H Huy Khoi, TS Phan Th H Dng,

TS Nguyn Chu Gia Vng m trch.


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KHM PH MT S TNH CHT CADY S TRUY HI TUYN TNH CP HAI

o Hong NhHS chuyn Ton kha 2009 - 2012

Trc ht, bc vo chuyn ny, chng ta cn bit th no l dy s truy hi tuyntnh cp hai.

nh ngha 1. Dy s truy hi tuyn tnh cp hai l dy s c dng nh sau

un+2= aun+1+bun,

trong a, b l cc s thc khc0.

Dy s ny tuy kh n gin nhng ng sau n l nhng b n th v ang ch ichng ta. No, chng ta hy bt u cuc hnh trnh.

M u cho cuc khm ph cc b n, chng ta hy cng nhau tm hiu mt tnh chtth v sau.

Tnh cht 1. Cho dy s{un}tha mnun+2= aun+1+ bun vi min N. Khi ta c hng ng thc sau

un+2unu2n+1= (b)n

1(u3u1u22), n N

. (1)

Chng minh.Trc ht, ta s chng minh

un+2unu2n+1=b(un+1un1u

2n). (2)

Tht vy, ta c

un+2unu2n+1+b(un+1un1u

2n) =un(un+2bun)un+1(un+1bun1)

=unaun+1un+1aun= 0.

Vy(2)c chng minh. T y, bng cch p dng lin tip (2),ta d dng thu chng ng thc (1)nh trn.

By gi, trong (1)cho b= 1,ta c

un+2unu2n+1=u3u1u

2

2. (3)

Gi sun= 0vi mi n 1.Khi , t c= u3u1u22,t(3)ta suy ra

un+2= u2n+1+c

un.

T ta rt ra c tnh cht nh sau.

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18 Chuyn Ton hc s 9

Tnh cht 2. Cho dy{un}c xc nh biu1=m, u2= p, u3=q (mpq= 0) v

un+2= u2n+1+c

un, n 1,

trong c= mqp2.Khi , ta c{un}chnh l dy s truy hi tuyn tnh cp 2 dng

un+2= aun+1un,

via= q+m

p .

Chng minh.T gi thit, ta suy ra

c= un+2unu2n+1. (4)

Thay nbi n1,ta cc= un+1un1u2n. (5)

T (4) v (5), ta c un+2unu2n+1= un+1un1u2n,hay

un(un+2+un) =un+1(un+1+un1),

un+2+unun+1

= un+1+un1

un.

Thay nln lt bi n1, n2, . . . , 2,ta c

un+2+unun+1

=un+1+un1

un= =

u3+u1u2

=q+m

p =a.

T suy ra un+2=aun+1 un.Tnh cht 2 c chng minh.

Cc tnh cht 1, 2 thng c s dng cho cc dng ton m dy s c xc nhnh trn. Sau y l mt s v d.

V d 1 (VMO 1999). Cho dy s{an}c xc nh bi

a1= 1, a2= 2, an+2= 3an+1an, n 1.

Chng minh rng

an+2+an2 +a2n+1

an, n 1.

Li gii.Ta d dng chng minh c dy s trn l dy tng, suy ra an >0 vi min.S dng tnh cht 1, ta c

an+2= a2n+1+ 1

an.

T y, s dng bt ng thc AM-GM cho hai s dng, ta c

an+2+an =a2n+1+ 1

an+an=

a2n+1an

+

1

an+an

a2n+1an

+ 2.

Bi ton c chng minh.


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Khm ph mt s tnh cht ca dy s truy hi tuyn tnh cp hai 19

V d 2 (Bulgaria 1978). Cho dy s{un}tha mn ng thi cc tnh cht sau

1. un = 0, n= 1, 2, . . .

2. u1, u2 Z.

3. u21+u

22+a

u1u2 Z.

4. un+2= u2n+1+a

un, n= 1, 2, . . . .

Chng minh rng dy{un}ch cha ton s nguyn.

Li gii.S dng tnh cht 2, ta c

un+2=kun+1un, n 1, (6)

vik= u3+u1

u2.

Do u3= u2

2+a

u1nn ta c

u3+u1u2

= u2

1+u2

2+a

u1u2 Z,suy ra k Z.M theo gi thit

th u1, u2 Znn t(6),ta d dng suy ra un Z, n 1.

Nhn xt.V d trn cho ta mt tiu chun nh gi dy s cc s hng ca dy

{un}vi un+2= u2n+1+a

un c l cc s nguyn hay khng.

By gi, chng ta hy tip tc xem xt dy s {un}tha mn

u1=a, u2= b, un+2=dun+1un.

R rng, nu a, b u l cc s nguyn v d cng l s nguyn th hin nhin mi shng ca dy {un} lun lun l s nguyn. Vy, nu d khng l s nguyn th d phitha mn iu kin g mi s hng ca dy u l cc s nguyn? Chnh cu hi ny a ta n vi bi ton sau.

Bi ton 1. Cho dy s{un}tha mn

u1= a, u2=b, un+2= dun+1un, n 1,

trong a, bl cc s nguyn khc0vdl mt s thc khc0.Tm mi gi tr cad mi s hng ca dy u l cc s nguyn.

By gi chng ta s cng nhau tm hiu li gii ca bi ton ny.

Li gii.Nu dl mt s v t th r rng u3= bd acng l s v t v a, bl cc s

nguyn. Do ta ch cn xt dtrong tp s hu t.


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20 Chuyn Ton hc s 9

Gi s d l s hu t tha mn yu cu bi ton. V d l s hu t nn ta c th td=

p

q vip, ql cc s nguyn khc 0 v(p, q) = 1.T un+2=

p

qun+1un,hay

q(un+2+un) =pun+1. (7)

Vi n = 1th (7) tr thnh q(u3+ u1) =pu2...q,suy ra u2

...q(do(p, q) = 1). Tng t,

bng cch cho nnhn gi tr ln lt bng 3, 4,ta cng c u4... qvu5

... q. T chon= 2,ta suy ra

pu3=q(u2+u4)...q2.

M (p, q) = 1nn(p, q2) = 1,do t trn suy ra u3...q2.Do vy

pu4=q(u3+u5)...q2.

V (p, q2) = 1nn u4...q2.C lm nh vy ta rt ra c nhn xt l

un...qk1, n k 1. (8)

iu ny ta c th chng minh bng quy np theo k. Tht vy, d thy (8) ng vik= 1.Gi s n ng ti k, tc l

un...qk1, n k 1.

Khi t (7) ta suy raun+1 ...qk,tc lun ...qk,n k + 1.Nhn xt c chng minh.

By gi, s dng tnh cht 1, ta c

un+2un u2n+1=c,

vic= u3u1u22 l hng s.

T y kt hp vi (8), ta suy ra c...q2k vikln ty .

Do c l mt hng s nn nu c = 0th qch c th bng 1 (v nu qkhc 1, ta s cq2k > ckhi ktng n mt gi tr no ), khi dl mt s nguyn (v mu bng 1).

Xt c= 0,khi ta cun+2un = u

2n+1. (9)

D thy, v a, bl cc s nguyn khc 0cho nn ta lun chng minh c un = 0vimi s nguyn dng n(c th quy np t (9)). T (9) suy ra

un+2

un+1=

un+1

un, n 1.

Suy raun+1

un=

un

un1= =

u2

u1=

b

a

=r, n 1.


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Do un+1= run, n 1.R rng, lc ny dy s {un}l cp s nhn, cho nn

un+1= rna. (10)

n y, ta vit li rdi dng phn s ti gin r = uv

vi u, v Zv (u, v) = 1.Do

unl s nguyn vi mi nnn

un+1= un

vna Z.

T suy ra a... vn (do (un, vn) = 1). M a li l mt hng s nn v ch c th nhn

gi tr l 1,do r Zhay b...a. Ta li c

d= u3+u1

u2=

ar2 +a

ar =

r2 + 1

r =

a2 +b2

ab .

Th li gi tr ca d,ta thy mi s hng ca dy u nguyn. T ta rt ra kt lun

Nu bkhng chia ht cho ath dl s nguyn.

Nu bchia ht cho ath dl mt s nguyn hoc d= a2 +b2

ab .

Chnh bi ton trn cho ta mt tnh cht tip theo.

Tnh cht 3. Cho dy s{un}tha mn

u1= a, u2=b, un+2= dun+1un, n 1,

via, b l cc s nguyn vab= 0. Khi , mi s hng ca dy u l cc s nguynkhi v ch khi ta c mt trong cc iu sau

d l mt s nguyn nubkhng chia ht cho a;

dl mt s nguyn hocd= a2 +b2

ab nubchia ht cho a.

Khng dng li ti y, chng ta hy tip tc cng nhau khm ph tip, vn cn nhiu

b n ng sau ang ch i chng ta.Xt dy s {un}tha mn u1= m, u2=p v

un+2= aun+1un. (11)

Theo tnh cht 1 th dy s trn lun c th bin i v dng

un+2unu2n+1=u3u1u

2

2=c. (12)

No, by gi chng ta hy nhn tht k (11) v (12), gia chng liu c mi lin h no

khng? Cc bn nhn ra cha no? Chng ta cng xem nh!


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22 Chuyn Ton hc s 9

T (11) v (12), ta suy ra c

un+2+un= aun+1un+2un = c+u

2

n+1

.

Do un+2vunchnh l nghim ca phng trnh bc hai

X2 aun+1X+ u2n+1+c= 0,

viXl n s, un+1chnh l tham s.

T ta rt ra c tnh cht tip theo.


u1= m, u2=p, un+2= aun+1un.

Khi , un+2 vun chnh l nghim ca phng trnh bc hai

X2 aun+1X+ u2n+1+c= 0, (13)

viX l n s, un+1l tham s vc= u3u1u22.

C l, n by gi cc bn s t hi, tnh cht 4 c th suy ra c iu g na?

By gi, ta hy xem, nu mi s hng ca dy {un} u l cc s nguyn v gi s acng l s nguyn, th r rng phng trnh (13) c hai nghim nguyn. M iu kin

cn mt phng trnh bc hai c cc h s nguyn c nghim nguyn l bit thc ca n phi l s chnh phng. M

=a2u2n+14(u2n+1+c) = (a

2 4)u2n+14c.

Do ta c (a2 4)u2n+14cl mt s chnh phng. Ta rt ra c tnh cht sau.


un+2=aun+1un, n 1.

Gi s mi s hng ca dy u l s nguyn vacng l mt s nguyn. Khi ta c

(a2 4)u2n+14cl mt s chnh phng(c= u3u1u22).

V d 3. Cho dy s{un}c xc nh nh sau

u1= 5, u2= 16, un+2= 1994un+1un, n 1.

Chng minh rng3976032u22508159239 l mt s chnh phng.

Li gii.Ta c 3976032 = 19942 4, u3= 1994165 = 31899v

159239 = 318995162.

Do theo tnh cht 5 th 3976032u2

2508159239l mt s chnh phng.


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V d 4 (VMO 1997). Cho dy s nguyn{an}c xc nh bia0= 1, a1= 45 v

an+2= 45an+17an,

vi min= 0, 1, 2, . . .

(a) Tm s c dng caa2n+1 anan+2 theo n.

(b) Chng minh rng vi minth1997a2n+ 4 7n+1 l s chnh phng.

Li gii.(a) Ta c a2= 452 71 = 2018.S dng tnh cht 1, ta c

an+2ana2n+1= 7

n(a2a0a2

1),

suy ra

a2

n+1an+2an = 7n+1

.Do 7l s nguyn t nn s cc c nguyn dng ca a2n+1anan+2l n+ 2 s.

(b)Ta s dng t tng ca cc tnh cht trn gii quyt cu hi ny. T gi thitan+2= 45an+17anva2n+1anan+2= 7

n+1 (theo tnh cht 1), ta suy raan+2+ 7an= 45an+1

(7an)an+2= 7a2n+17

n+2.

Do 7anvan+2l cc nghim ca phng trnh bc hai

X2 45an+1X+ 7a2n+17

n+2 = 0.

V 7an, an+2l cc s nguyn nn phng trnh trn c hai nghim nguyn, suy ra bitthc ca n l s chnh phng. M

= (45an+1)2 4(7a2n+17

n+2) = 1997a2n+1+ 4 7n+2,

nn ta c1997a2n+1 + 4 7n+2 l s chnh phng. Hn na, d thy 1997a20 + 4 7

1 = 452.Do vy ta c 1997a2n+ 4 7

n+1 l s chnh phng vi mi s t nhin n.

Nhn xt.Qua v d trn, ta thy tnh cht 4 v 5 c th tng qut cho trng hpun+2=pun+1+qun,

viu1, u2l cc s nguyn v p, qcng l cc s nguyn.

Khi un+2 qun=pun+1v un+2(qun) =qu2n+1+ (q)ncvi c = u3u1 u22.T

ta c un+2vqunl hai nghim ca phng trnh bc hai

X2 pun+1X qu2n+1+ (q)

nc= 0.

V do (p2

+ 4q)u2

n+14(q)n

cl mt s chnh phng.


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24 Chuyn Ton hc s 9

By gi, chng ta s tip tc khm ph tip, liu t tnh cht 4, ta c th khai thcc nhng iu th v no na? Cc bn hy cng ti th gii phng trnh (13). Theocng thc nghim ca phng trnh bc hai, ta suy ra

X1, 2 =aun+1

(a2 4)u2n+14c

2 .

Gi sun+2> un,khi ta c mt dy s truy hi c dng

un+2= a

2un+1+

1

2

(a2 4)u2n+14c.

Nh vy, ta c th rt ra c tnh cht 6 nh sau.

Tnh cht 6. Mi dy s{un}tha mn

u1= m, un+2=

a

2 un+1+

1

2

(a2

4)u2

n+1+ 4c, (ac= 0)

lun c th a v dy s truy hi tuyn tnh cp hai c dng nh sau

un+2= aun+1un.

Li gii.Chuyn a

2un+1sang v tri v bnh phng hai v, ta c

u2n+2aun+2un+1+a2

4u2n+1=

1

4

(a2 4)u2n+1+ 4c

,

u2n+2aun+2un+1+u2n+1c = 0.

Thay nbi n1,ta cu2n+1aun+1un+u

2nc = 0.

T y suy ra un+2vunl cc nghim ca phng trnh bc hai

X2 aun+1X+ u2n+1c= 0.

p dng nh l Viette, ta c un+2+ un =aun+1,hay un+2=aun+1un.Tnh cht 6c chng minh.

V d 5. Cho a, b, cl ba s nguyn tha mna2 =b + 1v dy s{un}c xc nhnh sau

u0= 0, un+1= aun+

bu2

n+c2

, n= 0, 1, 2, . . .Chng minh rng mi s hng ca dy trn u l s nguyn.

Li gii.Ta c

un+1=aun+

bun+c2 =aun+

(a2 1)u2n+c2 =

2a

2un+

1

2

(4a2 4)u2n+ 4c

2.

Theo tnh cht 6, ta suy raun+2= 2aun+1un.

V u0= 0, u1=|c| Znn t trn ta c unl s nguyn vi mi s t nhin n.

T tnh cht 3 v tnh cht 6, ta c th suy ra mt tnh cht nh sau.


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Tnh cht 7. Xt dy s{un}tha mnu1, u2 Zv

un+1= aun+ bu2n+c.

Nu c l mt s nguyn v tn ti mt sd sao cho a= d

2 vb=

d2 4

4 vidl mt

s nguyn (nuu2 chia ht hoc khng chia ht cho u1) hocd= u2

1+u22

u1u2(nuu2 chia

ht cho u1) th mi s hng ca dy u l nguyn.

y c th c xem nh mt tiu chun nh gi dy s dng trn l 1 dy nguyn.

Trc khi kt thc chuyn , chng ta hy cng nhau xem xt mt bi ton sau.

Bi ton 2 (Bulgaria 1987). Xt dy s{xn}c xc nh bix1=x2= 1v

xn+2= 14xn+1xn4, n 1.

Chng minh rng vi min 1,ta cxn l bnh phng ca mt s nguyn.

Li gii. gii thnh cng bi ton ny, trc ht chng ta cn phi nghin cu ccgi tr ca cc s hng u tin ca dy v sau rt ra nhn xt v mi quan h cachng. Th vi gi tr u, ta c

x1= 12, x2= 1

2, x3= 32, x4= 11

2, x5= 412, x6= 153

2, . . .

Hy quan st v , cc s trn hon ton c quy lut. Ta nhn thy3 = 411, 11 = 431, 41 = 4113, 153 = 44111, . . .

T cc yu t trn, cc s 1, 1, 3, 11,41, 153, . . .u c hnh thnh da vo dy sau

y1= 1, y2= 1, yn+2= 4un+1yn, n 1.

By gi, vic ca chng ta l chng minh rng

xn = y2n. (14)

Cch chng minh n gin nht chnh l s dng quy np: D thy (14) ng vin = 1,2.Gi s n ng vi n= 1, 2, . . . , k,tc l

xn=y2n, n= 1, 2, . . . , k.

Ta cn chng minh n cng ng vi n= k + 1.Ta c

y2k+1= (4ykyk1)2 = 16y2k8ykyk1+y

2k1

= 14xk xk14 + (2y2k8ykyk1+ 2y

2k1+ 4)

=xk+1+ 2y2k2(4yk yk1)yk1+ 4

=xk+1+ 2y2

k2yk+1yk1+ 4.


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26 Chuyn Ton hc s 9

S dng tnh cht 1, ta cyk1yk+1y

2k= 2.

T suy ra y2k+1= xk+1.Vy ta c iu phi chng minh.

Nhn xt. Ta hon ton c th tng qut bi ton trn. Xt dy s {un} tha mnun+2=aun+1 unvc= u1u3u22,ta c

un+2un u2n+1=c.

Xt tip dy s {vn}sao cho vn=u2nvi mi n,ta s c

vn+2= u2n+2=a

2u2n+12aun+1un+u2n=a

2u2n+12un(aun+1un)u2n

=a2u2n+1u2n2unun+2= (a

2 2)u2n+12(unun+2u2n+1)u

2n

= (a2 2)vn+1vn2c.

Suy ra vn+2= (a2 2)vn+1vn2c.

Mt khc, bng cch s dng quy np, ta cng d dng chng minh c iu ngcli ng, tc l: Nu hai dy {un}, {vn}tha mn v1= u21, v2=u

22

, c= u1u3 u22v

un+2=aun+1un, vn+2= (a2 2)vn+1vn2c,

th ta c vn=u2nvi mi n.

V nh vy, ta c bi ton tng qut nh sau: Cho hai dy s{un} v{vn} tha mnv1= u

21

, v2=u2, c= u1u3u

22vun+2=aun+1 un. Khi

vn+2= (a2 2)vn+1vn2c

khi v ch khivn= u2n vi mi s nguyn dngn.

T cch chng minh nh trn, ta c th rt ra c phng php gii cc bi ton vdy s c lin quan n s chnh phng. Vi t tng nh trn, ta c th gii c ccbi ton tng t sau.

Bi ton 3. Cho dy s{yn}tha mny1= y2= 1v

yn+2= (4k5)yn+1yn+ 4 2k, n 1.

Tmk Zsao cho mi s hng ca dy u l s chnh phng.

(Ta c th chng minh c dy s tha mn bi ton trn c dng nh sau

yn+2= 7yn+1yn2.

T , p dng cch lm bi ton 2 th c iu phi chng minh.)


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Bi ton 4 (Canada 1988). Cho hai dy s{xn}, {yn} xc nh bi x0 = 0, x1 = 1,xn+1= 4xnxn1 vy0= 1, y1= 2, yn+1= 4ynyn1.Chng minh rng

y2n

= 3x2n

+ 1, n N.

Bi ton 5. Cho dy{un} tha mnu1=k, u2=k + 1 vik l s nguyn bt k v

un+1=un(un1) + 2, n= 2, 3, . . .

Chng minh rng sA= (u21+ 1)(u22+ 1) (u

22005+ 1)1 l mt s chnh phng.

(Gi , bng quy np ta chng minh c An = (un+11)2 vi

An= (u2

1+ 1)(u2

2+ 1) (u2n+ 1)1.)

Sau y l mt s bi tp dnh cho bn c.

Bi tp 1. Cho dy s {un}tha mn u1= 1, u2=1v

un=un12un2, n= 3, 4, . . .

Xt dy {vn}nh sau

vn = 2n+1 7u2n1, n= 2, 3, . . .

Chng minh rng mi s hng ca dy {vn}u l s chnh phng.

Bi tp 2. Cho dy s {un}tha mn u1= 2, u2= 3v

un = nun1(n2)un22n+ 4

vi mi n= 3, 4, . . .

(a) Tm n |un2007|c gi tr nh nht.

(b) Tm s d khi chia u2007 cho2006.

Bi tp 3 (VMO 1998). Cho dy s {an}c xc nh bi a0= 20, a1= 100v

an+2= 4an+1+ 5an+ 20, n N.

Tm s nguyn dng hnh nht tha mn iu kin an+h an chia ht cho 1998vimi s t nhin n.

Bi tp 4 (TST Vit Nam 1993). Gi (n) l phi hm Euler. Tm tt c cc snguyn dng k > 1 tha mn iu kin: vi a l s nguyn > 1 bt k, t x0 = a,xn+1=k(xn)vi n= 0, 1, 2, . . .th {xn}lun b chn.

Bi tp 5 (Ba Lan 2002). Cho k N v dy s {an}c xc nh bi

a1=k + 1, an+1= a2nkan+k, n N

.

Chng minh rng vi mi m, n N, m=n, ta c (an, am) = 1.


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28 Chuyn Ton hc s 9

Bi tp 6. Cho s thc rv dy s {xn}c xc nh bi

x0= 0, x1= 1, xn+2=rxn+1 xn, n N.

Chng minh rngx1+x3+ +x2m1= x

2m, m N

.

Ti liu tham kho

[1] Phan Huy Khi,S hc v dy s, Nh xut bn Gio Dc, 2009.

[2] Nguyn Vn Mu (ch bin), Chuyn chn lc dy s v p dng, Nh xutbn Gio Dc, 2008.

[3] Cc ti liu su tm t internet.


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NH L THNG D TRUNG HOA

Phm Hy Hiu

HS chuyn Ton kha 2007 - 2010

Lc dn. Xa Trung Quc, c mt v tng khng nhng c ti thao lcchin trng m cn c nhng hiu bit su sc v Ton hc tn l Hn Tn.Tng truyn rng khi im qun, ng ra lnh cho qun i ca mnh xp thnhnhng hng 5 ngi, 7 ngi ri 11 ngi, sau da vo qun s d cn lisau mi ln xp hng cng vi vic c lng qun s ca mnh trong khong baonhiu, ng suy ra c qun s chnh xc ca mnh. Nhng g Hn Tn lm c cc nh Ton hc Trung Quc v th gii ghi nhn li thnh nh l Thng

d Trung Hoa. y thc s l mt kt qu p v c nhiu ng dng trong vicgii quyt cc bi ton v L thuyt s v c nhng vn cao cp hn ca Tonhc. Bi vit ny hy vng trao i vi cc bn v nh l ny.

1 Nhn li nh l Thng d Trung Hoa

nh l 1(nh l Thng d Trung Hoa). Chok l mt s nguyn dng, m1, m2, . . . ,mk l cc s nguyn i mt nguyn t cng nhau v a1, a2, . . . , ak l cc s nguynbt k. Khi h phng trnh ng d

x a1 (modm1)x a2 (modm2)

. . .

x ak (modmk)

c nghim duy nht modulo m1m2 mk.

Chng minh 1. T pht biu ca nh l, ta thy vic chng minh cn phi tri quahai bc nh sau:

(1) Chng minh s tn ti ca nghim xmodulo m1m2 mk.

(2) Chng minh tnh duy nht ca nghim theo modulo m1m2 mk.

(1) chng minh s tn ti nghim ca h ng d, ta s s dng nguyn l quy np.Gi s kt lun ca nh l ng vi 1, 2, . . . , k.Xt k+ 1s nguyn m1, m2, . . . ,mk+1 i mt nguyn t cng nhau. t mk = mkmk+1. V (mk, mk+1) = 1 nn theogi thit quy np, h ng d

x ak (modmk)

x ak+1 (modmk+1)

29


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30 Chuyn Ton hc s 9

c nghim duy nht x ak (mod m

k). Li theo gi thit quy np th v m1, m2, . . . ,mk l ks nguyn i mt nguyn tt cng nhau nn h ng d

x a1 (modm1)

x a2 (modm2)

. . .

x ak (modm

k)

c nghim duy nht theo modulo m1m2 mk nn nh l cng ng vi k+ 1.

Cui cng ta ch cn chng minh nh l trong trng hp k = 2. lm c iu ,ta cn n b sau.

B 1 (B Bezut). Num, n l hai s nguyn tha mn(m, n) = 1 th tn ticc s nguynu, v tha mn

mu+nv = 1.

Chng minh. Xt cc s c dng {1 mu | u Z}, v (m, n) = 1nn tn ti u sao

chon| 1 mu.Khi ta ch cn chn v=1 mu

n .B c chng minh.

Vo bi. Theo b Bezut suy ra rng tn ti cc s nguyn u, v sao cho

m2v m1u= 1.

t r= (a1 a2)u, s= (a1 a2)v th

m2s m1r= a1 a2 a1+m1r= a2+m2s.

t x= a1+m1r= a2+m2s,ta cx a1 (modm1)

x a2 (modm2)

Vyxl nghim ca h ng d vi k= 2.

(2)By gi ta chng minh tnh duy nht ca nghim. Tht vy gi s tn ti x sao cho

x x (mod m1)

x x (mod m2)

. . .

x x (mod mk)

Th th do gi thit m1, m2, . . . , mk i mt nguyn t cng nhau nn ta c

x x (mod m1m2 mk).

nh l c chng minh hon ton.

C rt nhiu cch chng minh nh l Thng d Trung Hoa. Sau y ta n vi mtchng minh khc cho phn tn ti nghim (tc phn (1) trn) ca nh l ny theohng xy dng v khng cn s dng nguyn l quy np.


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nh l Thng d Trung Hoa 31

Chng minh 2. Do gi thit m1, m2, . . . , mk i mt nguyn t cng nhau nn

gcd

m

mi, mi

= 1 (m = m1m2 mk) vi mi i = 1, 2, . . . , k. Do vy, tng t vi

cch l lun trong chng minh ca b Bezut, ta suy ra rng vi mi i u tn ti bisao chom

mibi1 (mod mi), i= 1, 2, . . . , k.

T m

mibiaiai (mod mi), i= 1, 2, . . . , k.

Xt s nguyn x0=k

i=1

m

mibiai vi mi i= 1, 2, . . . , k,ta c

x0=

ki=1

m

mi biai (mod mi)

m

mibiai (mod mi)

ai (mod mi) (v m

mibi1 (modai)).

Vyx0 l nghim ca h ng d trong bi.

Nghin cu hai chng minh trn, ta nhn thy chng u da trn t tng ca b Bezut. Thc t, b ny c mt h qu quan trng nh sau.

H qu 1.1. Nu m, n l cc s nguyn v (m, n) = d th vi mi s nguyn h md| h th tn ti cc s nguynu, v tha mn

h= mu+nv.

T h qu ny ta suy ra m rng sau y ca nh l Thng d Trung Hoa.

nh l 2 (M rng nh l Thng d Trung Hoa). Cho k l mt s nguyn dng,m1, m2, . . . , mk l cc s nguyn bt k va1, a2, . . . , ak l cc s nguyn. Khi hphng trnh ng d

x a1 (modm1)

x a2 (modm2)

. . .

x ak (modmk)

c nghim modulo lcm (m1, m2, . . . , mk) khi v ch khi

ai aj (mod (mi, mj)), 1 i < j k.

M rng ny hon ton c th c chng minh bng phng php tng t ha tc hai cch chng minh ca nh l lc u, ch khc ch s dng h qu ca b Bezut thay v chnh b ny.

Khng ch trong L thuyt s m nh l Thng d Trung Hoa cn c cc m rngtng t trong L thuyt nhm, L thuyt vnh, trng, ideal, . . . , nhng iu vtqu tm quan tm ca bi vit ny.


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32 Chuyn Ton hc s 9

2 nh l Thng d Trung Hoa trong gii ton ph thng

Trong mc ny ta tm hiu cc bi ton ng dng nh l Thng d Trung Hoa. Trc

ht ta n vi ba bi ton c bn u tin mang tnh kinh in cho k thut s dngnh l thng d Trung Hoa.

Bi ton 1. Chng minh rng vi mi s nguyn dng k ln ty u tn ti k snguyn dng lin tip gm ton hp s.

Li gii. Gi p1 < p2 < < pk l cc s nguyn t bt k. Theo nh l Thng dTrung Hoa th tn ti s nguyn dng nln ty sao cho

n 1 (modp1)

n 2 (modp2)

. . .n k (modpk)

Vy vi mi i= 1, 2, . . . , kth pi |n +i, hn na c th chn n > pk,do n+i ul hp s.

Bi ton 2 (Problem E27, PEN, Hojoo Lee). Chng minh rng vi mi s nguyndngk ln ty u tn tik s nguyn dng lin tip khng l ly tha ca mt snguyn dng no khc.

Li gii. Gi p1 < p2 < < pk l cc s nguyn t bt k. Theo nh l Thng d

Trung Hoa th tn ti s nguyn dng nln ty sao cho

n p1 1 (modp21)

n p2 2 (modp22)

. . .

n pk k (modp2k)

Vy vi mi i= 1, 2, . . . , k th pi n+i, hn na ta c th chn n > pk, do n+iu l hp s. Li do l lun trn, trong biu din c s ca n + ic tha spinn n + iu khng phi l ly tha ca mt s nguyn no.

Bi ton 3(VMO 2008). Chom = 20072008.Hi c tt c bao nhiu s t nhinn < msao cho

m| n(2n+ 1)(5n+ 2)?

Li gii.Xt phn tch tiu chun

m= 20072008 = 34016 2232008.

D dng kim tra thy gcd(i, j) < 3 vi mi i, j {n, 2n+ 1, 5n+ 2} nn ta cm| n(2n+ 1)(5n+ 2) khi v ch khi xy ra mt trong cc trng hp sau:

1. m| n.


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2. m| 2n+ 1.

3. m| 5n+ 2.

4. 34016 |n v2232008 |2n+ 1.

5. 34016 |n v2232008 |5n+ 2.

6. 34016 |2n+ 1 v2232008 |5n+ 2.

7. 34016 |2n+ 1 v2232008 |n.

8. 34016 |5n+ 2 v2232008 |n.

9. 34016 |5n+ 2 v2232008 |2n+ 1.

Trong mi trng hp trn, theo nh l Thng d Trung Hoa, tn ti duy nht mtgi tr n modulo m tha mn trng hp y. Do vy c tt c 9 s t nhinntha mnyu cu bi.

Thc cht bi ton VMO 2008 ch l trng hp ring ca bi ton tng qut sau.

Bi ton 4. Cho s nguyn dngnc phn tch tiu chun

n= pa11 pa22 p

ass .

Xt a thcP(x) c h s nguyn. Nghimx0 ca phng trnh ng d

P(x) 0 (mod n) (1)

l lp ng d x0 {0, 1, 2, . . . , n 1} tha mn P(x0) 0 (mod n). Khi , iukin cn v phng trnh(1) c nghim l vi mi i= 1, 2, . . . , s, phng trnhP(x) 0 (mod paii ) c nghim. Hn na, nu vi mi i = 1, 2, . . . , s, phng trnhP(x) 0 (mod paii ) c ri nghim modulo p

aii th phng trnh (1) c r = r1r2 rs

nghim modulo n.

Li gii. Nu mt trong cc phng trnh P(x) 0 (mod paii ) v nghim th hinnhin (1) v nghim. Ngc li, gi s vi mi i= 1, 2, . . . , s, phng trnh P(x) 0(mod paii )c ri nghim l xi1, xi2, . . . , xiri .Theo nh l Thng d Trung Hoa, vi micch chn b (x1j1 , x2j2 , . . . , xsjs),h ng d

x x1j1 (modpa11 )

x x2j2 (modpa22 )

. . .

x xsjs (modpass )

c nghim duy nht x(1j1 ,2j2 , ..., sjs) theo modulo n. R rng nghim ny l nghim ca

phng trnh (1). Vy (1) c nghim.

Hn na s nghim ca (1) bng s cch chn cc b (j1, j2, . . . , js)t cc tp c di tng ng l r

ivii= 1, 2, . . . , s.Vy nn r= r

1r2

rs

.


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34 Chuyn Ton hc s 9

Bi ton trn khng ch c ngha nh mt bi ton tng qut. Thc t, n cho phpta gii hn s lng modulo cn kho st trong vic gii cc phng trnh dng nh (1).Tht vy, t kt qu ca bi ton ny, ta ch cn xt trng hp modulo nca phng

trnh ng d l ly tha ca mt s nguyn t. Kt hp vi b Hensel, ta cn cth gii hn bi ton v trng hp modulo n ca bi ton l mt s nguyn t. Lcny, cc nh l ca L thuyt ng d v s nguyn t nh nh l nh Fermat, s tnti ca cn nguyn thy, . . . c th p dng c. Ni chung, bi ton trn c nghaln trong vic nghin cu cc phng trnh ng d dng (1).

Qua cc bi ton trn, c l cc bn nm c t tng c bn ca cc bi ton cnn nh l Thng d Trung Hoa. Cc bi ton c trnh by tip theo s i hi skho lo hn trong vic s dng nh l Thng d Trung Hoa cng nh cc kin thcS hc khc.

Bi ton 5 (Taiwan TST 2002). Trong li im nguyn ca mt phng ta Oxy,mt im A vi ta (x0, y0) Z2 c gi l nhn thy t O nu on thng OAkhng cha im nguyn no khc ngoi O v A. Chng minh rng vi mi n nguyndng ln ty , tn ti hnh vungn nc cc nh nguyn, hn na tt c cc imnguyn nm bn trong v trn bin ca hnh vung u khng nhn thy c tO.

Li gii.D thy rngiu kin cn v A(xA, yA)nhn thy c tOl

gcd(xA, yA) = 1.

gii quyt bi ton, ta s xy dng mt hnh vung n n vi n nguyn dng ty sao cho vi mi im nguyn (x, y)nm trong hoc trn bin hnh vung u khngth nhn thy c tO. Tht vy, chn pij l cc s nguyn t i mt khc nhau vi0 i, jn. Xt hai h ng d

x 0 (modp01p02 p0n)

x+ 1 0 (modp11p12 p1n)

x+ 2 0 (modp21p22 p2n)

. . .

x+n 0 (modpn1pn2 pnn)

v

y 0 (modp10p20 pn0)

y+ 1 0 (modp11p21 pn1)

y+ 2 0 (modp12p22 pn2)

. . .

y+n 0 (modp1np2n pnn)

.

Theo nh l Thng d Trung Hoa th tn ti (x0, y0) tha mn hai h ng d trn.Khi , r rng gcd(x0+ i, y0+ i)> 1 vi mi i, j = 0,1,2, . . . , n.iu c ngha lmi im nm trong hoc trn bin hnh vung n nxc nh bi im pha di bntri l(x0, y0)u khng th nhn thy c tO. Bi ton c chng minh.

Nhn xt.Yu cu ca bi ton trn khin ta phi la chn cc b s d v cc modulonguyn t thch hp trn mt din rng hn. iu ny khin cho vic p dng nh lThng d Trung Hoa tr nn khng n gin nh ba bi ton lc u na.

Bi ton 6 (Problem E30, PEN, Hojoo Lee). Cho n l s nguyn dng l vn >3.Gik, tl cc s nguyn dng nh nht sao cho kn + 1vtnu l s chnh phng.Chng minh rng iu kin cn v n l s nguyn t lmin{k, t} >

n

4.


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Li gii. Trc ht ta chng minh iu kin cn. Gi snl mt s nguyn t th v

tnl s chnh phng v n| tn nn n2 |tn, suy ra n| t, t t n > n

4.Hn na, t

a2 =kn + 1th a2 1 (mod n),suy raa 1 (mod n).Nhng v a >1 nn a n 1.

T kn+ 1 (n 1)2 k n 2 k > n4

(v n > 3 nn n 2 > n4

). Vy iu

kin cn c chng minh.

Ngc li, gi smin{k, t} > n

4.Ta xt hai trng hp.

Trng hp 1.n ch c mt c nguyn t duy nht. Do n l nnn = pa vip 3.Nu a chn, ta ly t = 1 AB + AC.

A

B C

D

E

O


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Rn luyn k nng gii cc bi ton Hnh hc phng 43

Khi ng trc bi ton ny, ta thng ngh n cc phng php sau.

S dng phng php ta : Dng h trc ta vi B(1, 0), C(1, 0)vA(a, b).Tnh ta im D ri dng bt ng thc.

K thm cc ng ph: Dng DE AB (nh hnh v trn) ri chng minhrng AC= DE, BE= ADa v AD + B E > AB + DE.

S dng nh l quen thuc:Dng nh l Ptolemy c

AD BC= AB CD + AC BD,

m BC < BD + CD= 2BD nn2AD > AB + AC.

S dng phng php lng gic:Tnh ton trc tip ta c

2AD= 2R sin

B +

A

2

+ 2R sin

C+

A

2

= 4R cos

B C2

,

suy ra AB + AC= 2R(sin B + sin C) = 4R sinB + C

2 cos

B C2

< 2AD.

Nh vy: Vi mt bi ban u, ta c c nhiu tng ring m t cho tanhng hng gii khc nhau; mi hng u c nhng im mnh, im yu no nhng chng u a ta n iu phi chng minh.

V d 2. Cho tam gic ABCc phn gicAD. Trn onAD ly hai im E, F saocho ABE= DBF. Chng minh rngACE= DCF.

???

A

B CD

E

F

P

N

M

A

CB D

F

E

Ta c th gii bi ton trn nh sau: Gi Ml im i xng vi F qua BC, N, P lnlt l im i xng vi Equa AC, AB.Chng minhBE M= BP F (c.g.c), suyraCEM= CN F(c.c.c). T d dng suy ra ACE= DCF.

Cch gii l nh vy, tuy nhin vn t ra y l:

Ti sao li bit ly cc im i xng?


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44 Chuyn Ton hc s 9

Ti sao li bit xt cc tam gic bng nhau? Nu khng dng ng ph th gii c khng?

Trong phn sau chng ta s cng tm cch gii quyt cc vn trn. Li gii ca cc vd s c trnh by ch yu l da trn hng suy ngh chnh, ch trng phn tch ccbc lp lun ch khng i su vo xt cc trng hp ca hnh v c th xy ra nhmhn ch s phc tp. D vy trn thc t, khi gii cc bi ton HHP, chng ta nn ch iu ny, nn xt ht cc trng hp (v tr cc im, cc tia; phn gic trong, ngoi;tam gic cn, khng cn; ng trn thc s v suy bin, . . . ) m bo li gii cy v chnh xc!

2 Mt s phng php rn luyn t duy hnh hc v nng cao k nnggii ton HHP

2.1 La chn cng c thch hp gii mt bi ton HHP

Chng ta hy th ngm ngh li, khi ang l hc sinh THPT nh hin nay, chng ta bitc ht thy bao nhiu phng php gii mt bi ton HHP. C th chng ta bit nhiu nhl, b nhng cng cha th gi l mt phng php theo ngha tng qut. y, ta nin phng php l nh hng, l t tng chnh ca li gii; gii bng cch no ch cha isu vo vic gii nh th no. Xin nu mt s phng php c bn sau

Phng php hnh hc thun ty (quan h song song, vung gc; tam gic ng dng,bng nhau; tnh cht ca tam gic, ng trn; cc nh l hnh hc quen thuc; cc

php bin hnh, ...).

Phng php lng gic (a yu t trong bi v lng gic ca cc gc v bin i). Phng php vector (dng vector trong chng minh tnh cht hnh hc hoc dng mt

h vector n v gii bi ton).

Phng php i s (a cc yu t trong bi v di cnh v bin i). Phng php ta (a gi thit cho vo mt h trc ta v tm ta im,

phng trnh ng thng, ng trn lin quan).

Trong , d thy rng mc t duy hnh hc c th hin gim dn qua th t cc phngphp trn. Nu chng ta l mt hc sinh cha gii HHP th thng vi cc bi ton c githit thun li th lp tc s dng ta , iu tt nhin c ch cho k nng tnh ton,bin i i s ca chng ta nhng ni chung khng c li cho vic rn luyn t duy hnh hc.V a s cc bi ton hnh kh c th s dng phng php ny, ch cn mt ng trn hocmt tm ng trn ni tip khin cho vic dng phng php ta tht kh khn ri.Th nhng khng phi ni vy m ta li qun i phng php c. C vi bn kh ni dung ny th li khng thch s dng ta v c i tm mt cch gii thun ty cho n.Cng vic ny khng phi lc no cng ng, nht l i vi cc k thi HSG c thi gian gprt v s lng bi ton cn gii c li tng i nhiu.

Chng ta hy th ni v mt bi ton n gin sau.


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V d 3. Cho on thngAB c nh v ng thngd c nh song song viAB. imCdi ng trnd. Tm qu tch trc tm tam gicABC.

A B

C

H

O

Phn tch.Mt s bn thy bi ton ny c gi thit tht n gin, ch c on thng cnh, mt im di ng trn ng thng song song ri tm trc tm; thm na, bi ton nyc v quen thuc nn h ch v hnh ra v c gng k ng ph gii. Th nhng, chc chncc bn y s kh m tm c mt li gii hnh hc thun ty cho bi ton ny khi m trnthc t qu tch ca Hl mt ng parabol!

Nu khng cn thn v hnh trc nhiu ln d on qu tch chc chn rng y khngcn l mt qu tch ng thng, ng cong thng thng m m mn i tm khng ngcch s khng i n kt qu mun c. Bi ton ny khng kh nhng nu khng la chnng cng c th khng th nhanh chng thnh cng trong vic gii n c.

Li gii.Trong mt phng ta Oxy,xt A(1, 0), B(1, 0)v ng thng dc phngtrnh y = a, a = 0,do Cdi ng trn nn c ta l C(m, a), m R.Ta s tm ta trc tm ca tam gic ABC.Phng trnh ng cao ca tam gic ng vi nh C l

x= m.

Phng trnh ng cao ng vi nh A l (m 1)(x + 1) + ay= 0,hay

(m 1)x + ay+ m 1 = 0.

T y suy ra ta trc tm ca tam gic AB Cl nghim ca h

(m 1)x + ay+ m 1 = 0x= m

.

Gii h ny ta tm c xH=mvyH= 1 m2

a .Suy ra y

H=

1 x2H

a .

Vy qu tch ca Hl parabol c phng trnh y=1 x2

a .

V d 4. Cho tam gicABCc cnhB Cc nh, Adi ng trong mt phng. GiG, H lnlt l trng tm, trc tm ca tam gic. Bit rng onGHctB Cti trung im caGH,tm qu tch caA.


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46 Chuyn Ton hc s 9

A

B C

G

H

Phn tch.Ta thy gi thit ca bi ton khng phc tp nhng iu kin GHct BC titrung im ca GHqu tht hi kh vn dng; ta cng c th hiu n gin hn l trungim ca GHthuc BCnhng vy th cng khng em li nhiu gi cho li gii bi ton.

V nu ng trc nhng bi ton c gi thit n gin nhng kh vn dng nh th th hyth ngh n phng php ta . Khi , d cc tnh cht hnh hc cha c th hin y nhng cc iu kin hnh hc th s c m bo cht ch hn.

Cng tin hnh la chn mt h trc ta thch hp tng t nh trn ri tnh ta ccimG, Hv vit phng trnh cc ng thng cn thit, t vo iu kin ca bi ton, tas tm c qu tch ca im Achnh l mt ng hypebol. Cc bn th gii li bi tonny vi vic gi nguyn cc gi thit ban u, ch thay trc tm Hbng tm ng trn ngoitip O,cc cng vic ni chung cng c tin hnh tng t nhng d vy ta cng c thmmt khm ph mi. V nu c, hy gii li hai bi ton va ri bng phng php hnh hcthun ty da trn nh ngha cc ng conic, tm tiu im v ng chun ca chng! yl mt vn kh th v nhng cng kh kh!

Ta hy so snh hai phng php gii bi ton sau rt ra tm quan trng ca vic la chnphng php ph hp gii cc bi ton HHP.

V d 5. Cho tam gicABC. V pha ngoi tam gicABCdng cc imD, E, F sao chocc tam gicBCD, C AE, ABFl cc tam gic u. Chng minh hai tam gicAB CvDEFc cng trng tm.

Li gii 1. S dng phng php vector(kh nh nhng v khng cn tn nhiu thi gian ngh ra cch gii ny).

Gi M, N, Pln lt l trung im ca BC, CA, AB.Ta c

AD+ BE+ CF = AM+ M D+ BN+ N E+ CP+P F=

AM+BN+

CP

+

M D+N E+

P F

.

D thyAM+

BN+

CP =

1

2

AB+

AC

+

1

2

BA +

BC

+

1

2

CA +

CB

=0

vM D+

N E+

P F =

0 theo nh l con nhm nn

AD+

BE+

CF =

0.

Vy hai tam gic ABCvDEFc cng trng tm.

Li gii 2. S dng hnh hc phng thun ty(dng nhiu ng ph, hng suy ngh hi

thiu t nhin v i hi c kinh nghim v cc bi ton c gi thit tng t nh th ny).


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A

B CM

NP

D

Q

F

E

I

G

Gi Il trung im EF v Ql im i xng vi D qua BC, khi BC Qcng l tamgic u. Ta thy php quay tm B gc quay 60 bin C thnh Q, bin A thnh F nnABC= FBQ.Tng tABC= EQC.T y ta c

F BQ =

EQC.

Suy raF Q= AC=AE, QE= AB = AFv t gic AEQFl hnh bnh hnh. Do Ichnhl trung im ca AQ,m Ml trung im ca QDnn IMchnh l ng trung bnh ca

tam gic QAD,suy ra I M= 1

2ADv I M AD.

Gi Gl giao im ca AMvIDth theo nh l Thals

GM

GA =

GI

GD =

IM

AD =

1

2.

Hn naGcng thuc hai trung tuyn ca tam gic ABCvDEFnn n chnh l trng tmchung ca hai tam gic ABCvDEF.T y ta c iu phi chng minh.

Trong vic gii cc bi ton bng phng php ta , ta cng cn ch n vic chn cch trc ta hp l: ta cc im, phng trnh ng thng cn vit n gin; c nhiulin h vi cc im cho trong gi thit, tn dng c cc yu t ng song song, vunggc, trung im do hnh cn dng n gin, . . .Chng hn chng ta c bi ton sau.

V d 6. Cho tam gic ABC c D l trung im ca cnh BC. Gi d l ng thng quaD v vung gc vi ng thngAD. Trn ng thng d ly mt imMbt k. GiE, Fln lt l trung im ca cc on thngMB, M C.ng thng quaEvung gc vid ctng thngAB tiP, ng thng quaFvung gc vidct ng thngAC tiQ.Chngminh rng ng thng quaM, vung gc vi ng thngP Q lun i qua mt im c nh

khiMdi ng trn ng thngd.


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48 Chuyn Ton hc s 9

Phn tch. Ta thy trong bi ny cc gi thit a ra ch xoay quanh cc yu t nhtrung im, ng vung gc, on thng, . . .nhng v c hi nhiu yu t nh vy nn viclin kt chng li v m bo s dng c tt c cc gi thit qu l iu khng d dng.

Chng ta c mt li gii bng cch s dng phng php hnh hc thun ty nh kin thctrc ng phng nh sau nhng n hi phc tp v cn phi k nhiu ng ph.

Li gii.

A

B

CD

M

EF

Q

P

HH

K

K

d

I

d

GiH, Kln lt l hnh chiu caB , Cln ng thng d.Do D l trung im ca B CnnDH=DK,suy ra AD l trung trc ca HK,do AH=AK.

Gi()l ng trn tm A i qua HvK. Gi H, K ln lt l cc im i xng vi H,Kqua cc ng thng AB, AC.Khi d thy H, K thuc ().Gi s cc ng thngHH, KK ct nhau ti Ith Il im c nh. ()Ta c P E BH(cng vung gc vi d) m P Ei qua trung im ca M B nn cng quatrung im ca M H,suy ra P El trung trc ca M Hv v th P H=P M.

Gi(1)l ng trn tmPi quaHvM,do tnh i xng nnH cng thuc(1).Honton tng t, ta cng c QFl trung trc ca M K;nu gi(2)l ng trn tm Qi quaKvM th K thuc(2).Ta li c

(), (1)ct nhau tai H, H nn H H l trc ng phng ca (), (1). (), (2)ct nhau tai K, K nn K K l trc ng phng ca (), (2).

Mt khc, M cng thuc (1), (2)v P, Qln lt l tm ca (1), (2)nn ng thngd qua M, vung gc vi P Qchnh l trc ng phng ca (1), (2).T suy ra HH,KK, d ng quy ti tm ng phng ca ba ng trn (), (1), (2). ()T()v()suy ra d i qua Il im c nh.Vy ng thng qua M, vung gc vi ng thng P Qlun i qua mt im c nh khi

Mdi ng trn ng thng d. Ta c iu phi chng minh.


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Nhn xt.Ta c th s dng phng php ta gii nh nhng bi ton trn v vic xcnh ta trung im v vit phng trnh ng vung gc cho cc biu thc n gin, cng chnh l p n chnh thc ca thi HSG quc gia ny. Th nhng, cng khng phi

cch chn trc ta no cng cho ta mt li gii nhanh gn. Nu chn h trc ta gcDv trc honh trng vi BCtheo suy ngh thng thng th li gii s di v phc tp hnso vi chn gc ta l D v trc honh l ng thng d. Cc bn hy th vi cch ny sthy ngay s khc bit !

Qua cc v d va nu, ta thy rngvic la chn cng c thch hp gii cc bi ton hnhhc cng l mt yu t quan trng c th i n kt qu mt cch n gin v ngn gnhn, nhiu khi cng l cch duy nht c th gii quyt c vn .

2.2 V vic tn dng gi thit ca bi

Trong mt bi ton thng thng, cc gi thit a ra, d t hay nhiu, d gin tip hay trc

tip, th trong bt c li gii no ca bi ton u c tn dng. Mt bi ton cng c tgi thit th ni chung vic s dng chng cng n gin bi khng phi d dng g cho vica hng lot gi thit, yu t, cc quan h hnh hc vo li gii ca mnh. Mi gi thit ara u c mc ch v tm quan trng nht nh; nhim v ca chng ta l xc nh xem cino l quan trng nht v lm sao tn dng v lin kt tt c vo trong li gii bi tonca mnh!

Trc ht, ta hy t cu hi: gi thit ni ln iu g?, chng hn cho gi thit: tamgicABC cM, N, Pl trung im cc cnh, iu gi cho ta suy ngh rng

Cc cnh caM N Psong song v bng na cc cnh caABCtng ng;

Tam gic M N Png dng vi tam gic ABCvi t s ng dng l 12

;

Din tch tam gic M N Pbng 14

din tch tam gic AB C;

Php v t tm G trng tm tam gic ABCvi t s12

bin tam gic ABC cho

thnh tam gic M N P;

Hai tam gic ny c cng trng tm; ng trn ngoi tip tam gic M N Pchnh l ng trn Euler nn n cng i qua

chn cc ng cao v trung im cc on ni trc tm v nh ca tam gic ABC; Trc tm ca tam gic M N Pcng l tm ng trn ngoi tip ca tam gic ABC, . . .

C tht nhiu suy ngh t mt gi thit v nu ta b st mt trong s chng th c th khnggii c bi ton v chnh l cha kha vn (tt nhin cng khng phi dng ht cc). Chng ta cng c c nhiu lin tng khi kin thc hnh hc ca chng ta cng nhiu vkinh nghim cng su sc, iu i hi ta cn lm 1 s lng nht nh cc bi ton HHP.

Tip theo ta li hi:vy nu cha c nhiu kinh nghim th sao?, tt nhin cng c mt cchnh ny gip ta c th thy trc quan hn gi thit . Chng ta hy th i tm cch dngcc gi thit bng thc v compa, nht l vi cc gi thit c phn phc tp, iu ny

nhiu lc cng rt c ch. Chng ta th tm hiu r iu qua bi ton sau.


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50 Chuyn Ton hc s 9

V d 7. Cho tam gicABC cK l im nm trong tam gic v tha

KAB = KB C= KCA.

GiD, E, F ln lt l tm ng trn ngoi tip cc tam gicKBC, KCA, KAB. GiM,N, P ln lt l giao im ca BC, FD; CA, DE; AB, EF. Chng minh rng cc tam gicABC, DEF, MNPng dng vi nhau.

Phn tch.Ta thy im Kcho nh trn l mt gi thit quen thuc (im Brocard) nhngni chung cc tnh cht ta bit v n khng phc v nhiu cho iu cn chng minh y.Nu nh ta v 1 hnh n iu nh bn di th vic gii v nh hng cho bi ton s khngn gin. Ta s th dng php dng hnh xc nh im K trong gi thit bng thc vcompa xem th n c tnh cht g c bit khng. Ta d dng c c php dng hnh sau.

Dng trung trc ca on AB v ng thng vung gc vi BC ti B, gi F l giaoim ca hai ng thng trn.

Dng ng trn tm Fbn knh F A. Tng t, dng im El giao im ca trung trc ACv ng thng vung gc vi

ACti A.

Dng ng trn tm E, bn knh EA. Giao im ca hai ng trn trn chnh l im Kcn tm.

A

B C

F

D

E

P

M

NK

T vic tm cch dng cho im K, ta cng c thm trn hnh mt s ng ph cn thit,bi ton r rng hn nhiu. Vi nhng gi c c t hnh v ta va dng, c th giiquyt c bi ton ny theo cch nh sau.


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Hng dn gii.

A

B C

F

D

E

P

M

NK

Chng minh AK EF, BK DE. Chng minh AKB + DF E= 180. Chng minh AKB + ABC= 180. Suy raABC EF D(g.g). Suy ra t gic B M PFni tip v M P EF. Chng minh M P N= FED. Chng minhM P N F ED (g.g).

T suy ra iu phi chng minh.

Cn i vi cc bi ton m hnh v khng th dng c bng thc v compa th sao, chnghn nh nh l Morley: Cho tam gic ABC. Cc ng chia ba cc gc ca tam gic ctnhau ti cc imM, N, P. Khi ta c M N P l tam gic u.

Ta bit rng vic chia ba mt gc khng th gii c bng thc v compa nn cch tm gi t vic dng hnh khng th thc hin c. V c l v vy m n sau hn 50 nm xuthin, bi ton ni ting ny mi c mt li gii HHP thun ty rt p v hon chnh. Nhng l cu chuyn ca nhng bi ton ni ting th gii; trn thc t, nu cn thit, chng talun c th dng cch dng hnh ny cho vic tm gi cho bi ton v tn dng c gi

thit ca bi.


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52 Chuyn Ton hc s 9

2.3 V vic rt ngn con ng i t gi thit n kt lun

Cng tng t phn trn, ta cng t cc cu hi: kt lun t u m ra?,nhng iu c lin h g n gi thit ca chng ta c?. Chng ta cng tin hnh i ngc ln t iu

cn chng minh, tm ra cc iu cn phi c c c kt lun.V d 8. Cho tam gicABCc ng cao BD vCEct nhau tiH. M l trung im caBC, Nl giao im caDEvBC. Chng minh rngN Hvung gc viAM.

A

BC

F

D

EH

MN

K

I

Phn tch.T gi thit ta d dng thy rng t gic DEF Mni tip trong ng trn Eulerca tam gic AB Cnn

N E N D= N F N M .Mt khc D, Enm trn ng trn ng knh AH;cn F, Mnm trn ng trn ngknh H Mnn Nnm trn trc ng phng ca ng trn ng knh M Hv ng trnng knh AH.n y ta cha c ngay kt qu N H AMc.

Ta thy thiu mt vi yu t trong hnh, mt yu t no cn c kt ni cc iu ta vaphn tch c t gi thit n kt lun ca bi, yu t va phi m bo rng c linquan nN Htrong cc phng tch trn, va m bo rng c lin h n on AM.V vicchn hai im ph dng thm l trung im AHv HM (I l trung im AH, K l trungim H M) cng l iu t nhin v khi I Kl ng trung bnh ca tam gic HAM, IvKcng l tm ca cc ng trn va nu trn nn trc ng phng N Hvung gc ving ni hai tm .

V d 9. Cho tam gic ABC c O nm trong tam gic. Cc tia AO, BO, CO ct cc cnhi din ln lt tiM, N, P.QuaO k ng thng song song viB CctMN, MPln lttiH, K.Chng minh rngOH=OK.

A

B CM

N

OP

K H


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Phn tch. Ta thy cc gi thit trong bi cho rt thong nhng kt lun c c cngtht th v. R rng, O l im nm bt k trong tam gic th khng th c mt tnh chtno c bit c th khai thc; do , ta s i vo phn tch cc t s c c t ng thng

song song k. Nu chng ta quen vi cc bi ton v t s ny th ta thy c mt s cngc h tr cho chng ta nh t s din tch, t s ng dng, nh l Thals, nh l Menelaus,nh l Cva, . . .Trc tin, ng thng song song trong bi gi cho ta s dng nh lThales a cc on thng OH, OKv cc on thng d gii quyt hn. Ta c

OHBM

= ON

BN,suy ra OH=

ON

BN BM .

OKCM

= OP

CP,suy ra OK=

OP

CP CM.

Do , mun c OH=OKth ON

BN BM= OP

CP CM, hay

ON

BNCP

OP =

CM

BM.

Nu c bin i cc t s ny tip tc th dn dn, ta s b ng nhn vi kt lun c sn; thayvo , ta s a cc t s on thng ny v t s din tch cc tam gic.

Hai tam gic c cng cnh y th t s din tch bng t s chiu cao. Hai tam gic c cng chiu cao th t s din tch bng t s cnh y.

Ta c th d dng thay cc t s trn c lin quan trn nh sau.ON

BN =

SAON

SABN=

SCON

SCBN=

SAON+ SCONSABN+ SCBN

= SAOC

SABC,

OP

CP =

SAOB

SABC.

Do ONBN

C POP

= SAOC

SABC SABC

SAOB=

SAOC

SAOB=

CM

BM.

n y kt lun hon ton r rng.

V d 10. Cho tam gic ABCnhn c ng cao AD tha AD = BC. Gi H l trc tmtam gic, M vNln lt l trung im caBC vAD. Chng minh rngHN =H M.

B C

A

D

E

H

M

N


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54 Chuyn Ton hc s 9

Phn tch.Ta thy trong gi thit ca bi ton c hai iu ng ch l: ng cao ADca tam gic ABCbng B CvM, Nli chnh l trung im ca hai cnh y. Ta c th suyngh rng:

Din tch tam gic ABC l SABC=12

AD BC= BC22

.

S dng lng gic, ta c BC= BD +CD= AD cot B +AD cot C=AD(cot B +cot C),t suy ra cot B+ cot C= 1vcot A + cot B cot C= 1.

T gic ANMCc hai cnh i AN vCMbng nhau v ng thng qua hai cnh vung gc vi nhau nn c th c mt s tnh cht c bit.

DoM , Nu l trung im ca BC, ADnn nu ta v cc ng trn ng knh B C,AD th M, Nchnh l tm ca ca cc ng trn ny. Hn na, AD = BCnn haing trn ny bng nhau v M, Ni xng nhau qua dy chung . . .

V cn nhiu iu c th suy lun ra t gi thit , nhng mc ch ca ta l tm mt cchgii hp l v n gin. Ta thy suy lun th 4 trn c th s dng c do c xut hins i xng gia cc yu t v gi thit c s dng mt cch trit hn. Do , ta thi theo con ng bng cch dng thm hai ng trn. n y, c v nh gi thit trctmHcha c dng n nhng vn cha c mt con ng r rng ch cho ta cch s dngn. Ta hy dng vic phn tch gi thit li v xem n kt lun: chng minhHM=H N.

Kt lun ny cng c th c c t nhiu hng, chng hn nh t t s gia cc cnh, t haitam gic bng nhau, t hai h thc lng gic bng nhau, hay t 1 php bin hnh no .

Tt nhin, vi cc i hi cn thit i n kt lun , ta c th hnh thnh nhiu tng

cho li gii nhng do chn cch dng ng trn nn ta th bm theo tnh i xng cahai ng trn. Mun c H M=H N th Hphi nm trn trung trc ca M N, m M, Ni xng nhau qua dy chung nn Hphi nm trn dy chung ! n y, ta thy c th gn lin kt c cc d kin. Ta tin thm mt cht na! Nh vy, mun c dy chung thphi gi tn hai giao im ca hai ng trn, nhng trn thc t hai giao im nm quri rc, kh m chng minh chng v Hthng hng. Ta s khng chn cch ny. Th nhndy chung mt phng din khc, khng phi l im chung ca hai ng trn thunty na m l trc ng phng ca hai ng trn, nh th mun Hnm trn th Hphic cng phng tch n hai ng trn. Nhng phng tch c d dng tnh c khng?Vi ng trn ng knh AD th qu n gin, chnh l HA HD;cn vi ng trnng knh BCth cha c, ta c th v qua Hmt dy cung ca ng trn ny gn vi

mt u mt l B hoc C,ta th v dy B Ev phng tch c c l H B HE, gi ch cnchng minh H A HD= H B HEna l xong!Hn na, nu ta v nh th th B Ephi vung gc vi ACdoHl trc tm tam gic; m Ethuc ng trn ng knh BCnn BEvung gc vi EC.Do , ha ra A, E, Cthnghng hayEchnh l chn ng cao ca tam gic ABC,cng viHl trc tm th ng thccn c l HA HD= H B HEkhng c kh khn g na. V cc mc xch trn c nilin, bi ton c gii quyt. Vic phi lm cn li ch l trnh by li gii m thi.

R rng bi ton ny khng qu kh v vn cn nhiu cch gii khc cho n na m chng tac th thy ngay rng ta cng l mt cch tt. Th nhng, nu c thi gian, chng ta

hy phn tch bi ton t t tm c mt li gii hnh hc thun ty tht p nh trn!


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C th ni trc nhng bi ton HHP kh, cc cng vic phn tch bi ton nh trn lrt cn thit. l cch chng ta m mn, d tm cch gii bi ton, cch c c pcm tnhng yu t cho trc thng qua vic kt ni cc mc xch lin h gia chng.

Ta hiu mc xch y c th l mt bc xung dng, mt du suy ra(),tng ng(), mt php bin i, . . .no ; tt nhin l khng d dng g m ta c c chng.Chng ta phn tch c cng nhiu iu t gi thit v kt lun cng tt, bi c nh th thvic dng nhng lin tng, nhng phn on, nhng kinh nghim cho vic vit tip nhngmc xch quan trng vo gia bi nhm hon chnh li gii s d dng hn. chnh l tmquan trng ca vicrt ngn con ng i t gi thit n kt lun.

2.4 Dng thm yu t ph trong cc bi ton hnh hc

Ta thy a s cc v d trn u c a thm cc yu t ph vo, c th l mt giao im,mt trung im, chn ng vung gc, ng thng song song hay thm ch l c mt ngtrn. Yu t ph chnh l cu ni gia gi thit v kt lun, n lin kt cc yu t ri rc csn li v gip tn dng trit cng nh pht trin gi thit cho thnh nhiu kt qu,cui cng i n c kt lun. Nu khng c chng, ta c th gii bi ton rt kh khn hockhng th gii c.

C th ni rng nu mt hc sinh bit cch k ng ph trong vic gii cc bi HHP th khng th no l mt hc sinh km phn ny c. Mun k c ng ph, i hichng ta phi c s quan st, nh gi vn tt; c mt kinh nghim su sc v kh nngphn tch, sng to mc nht nh.

Vic gi tn cho mt im cha c tn trong hnh v trn thc t cng l mt chuyn khngn gin d im c sn ni chi n vic dng thm mt hoc nhiu yu t ph, nhng

ci khng h c trc .iu ny cng khng kh hiu v khi lm cc bi ton i s Gii tch, chng ta thngquen vi cc lp lun logic c sn, mi th xut hin u phi c mt l do r rng. Cn HHPth khng phi nh vy, nu c cng nhc cho rng mt ng ph no mun k c ucn phi c mt lp lun logic no cho n th kh m thc hin c cng vic ny bi trnthc t, nhiu khi ta k mt ng ph m khng c mt l do xc ng!

Do trong phn ny ta s suy ngh thm v vic k ng ph v vai tr quan trng cakinh nghim qua mt qu trnh rn luyn lu di gii ton HHP bng cch k thm ngph. Ta xt bi ton sau.

V d 11. Cho t gic ABCD ni tip (O, R) c M, N ln lt l giao im ca cc cpcnh i. Chng minh rng

OM ON =R2.Phn tch.Khi gii bi ton ny, chc chn cc bn cng s m mn bin i tch v hngca hai vector v tri i n kt qu nhng cui cng cng s b ng nhn hoc cng lccng phc tp thm. Do , vic dng thm mt yu t ph s l iu tt yu. Chng ta nglm tng bi hnh thc n gin ca bi ton ny!

Vic dng ng ph di y c th l kh vi mt s bn nhng nu chng ta quen vibi ton sau th mi chuyn s tr nn n gin hn rt nhiu: Cho t gic ABCD ni tip(O)cM, Nln lt l giao im caAD, BC vAB, CD.Chng minh rng

M A M B+ N A N D= M N2

.


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56 Chuyn Ton hc s 9

O

A

B

CD

N

M

P

Ta s gii bi ton ny xem nh mt b v p dng n vo bi ton cho: Gi P l giaoim ca ng trn ngoi tip tam gic BC M viM N. Ta thy B PM Cr rng l mt tgic ni tip nn BP M= BC D= NAB,suy ra t gic ANBPcng ni tip. Theo tnhcht phng tch, ta c M A M B= M P MN, NA N D= N P N M. T suy ra

M A M B+ N A N D= M N(M P+ N P) =M N2.

Ta s quay tr li bi ton cho, bin i biu thc cn chng minh mt cht cho vn c r rng hn (gi sMl giao im ca AB v CD, Nl giao im ca AD v B C)

OM ON=R2

OM2

+ ON2

M N2

= 2R2

.

p dng b trn, thay M A M B+ N A N D= M N2 vo biu thc trn

OM2 + ON2 (M A M B+ N A N D) = 2R2.

Nhng iu ny l ng do theo tnh cht phng tch

M A M B= OM2 R2, N A N D= ON2 R2.

T , ta gii thnh cng bi ton.

Th ngh nu khng c s h tr ca b trn th vic k ng trn ngoi tip tam gicBC Mri i chng minh tun t nh trn qu l chuyn khng n gin. V phi cng nhnrng kinh nghim gii ton HHP th hin trong bi ny khng t! Ta tip tc phn tch mtv d khc.

V d 12. Trong mt phng cho hai im A, B c nh (A khc B). Mt im Cdi ngtrong mt phng sao cho gc ACB = khng i (0 < < 180). ng trn tm I nitip tam gicABCv tip xc vi cc cnhAB, BC, CA ln lt tiD, E, F. ng thngAI, BIln lt ct ng thngEF tiM, N. Chng minh rng

(a) onM Nc di khng i.

(b) ng trn ngoi tip tam gicDM N lun i qua mt im c nh.


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C

A B

I

D

E

F

N

M

P

Phn tch.Nu bi ny ta c k gi thit th s thy rng cc imMvNxc nh nhtrn xut hin trong nhiu bi ton quen thuc trc m yu cu ca ch dng li

vic chng minh cc tam gic MBC, NBCvung. Nu bit iu ny, ta s chng minh likt qu v s dng vo vic gii bi ton cho nh mt b (trong bi ny khng xt

cc v tr c th c ca M, N). Ta thy M EB = CEF = 180 C

2 v

M IB= IAB+ IBA =ABC+ ACB

2 =

180 C2

.

Do M EB = MIB.T y suy ra t gic EMBIni tip v IM B = IEB = 90,suy ra tam gic AM Bvung M. Tng t, ta cng c tam gic N ABvung ti N .

p dng iu ny vo bi ton, ta c t gic ANMBni tip ng trn ng knh AB.

Suy raAIB

NIM,t ta c AB

M N =

IA

N I,suy ra

M N=AB I NIA

=AB sinN AI=AB sin

90 CAB+ CBA

2

=AB sinC

2 =AB sin

2 =const.

Hn na, ta thy

M DN = IDN+ IDM= 2

90 CAB+ CBA

2

= C.

GiPl trung im caAB th Pchnh l tm ng trn ngoi tip t gicANMB,suy ra

M P N= M P A N P A= 2(M BA N BA) = 2(90 M AB N BA)

= 2

90 CAB+ CBA

2

= C.

Do M P N = MDN,suy ra t gic MNDPni tip hay ng trn ngoi tip tam gicDM Nlun i qua Pc nh. y chnh l iu phi chng minh.

Bi ton ny vn cn nhiu cch gii khc nhng c l cch ny n gin, ngn gn hn c.

Mt s bi ton cng c th gii c bng nhiu cch dng ng ph v nu chng ta cngc nhiu cng c h tr nh nhng b , nh l quen thuc th vic dng hnh s n gin

v li gii s nh nhng hn, chng ta hy xt vic chng minh nh l Pascal sau y.


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58 Chuyn Ton hc s 9

V d 13. Cho lc gicABCDEFni tip ng trn(O)c M, N, P ln lt l giao imcaAB, DE; BC, EF; CD, FA. Chng minh rngM, N, Pthng hng.

Phn tch. Vic chng minh nh l ny qu quen thuc bng cch s dng nh l

Menelaus thun v o cho cc tam gic. Cch tng i ngn gn v khng k nhiung ph. Nhng nu nh ta khng bit trc nh l Menelaus v s dng mt cch chngminh khc th mi cc bn hy theo di li gii sau y vi vic k thm hai ng trn ph.

O

D

C

B

E

F

A

MN

PH

Gi I

l giao im ca ng trn ngoi tip tam gic BDM

v FDQ.

Ta s chng minhrng c bn im M, N, P, Ithng hng bng cch chng minh tng b ba im thng hng.(Vic ngh ra hai ng trn ph ny c th xut pht t mt bi ton quen thuc l: Choba ng trn (1), (2), (3) cng i quaD. (1) ct (2) ti A, (2) ct (3) tiB, (3) ct (1) tiC(A, B, C khcD). ViMbt k nm trn(1),giP, Ql giao im caM Avi(2), M Cvi(3). Chng minh rngP Qi quaB.)

Tht vy, t cc t gic ni tip BDIM, FDIP,ta c

DI M+ DIP = DBA + DF A= 180,

suy raM, I, Pthng hng. Tip theo ta s chng minh rngM, N, Ithng hng. Ta c t gic

BDIM

ni tip nnBI M

=BDM

= 180

BDE

= 180

1

2s(BAE

) =

1

2s(BDE

).


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Mt khc, li c

BI F = BI D+ F ID= BM D+ F P D

=

1

2[(s(AF E) s(BC D)) + (s(ABC) s(DEF))]=

1

2[(s(AF) +s(AB)) (s(DC) +s(DE))]

=1

2(s(BAF) s(CDE)) = BNF,

nn t gic BNIFni tip. Do

BI N= BF N =1

2s(BDE) = BIM,

suy ra M, N, Ithng hng. Tng tN, I, Pthng hng.

Vy ta c M, N, P thng hng (pcm).Qua cc v d trn, ta thy rng vic dng ng ph l cng vic i hi phi c qu trnhrn luyn v tch ly kinh nghim lu di. C th ni khi chng ta k thnh cng cng ph gii mt bi ton no chnh l lc chng ta c mt bc tin di trong vichc tp HHP.

2.5 V vic hc tp v rn luyn HHP mc nng cao

C khi no chng ta t cu hi: Ti sao ngi ta li c th ngh ra c mt bi ton haynh vy nh?. Thng thng, chng ta gii c mt bi ton vi li gii tht hay v p rigc n li m khng dnh thi gian tm hiu thm nhng iu l th ng sau n hay thmch l a ra c mt bi ton mi t bi ton c . Vic tm ti nh th s gip chng tach ng hn cc bi ton v pht trin k nng HHP tt hn.

Khi chng ta tm ti sng tc ra cc bi ton mi chnh l lc chng ta i trn con ngm nhng ngi ra i v tm hiu xem h lm th no c c bi ton nh vy.Thng thng cc bi ton HHP t ra di dng che giu cc vn v cng vic ca chngta l ln m theo cc gi thit c sn gii.

Vic che giu cng hay khi m mt s im v ng trong hnh b xa i m yu cu ca biton li khng b nh hng, ngi gii cc bi nh vy phi khi phc li cc im thngqua cch k cc yu t ph; cng c th l vic bin i cc yu t trong bi, thm cc ng

mi che giu bn cht vn .V vic t ngh ra cc bi ton HHP hoc pht trin t mt bi ton c l mt vic lm rtc ch cho chng ta khi m ta tr thnh th sinh trong k thi no , i mt vi mt bi tonHHP kh, khng ri vo hon cnh b ng v lng tng.

Ta th xem cc bi ton sau y.

V d 14. Cho tam gicABCnhn c A l gc ln nht, ni tip ng trn (O) v ngoitip ng trn(I), Hl trc tm. Trung tuyn nhIca tam gicIOH ct(I) tiP. GiM, Nln lt l trung im caAB, AC. Chng minh rng

BAC=

M PN .


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60 Chuyn Ton hc s 9

A

B C

OI

H

E

M N

Phn tch.Ta thy gi thit ca bi ton khng qu phc tp nhng cc yu t ri rc tronghnh cng nh vic dng hnh phc tp c th khin ta kh tm ra li gii. Thc ra, bi tonny pht trin t nh l ng trn Euler tip xc trong vi ng trn ni tip.Cc vn b che lp l

Trung im ca on OHchnh l tm ng trn Euler. Giao im Pchnh l tip im ca ng trn Euler vi ng trn ni tip nn tt

nhin n s thuc ng trn Euler.

ng trn Euler i qua trung im cc cnh nn nu gi Q l trung im BC thM N P Qni tip v M P N = MQN.

DoM, N, Ql trung im cc cnh nn BAC= MQN.T ta d dng i n ligii cho bi ton.

Nu chng ta cha quen ln m theo con ng ca ngi cho tm ra li gii th bi tontrn qu tht khng n gin cht no, bt k l chng ta c nng khiu HHP hay khng.

Chng hn bn l ngi cho , bn c sn mt bi ton chng minh cc imM, N, P no cng nm trn ng thngd,bn mun bi ton ny kh hn v bn s rt d dng ngh

ra rng nu A l mt im no nm ngoi ng thng d th trc tm ca ba tam gicAMN, ANP, AP Mcng thng hng (cng nm trn ng thng quaAvung gc vid)vnh th, bn cng tch ly c thm mt kinh nghim cho vic chng minh ba im thnghng. Th hi nu l mt ngi i tm li gii bi ton th vic nhn ra cch chng minh c d dng khng?

V d 15. Cho tam gicABCni tip trong(O) c A l gc ln nht. Trung trc ca AB,ACct cnhBC ln lt tiD, E. ng thngAD, AEct(O) ln lt tiM, N. GiKl giao im ca BM v CN; d l ng thng i xng vi phn gic gc DAEqua phngic gcBAC. ng thngOK ctdtiI.Chng minh rng

BI C=

DOE.


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O

A

B CE D

N

MK

I

Phn tch.Chc hn chng ta tng nghe n bi ton sau: Cho t gicABCDtha mn

DAB = ABC= BCD.Chng minh rng ng thng Euler caABCi quaD.Bi ton ny l mt bi ton kh nhng n hu nh kh quen thuc vi nhiu cch gii.Tng chng bi ton ny v v d trn khng c lin h g nhng thc ra v d trn l mtpht trin ca bi ton va nu vi vic che lp v b sung thm hng lot vn . Nu chabit n n th v d ny qu l mt bi ton rt kh. Chng ta th chng minh xem t gicABKCtrong hnh v c tnh cht ba gc bng nhau khng, r rng iu l ng. Khi a kt qu trn vo th ng thng Euler ca tam gic AB Ci quaKhay ngc liOKsi qua trc tm tam gic ABC.ng thng dtrong bi thc cht l ng cao ca tamgicAB CvIchnh l trc tm. y l cc yu t b che lp i, nu chng ta khng tinhnh tng bc khai thc gi thit th kh c th thy c iu ny. n y th r rngBI

OE, CI

OD nn BI C= DOEl ng. Vn c gii quyt!

Ta s phn tch thm mt v d na thy r vai tr ca kinh nghim tch ly c ca bnthn trong vic gii cc bi ton HHP.

V d 16. Cho hai ng trn (O), (O) ct nhau ti A v B. Gi CC l tip tuyn chung(gnAhn) ca hai ng trn, C (O), C (O).GiD, D ln lt l hnh chiu caC,C trn ng thngOO. Gi sAD ct(O) tiE, AD ct(O) tiE. Chng minhE, B,E thng hng.

O O

A

B

K

I

C

C

D D

E

E

FF


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62 Chuyn Ton hc s 9

Phn tch.y l mt bi ton hnh hc ca k thi HSG quc gia v cc li gii ca n nichung u mang nhiu tnh cht ca php v t. p n chnh thc cng l mt li ngn gnv p. D vy, nu cc bn nhiu ln gii cc bi ton v hai ng trn ct nhau cng

vi tip tuyn ca n th s nhiu kinh nghim v dng ny v s c th dng kinh nghim nh nhng b gii bi ton ny mt cch n tng hn. Hy suy ngh v cch gii sau.

Gi F, F ln lt l giao im khc Aca cc ng thng AO vi (O), AO vi (O).DoAF, AF ln lt l ng knh ca cc ng trn (O), (O)tng ng nn

ABF = ABF = 90,

suy ra F, B, F thng hng. Gi R, R ln lt l bn knh ca hai ng trn (O), (O).Tas chng minh rng

OAD = OAD. ()Tht vy, gi Il giao im ca C C viOO vKl giao im ca I Avi(O).D dng thy

rngIchnh l tm v t ca hai ng trn. Do IO

IO =

R

R =

OA

OK =

IA

IK =

IC

IC =

ID

ID.

Suy ra AD KD hay IAD = IKD.Mt khc, do AC KCnn

IAC = IKC= ICA,

suy raIAC ICAv IAIC

= IC

IA.T y ta c IA2 =I C IC.

Do t gic CCDOni tip nn IC IC =I D IO.Kt hp vi trn ta c I A2 =I D IO,hay IAID

= IOIA

,t suy raIAD IOA.Do IAD = IOA,m IAD = IKDnn IKD = IOA,suy ra t gic ADOKni tip. T y ta c

OAD = OKD.

Hn na do OK OA, DK DAnn

OAD = OKD.

Suy ra OAD = OAD,()c chng minh.p dng vo bi ton, theo tnh cht ca gc ni tip ta c

OAD = FBE, OAD = FBE.

Do , kt hp vi (), ta c F BE = FBE.M F, B, F thng hng nn theo tnhcht ca gc i nh, ta cng c E, B, E thng hng. Ta c iu phi chng minh.

Bn cnh , ta cng cn phi nhc n mt s cng c gi l cao cp gii cc bi tonHHP nh: gc nh hng, di i s, tch c hng v din tch i s, phng tch vtrc ng phng, hng im iu ha, cc v i cc, php nghch o v ng dng, nhl Carno, Michael, ... Cng tng t nh nhng iu gi l kinh nghim hay b trn,nhng cng c ny c th gip ta gii quyt nhanh gn v d dng nhiu bi ton kh m nu

s dng cng c thng thng th li gii s di dng v phc tp; c nhiu khi ta khng


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kh nng nhn ra mt li gii kiu nh th. Th nhng, mun p dng nh l no vo vicgii mt bi ton qu l iu khng n gin khi m s lng nh l c sn tng i ln vcng kh khn hn khi c trng ca nh l cha th hin bt c mt no ca bi ton,

chng ta phi ln m theo gi thit t ra cc yu cu cn c nhm i n kt lun v cth bt cht mt nh l no s xut hin h tr cho ta.

V d 17. Cho tam gicABC. Dng pha ngoi tam gicABCcc tam gic u v giM,N, P l tm ca chng. Chng minh rng tam gicM N P u.

Ta thy y l ni dung ca nh l Napolon vi cch chng minh quen thuc l dng thmcc ng trn ngoi tip cc tam gic u c ri gi tn giao im ca chng. Sau y, tas cng xem hai cch chng minh khc na v a ra nhn xt so snh.

Li gii 1.S dng phng php thng thng.

A

B C

P

N

M

Q

Dng im Q khc pha M so vi N P sao cho QP A = M P B v P Q = M Q. Ta cAP Q= BP M(c.g.c), suy ra

P AQ= M BP = ABC+ 60, AQ= BM=C M.

Do

N AQ= 360 (P AQ + P AN) = 360 (ABC+ 60 + CAB+ 60)= ACB+ 60 = MCN.

Suy raAQN =CM N(c.g.c) v N Q = N M. M P Q = P MdoAP Q =BP MnnP Nl trung trc ca M Q,tc l M , Qi xng nhau qua P Nhay

QP N=

M PN ,

QN P =

MNP.


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64 Chuyn Ton hc s 9

Mt khc, ta c QP M = AP M+ QP A= AP M+ M P B= AP B = 120 v tng

tQNM= 120 nn M P N= M N P =120

2 = 60,suy ra tam gic M N Pu.

Li gii 2.S dng php quay vector.

A

B C

C

B

P

N

M

A

Gi A, B, C l nh ca cc tam gic u tng ng dng trn cc on BC, CA, AB. V

cc imM, Nln lt l trng tm ca BAC, ACB nn M N= 13

BA +

AC+

CB

.

Tng t, ta cng cM P =

1

3

BC +

AB+

CA

.

Xt php quay vector gc quay l 3 ,ta c

Q3

M N

=

1

3Q

3

BA +

AC+

CB

=

1

3

Q

3

BA

+ Q

3

AC

+ Q

3

CB

=

1

3

BC +

AB+

CA

=M P .

Suy ra tam gic M N Pu.

Ta thy rng php quay vector s dng trong bi ton gip hn ch nhiu lp lun hnhhc phc tp v cho ta mt li gii ht sc nh nhng, vn l chng ta phi bit cn c

vo cc c trng ca bi ton vn dng cho ph hp v chnh xc. Ni chung ba cng c


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sau:vector (tng ng vi on thng), gc nh hng (tng ng vi gc), din tch i sv tch ngoi (tng ng vi din tch)l ba cng c mnh, pht trin t cc yu t hnh hcc bn. Chng ta nn tm hiu thm v chng c thm c nhng s tr gip rt tt khi

ng trc mt bi ton HHP no m cc phng php hnh hc thun ty khc dngnh khng cn tc dng na.

Tip tc ni v vic nghin cu ra cc bi ton HHP mi, chng ta thy mt iu rng: munt ngh ra mt bi HHP hon ton c lp vi cc bi c qu l chuyn khng n gin; tavn c th s dng nhng s tng t gia cc yu t ng v im to ra cc bi ton co.

Chng hn nh trong bi ton trn, cc bn c th t hi nu nh khng dng cc tam gicu pha ngoi tam gic m dng v pha ngc li th kt qu trn s ra sao, nh l c cnng hay khng. Vn cn nhiu v d v nhng pht hin ny nh:

Tam gic c hai ng trung tuyn, ng cao bng nhau l tam gic cn; vy phngic th sao? Giao im cc ng chia ba pha trong cc gc ca mt tam gic l mt tam gic l

tam gic u (nh l Morley); vy th chia ba pha ngoi th sao?

Ta c bi ton quen thuc l:Cho tam gicAB Cni tip(O),c trng tmG.Gi scc tiaGA, GB, GC ct(O) ln lt tiA, B, C.Chng minh rng

GA + GB+ GC GA + GB + GC;vy th nu thay G bi trc tm hay tm ng trn ni tip th sao?

Qu tch trc tm Hca tam gic ABCc BCc nh v

BACkhng i l ngtrn, ca trng tm G cng l ng trn; vy ca tm ng trn ni tip l g?

Nu bit c trung im cc cnh c th dng c cc nh tam gic, bit c chnng cao c th dng c cc nh tam gic; vy nu bit chn cc ng phn gicth c dng c khng v dng nh th no?

Chng hn t vn cui va nu trn, chng ta c c mt bi ton sau:Cho tam gicABCni tip ng trn tmO c cc phn gicAD, BE, CFng quy I.Gix, y, z lnlt l cc tip tuyn ca(O) song song vi cc on thngEF, FD, DE.Gi sxcty tiP, y ctz tiM, z ctxtiN. GiH, K, L l chn ng phn gic k t gcM, N, P catam gicMNP. Chng minh rng

(a) Cc on thngMD, NE, PFng quy. Gi im lR.

(b) Cc on thngHD, KE, LFng quy. Gi im lS.

(c) Ba imR, S, O thng hng. Gi ng thng qua cc im ny ld.

(d) ng thngdi qua tm ng trn ni tip ca bn tam gicABC, DEF, MNP vHKL.

Cn rt nhiu iu rt gn gi, quen thuc m chng ta cha tm hiu nhiu v chng cth pht hin thm nhng s th v cng nh rn luyn cho mnh k nng gii ton HHP. Tisao chng ta khng th bt u ngay vi mt bi HHP n gin no i tm n nhng

iu th v?


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66 Chuyn Ton hc s 9

Tm li,mun hc tt mn HHP, chng ta cn phi c mt qu trnh rn luyn y cngvi mt cch hc tp ph hp. Chng ta nn rn luyn t duy hnh hc ca mnh t nhiu dngton v nn tp trung vo cc cng c chnh; ng i qu su vo mt phng php, mt cng

c h tr c bit no . Ta hc tht nhiu nh l, b nhng khi cn chng ta khng thno nh ht chng v khng bit la chn cng c no cho ph hp gii quyt. Chng tacng nn bit rng cc bi ton HHP trong cc k thi thng khng gii da trn mt b ,nh l kh no nh gi k nng nh, thuc bi m ch dng cc cng c thng thng,quen thuc c th nh gi c kh nng t duy, lp lun ca hc sinh. Hy trang b chomnh nhng th cn thit c th i u vi cc bi ton HHP kh khn pha trc; ttnhin, mt hnh trang tt l mt hnh trang y v gn gng, c th s dng c trongnhiu tnh hung ch khng phi mt hnh trang qu cng knh, qu phc tp khi cn dngmt th no khng bit tm kim u, ...

Chc cc bn c th rn luyn tt v thnh cng b mn HHP tht th v v hp dn ny!

3 Cc bi ton rn luyn

Sau y xin mi cc bn hy tham kho th 16 bi ton trong k thi chn i tuyn quc giaca Vit Nam d thi IMO di y v hy th vn dng cc hng va ri tm cch giiquyt chng. y u l cc bi ton rt hay v cng rt kh! (Cc bi ton c sp xpmt cch tng i t d n kh)

Bi tp 1( TST 2000). Hai ng trn (C1)v(C2)ct nhau ti hai im P vQ. Tiptuyn chung ca hai ng trn gn PhnQtip xc vi (C1)ti Av tip xc vi (C2)tiB. Cc tip tuyn ca (C1), (C2)k tPct ng trn kia ln lt ti EvF (E, FkhcP). Gi H, Kln lt l cc im nm trn cc ng thng AF, BEsao cho AH=AP v

BK=BP.Chng minh rng nm im A, H, Q, K, B cng thuc mt ng trn.Bi tp 2 ( TST 2003). Trn cc cnh ca tam gic ABCly cc im M1, N1, P1 saocho cc on M M1, N N1, P P1 chia i chu vi tam gic, trong M, N, Pln lt l trungim ca cc on BC, CA, AB.Chng minh rng

(a) Cc ng thng M M1, N N1, P P1ng quy ti mt im. Gi im l K.

(b) Trong cc t s KA

BC,

KB

CA,

KC

ABc t nht mt t s khng nh hn

13

.

Bi tp 3( TST 2006). Cho tam gic AB CcHl trc tm. ng phn gic ngoi cagc BH Cct cc cnh AB, AC ln lt ti D vE. ng phn gic trong ca gc BAC

ct ng trn ngoi tip tam gic ADEti im K.Chng minh rng ng thng HKiqua trung im ca on B C.

Bi tp 4( TST 2006). Trong mt phng cho gc xOy.Gi M , Nln lt l hai im lnlt nm trn cc tia Ox, Oy.Gi dl ng phn gic gc ngoi ca gc xOy vIl giaoim ca trung trc M Nvi ng thng d. Gi P, Ql hai im phn bit nm trn ngthng d sao cho IM=I N=I P =I Q,gi sKl giao im ca M QvN P.

(a) Chng minh rng Knm trn mt ng thng c nh.

(b) Gi d1l ng thng vung gc vi I MtiMvd2l ng thng vung gc vi I Nti N .Gi s cc ng thng d1, d2ct ng thngd ti E, F.Chng minh rng cc

ng thng EN, FMvOKng quy.


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Bi tp 5 ( TST 2009). Cho tam gic nhn ABCni tip ng trn (O).Gi A1, B1,C1 vA2, B2, C2ln lt l cc chn ng cao ca tam gic ABCh t cc nh A, B, Cv cc im i xng vi A1, B1, C1qua trung im ca cc cnh BC, CA, AB.Gi A3, B3,

C3 ln lt l cc giao im ca ng trn ngoi tip cc tam gic AB2C2, BC2A2, CA2B2vi ng trn (O).Chng minh rng A1A3, B1B3, C1C3ng quy.

Bi tp 6( TST 2001). Trong mt phng cho hai ng trn ct nhau ti hai im A, B.Gi P Tl mt trong hai tip tuyn chung ca hai ng trn trong P, Tl cc tip im.Tip tuyn ti PvTca ng trn ngoi tip tam gic AP Tct nhau tiS.Gi Hl imi xng vi B qua ng thng P T.Chng minh rng cc im A, S, Hthng hng.

Bi tp 7( TST 1999). Cho tam gic AB Cni tip trong ng trn .Mt ng trntip xc vi cc cnh AB, ACv tip xc trong vi ng trn ln lt ti cc im M1,N1, P1.Cc im M2, N2, P2 vM3, N3, P3 xc nh mt cch tng t. Chng minh rngcc on thng M1N1, M2N2, M3N3ct nhau ti trung im mi ng.

Bi tp 8 ( TST 1995). Cho tam gic ABCv im Mnm trong tam gic. Gi A, B,C ln lt l nh ca cc im A, B, Cqua php i xng tm M .

(a) Chng minh rng tn ti duy nht mt im imPtrong mt phng cch u hai umt ca cc on thng AB , BC, CA.

(b) Gi D l trung im ca on AB. Chng minh rng khi Mthay i trong tam gicABCv khng trng vi D th ng trn ngoi tip tam gic MNP,trong N lgiao im ca DMvAP,lun i qua mt im c nh.

Bi tp 9 ( TST 2004). Trong mt phng cho hai ng trn (O1), (O2)ct nhau ti AvB. Cc tip tuyn ti A, B ca ng trn (O1)ct nhau ti K.Xt mt im Mkhngtrng vi A, B nm trn ng trn (O1).Gi Pl giao im th hai ca ng thng M Avi ng trn (O2).Gi Cl giao im th hai ca ng thng M Kvi ng trn (O1).Gi Ql giao im th hai ca ng thng CAvi ng trn (O2).Chng minh rng

(a) Trung im ca on thng P Qnm trn ng thng M C.

(b) ng thng P Qlun i qua mt im c nh khi Mdi ng trn(O1).

Bi tp 10 ( TST 2003). Cho tam gic ABCc Ol tm ng trn ngoi tip. Gi H,K, Lln lt l chn cc ng vung gc k t cc nh A, B, Cca tam gic ABC.Gi

A0, B0, C0ln lt l trung im ca cc ng cao AH, BK, CL.ng trn ni tip tmIca tam gic ABCtip xc vi cc on BC, CA, AB ln lt ti D, E, F. Chng minhrngA0D, B0E, C0Fcng i qua mt im v im nm trn ng thng OI .

Bi tp 11( TST 2006). Cho tam gic AB Cl tam gic nhn, khng cn, ni tip trongng trn tm O bn knh R.Mt ng thng d thay i sao cho d lun vung gc vi OAv lun ct cc tia AB, AC.Gi M, Nln lt l giao im ca ng thng dv cc onAB, AC. Gi s cc ng thng BN v CNct nhau ti K; gi s ng thng AK ctng thng B C.

(a) Gi Pl giao ca ng thng AKv ng thng BC. Chng minh rng ng trn

ngoi tip ca tam gic M N Plun i qua mt im c nh khi dthay i.


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68 Chuyn Ton hc s 9

(b) Gi Hl trc tm ca tam gic AMN.t B C= a vl l khong cch t im A nng thng HK.Chng minh rng ng thng HKlun i qua trc tm ca tamgicABC.T suy ra l

4R2 a2.ng thc xy ra khi v ch khi no?

Bi tp 12 ( TST 2009). Cho ng trn (O)ng knh AB vMl mt im bt knm trong (O), Mkhng nm trn on thng AB. Gi Nl giao im ca phn gic tronggc Mca tam gic AM B vi ng trn (O).ng phn gic ngoi gc AM B ct ccng thng NA, NB ln lt ti P, Q.ng thng M Act ng trn ng knh N Qti R,ng thng M Bct ng trn ng knh N P ti SvR, Skhc M. Chng minhrng ng trung tuyn ng vi nh Nca tam gic N RSlun i qua mt im c nh khiMdi ng pha trong ng trn.

Bi tp 13 ( TST 2005). Cho tam gic AB Cc(I)v(O)ln lt l cc ng trn nitip, ngoi tip tam gic. Gi D, E, Fln lt l tip im ca ng trn (I)trn cc cnhBC, CA, AB.Gi

A,

B,

Cln 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

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