S ha bi Trung tm Hc liu i hc Thi Nguynhttp://www.lrc-tnu.edu.vn 1

I HC THI NGUYN TRNG I HC S PHM ----------------------------- L TH THU H RN LUYN K NNG GII TON CHO HC SINHBNG PHNG PHP VCT TRONG CHNG TRNH HNH HC 10 (CHNG I, II - HNH HC 10 - SCH GIO KHOA NNG CAO ) LUN VN THC S KHOA HC GIO DC THI NGUYN -2007 www.VNMATH.com S ha bi Trung tm Hc liu i hc Thi Nguynhttp://www.lrc-tnu.edu.vn 2 I HC THI NGUYN TRNG I HC S PHM ------------------------- L TH THU H RN LUYN K NNG GII TON CHO HC SINHBNG PHNG PHP VCT TRONG CHNG TRNH HNH HC 10 (CHNG I, II - HNH HC 10 - SCH GIO KHOA NNG CAO ) Chuyn ngnh: L lun v phng php dy hc ton M s: 60.14.10 L LU U N N V V N N T TH H C C S S K KH HO OA A H H C C G GI I O O D D C C Ngi hng dn khoa hc : TS. NGUYN NGC UY THI NGUYN - 2007 www.VNMATH.com S ha bi Trung tm Hc liu i hc Thi Nguynhttp://www.lrc-tnu.edu.vn 3 Li cm n Em xin by t lng bit n su sc n thy gio - TS. Nguyn Ngc Uy, ngi tn tnh hng dn, gip em trong sut qa trnh thc hin ti. Em xin chn thnh cm n cc thy c gio trong t : Phng php ging dy ton, Khoa Ton - Tin trng i Hc S Phm H Ni, cc thy c gio trong khoa Ton- Tin Trng i Hc S Phm - i Hc Thi Nguyn trong sut qu trnh hc tp v nghin cu. Em xin trn trng cm n Ban gim hiu, Khoa sau i hc trng i Hc S Phm-iHc Thi Nguyn to mi iu kin thun li em hon thnh lun vn. Xin chn thnh cm n Ban gim hiu v cc bn ng nghip trng THPT Bm Sn - Thanh Ha ng vin, gip ti hon thnh nhim v hc tp v nghin cu ca mnh. Thi Nguyn, thng 9 nm 2007 L Th Thu H www.VNMATH.com S ha bi Trung tm Hc liu i hc Thi Nguynhttp://www.lrc-tnu.edu.vn 4 NHNG CM T VIT TT TRONG LUN VN Hc sinhHS Hnh hcHH Phng php vctPPVT Sch gio khoaSGK Sch bi tpSBT Trung hc ph thngTHPT www.VNMATH.com S ha bi Trung tm Hc liu i hc Thi Nguynhttp://www.lrc-tnu.edu.vn 5 MC LC

Trang M U .............................................................................................................. 1 Chng1.CSLLUNVTHCTINTRONGVICDYHC GII BI TP BNG PPVT ......................................................................... 4 1.1 L lun v dy hc gii bi tp ton........................................................... 4 1.1.1 Mc ch, vai tr, ngha ca bi tp ton trong trng ph thng......... 4 1.1.2 V tr v chc nng ca bi tp ton......................................................... 5 1.1.3 Dy hc phng php gii bi ton......................................................... 6 1.1.4 Bi dng nng lc gii ton................................................................. 10 1.2K nng gii ton v vn rn luyn k nng gii ton cho hc sinh.... 13 1.2.1 K nng ................................................................................................... 13 1.2.2 K nng gii ton .................................................................................... 14 1.2.3 c im ca k nng ............................................................................. 14 1.2.4 S hnh thnh k nng............................................................................ 15 1.2.5Mtsknngcbntrongquytrnhgiibitonbngphng php vct.............................................................................................. 17 1.2.5.1 Din t quan h hnh hc bng ngn ng vc t ............................ 17 1.2.5.2 Phn tch 1 vc t thnh mt t hp vct...................................... 18 1.2.5.3K nng bit cch ghp 1 s vct trong 1 t hp vct ................ 20 1.2.5.4 Bit khi qut ha 1 s nhng kt qu vn dng vo bi ton tng qut hn .................................................................................... 21 1.3 Ni dung chng trnh HH10-SGK nng cao ........................................... 21 1.3.1 Nhim v ca HH10-SGK nng cao ....................................................... 21 1.3.2 Nhng ch khi ging dy HH10-SGK nng cao ................................. 22 1.3.3McchyucucaPPVTtrongchngtrnhHH10-SGK nng cao ............................................................................................ 25 1.4Nhng kh khn sai lm ca hc sinh lp 10 khi gii ton hnh hc phng bng PPVT .................................................................................... 26 www.VNMATH.com S ha bi Trung tm Hc liu i hc Thi Nguynhttp://www.lrc-tnu.edu.vn 6 1.4.1NhngiucnlukhigingdyvcttrongHH10-SGK nng cao ......................................................................................... 26 1.4.2 Nhng kh khn sai lm ca hc sinh lp 10 khi gii ton hnh hc phng bng PPVT .................................................................................... 28 1.5 Kt lun chng 1 ...................................................................................... 32 Chng2.XYDNGHTHNGBITPHNHHC10THEO HNG RN LUYN K NNG GII TON BNG PPVT .......... 33 2.1NhngkinthccbnvvcttrongchngtrnhHH10-SGK nng cao.................................................................................................... 34 2.2Quy trnh bn bc gii bi ton hnh hc bng PPVT ............................ 37 2.3 H thng bi tp...................................................................................... 40 2.3.1 Nhng kin thc b tr xy dng h thng bi tp ............................ 40 2.3.2 Nhng dng s phm khi xy dng h thng bi tp .......................... 46 2.3.3 Chng minh 3 im thng hng............................................................. 46 2.3.4 Chng minh hai ng thng vung gc ................................................ 60 2.3.5 Chng minh ng thc vct .................................................................. 72 2.3.6 Cc bi ton tm tp hp im............................................................... 81 2.3.7 ng dng ca vct vo i s ............................................................... 93 2.4 Kt lun chng 2 ...................................................................................... 96 Chng 3. TH NGHIM S PHM ............................................................... 97 3.1 Mc ch th nghim s phm ................................................................... 973.2 Ni dung th nghim ................................................................................. 973.3 T chc th nghim ................................................................................. 1103.3.1 Chn lp th nghim ............................................................................. 1103.3.2 Tin trnh th nghim........................................................................... 110 3.4 nh gi kt qu th nghim ................................................................... 110 3.5 Kt lun chng 3 .................................................................................... 114 KT LUN CHUNG ........................................................................................ 115 TI LIU THAM KHO .................................................................................. 116

www.VNMATH.com S ha bi Trung tm Hc liu i hc Thi Nguynhttp://www.lrc-tnu.edu.vn 1 M U 1.L do chn ti Tronggiaionhinnay,khikhoahccngnghcnhngbctin nhyvt,vicotonhngconngikhngchnmvngkinthcm cn c nng lc sng to, c ngha quan trng i vi tim lc khoa hc k thut ca t nc. NghquythinghlnthIVBanchphnhtrungngngCng Sn Vit Nam (kha VII, 1993) ch r: Mctiugiodc-oto phihungvootonhngcon ngi lao ng, t ch, sng to,c nng lc gii quyt nhng vn thng gp, qua m gp phn tch cc thc hin mc tiu ln ca t nc l dn giu, nc mnh, x hi cng bng, dn ch, vn minh. NghquythinghlnthIIBanchphnhtrungngngCng Sn Vit Nam (kha VIII, 1997), tip tc khng nh: Phiimiphngphpgiodcoto,khcphclitruynth mt chiu, rn luyn thnh np t duy sng to ca ngi hc. Tng bc p dngphngphptintinvphngtinhinivoqutrnhdyhc, mboiukinvthigianthc,tnghincuchohcsinh,nhtl sinh vin i hc. Nh vy, quan im chung v i mi phng php dy hc khng nh, ct li ca vic i mi phng php dy hc mn ton trng THPT llmchohcsinhhctptchcc,chng,chnglithiquenhctp th ng. Trong vic i mi phng php dy hc mn ton trng THPT, vic dy gii bi tp ton trng ph thng c vai tr quan trng v: .Dy ton trng ph thng l dy hot ng ton hc. Vic gii ton l hnh thc ch yu ca hot ng ton hc, gip hc sinh pht trin t duy, www.VNMATH.com S ha bi Trung tm Hc liu i hc Thi Nguynhttp://www.lrc-tnu.edu.vn 2 tnh sng to. Hot ng gii bi tp ton l iu kin thc hin ccmc ch dy hc ton trng ph thng. Dy gii bi tp ton cho hc sinh c tc dng pht huy tnh ch ng sng to, pht trin t duy, gy hng th cho hctpchohcsinh,yucuhcsinhcknngvndngkinthcvo tnh hung mi, c kh nng pht hin v gii quyt vn , c nng lc c lp suy ngh, sng to trong t duy v bit la trn phng php t hc ti u. Thctindyhcchothy:Vicsdngphngphpvcttrong nghincuhnhhc,hcsinhcthmnhngcngcmidint,suy lun gii ton, trnh c nh hng khng c li ca trc gic. y cng ldptt hcsinhlmquenvi ngnngtonhccaocp.Th nhng vic s dng khng thnh tho phng php trn lm hc sinh gp nhiu kh khn v lng tng, hn ch ti kt qu hc tp. Vi nhng l do trn,ti chn tinghin cu l"Rn luyn k nng gii ton cho hc sinh bng phng php vct, trong chng trnh hnh hc 10 (Chng I,II - Hnh hc 10 - Sch gio khoa nng cao ).2. Gi thuyt khoa hc Nuhngdnhcsinhcchtmligiibitontheo4bctrong lc ca Plya v xy dng c h thng bi tp nhm rn luyn k nng gii ton cho hc sinh bng PPVT trong chng trnh hnh hc 10, ng thi cccbinphpsphmphhpthsgpphnphttrinnnglcgii ton cho hc sinh. Gip hc sinh khc su kin thc hc, pht huy tnh ch ng, tnh tch cc trong vic tip thu kin thc mi gp phn nng cao cht lng dy v hc trng THPT. 3. Mc ch nghin cu Nghin cu vic vn dng bn bc gii bi tp ton theo lc ca Plyavo gii bi tp theo PPVT, nhm rn luyn k nng gii bi ton hnh hcphngbngPPVT,quaphttrinnnglcgiitonchohcsinh. www.VNMATH.com S ha bi Trung tm Hc liu i hc Thi Nguynhttp://www.lrc-tnu.edu.vn 3 ngthixutmtsbinphpdyhcnhmnngcaonnglcgii ton cho hc sinh THPT. 4. Nhim v nghin cu- Nghin cu c s l lun v thc tin ca vn c nghin cu.-Xy dng h thng bi tp nhm rn luyn k nng gii ton cho hc sinh bng PPVT trong chng trnh hnh hc 10,gp phn i mi phng php dy v hc tp trng ph thng.- Th nghim s phm kim nghim tnh kh thi ca ti.5. Phng php nghin cu - Phng php nghin cu l lun:+Nghincumtstiliuvllundyhc,giodchc,tml hc, nghin cu SGK ca chng trnh THPT, cc gio trnh v phng php ging dy ton.+ Nghin cu sch bo, tp ch lin quan n dy v hc hnh hc phng bng PPVT. - Phng php nghin cu thc tin: + Tng kt kinh nghim qu trnh cng tc ca bn thn, hc tp v tip thu kinh nghim cang nghip. Trao i trc tip vi hc sinh,gio vin ging dy tm ra nhng kh khn vng mc ca hc sinh khi gii bi tp v ch ny v tm bin php khc phc. - Phng php th nghim s phm.6. B cc lun vnM u. Chng1.Csllunvthctintrongvicdyhcgiibitp bng PPVT. Chng 2. Xy dng h thng bi tp hnh hc 10 theo hng rn luyn k nng gii ton bng PPVT. Chng 3. Th nghim s phm Kt lun. Ti liu tham kho www.VNMATH.com S ha bi Trung tm Hc liu i hc Thi Nguynhttp://www.lrc-tnu.edu.vn 4 CHNG 1.C S L LUN V THC TIN TRONG VIC DY HC GII BI TP BNG PPVT 1.1L lun v dy hc gii bi tp ton. 1.1.1 Mc ch, vai tr, ngha ca bi tp ton trong trng ph thng Plya cho rng Trong ton hc, nm vng b mn ton quan trng hn rt nhiu so vi mt kin thc thun ty m ta c th b sung nhmt cun sch tra cu thch hp. V vy c trong trng trung hc cng nh trong cc trng chuyn nghip, ta khng ch truyn th cho hc sinh nhng kin thc nht nh, m quan trng hn nhiu l phi dy cho h n mt mc no nng vng mn hc. Vy th no l mun nm vng mn ton ? l bit gii ton [25, tr.82]. a. Mc ch: Mt trong nhng mc ch dy ton trng ph thng l:Pht trin hc sinh nhng nng lc v phm cht tr tu, gip hc sinh bin nhng tri thc khoa hc ca nhn loi c tip thu thnh kin thc ca bn thn, thnh cng c nhn thc v hnh ng ng n trong cc lnh vc hot ng cng nh trong hc tp hin nay v sau ny. Lmchohcsinhnmcmtcchchnhxc,vngchcvch thng nhng kin thc v k nng ton hc ph thngc bn, hin i, ph hp vi thc tin v c nng lc vn dng nhng tri thc vo nhng tnh hungcth,voisng,volaongsnxut,vovichctpccb mn khoa hc khc.b.Vaitr:Tonhccvaitrlntrongisng,trongkhoahcv cng ngh hin i, kin thc ton hc l cng c hc sinh hc tt cc mn hc khc, gip hc sinh hot ng c hiu qu trong mi lnh vc. Cc-Mc ni Mt khoa hc ch thc s pht trin nu n c th s dng c phng php ca ton hc[5, tr.5]. www.VNMATH.com S ha bi Trung tm Hc liu i hc Thi Nguynhttp://www.lrc-tnu.edu.vn 5 Mn ton c kh nng to ln gip hc sinh pht trin cc nng lc tr tu nh:phntch,tnghp,sosnh,cbitha,khiqutha...Rnluyn nhng phm cht, c tnh ca ngi lao ng mi nh: tnh cn thn, chnh xc, tnh k lut, khoa hc, sng to.... c. ngha: trng ph thng gii bi tp ton l hnh thc tt nht cng c, h thnghakinthcvrnluynknng,lmthnhthcvndngkin thc hc vo nhng vn c th, vo thc t, vo nhng vn mi, l hnhthcttnhtgiovinkimtravnnglc,vmctipthuv kh nng vn dng kin thc hc.Vicgiibitptonctcdnglntrongvicgyhngthhctp chohcsinhnhmphttrintrtuvgpphngiodc,rnluyncon ngi hc sinh v nhiu mt.Vicgiimtbitoncthkhngnhngnhmmtdng nnht no m thng bao hm ngha nhiu mt nh nu trn.1.1.2 V tr v chc nng ca bi tp tona. V tr: " trung ph thng, dy ton l dy hot ng ton hc. i vihcsinhcthxemgiitonlhnhthcchyucahotngton hc. Cc bi tp ton trng ph thng l mt phng tin rt c hiu qu v khng th thay th c trong vic gip hc sinh nm vng tri thc, pht trin t duy, hnh thnh k nng k xo, ng dng ton hc vo thc tin. Hot ng gii bi tp ton l iu kin thc hin tt cc nhim v dy hc ton trng ph thng. V vy, t chc c hiu qu vic dy gii bi tp ton hc c vai tr quyt nh i vi cht lng dy hc ton.[13, tr.201]. b. Cc chc nng ca bi tp ton. Mi bi tp ton t ra mt thi im no ca qu trnh dy hc u changmtcchtngminhhayntngnhngchcnngkhcnhau. Cc chc nng l: www.VNMATH.com S ha bi Trung tm Hc liu i hc Thi Nguynhttp://www.lrc-tnu.edu.vn 6 - Chc nng dy hc. - Chc nng gio dc.- Chc nng pht trin. - Chc nng kim tra. Cc chc nng u hng ti vic thc hin cc mc ch dy hc: - Chc nng dy hc: Bi tp ton nhm hnh thnh cng c cho hc sinh nhng tri thc, k nng, k xo cc giai on khc nhau ca qu trnh dy hc. -Chcnnggiodc:Bitptonnhmhnhthnhchohcsinhth gii quan duy vt bin chng, hng th hc tp, sng to, c nin tin v phm cht o c ca ngi lao ng mi.- Chc nng pht trin: Bi tp ton nhm pht trin nng lc t duy cho hc sinh, c bit l rn luyn nhng thao tc tr thnh thnh nhng phm cht ca t duy khoa hc. - Chc nng kim tra: Bi tp ton nhm nh gi mc kt qu dy v hc,nhgikhnngclphcton,khnngtipthu,vndngkin thc v trnh pht trin ca hc sinh. Hiu qu ca vic dy ton trng ph thng phn ln ph thuc vo vic khai thc v thc hin mt cch y cc chc nng c th c ca cc tcgivitschgiokhoacdngavochngtrnh.Ngigio vin phi c nhim v khm ph v thc hin dng ca tc gi bng nng lc s phm ca mnh. 1.1.3 Dy hc phng php gii bi ton. Trongmntontrngphthngcnhiubitonchachoc khng c thut gii v cng khng c mt thut gii tng qut no gii tt cccbiton.Chngtachcththngquavicdyhcgiimtsbi ton c th m dn dntruyn th cho hc sinh cchthc, kinh nghim trong vic suy ngh, tm ti li gii cho mi bi ton. www.VNMATH.com S ha bi Trung tm Hc liu i hc Thi Nguynhttp://www.lrc-tnu.edu.vn 7 Dy hc gii bi tp ton khng c ngha l gio vin cung cpcho hc sinh li gii bi ton. Bit li gii ca bi ton khng quan trng bng lm th nogiicbiton.lmtnghngthhctpcahcsinh,pht trintduy,thygiophihnhthnhchohcsinhmtquytrnhchung, phng php tm li gii cho mt bi ton. Theo Plya, phng php tm li gii cho mt bi ton thng c tin hnh theo 4 bc sau:Bc 1: Tm hiu ni dung bi ton.giicmtbiton,trchtphihiubitonvchng th vi vic gii bi ton . V th ngi gio vin phi ch gi ng c, kch thch tr t m, hng th cho hc sinh v gip cc em tm hiu bi ton mt cch tng qut. Tip theo phi phn tch bi ton cho: -u l n, u l d kin. -V hnh, s dng cc k hiu thch hp (nu cn). -Phnbitccthnh phn khcnhauca iu kin,c th dintcc iu kin di dng cng thc ton hc c khng? Bc 2: Xy dng chng trnh gii. Phi phn tch bi ton cho thnh nhiu bi ton n gin hn. Phi huy ng nhng kin thc hc( nh ngha, nh l, quy tc...) c lin quan nnhngiukin,nhngquanhtrongtonrilachntrongs nhngkinthcgngihncvidkincabitonrimmm,d on kt qu. Xt vi kh nng c th xy ra, k c trng hp c bit. Sau , xt mt bi ton tng t hoc khi qut ha bi ton cho[13, tr.210]. Bc 3: Thc hin chng trnh gii. Bc 4: Kim tra v nghin cu li gii. - Kim tra li kt qu, xem li cc lp lun trong qu trnh gii. - Nhn li ton b cc bc gii, rt ra tri thc phng php gii mt loi bi ton no . www.VNMATH.com S ha bi Trung tm Hc liu i hc Thi Nguynhttp://www.lrc-tnu.edu.vn 8 B A C M A B C O B -Tm thm cc cch gii khc (nu c th ). -Khai thc kt qu c th c ca bi ton. - xut bi ton tng t, bi ton c bit hoc khi qut ha bi ton. Cngvickimtraligiicamtbitoncnghaquantrng. Trong nhiu trng hp, s kt thc ca bi ton ny li m u cho mt bi ton khc. V vy "Cn phi luyn tp cho hc sinh c mt thi quen kim tra li bi ton, xt xem c sai lm hay thiu st g khng, nht l nhng bi ton c t iu kin hoc bi ton i hi phi bin lun. Vic kim tra li li gii yu cu hc sinh thc hin mt cch thng xuyn [13, tr.212].Sau y l v d s dng 4 bc gii bi ton ca Polya chng minhV d: (Bi 89-tr 52- SBT HH10 - Nng cao )Cho im M nm trong ng trn (O) ngai tip tam gic ABC. K cc ngthngMA,MB,MC,chngctngtrnlnltA,B, C.Chng minh rng ( )( )232 2. ., , ,MC MB MAMO RSSABCC B A=(*) Gii: Bc 1: Tm hiu bi tonGv: Nhn xt 2 v ca ng thc (*) Hs: -V tri cha cc yu t din tch SA,B,C,, SAB C. -Vphichaccyutv HM/(O);vtch di cc cnh MA, MB, MC. Ta c: HM/(O)=2 2 ' ' '. . .. R MO MC MC MB MB MA MA = = =Bc 2: Xy dng chng trnh giiGv:binivtrithnhvphi,phis dng cng thc tnh din tch tam gic no chuyn dn t yu t din tch sang yu t di ?www.VNMATH.com S ha bi Trung tm Hc liu i hc Thi Nguynhttp://www.lrc-tnu.edu.vn 9 Hs: SABC= RCA BC AB4. .; SABC = RA C C B B A4' ' '. ' '. '; Gv:chuyndntyutdicccnhcatamgicABC,tam gic ABC v di cnh MA, MB, MC, H M/(O) th phi lm g ? Hs: Phi tm mi lin h gia chng bng cch xt cc tam gic ng dng: MAB A ~ MB MAMA MAMBMAABB AA MB.' . ' ' '' ' = = A(MA.MA=H M/(O) = R2- MO2 ) Lm tng t vi CAA CBCC B ' ',' ', khi (*) c chng minh. Bc 3: Trnh by li gii-Hs: SABC = RCA BC ABSRA C C B B AABC4. .;4' ' '. ' '. '=CA BC ABA C C B B ASSABCC B A. .' ' '. ' '. '' ' '= (**) Mtkhc:MAB A ~ ' ' A MB A nn: MB MAMO RMB MAMA MAMBMAABB A. MA.MB .' . ' ' '2 2M/(O) = = = = Tng t MA MCMO RCAA CMC MBMO RBCC B.' ';.' '2 2 2 2== ( ***)Thay (***) vo (**) ta c iu phi chng minh. Bc 4: Kim tra v nghin cu li gii. Gv: Bi ton trn cn cch gii no khc khng ? Hs:Cthchngminhvphibngvtribngcchsdngcng thc tnh HM/(O), s dng cc tam gic ng dng chuyn dn t yu t dicccnh,HM/(O)vyutdintchtamgicABCvdintchtam gic ABC. www.VNMATH.com S ha bi Trung tm Hc liu i hc Thi Nguynhttp://www.lrc-tnu.edu.vn 10 Vdnycungcpchohcsinhmtsknngvndngcng thc tnh phng tch ca mt im i vi mt ng trn v lm bi tp hnh hc. 1.1.4 Bi dng nng lc gii ton. Bi tp ton nhmpht trintduycho hcsinh,cbitl rn luyn cc thao tc tr tu. V vy, trong qu trnh dy hc ngi thy gio phi ch trngbidngnnglcgiitonchohcsinh.Nnglcgiitonlkh nng thc hin 4 bc trong phng php tm li gii bi ton ca Plya. Rn luyn nng lc gii ton cho hc sinh chnh l rn luyn cho h kh nng thc hin bn bctheo phng php tm li gii bi ton ca Plya. iu ny cng ph hp vi phng php dy hc pht hin v gii quyt vn theoxuhngimiphngphpdyhccanngiodcncta hin nay. Mtimngchnal:"Trongqutrnhgiibitpton,cn khuynkhchhcsinhtmnhiucchgiichomtbiton.Micchgii udavomtscimnocadkin,chonntmcnhiu cch gii l luyn tp cho hc sinh bit cch nhn nhn mt vn theo nhiu kha cnh khcnhau, iu rt b ch cho vic pht trin nng lc t duy. Mtkhc,tmcnhiucchgiithstmccchgiihaynht,p nht...[13, tr.214]. V d 1:Cho 6 im A, B, C, D, E, F. chng minh rng:CD BF AE CF BE AD + + = + + (1) gii bi ton ny, hc sinh thng ngh n cch dng cc php ton v vc t chng minh v phi bng v tri v c li gii nh sau: Li gii 1: Ta c (1)AD AE CF CD BF BE + =

EF DF ED = + EF EF = www.VNMATH.com S ha bi Trung tm Hc liu i hc Thi Nguynhttp://www.lrc-tnu.edu.vn 11 Vy ng thc (1) c chng minhLi gii 2: Bin i v triDF CD FE BF ED AE CF BE AD + + + + + = + += DF FE ED CD BF AE + + + + +=CD BF AE + +(VO FF DF FD DF FE ED = = + = + + ) Li gii 3: Bin i v phi:FD EF DE CF BE AD FD CF EF BE DE AD CD BF AE + + + + + = + + + + + = + + =CF BE AD + + (VO FD EF DE = + + ) Nhnxt:Trong3ligiitrnchothyligiithnhtlngin nht, ch cn bin i ng thc vct cn chng minh tng ng vi mt ng thc vct c cng nhn l ng. V d 2: Cho tam gic ABC. Gi M, N, P l nhng im c xc nh nh sau:PB PA NA NC MC MB 3 , 3 , 3 = = = .ChngminhhaitamgicABCvtam gic MNP c cngtrng tm. gii bi ton ny hc sinh thng ngh n cch chng minh tnh cht trng tm ca tam gic v c li gii nh sau:Li gii 1: Gi S, Q, R ln lt l trung im ca BC, CA v AB P N Q A B C S MR QS RB BP PB PACQ AN NA NCSC CM MC MB= = == == =333 www.VNMATH.com S ha bi Trung tm Hc liu i hc Thi Nguynhttp://www.lrc-tnu.edu.vn 12 Gi G l trng tm ca tam gic ABC th:O GC GB GA = + + . Ta c:( ) ( )GM GN GP GC CM GAAN GB BPGA GB GC SC CQ QSO O O+ + = + + ++ + += + + + + += + = V G l trng tm ca tam gic MNP Li gii 2:-Gi G l trng tm ca tam gic ABC th O GC GB GA = + + -Gi G l trng tm ca tam gic MNP thO P G N G M G = + + ' ' 'Ta c: ( ) ( ) ( )( )'21' ' ' ' 3' '' '' 'G GO O BC AB CA OMG PG NG CM BP AN GC GB GA GGMG CM GC GGPG BP GB GGNG AN GA GG = + + + + =+ + + + + + + + = + + =+ + =+ + = Vy tam gic ABC, tam gic MNP c cng trng tm. Li gii 3: Gi G l trng tm ca tam gic MNP thO GP GN GM = + +Ta c:MC GM PB GP NA GN GC GB GA + + + + + = + + = ( ) ( ) BA CB AC GM GP GN + + + + +21 =O O O = + .21 Suy ra G l trng tm l tam gic ABC. Nhn xt: Trong 3 li gii nu trn, li gii th 3 l ngn gn nht v t nhin nht, v n vn dng tnh cht trng tm ca tam gic chng minh. www.VNMATH.com S ha bi Trung tm Hc liu i hc Thi Nguynhttp://www.lrc-tnu.edu.vn 13 TrongqutrnhtmligiibitontheobnggicaPlyartc hiu qu, n t hc sinh trc nhng ngh tch cc, chng hn nh: - Bn gp bi ton ny ln no cha ? Hay bn gp bi ton ny dng hi khc ? - Bn c bit bi ton no c lin quan khng ? C th dng nh l hay cng thc no gii n ? - C th s dng kt qu ca bi ton khc vo vic gii bi ton ny hay khng? c th a ra mt bi ton tng t hoc mt bi ton tng quthn bi ton cho khng ?... 1.2Knnggiitonvvnrnluynknnggiitoncho hc sinh1.2.1 K nng Knnglkhnngvndngtrithckhoahcvothctin.Trong , kh nng chiu l:sc c (v mtmtno) thchinmt vic g[3, tr.548]. Theotmlhc,knnglkhnngthchinchiuqumthnh ng no theo mt mc ch trong nhng iu kin xc nh. Nu tm thi tch tri thc v k nng xem xt ring tng cc tri thc thuc phm vi nhn thc, thuc v kh nng bit, cn k nng thuc phm vi hnh ng, thuc v kh nng bit lm. Ccnhgiodchcchorng:Mikinthcbaogmmtphnl thng tin kin thc thun ty v mt phn l k nng. K nng l mt ngh thut, l kh nng vn dng nhng hiu bit mi ngi t c mc ch. K nng cn c th c c trng nh mt thi quen nht nh v cui cng k nng l kh nng lm vic c phng php. Trongtonhc,knnglkhnnggiiccbiton,thchincc chng minh nhn c. K nng trong ton hc quan trng hn nhiu so vi kin thc thun ty, so vi thng tin trn.[25, tr.99]. www.VNMATH.com S ha bi Trung tm Hc liu i hc Thi Nguynhttp://www.lrc-tnu.edu.vn 14 Trong thc t dy hc cho thy, hc sinh thng gp kh khn khi vn dngkinthcvogiiquytccbitpcthldo:hcsinhkhngnm vng kin thc cc khi nim, nh l, qui tc,khng tr thnh c s ca k nng. Munhnh thnh c k nng, c bit l k nng gii ton cho hc sinh, ngi thy gio cn phi t chc cho hc sinh hc ton trong hot ng v bng hot ng t gic, tch cc, sng to hc sinh c th nm vng tri thc,cknngvsnsngvndngvothctin.Gpphnthchin nguynlcanhtrngphthngl:Hciivihnh,giodckt hp vi lao ng sn xut, nh trng gn lin vi x hi. 1.2.2 K nng gii ton. K nng gii ton l kh nng vn dng cc tri thc ton hc gii cc bi tp ton (bng suy lun, chng minh)[5, tr.12]. thchinttmntontrongtrngTHPT,mttrongnhngyu cu c t ra l: V tri thc v k nng, cn ch nhng tri thc, phng php c bit l tri thc c tnh cht thut ton v nhng k nng tng ng. Chng hn: tri thc v k nng gii bi ton bng cch lp phng trnh, tri thc v k nng chng minh ton hc, k nng hot ng v t duy hm...[13, tr.41]. Cn ch l ty theo ni dung kin thc ton hc m c nhng yu cu rn luyn k nng khc nhau. 1.2.3 c im ca k nng. Khi nim k nng trnh by trn cha ng nhng c im sau: - Bt c k nng no cng phi da trn c s l thuyt l kin thc. Bi v, cu trc ca k nng l: hiu mc ch - bit cch thc i n kt qu - hiu nhng iu kin trin khai cch thc . - Kin thc l c s ca k nng, khi kin thc phn nh y cc thuc tnh bn cht ca i tng, c th nghim trong thc tin v tn ti www.VNMATH.com S ha bi Trung tm Hc liu i hc Thi Nguynhttp://www.lrc-tnu.edu.vn 15 trong thc vi t cch l cng c ca hnh ng. Cng vi vai tr c s ca trithc,cn thyr tmquantrngca knng.Bi v:Mntonlmn hc cng c c c dim v v tr c bit trong vic thc hin nhim v pht trin nhn cch trong trng ph thng.[13, tr.29].V vy, cn hng mnh vo vic vn dng nhng tri thc v rn luyn k nng, v k nng ch c th c hnh thnh v pht trin trong hot ng. -K nng gii ton phi da trn c s tri thc ton hc, bao gm: kin thc, k nng, phng php. 1.2.4 S hnh thnh k nng S hnh thnh k nng l lm cho hc sinh nm vng mt h thng phc tpccthaotcnhmbinivlmsngtnhngthngtinchang trong cc bi tp. V vy, mun hnh thnh k nng cho hc sinh, ch yu l k nng hc tp v k nng gii ton, ngi thy gio cn phi:-Gip hc sinh hnh thnh mt ng li chung (khi qut ) gii quyt cc i tng, cc bi tp cng loi.-Xc lp c mi lin h gia nhng bi tp khi qut v cc kin thc tng ng. Vd:Khirnluynknngchngminhngthcvct,cnch gip hc sinh nhn ra mi quan h gia v phi v v tri ca ng thc cn chng minh. Chng hn:1/ Cho 2 im A, B v hai s thc| o,sao choO = + | oa.Chng minh tn ti duy nht im I sao choO IB IA = + | o.b.Chng minh vi mi im M ta lun c:( ) MI MB MA . | o | o + = +2/ Cho t gic ABCD c M, N, P, Q theo th t l cc trung im ca AD, BC, DB, AC. Chng minh: www.VNMATH.com S ha bi Trung tm Hc liu i hc Thi Nguynhttp://www.lrc-tnu.edu.vn 16 a.( ) DC AB MN + =21 b.( ) DC AB QP =21 Nhngbitondngnygiphcsinhcngcknngsdngcc tnh cht ca vc t, php cng vc t, php tr vc t, php nhn vc t vi mt s thc, cc quy tc nh quy tc 3 im, quy tc trung im... Do c im, vai tr v v tr ca mn ton trong nh trng ph thng, theo l lun dy hc mn ton cn ch :Trongkhidyhcmntoncnquantmrnluynchohcsinh nhng k nng trn nhng bnh din khc nhau l:-K nng vn dng tri thc trong ni b mn ton-K nng vn dng tri thc ton hc vo nhng mn hc khc-K nng vn dng tri thc vo i sng[12, tr.19].Theo quan im trn, truyn th tri thc, rn luyn k nng l nhim v quan trng hng u ca b mn ton trong nh trng ph thng.Rnluynknngtonhcvknngvndngtonhcvothc tin m trc tin l k nng gii ton cn t c cc yu cu sau: 1/Giphcsinhhnhthnhnmvngnhngmchkinthccbn xuynsutchngtrnhphthng.Trongmntoncthkticckin thc sau:- Cc h thng s. - Hm s v nh x. - Phng trnh v bt phng trnh. - nh ngha v chng minh ton hc. - ng dng ton hc. 2/ Gip hc sinh pht trin cc nng lc tr tu, c th l: - T duy logic v ngn ng chnh xc, trong c t duy thut ton. www.VNMATH.com S ha bi Trung tm Hc liu i hc Thi Nguynhttp://www.lrc-tnu.edu.vn 17 - Kh nng suy on, t duy tru tng v tr tng tng khng gian.- Nhng thao tc t duy nh phn tch, tng hp, khi qut ha. - Cc phm cht tr tu nh t duy c lp, t duy linh hot v sng to. 3/ Coi trng vic rn luyn k nng tnh ton trong tt c gi hc ton, gn vi vic rn luyn cc k nng thc hnh nh tnh ton, bin i, v hnh, v th. 4/ Gip hc sinh rn luyn phm cht ca ngi lao ng mi nh: Tnh cn thn, chnh xc, kin tr, thi quen t kim tra nhng sai lm c th gp. 1.2.5Mtsknngcbntrongquytrnhgiibitonbng phng php vctKnnggiibitpton,cbitvgiitonvctbaogmmth thngccthaotctrtuvthchnhvndngtrithc(kinthc, phngphp)vovicgiiccbitpkhcnhautcmtsyucu ca ch gii bi tp v vct trong chng trnh Hnh Hc 10. Trong quy trnh gii 1 bi tp ton bng phng php vc t, c nhng k nng c bn sau: - Chuyn bi ton sang ngn ng vc t. - Phn tch 1 vc t thnh mt t hp vc t. - K nng ghp 1 s vct trong 1 t hp vct. - Khi qut ha 1 s nhng kt qu vn dng vo bi ton tng qut hn. *ylnhngkhumuchttrongphngphpgiitonbng cng c vct. 1.2.5.1 Din t quan h hnh hc bng ngn ng vc t - Cn rn luyn cho hc sinh k nng chuyn tng ng nhng quan hhnhhctcchnithngthngsangdngvctcthvndng cng c vct vo gii ton.www.VNMATH.com S ha bi Trung tm Hc liu i hc Thi Nguynhttp://www.lrc-tnu.edu.vn 18 V d: T quan h hnh hc "Ba im A, B, C thng hng c din t bng kin thc vc t l: OB m OA k OC BC k AC AC k AB + = = = , ; vi O ty v k+m = 1. -TquanhhnhhcHaiimB,Ctrngnhaucdintbng kin thc vct lAC AB = . - T quan h hnh hc "Hai ng thng song song AB// CDc din t bng kin thc vc t lAB kCD = . - Tquan h hnh hc "im M chia on thng AB theo t s k=1 c din t bng kin thc vc t lMB k MA = . - T quan h hnh hc "AM l trung tuyn caA ABCc din t bng kin thc vc t lAM AC AB 2 = + . - T quan h hnh hc "G l trng tmA ABC c din t bng kin thc vc t lO GC GB GA = + + . -Tquanhhnhhc"HaingthngvunggcAB CDc din t bng kin thc vc t lO CD AB = . ... Nhvy,vicchuynbitonsangngnngvctlimxutpht trong vic s dng cng c vct gii ton. 1.2.5.2 Phn tch 1 vc t thnh mt t hp vctMtkhumuchtkhcnamtacnrnluynchohcsinhlk nng phn tch 1 vct thnh 1 t hp vct ca nhng vct khc, ch yu l phn tch 1 vct thnh tng 2 vct hoc thnh hiu hai vct. * Phng php 1: Vn dng quy tc hnh bnh hnh. Vd:ChotamgicABC,Ilimbtktrongtamgic.Chng minh rng: 0 = + + IC S IB S IA SIAB ICA IBC Hng dn gii: www.VNMATH.com S ha bi Trung tm Hc liu i hc Thi Nguynhttp://www.lrc-tnu.edu.vn 19 Phn tch ICtheoIB IA,bng quy tc hnh bnh hnh. Gi giao im ca cc tia AI, BI, CI vi BC, CA, AB ln lt l A1, B1, C1. Dng hnh bnh hnh IACB, ta c: .IB IA IB IA IC | o + = + = ' '. IABIBCSSAMCHA BC BIAIA = = = =11'oTng t:IABIACSS = |Vy IBSSIASSICIABICIABIBC = 0 = + + IC S IB S IA SIAB IAC IBC * Phng php 2: Phng php xen im (vn dng quy tc ba im). V d1: Cho tam gic ABC vi trng tm G. Chng minh rng vi im O bt k, ta c( ) OC OB OA OG + + =31

-Phn tch: T vc tOG , xut hin cc vc t c im cui l A, B, C, ta dng quy tc tam gic xen im A, B, C vo v c cch phn tch vct di y: CG OC OGBG OB OGAG OA OG+ =+ =+ =

T cng theo tng v ri lp lun ri suy ra iu cn chng minh. V d 2: Cho bn im M, A, B, C ty . Chng minh rng: . . . . O AB MC CA MB BC MA = + +-Phn tch: c mt tng bng khng, ta c th chn php bin i lmxuthincccpgitrinhau.Munvy,cnvndngcchphn A BC A B A1 B1 M H C1 I www.VNMATH.com S ha bi Trung tm Hc liu i hc Thi Nguynhttp://www.lrc-tnu.edu.vn 20 tch mi vct thnh mt hiu, im gc c th chn ty , song khi di dng, ta chn im gc ny l M. ( )( )( ) MC MA MC MB MA MB MC AB MCMB MC MB MA MC MA MB CA MBMB MA MC MA MB MC MA BC MA. . .. . .. . . = = = = = = T c th d dng i n iu phi chng minh. 1.2.5.3K nng ghp 1 s vct trong 1 t hp vct V d: Cho t gic ABCD. Gi M, N, I ln lt l trung im ca AB, CD, MN. Ta bit rng: O ID IC IB IA = + + +t t hp vct:: v ID IC IB IA = + + +-Nu nhnvdi dng:( )( ) IF IE IC IB ID IA v 2 2 + = + + + =(E, F l trung im ca AB, CD ) R rng, nu nhn mt t hp vct theo tng nhm ta c c nhiu kt qu th v. A B M E I F D N C Ta c kt qu E, I, F thng hng. - Nu nhnvdi dng: ( )( ) IQ IP ID IB IC IA v 2 2 + = + + + =(P, Q l trung im ca AC, BD) Ta c P, I, Q thng hng. -Nu nhnvdi dng: ( ) ID IG ID IC IB IA v + = + + + = 3(G l trng tm tam gic ABC) ta c G, I, D thng hng. Tngt,sdnnccon niminhtgicABCDvtrng tmtamgictobibanhcnli ng quy. D A B M I G N C www.VNMATH.com S ha bi Trung tm Hc liu i hc Thi Nguynhttp://www.lrc-tnu.edu.vn 21 1.2.5.4 Khi qut ha 1 s nhng kt qu vn dng vo bi ton tng qut hn Thngquavichcsinhvndngnhngkinthcvvctchng minhmtstnhchttronghnhhc,tnhchtcatrungimonthng, ca trng tm tam gic..., ngi thy gio cn tn dng nhng c hi cho hcsinhcrnluynvphntch,tnghp,khiqutha...,chnghn gip hc sinh khi qut ha nhng s kin sau y: -Trung im O ca on thng AB:O OB OA = +-Trng tm G ca tam gic ABC:O GC GB GA = + + . -Tm O ca hnh bnh hnh ABCD:O OD OC OB OA = + + + . -TrungimOcaonthngnicctrungimcahaingcho hoc ca hai cnh i din ca t gic ABCD:O OD OC OB OA = + + + . Cn cho hc sinh pht hin s tng t gia cc s kin tng t trn, t cthcmtcchnhnkhiqutvnhngkinthcvcttngng. Thtranhngbitontrnulnhngtrnghpcthcatnhcht chung v trng tm ca mt h n im trong mt phng. 1.3 Ni dung chng trnh HH10-SGK nng cao 1.3.1 Nhim v ca HH10-SGK nng cao Mn ton THPT c nhim v cung cp cho hc sinh nhng kin thc, k nng,phngphptonhcphthngcbn,thitthc,gpphnquan trngvovicphttrinnnglctrtu,hnhthnhkhnngsuylunc trng ca ton hc, cn thit cho cuc sng, gp phn hnh thnh v pht trin cc phm cht, phong cch lao ng khoa hc, bit hp tc lao ng, c ch vthiquenthcthngxuyn.Mntontocshcsinhtiptc hclnihc,caong,trunghcchuynnghip,hcnghhocivo cuc sng lao ng theo nh hng ca Ban khoa hc t nhin. Chng trnh HH10-SGK nng cao m nhn cc nhim v c bn sau: www.VNMATH.com S ha bi Trung tm Hc liu i hc Thi Nguynhttp://www.lrc-tnu.edu.vn 22 1 / B sung thm mt s kin thc v hnh hc phng v c bit b sung thm hai phng php mi: l phng php vct v phng php ta . -Vctlmtkhinimquantrng,hcsinhcnnmvngcth hc tip ton b chng trnh hnh hc bc THPT. N cng l c s trnh by phng php ta trn mt phng. Ngoi racckin thc v vcts c p dng trong vt l nh: vn tng hp lc, phn tch mt lc theo hai thnh phn, cng sinh ra bi mt lc... -Phngphptatrnmtphngctrnhbydatrncckinthcvvctvccphptonvct.Phngphpnygipchohcsinh i s hacc kin thc c v hnh hc, v t c th gii quyt cc bi ton hnh hc bng thun ty tnh ton. Phngphptacncsdngbcutmhiucctnh cht ca ba ng Cnc. 2/Tiptcrnluynvphttrintduylogc,trtngtngkhng gian,vknngvndngkinthchnhhcvovicgiiton,vohot ng thc tin, vo vic hc tp cc b mn khc. 1.3.2 Nhng ch khi ging dy HH10-SGK nng cao. -Trckiatheocchgingdyc,SGKchnthunlmttiliu khoa hc dng cho gio vin. Ni dung cc tit dy thng c vit c ng, ging nh mt bi bo vit trn cc tp tr ton hc: u tin l nu nh ngha ca mt khi nim mi, sau l cc tnh cht v chng minh, ri cc nh l v chng minh, cui cng l cc v d hoc cc bi ton. -Trongtthayischnm2006-2007,schgiokhoacgnggp phn vo vic ci tin phng php ging dy ca thy v phng php hc ca tr. V ni dung kin thc, chng trnh mi c nhng thay i nh sau: 1.C gng gimnhphnl thuyt, khngi hi phi chnh xcmt cch hon ho. Nhng chng minh rm r, rc ri th c th b qua v thay www.VNMATH.com S ha bi Trung tm Hc liu i hc Thi Nguynhttp://www.lrc-tnu.edu.vn 23 bng nhng kim chng hoc nhng minh ha n gin (V d: Cc tnhcht ca tch vct vi mt s hoc tch v hng ca hai vct...) Nhng vn l thuyt qu i su, khng cn thit th cng quyt gt b. 2. Tng cng phn luyn tp v thc hnh.Cc bi tp phn ln nhm mc ch cng c nhng kin thc c bn, nhm rn luyn k nng tnh ton khngquphctp,vcchtrngnccbitonthctin.Khngch trngnccbitpkh,phctp,hocccbitpphidngnhiumo mc mi gii c. 3. Tng cng tnhthc t, ch trng p dng vo thc t i sng. Vi tinh thn trn, ni dung HH10-SGK nng cao c trnh by theo tng sau y: - Sch gio khoa phi l ti liu dng cho c thy gio v hc sinh phi trnhbyvhngdnnhthnochonukhngcthygio,hc sinh cng c th t hc c, tuy nhin l kh khn v vt v hnSch gio khoa c thng gii thiu mt khi nimmi bng mt nh ngha c tnh cht p t. V d: Khi nim "Vct l hon ton mi i vi hc sinh, c nh ngha: "L mt on thng nh hng, ngha l c phn bit im u v im cui. Khi ging dy, gio vin lun lun tm cch dn dt mt cch hp l, lm cho hc sinh thy c rng khi nim c xut hin mt cch t nhin, ch khng phi l ci g t trn tri ri xung, hay t trong cc nh ton hc bt ra. khc phc iu ny, SGK mi a thm phndndthcsinhcthccn.Vd:ankhinim vct, SGK mi lin h n vt l ni n cc i lng v hng v cc i lng c hng. - SGK gip thy gio to iu kin cho hc sinh suy ngh v hot ng, trnh tnh trng hc sinh ch nghe v ghi chp. Bi vy, SGK a vo mt h thng cc cu hi v cc hot ng.Cc cu hi nhm gip hc sinh nh www.VNMATH.com S ha bi Trung tm Hc liu i hc Thi Nguynhttp://www.lrc-tnu.edu.vn 24 limtkinthcnohocgi,hocnhhngchonhngsuy ngh ca hc sinh, cc cu hi ni chung l d, v th khng nn a cu tr li trong SGK. Cchotngihihcsinhphilmvic,phitnhtonin mt kt qu no . i vi nhng chng minh hoc tnh ton khng qu kh, mt vi bc hot ng ca hc sinh c th thay th cho li gii ca thy gio. Ty tnh hnh lp v trnh hc sinh, t chc cc hot ng c th c nhiu cch: C th l mi hc sinh t lm vic theo hng dn ca hat ng, thy kim tra cc kt quv tng kt, cng c th hc sinh lm vic theo tng nhm hai ngi, nhiu ngi, cng c th t chc tholun chung trong lp. - SGK gim nh phn l thuyt, ch yu l gim nh cc chng minh ca cctnhchthocnhl.Cctnhchtvnhlnynhiulcrthin nhin, hon ton c th thy c bng trc gic, nhng thc ra chng minh chng li khng n gin. Vd:Vicchngminhtnhchtphpnhnvctvimts () ( )a l k a l k . . = kh phc tp v di dng m khng mang li li ch g nhiu. V vy SGK khng trnh by chng minh m ch nu ra mt s trng hp c th kim chng. Ngoi ra, nu mt tnh cht no qu hin nhin SGK cng khng a ra, v nu lm nh vy, i khi li gy thc mc cho hc sinh.V d v vc t i:Sau khi nh ngha vc t i SGK dn ra cu hi hc sinh c ngay nhn xt: nu cho vc tABthO BA AB = + , vyBAchnh l vct i ca vctAB . T i n kt lun mi vct u c vct i, m khng ni g n tnh duy nht ca vc t i, xem nh hin nhin. -SGKlnnycgnglinhthcttrongtrnghpcth.Chng hn,trongphnvctcthathmnhngngdngtrongvtl:Tng hp lc, phn tch lc, cng sinh ra bi mt lc, phn gii tam gic c th a www.VNMATH.com S ha bi Trung tm Hc liu i hc Thi Nguynhttp://www.lrc-tnu.edu.vn 25 vo cc bi ton o c trn hin trng. V d khc: Khi ni n ng elp, parabol v hybebol th trong bi c thm, sch nu nhiu p dng thc t ca cc ng ny. Nu khng lm nh vy, hc sinh ch bit v l thuyt c cc ng nh th cn khng bit n c tn ti trong thc t hay khng. 1.3.3McchyucucaPPVTtrongchngtrnhHH10-SGK nng cao Trongchngtrnhhnhhclp10hcsinhchcvvct,cc php tontrn vct, cctnh cht cbn ca tch v hng v nhng ng dng ca chng, c bit l nhng h thc quan trng trong tam gic: nh l Csin,nhlSin,cngthctrungtuyn,cccngthctnhdintchtam gic...hc sinh phi bit tn dng cc kin thc c bn ni trn gii mt sbi ton hnh hc v bi ton thc t. Ccyucuivihcsinhvkinthccbnvknngcbn trong chng I, II- SGK HH10 nng cao l:-V kin thcc bn: nm ckhi nim vct, hai vctbng nhau, hai vct i nhau, vct khng, quy tc ba im, quy tc hnh bnh hnh, quy tctrungim,nhnghavtnhchtcaphpcng,phptr,phpnhn vct vi s thc, tch v hng ca hai vct. -V k nng c bn: bit dng mt vct bngvct cho trc, bit lp lun hai vct bng nhau, vn dng quy tc hnh bnh hnh, quy tc ba im dng vct tng v gii mt s bi ton, bit xc nh s thc k i vi hai vct cng phngb a,sao choa k b . = , vn dng tnh cht c bn ca tch v hng,cbitxcnhiukincnvcahaivct(khcvct khng) vung gc vi nhau, vn dng tng hp kin thc v vct nghin cumtsquanhhnhhcnh:tnhthnghngcabaim,trungim ca on thng, trng tm ca tam gic, giao im hai ng cho ca hnh bnh hnh... www.VNMATH.com S ha bi Trung tm Hc liu i hc Thi Nguynhttp://www.lrc-tnu.edu.vn 26 1.4Nhngkhkhnsai lmcahc sinhlp10khigii tonhnh hc phng bng PPVT 1.4.1NhngiucnlukhigingdyvcttrongHH10-SGK nng cao Ngaytchngutin,chngtatrnhbychohcsinhcckhi nimhontonmi: lvct,cc php tontrnvctv h trc ta ccvunggc.Cckhinimnycsdngtrongtonbnidung ca hnh hc 10. iu quan trng l gio vin cn lm cho hc sinh hiu r v nm c vvctcng vinhngkhinimclinquannh scng phng,khc phng, cng hng, ngc hng ca hai vct, s bng nhau ca hai vct vnhnghavctkhng,cngnhngquycringchovctkhng. Thngquaccvd,phnvd,giovincnlmchohcsinhhiur nhngkhinincbncnhnghahocgiithiubngccnh ngha c tnh cht m t. Cn phi ly nhng hnh nh trong thc t minh ha cc khi nim c cp trong SGK. Sau khi dy cc khi nim mi, gio vin cn phi c k hoch kim tra li xem hc sinh camnh r v nm chc kin thc va hc hay cha ? - Khi hc cc php ton vvct, hc sinh thng so snh vi cc php toncng,tr,nhn,chia,ccs.Do,giovincnkhngnhhc sinhbitrngivitphpccvct,khngcphpchiavctchomt vct. y ch c khi nim t s ca hai vct cng phng l mt s thc k. Khi nim ny c lin qua n khi nim php nhn mt s vi mt vct. V d: Ta cb k a . =nn c th vit bak = hcsinhcthsdngPPVTgiitonhnhhcphngthvic chuyn ngn ng hnh hc sang ngn ng vct v ngc li phi thnh tho. www.VNMATH.com S ha bi Trung tm Hc liu i hc Thi Nguynhttp://www.lrc-tnu.edu.vn 27 Do , trong khi dy, gio vin philin h nhng s kin hnh hc m hc sinhchclpdivinhngiuanghc,tdintchng bng ngn ngvct v ngc li. V d: Khi nim I l trung im ca on thng AB Th c th c din t bng ngn ngvct "I l im tha mnO IB IA = + , Hay hai ng thng AB v CD vung gc vi nhau th c th niO CD AB = . ,... Gio vin cn lm cho hc sinh bit cch phn tch mt vct thnh tng ca 2 hay nhiu vct ty thuc vo mc ch ca vic phn tch . V d:OB AO AB + =vi O l mt im ty . AN AM AB + =vi AMBN l mt hnh bnh hnh. KB HK IH AI AB + + + =vi I, H, K l cc im ty . hc sinh bit vndngtnhchtgiao honv tnhcht kthpca phpcngvcttrongkhitnhtonhocbinimththcvvctv dng cn chng minh, trc ht gio vin cn cho hc sinh lm quen vi vic binimtvct thnhhiucahaivctvsau thuc hinphp bin i ngc li. V d:Chng minh rng vi 4 im A, B, C, D bt k ta lun c h thc CB AD CD AB + = +Ta ly mt im O ty ri bin i a v cc vct c im u l O. Ta c:CB AD OC OB OA OD OC OD OA OB CD AB + = + = + = + ) ( ) ( ) ( ) ( Cch khc: Ta c th bin i nh sau:i vi 4 im A, B, C, D ta lun c h thcO DA CD BC AB = + + + Do AD CB DA BC CD AB + = = +SGK mi a vo mt h thng cu hi v cc hot ng nhm gip gio vin to iu kin cho hc sinh suy ngh v hot ng. Tt nhin, cc ni dung ny u mang tnh cht gi gio vin tham kho khi son bi v ln www.VNMATH.com S ha bi Trung tm Hc liu i hc Thi Nguynhttp://www.lrc-tnu.edu.vn 28 lp.Vnquantrnglcnphitoiukinhcsinhcsuyngh, pht huy tnh sng to ch ng chim lnh c kin thc, hnh thnh c k nng c bn tip thu ni dung cc bi ging mt cch tch cc y hng th. 1.4.2 Nhng kh khn sai lm ca hc sinh lp 10 khi gii ton hnh hc phng bng PPVT PPVT c nhiu tin li trong vic gii cc bi tp hnh hc. Tuy vy, khi s dng phng php ny hc sinh vn gp phi mt s kh khn, v khng trnh khi nhng sai lm trong khi gii ton hnh hc lp 10. Kh khn th nht m hc sinh gp phi l ln u tin lm quen vi i tng mi l vct, cc php ton trn cc vct. Cc php ton trn cc vctli c nhiu tnh cht tng t nh i vi cc s m hc sinh hc trc,dovhcsinhchahiurbnchtcacckhinimvcc php ton nn d ng nhn, mc sai lm trong khi s dng PPVT. V d 1: Cho bn im A, B, C, D. Chng minh rngCB AD CD AB + = +Vi bi ton trn, nhiu hc sinh b nhm trong qu trnh lm bi, c hcsinhhiubitonnynhsau:ChobnimA,B,C,D.Chng minh rng: AB + CD =AD + CB. V hiu sai bi ton, dn n kh khn trong qu trnh tm li gii bi ton. Vd2:ChotamgicABCviAB=3,AC=5,BC=7.Tnh AB . AC ,tnh gc A, v gc gia hai ng thng AB v AC C hc sinh gii bi ton ny nh sau: Li gii 1:Ta c 15 5 . 3 . = = CD AB1..cos = =AC ABAC ABA . Vy s o ca gc A l O0, gc gia hai ng thng AB v AC l O0. www.VNMATH.com S ha bi Trung tm Hc liu i hc Thi Nguynhttp://www.lrc-tnu.edu.vn 29 Li gii 2: Ta c ( )21521.2 2 2 = + = BC AC AB AC AB = = =2115215..cosAC ABAC ABA gcAbng120o.Gcgiahai ng thng AB v AC l 120o Bi ny hc sinh trn gii sai do cha nm vng cc kin thc v vct, di ca vct v tch v hng ca hai vct. c bit c s nhm ln v cch xc nh gc gia 2 vct v gc gia hai ng thng. Li gii ng nhsau: Ta c ( )21521.2 2 2 = + = BC AC AB AC AB = = =2115215..cosAC ABAC ABAgc A bng 120o. Vy gc gia hai ng thng AB v AC l 600. KhkhnthhaikhisdngPPVTldothotlykhihnhnhtrc quan, hnh v nn kh tng tng, hiu bi ton mt cch hnh thc, khng hiu ht ngha hnh hc ca bi ton. V hc sinh c thi quen gii bi ton hnh hc l phi v hnh nn khi s dng PPVT gii mt s bi tp khng s dng hnh v, hc sinh gp nhiu kh khn lng tng. V d 3: Cho tam gic ABC c AB=a, AC=b. AD l phn gic trong ca tam gic ABC. im D chia on thng BC theo t s no?. C hc sinh gii bi ton ny nh sau: -p dng tnh cht ng phn gic ta c:DCbaDBbaACABDCDB= = = . Suy raDCbaDB =www.VNMATH.com S ha bi Trung tm Hc liu i hc Thi Nguynhttp://www.lrc-tnu.edu.vn 30 Phn tch sai lm:Hc sinh xc nh sai chiu ca vct. Hai vct DC DB,ngc hng nhau, do nu im D chia an thng BC theo t s k th k 0, n > 0. Mt khc, HS xc nh nhm: t t s ca 2 on thng '1BBnCB = suy ra ngay im B chia on thng BC theo t s 1-n, v cng lm tng t nh th i vi im A. -Li gii ng ca bi ton ny nhsau: V I nm trn AB v AB nn c cc s x v y sao cho: . ' (1 ). . (1 ) ' CI x CA x CB y CA y CB = + = + Hay b n y a y b x a m x . ). 1 ( . ). 1 ( . . + = +V hai vct, a bkhng cng phung nnmnnxy n xy mx= = =11) 1 ( 1 Vybmnnamnn mCI ).111 (1) 1 ( += = bmnm namnn m.1) 1 (.1) 1 (+ Hc sinh thng gp kh khn khi chuyn bi ton t ngn ng hnh hc thngthngsangngnnghnhhcvctvngcli.Vvycnrn luyn cho hc sinh k nng chuyn tng ng nhng quan h hnh hc t cchnithngthngsangdngvctcthvndngcngcvct trong gii ton. Vd5:ChotamgicABC.imKchiatrungtuynADtheots 13=KDAK. ng thng BK chia din tch tam gic ABC theo t s no? Nhnxt:Trongrakhngcbngdngvct,hcsinhslng tng khi chuyn sang dng vct v kh xc nh c cch gii bi tp ny lg.V vygiovincn phi gichoccembitsuyngh v lachn cch chuyn bi ton trn sang ngn ng vct. (V d: bitng thng www.VNMATH.com S ha bi Trung tm Hc liu i hc Thi Nguynhttp://www.lrc-tnu.edu.vn 32 BK chia din tch tam gicABC theo t s no th cn phi tmxem im F chia on thng AC theo t s no, vi F l giao im ca BK v AC) Nh vy, sau khi hc PPVT, hc sinh c trong tay thm mt cng c gii bi ton hnh hc. Khng th ni phng php no tt hn phng php no. V c nhng bi ton gii bng phng php ny th d, nhng li rt vt v khi gii bng phng php khc, thm ch cn khng gii ni. Do vic sdngphngphpno giiloibitonhnh hcnoth thunli l mt trong nhng vn kh khn i vi hc sinh. 1.5 Kt lun chng 1 nhhngimiphngphpdyhccanctahinnayl "Hot nghangi hcnhmmcch nngcao hiu qugio dc v oto.Vinidungtrnhbychng1:Dyhcphngphptm li gii bi ton, bidng nnglc giiton,rn luyn knng gii ton chohcsinhtathy:dyhcgiibitptonchohcsinhtrunghcph thng l rn luyn khnng tm li gii bi ton theo bn bc ca Plya. Trong thc t hin nay, k nng gii ton ca hc sinh trung hc ph thng cn nhiu hn ch.gpphnkhcphctnhtrng,trongchng2calunvn, chng ti s a ra 1 h thng bi tp hnh hc 10 gii bngPPVT v 1 s bin php s phm nhm rn luyn cho hc sinh kh nng tm li gii bi tp ton theo bn bc gi ca Plya. www.VNMATH.com S ha bi Trung tm Hc liu i hc Thi Nguynhttp://www.lrc-tnu.edu.vn 33 CHNG 2. XY DNG H THNG BI TP HNH HC 10 THEO HNG RN LUYN K NNG GII TON BNG PPVT Trongchng ny,strnh byccdng bitp hnh hc10 gii bng PPVT,mi dng bi tp c thng qua ccv d tiu biu phn tch li gii.Quaaracctrithcphngphphocnhngktlunsphm cho mi dng bi tp c th. Hthngbitptrongchngnynhmmcchrnluynchohc sinh k nng gii bi tp hnh hc 10 bng PPVT bao gm c nhng k nng gii ton ni chung v k nng gii ton vct ni ring th hin trong hai ni dung chnh sau y: - Rn luyn cch tm ng li gii bi ton. - Rn luyn kh nng gii ton. Tm ng li gii bi ton l khu quan trng trong qu trnh gii ton, yu cu hc sinh phi t cc d liu ca bi ton bao gm: gi thit, iu kin c trong bi ton xc nh:- Th loi bi ton. - Vch ra phng hng gii bi ton. - Tm c cng c v phng php thch hp gii bi ton. -Phthincmilinhctnhttyugiagithitvktlun, gia nhng iu cho v nhng iu bi ton i hi. Khi gii ton cn ch tm kim nhng bi ton c lin quan v xut ra nhng bi ton mi. Trong qu trnh tin hnh gii mt bi ton, hc sinh c th s dng cc phng php khc nhau, cc suy lun khc nhau v tm ra cc li gii khc nhau. V vy, tm c li gii hay cho mt bi ton, hc sinh cn phi kim tra v nghin cu k li gii. www.VNMATH.com S ha bi Trung tm Hc liu i hc Thi Nguynhttp://www.lrc-tnu.edu.vn 34 cbitivinhngbitonkhngcthutgii,ihihcsinh phi tch cc suy ngh tm ti li gii v c phng php suy lun hp l ng thi cn c kinh nghim trong vic s dng cc cng c gii ton. phhpvikhnngtipthucahcsinh,hthngbitpc aratdnkh.Cnhngbitpcbncthdngcccngthc, nhlhcchngminhvktqucanhngbitpnycthvn dng vo chng minh cc bi ton khc. C nhng bi tp phi s dng kin thctnghpnhmrnluynknng,khnngvndngkinthc,kh nng pht trin t duy cho hc sinh. 2.1 Nhng kin thc c bn v vct trong chng trnh HH10-SGK nng cao A- vct v cc php ton vct. 1. Vct v l mt on thng c hng trong ch r im u v im cui. VctABc im u l A, im cui l B c hng t A n B, c dil dionthngAB,ck hiulAB ,vcgi l ngthng AB. Ngi ta cn k hiu vct bng cc ch thng nha ,b ,x ,y .... 2. Hai vcta ,bc gi l cng phng nu gi ca chng song song hoctrngnhau.Nuhaivctcngphngthchngcthcnghng hoc ngc hng. Hai vctavbc gi l bng nhau, k hiu la =bnu chng cng hng v c di bng nhau. 3. Vi mi im A, ta giAA l vct khng. Vct khng c k hiu lO. Ta qui c vctO cng phng, cng hng vi bt k vct no v O =O. 4. Cho hai vct avb . Ly mt im A ty , vAB =a , BC =b . Khi vctAC c gi l tng ca hai vcta vb . Php ton tm tng ca hai vct c gi l php cng hai vct. www.VNMATH.com S ha bi Trung tm Hc liu i hc Thi Nguynhttp://www.lrc-tnu.edu.vn 35 5.Chohaivcta vb .Tagihiucahaivcta vb lvct ) ( b a + ckhiula - b .Phptontmhiucahaivcta vb cn c gi l php tr hai vctavb . 6. Tch ca vcta O =vi s k = O l mt vct k hiu l k acng hng vianu k > O, ngc hng vianu k < O v c di bng a k . . Ta qui c O.a =O,k. O=O. 7. Phn tch mt vct theo hai vct khng cng phng. a)nhngha:cho2vcta vb khngcngphng.Nuvctcc vit di dngb k a h c + =vi h, k l s thc no th ta ni rng vct cphn tch c theo2 vctavbkhng cng phng hoc vctcbiu th c qua hai vctavbkhng cng phng. b) nh l:Cho hai vctavbkhng cng phng. Khi mi vct x ucthphntchc(hocbiuthc)mtcchduynhtqua2 vct avb , ngha l c duy nht cp s h, k sao chob k a h x + = . 8. Cc quy tc cn nh khi thc hin thc cc php ton v vcta) Quy tc hnh bnh hnh: Nu ABCD l hnh bnh hnh th: AC AD AB = +b) Qui tc ba im: * BC AB AC + =(qui tc ba im i vi php cng vct) * CA CB AB =(qui tc v hiu vct) Vn dng qui tc ny c th biu th mt vct bt k thnh hiu ca hai vct c chung im u. B C AD www.VNMATH.com S ha bi Trung tm Hc liu i hc Thi Nguynhttp://www.lrc-tnu.edu.vn 36 O A B B B-Tch v hng9.nhngha:chohaivcta vb ukhcvctO.Tchvhng cahaivcta vb lmts,khiula . b cxcnhbicngthc ) , cos( . . . b a b a b a = . .Trng hp t nht mt trong hai vctavb bng vctO ta qui c a . b =O. . Nuavbu khc vctO ta ca .b =Ob a . . Khia =b ta ca . 2a a = l bnh phng v hng ca vcta . Ta c 22a a = . 10. Cng thc hnh chiu. ChohaivctOB OA, .GisBlhnhchiucaBtrnngthng OA.Tagivct' OB l hnhchiuca vctOB trn ngthngOA.Khi , ta c cng thc hnh chiu sau y:' . . OB OA OB OA = C-Ta ca im v ta ca vct trn mt phng. 11. Ta ca vct v ca im. Trong mt phng Oxy cho mt vctaty . Nuj y i x a + =th cp s (x;y) c gi l ta ca vctai vi h ta Oxy, k hiu la =(x;y) hay la (x;y). B OA B O A A1ijaA2 y x www.VNMATH.com S ha bi Trung tm Hc liu i hc Thi Nguynhttp://www.lrc-tnu.edu.vn 37 Ta ca im M l ta ca vctOM . Vi 2 im M(xM;yM) v N(xN;yN) thMN =(xN-xM;yN-yM) 12. Nu I l trung im ca on thng AB th: xI=2A Bx x +;yI=2B Ay y+ Nu G l trng tm ca tam gic ABC th: xG=3C B Ax x x + + ;yG=3C B Ay y y + + 13.KhongcchgiahaiimA(xA;yA),B(xB;yB)ctnhtheo cng thc: AB = 2 2) ( ) (A B A By y x x AB + =14. Gc gia hai vct. Cho hai vcta =(x;y) v) ' ; ' ( ' y x a = khcO

Cos( ' ; a a )=2 2 2 2' ' .' '' .' .y x y xyy xxa aa a+ ++= gii cc bi ton hnh hc bng PPVT, hc sinh cn nm vng nhng kinthccbntrn,bitvndnglinhhotvomibitoncth.Bit kt hp gia k nng tnh ton vi k nng bin i cc ng thc vct v cc kinthcvhnhhc,mihcsinhcncrnluynkhnngtmra ng li gii cho mi bi ton hnh hc bng PPVT s nu ra trong h thng bi tp sau y. 2.2Quy trnh bn bc gii bi ton hnh hc bng PPVT. lp10,hcsinhchcvvct,ccphptontrnvct(php cng,phptr,phpnhnvctvisthc,tchvhngcahaivct), saultrc,htrcta,tacaim,tacavctvmtvi ngdngngincaphngphpta.Tuyhcsinhchcchai phngphp:vctvta,phngphpchyuvnlphngphp vct. Bi v, cc h thc lng trong tam gic v trong ng trn oc xy www.VNMATH.com S ha bi Trung tm Hc liu i hc Thi Nguynhttp://www.lrc-tnu.edu.vn 38 dng nh vct cng cc php ton, c bit l tch v hng ca hai vct cnhnghatheomtngthcvct...giphcsinhsdng thnhthoPPVTgiccbiton,ivihcsinhlp10,trcht gio vin cn rn luyn cho hc sinh nm vng quy trnh bn bc gii bi ton bng PPVT. Quy trnh bn bc gii bi ton hnh hc bng PPVT. Bc 1: Chn cc vct c s. Bc2:Dngphngphpphntchvctvccphptonvct biu din, chuyn ngn ng t hnh hc thng thng sang ngn ng vct. Bc 3: Gii bi ton vct. Bc 4: Kt lun, nh gi kt qu. Giovincntndngccchirnluynchohcsinhkhnng thc hin 4 bc gii bi ton hnh hc bng PPVT thng qua cc bi tp, c th minh ha quy trnh 4 bc trn bng v d sau: Bi ton: Cho gc xOy v hai im di chuyn trn hai cnh ca gc. M thuc Ox, N thuc Oy, lun lun tha mn OM=2ON. Chng minh rng trung im I ca MN lun thuc mt ng thng c nh. Hng dn gii: Bc1: Ly im AeOx, BeOy sao cho OA=OB, v chn hai vct OB OA,lm hai vct c s. Mi vct trong bi ton u phn tch c c (hoc biu th c) qua hai vct ny. Bc 2: Gi thit cho OM=2ON, nn nuOB k ON = , thOA k OM 2 = . iu phi chng minh l I thuc mt ng thng c nh (d thy ng thng ny i qua O) tng ngv p OI = , vivl mt vct c nh no . Bc 3: Do I l trung im ca MN, nn ta c) 2 (21) (21OB OA k ON OM OI + = + =www.VNMATH.com S ha bi Trung tm Hc liu i hc Thi Nguynhttp://www.lrc-tnu.edu.vn 39 tv OB OA p k = + = 2 ,21, ta c i phi chng minh Bc 4: Nhn xt: NulyOA OA 2 ' = th ' v OA OB = + ngthngcnhi qua trung im AB. *Cth tngqut habiton theo 2 cch: -Thay cho gii thit OM=2ON bng OM= m.ON (m l mt hng s) -Thay cho kt lun: trung im I ca MN thuc mt ng thng c nh bng kt lun: mi im chia MN theo t s qpINIM=(p, q l hng s dng) u thuc mt ng thng c nh. Trong qu trnh hng dn hc sinh gii bi ton bng PPVT, gio vin cn ch n nhng tri thc phng php: bc1:Nnchnccvctcssaochoccvcttrongbiton phn tch theochngthun linht.Qua mibi ton hc sinhsthyvic chn cc vct c s nh th no. bc2:Cnrnluynchohcsinhchuyningnngmtcch thnhtho.Cchchuyninhthnotacththyquatngnhmbi ton s c trnh by i y. bc 3: Cn nm vng cc php ton vct. ng thi, thng qua cc bi tp c th, gio vin cn lm cho hc sinh hiu r c tnh u vit ca PPVT. c bit cc bi tp v tm tp hp im, ccbitp vchngminh3 imthng hng,chngminhhaing thng song song, hai ng thng vung gc,...l nhng dng ton c nhiu c hi lm r vn ny. O A A M B I N x y www.VNMATH.com S ha bi Trung tm Hc liu i hc Thi Nguynhttp://www.lrc-tnu.edu.vn 40 Tuynhin,khngphilcnocnglmtheo4bcnhtrn,khng phi lc no cng phn tch cc vct theo hai vct c s cho trc,m c th gii quyt bi ton mt cch linh hot. Theo chng ti thy, vic rn luyn cho hc sinh thng qua mt h thng bi tp c phn loi s em li hiu qu cao trong dy hc. 2.3 H thng bi tp2.3.1 Nhng kin thc b tr xy dng h thng bi tp. KinthctrongSGKarachalcctrithcphngphpy cho hc sinh. V tri thc phng php khng phi l tri thc tng minh di dng l thuyt(nh ngha, nh l,...)m cn c th hin di dng bi tp. Vytrongqutrnhgingdygiovincnphinhnmnhccbitpc bntrongSGKhocphibsungthmccbitp(vylcctrithc phng php gii cc bi tp sau ny). A - iu kin cn v d hai vct khng cng phng Bi ton 1: ( Bi 12- trang17 - SBT-HH10- nng cao) Chng minh rng hai vctavbcng phng khi v ch khi c cp s m, n khng ng thi bng 0 sao cho0 = + b n a m Hy pht biu diu kin cn v hai vct khng cng phng Gii:Nucma nb O + = vim= O,tacbmna = suyraa vbcng phng. Ngc li, gi s avb cng phng. . Nua =O th c th vit. ma Ob O + = vi m= O. .Nua = Othcsmsaochoa m b = tclma nb O + = ,trong n = -1 = O. *Vyiukincnva vb cngphnglccpsm,n khng ng thi bng khng sao choma nb O + = . www.VNMATH.com S ha bi Trung tm Hc liu i hc Thi Nguynhttp://www.lrc-tnu.edu.vn 41 T suy ra: iu kin cn v 2 vct avb khng cng phng l numa nb O + = thm = n = O. B-Tmtccahim{A1,A2,......An}ngvicchs {no o o ,...... ,2 1} (n>2) Bi ton 2: Cho hai im A, B phn bit v hai s| o,khng ng thi bng khng. Chng minh rng: a) NuO o | + =th khng tn ti im M sao choMA MB O o | + = . b) NuO o | + = th tn ti duy nht im M sao choMA MB O o | + = . Gii: a) Gi sO o | + =m c im M sao choMA MB O o | + = . MA MB O o o = ( ) . MA MB O BA O o o = = VBA O = nnO O o | = = : mu thun. Vy khng tn ti im M. b) Gi s O o | + = , ta cMA MB O o | + = ( )( )AM AB AM OAM AB AM ABo ||o | |o | + = + = =+ ng thc cui cng chng t s tn ti v duy nht ca im M, ng thi ch ra cch dng im M. Biton3:ChohaiimA,Bv2sthc| o, .Chngminh:nu O o | + = thvctMB MA v | o + = khngi,khngphthucvovtr im M. Gii: MB MA v | o + = = BA MB MA MB MA o o o o = = ) (l 1 vct khng i. Biton4:ChotamgicABCv3s | o , , khngngthibng khng. Chng minh rng: a) NuO o | + + = th tn ti duy nht im I sao choIA IB IC O o | + + = www.VNMATH.com S ha bi Trung tm Hc liu i hc Thi Nguynhttp://www.lrc-tnu.edu.vn 42 b) NuO o | + + =th khng tn ti im M sao cho:MA MB MC O o | + + = Gii: a) VO o | + + = ( ) ( ) ( ) O o | | o + + + + + =nn 1 trong 3 s: ) ( ), ( ), ( o | | o + + + khc khng. Chng hn( ) O o | + =theo bi ton 3b, tn ti im E sao cho: EA EB O o | + = khi : IA IB IC O o | + + = ( ) ( ) IE EA IE EB IC O o | + + + + = ( ) ( ) IE EA EB IC O o | o | + + + + = ( )IE IC O o | + + = (*) V( ) O o | + + = nn tn ti duy nht im I tha mn (*) b) Gi s tn ti im M tha mn ng thc cho v gi s, chng hn O o = .Ta c: MA MB MC O o | + + = ( ) MA MB MC O o | o | + + = ( ) ( ) MA MC MB MC O o | + = CA CB O CA CB|o |o + = = CA song songCB(mu thun). Vy khng tn ti im M. Nhn xt: Trong trng hpO o | + + = , vi im M ty ta c: = + + MC MB MA | o ) ( ) ( ) ( IC MI IB MI IA MI + + + + + | o= ) ( ) ( IC IB IA MI | o | o + + + + +=( MI ) | o + +Bng phng php quy np, ta c th chng minh c kt qu tng qut: www.VNMATH.com S ha bi Trung tm Hc liu i hc Thi Nguynhttp://www.lrc-tnu.edu.vn 43 - Cho n im A1,A2,......An v n s thc no o o ,...... ,2 1 sao cho:on = + + + o o o ......2 1. Khi tn ti duy nht im I sao cho: 1 1 2 2.........n nIA IA IA O o o o + + + = (1). imIgiltmtccahim{A1,A2,......An}ngvicchs {no o o ,...... ,2 1} (n>2). T (1), vi im M ty ta c: ( .........2 2 1 1= + + +n nMA MA MA o o o MIn) ......2 1o o o + + +Cng thc ny thng xuyn c s dng trong nhng bi ton c lin quan ti tm t c. Ta gi n l cng thc thu gn. Vin=3v13 2 1= = = o o o ,tathyyltnhchttrngtmcatam gic c trnh by di y. C- Tnh cht ca trung im. Bi ton 5: Nu M l trung im ca on thng AB thMA MB O + = Gii: Theo quy tc 3 im, ta cMA AM MM O + = = . Mt khc, v M l trung im ca AB nnMB AM = . VyMA MB O + = . Nhnxt:tnhchttrnltrnghpcbitcabiton2a,khi 1 = = | o . Bi ton 6: Chng minh rng I l trung im ca on thng AB khi v ch khi vi im M bt k, ta cMI MB MA 2 = +Gii: Vi im M bt k ta c:

IB MI MBIA MI MA+ =+ = Nh vyIB IA MI MB MA + + = + 2 .Ta bit rng I l trung im ca AB khi v ch khiIA IB O + = . Suy ra iu phi chng minh. M A B I www.VNMATH.com S ha bi Trung tm Hc liu i hc Thi Nguynhttp://www.lrc-tnu.edu.vn 44 Tnh cht trng tm ca tam gic Biton7:ChotamgicABC.ChngminhimGltrngtmca tam gic ABC khi v ch khiGA GB GC O + + = . Gii: Gi M l trung im cnh BC, ta c: GA GB GC O + + = . 2 GA GM O + = G thuc on AM v GA=2GM. G l trng tm ca tam gic ABC. Biton8:ChotamgicABCctrngtmG.Chngminhrngvi im M bt k, ta c:MG MC MB MA 3 = + + . Gii:. = + + MC MB MA G M +GA MG GB MG GC + + + + . =( ) 3 GA GB GC MG + + + = 3 3 O MG MG + = . ( V G l trng tm ca tam gic ABC GA GB GC O + + = .) D- iu kin cn v 3 im phn bit A, B, C thng hng. Bi ton 9: (Bi 15- tr7 -SBT-HH10- nng cao) Cho 3 im ABC. a)ChngminhrngnucmtimIvmtstnosaocho IC t IB t IA ) 1 ( + =th vi mi im I ta c:C I t B I t A I ' ) 1 ( ' ' + =b) Chng t rngIC t IB t IA ) 1 ( + = l iu kin cn v 3 im A, B, C thng hng. Gii:a) Theo gi thitIC t IB t IA ) 1 ( + = , th vi mi im I ta c' ' ) 1 ( ' ) ' ' )( 1 ( ) ' ' ( ' ' II C I t B I t C I II t B I II t A I II + + = + + + = +Suy raC I t B I t A I ' ) 1 ( ' ' + = . A B CM G A B C G M www.VNMATH.com S ha bi Trung tm Hc liu i hc Thi Nguynhttp://www.lrc-tnu.edu.vn 45 b)NutachnItrngviAthc(1 ) O t AB t AC = + ,liukin cn v 3 im A, B, C thng hng. E- Cng thc im chia. Bi ton 10: Choon thng AB, s thc k khc Ov 1. Ta ni M chia on AB theo t s k nuMB k MA = . Chng minh rng vi im C bt k ta c:CBkkCAkCM=1 11(*) Ta gi (*) l cng thc im chia. Gii: Ta cMB k MA = CM k CB k CM = CA CB k CA CM k = ) 1 ( CBkkCAkCM=1 11 F- Cng thc hnh chiu. Chohaivct. , OB OA GiBlhnhchiucaBtrnngthng OA.Chng minh rng:' . OB OA OB OA =Gii:Trng hp 1: NuB O A< 90o Th = OB OA. OA.OB.cosAOB = AO.OB = AO.OB.cosOo

=' .OB OATrng hp 2: Nu AOB > 900 Th= OB OA.OA.OB.cosAOB = -OA.OB.cosBOB = - OA.OB=OA.OB.cos1800 =' .OB OAO B BA B BOA www.VNMATH.com S ha bi Trung tm Hc liu i hc Thi Nguynhttp://www.lrc-tnu.edu.vn 46 Vct' OB gi lhnh chiu ca vc tOB trnngthngOA.Cng thc= OB OA. ' .OB OAgi l cng thc hnh chiu.2.3.2 Nhng dng s phm khi xy dng h thng bi tp * H thng bi tp di y c xy dng theo cu trc nh sau: -Bc1: a ra tri thc phng php cho mi dng bi tp. - Bc 2: a ra v d, v hng dn HS thc hin 4 bc theo phng php tm li gii bi ton ca Plya hoc theo 4 bc gii bi tp HH bng PPVT. -Bc 3: a ra h thng bi tp cho mi dng bi tp. -Bc 4: a ra li gii hoc ch dn cho h thng bi tp trn. *VicarahthngbitpphndngnhmgipHSckinh nghim gii ton v rn luyn cc k nng: - Chuyn bi ton sang ngn ng vc t.- Phn tch 1 vc t thnh mt t hp vc t. - K nng bit cch ghp 1 s vct trong 1 t hp vct. -Bit khi qutha 1s nhng kt qu vn dng vobi tontng qut hn.cbitbitvndngquytrnh4bcgiibitonhnhhcbng PPVT vo gii cc bi tp HH. *Giovincthsdnghthngbitpphndngnytrongcc tnh hung dy hc khc nhau nh: lm bi tp v nh, bi tp phn ha, dng bi dng HS kh gii, dng lm kim tra, kim tra trc nghimgp phn bi dng nng lc gii ton cho HS. 2.3.3 Chng minh 3 im thng hngi vi dng ton trn ta c th dng iu kin cng phng ca 2 vct gii ton.Vct bcng phng vi vcta( a = O) khi v ch khi c s k sao chob =kawww.VNMATH.com S ha bi Trung tm Hc liu i hc Thi Nguynhttp://www.lrc-tnu.edu.vn 47 *T ng dng vo dng ton: Cho 3 im A,B,C tha mn 1iu kin xc nh, chng minh rng A, B, C thng hng. Phng php: - Hy xc nh vctAC AB,- Ch ra rng 2 vct cng phng, ngha l hy ch ra s thc k sao choAC k AB =V d 1: (Bi 19- tr8- SBT-HH 10 nng cao) Cho tam gic ABC. Cc im M, N, P ln lt chia cc on thng AB, BC, CA theo cc t s ln lt l m, n, p (u khc 1) Chngminhrng:M,N,Pthnghngkhivchkhimnp=1(nhl Mnlaut) Hng dn gii: (Theo quy trnh 4 bc gii bi ton HH bng PPVT) Bc 1: GV Chn vct c s. HS: Chon hai vctCB CA, lm 2 vct c s. Mi vct xut hin trong bi ton u phn tch c theo 2 vct ny.Bc 2:GV: Cc im M,N,P ln lt chia ccon thng AB,BC,CA theo cc t sln lt l m, n, p (u khc 1) tng ng vi cc ng thc vct no ? HS:MB m MA = ;NC n NB = ;PA p PC =GV:iuphichngminhM,N,Pthnghngtngngving thc vct no phi xy ra ? HS: - ch ra s thc k sao choMN k MP =hoc - Vi im O bt k v t s thc t ta cOP t ON t OM ) 1 ( + =A B C P N M www.VNMATH.com S ha bi Trung tm Hc liu i hc Thi Nguynhttp://www.lrc-tnu.edu.vn 48 Bc 3: Ly im O no , ta cmOB m OAOM=1;nOC n OBON=1;pOA p OCOP=1 n gin tnh ton, ta chn im O trng vi im C khi ta c:mCB m CACM=1; nCBCN=1;pCA pCP=1 (1)T hai ng thc cui ca (1), ta cCN n CB ) 1 ( = ;CPppCA1 =V thay vo ng thc u ca (1) ta c) 1 (1m ppCM= CNmn mCP1) 1 ( T bi ton 9- iu kin cn v 3 im M, N, P thng hng l: ) 1 ( ) 1 ( 1 11) 1 () 1 (1m p n pm pnn mm pp = = mnp = 1 Bc4:VychotamgicABC.CcimM,N,Plnltchiacc on thng AB, BC, CA theo t s m, n, p th M, N, P thng hng khi v ch khi: mnp=1 Vd2:TrnngthngachoccimA1,B1,C1vtrnng thng b cho cc im A2, B2, C2 tha mn: 1 1 1 1C A k B A = ; 2 2 2 2C A k B A = (k 1 = )Gi s cc im Ao, Bo, Co trn A1A2 , B1B2, C1C2 sao cho 2 1 0 1A A l A A = ;

2 1 0 1B B l B B = . 2 1 0 1C C l C C =Chng minh 3 im Ao, Bo, Co thng hng.www.VNMATH.com S ha bi Trung tm Hc liu i hc Thi Nguynhttp://www.lrc-tnu.edu.vn 49 A1 A2 C1 C2 B1 B2 Ao Bo Co Hng dn gii:- Theo gi thit ta c: 0 1 2 1 1 2( )o oA A l A A l A A A A = = = l2 1A A l A Ao o2 1) 1 ( A A l A A lo o = 2 1) 1 ( A A l A A lo o= Tng t) 1 ( l2 1 0B B l B Bo=) 1 ( l2 1 0C C l C Co=.

==1 lo ld c Ao, Bo, Co thng hng. .Vi l0 = , l 1 =ta c + + =+ + =0 2 2 2 2 0 0 00 1 1 1 1 0 0 0B B B A A A B AB B B A A A B A

+ + = + + = 0 2 2 2 21 1 1 1) 1 ( ) 1 ( ) 1 ( ) 1 (B B l B A l A A l B A lBo B l B A l A A l B A lo o oo o o ( ) ( )o o o o o oB B l B B l B A l B A l A A l A A l B A2 1 2 2 1 1 2 1) 1 ( ) 1 ( ) 1 ( + + + + + = =( ) 0 ) 1 ( 02 2 1 1+ + + B A l B A l =| |2 2 1 1 2 2 1 1) 1 ( ) 1 ( C A l C A l k B A l B A l + = + Tng t: 2 2 1 1) 1 ( C A l C A l C Ao o+ = o o o oC A k B A = .Vy 3 im Ao, Bo, Co thng hng. www.VNMATH.com S ha bi Trung tm Hc liu i hc Thi Nguynhttp://www.lrc-tnu.edu.vn 50 Lu : Vi l =21 th Ao, Bo,Co ln lt l trung im ca A1A2 , B1B2, C1C2, lc ny hc sinh d dng chng minh c bi 36-tr11-SBT-HH10-nng cao: ChotgicABCD.Viskty,lyccimMvNsaocho AB k AM = vDC k DN = .TmtphpcctrungimIcaonthngMN khi k thay i. Hoc:Cho t gic ABCD. Trn cc cnh AB, CD ta ly cc im tng ng M, N sao cho DCDNABAM= . Chng minh rng trung im ca 3 on thng AD, BC, MN thng hng. V d 3:(Bi ton 3-tr21-SGK HH10-nng cao) Cho tam gic ABC c trc tm H, trng tm G v tm ng trn ngoi tip tm O. Chng minh 3 im O, G, H thng hng. Hng dn gii:Gi I l trung im ca BC. D thyOI AH 2 =nu tam gic ABC vung. Nu tam gic ABC khng vung, gi D l im i xng ca A qua O. Khi :BH song song DC( v cng vung gc vi AC) BD song song CH ( V cng vung gc vi AB) SuyraBDCHlhnhbnhhnh,doIl trungim ca HD. T OI AH 2 =* Ta c:AH OI OC OB = = + 2 nnOH AH OA OC OB OA = + = + +* Ta bitOG OC OB OA 3 = + +VyOG OH 3 = suyra3imO,G,Hthng hng(ngthngiqua3imnygil ng thng le ca tam gic ABC). A B C D O I H G www.VNMATH.com S ha bi Trung tm Hc liu i hc Thi Nguynhttp://www.lrc-tnu.edu.vn 51 Lu : Hc sinh phi c th vn dng cch chng minh bi ton trn vo gii cc bi ton sau: 1/ (Bi 38 - tr11-SBT- HH10 - nng cao ). Cho tam gic ABC c trc tm H v tmng trn trn ngoi tip O. Chng minh rng:a/OH OC OB OA = + +b/OH HC HB HA 2 = + +2/ (Bi tp 39- tr11 SBT - HH10 - nng cao ) Cho 3 dy cung song song AA1, BB1, CC1 ca hnh trn (O) chng minh rng trc tm ca 3 tam gic ABC1, BCA1 v ACB1 nm trn mt ng thng.Hng dn gii:GiH1,H2,H3lnltltrctmca tam gicABC1,BCA1vACB1theoktquv d 3, ta c:11OC OB OA OH + + =1 2OA OC OB OH + + =13OB OA OC OH + + =Suy ra: 1 1 1 1 1 2 2 1AA C C OA OA OC OC OH OH H H + = + = = 1 1 1 1 1 3 3 1BB C C OB OB OC OC OH OH H H + = + = =VccdycungAA1,BB1,CC1 songsongvinhaunn3vct 1 1 1, , CC BB AAcng phng. Do 2 vct 2 1H H v 3 1H H cng phng, hay 3 im H1, H2, H3 thng hng.V d 4: Cho tam gic ABC ng trn tm I ni tip tam gic ABC tip xcvicnhBCtiD.GiM,NlnltltrungimcaADvBC. Chng minh 3 im M, N, I thng hng. A B A1 B1 C1 C www.VNMATH.com S ha bi Trung tm Hc liu i hc Thi Nguynhttp://www.lrc-tnu.edu.vn 52 Hng dn gii: Ta c: ) (21IC IB IN + =) (21ID IA IM + =Ta c: o IC c IB b IA a = + +(*) ) ( ICacIBabIA + = (1) Mt khc: DB=P-b; DC=p-c. y p l na chu vi tam gic ABC. DCc pb pDBc pb pDCDB = = aIC b p IB c pc pb pICc pb pIBID) ( ) (1 + =++= (2) T (2) v (3) ta c: ) () ( ) ( ) (IC IBac b paIC c b p IB c b pID IA + = + = + INac b pIC IBac b pIM = + = ) )( (21 3 im A, B, C thng hng. Chng minh trn c sdng ng thc (*) l kt qu ca bi tp sau: Bi 37b-tr11-SBT HH10-nng cao Cho tam gicABC vi cccnh AB=c, BC=a,CA=b.GiIltmngtrnni tiptamgicABC.Chngminhrng o IC c IB b IA a = + + B c N A b C M I D a A B C M I www.VNMATH.com S ha bi Trung tm Hc liu i hc Thi Nguynhttp://www.lrc-tnu.edu.vn 53 Chng minh: .Gi CM l phn gic trong ca gc C. .V I l tm ng trn ni tip tam gicABC nn AI l phn gic ca tam gicACM. . Theo tnh cht ng phn gic ta c: ICACAMIMACAMICIM = =T ta c: ACb abcbb abcABb abb abcbbICAM ACAMAMAM ACACACAMICACAMAMAI++ ++++=+++=++=1 00 ) 1 ( :) ( ) (= + + =+ +++ +++ +++ ++ + +=+ +++ +=IC c IB b IA aICc b acIBc b abIAc b ac bSuyraIA ICc b acIA IBc b abACc b acABc b ab *H thng bi tp. Bi 1: (Bi 26- SBT HH10-Nng cao) ChoimOcnhvngthngdiquahaiimA,Bcnh. ChngminhrngimMthucngthngdkhivchkhicso sao cho:OB OA OM ) 1 ( o o + =Vi iu kin no ca oth M thuc on thng AB. Bi 2. Trn cc cnh ca tam gic ABC, ly cc im M, N, P sao cho: 3 6 2 MA MB NB NC PC PA O + = = + = . Hy biu thANquaAMvAP , t suy ra M, N, P thng hng. www.VNMATH.com S ha bi Trung tm Hc liu i hc Thi Nguynhttp://www.lrc-tnu.edu.vn 54 Bi 3. Cho tam gic ABC, gi D, I, N l cc im xc nh bi cc h thc: CN CI NB AN DC DB 2 , 3 , 0 2 3 = = = . Chng minh A, I, D thng hng. Bi 4. (Bi 20a-tr8-SBT HH10-Nng cao) Cho tam gic ABC, v cc im A1, B1, C1 ln lt nm trn cc ng thng BC, CA, AB. Gi A2, B2, C2 ln lut l cc dim i xng vi A1, B1, C1 qua trung im ca BC, CA, AB. Chng minh rng; a) Nu 3 im A1, B1, C1 thng hng th 3 im A2, B2, C2 cng th. b) Trng tm ca 3 tam gic ABC, A1B1C1, A2B2C2 thng hng. Bi 5. Cho tam gic ABC u, tm O. M bt k trong tam gic ABC v c hnh chiu xung 3 cnh BC, CA, AB tng ng l P, Q, R. Gi K l trng tm tam gic PQR. a) Chng minh: M, O, K thng hng. b) Cho N l mt dim ty trn BC. H NE, NF tng ng vunggc vi AC, AC. Chng minh N, J, O thng hng, vi J l trung im ca EF. Bi 6. Cho tam gic ABC, G l trng tm ca tam gic ABC. Qua im M ty trn mt phng tam gic ABC dng cc ng thng song song vi GA, GB, GC, chng tng ng ct BC, CA, AB ti A1, B1, C1. Chng minh M, G, G1 thng hng, vi G1 l trng tm tam gic A1B1C1. C nhn xt g v im G1? Bi 7.(Bi 28-tr24-SGK HH10-nng cao) Cho t gic ABCD. Chng minh rng: a) C mt im G duy nht sao cho 0 = + + + GD GC GB GA . im G nh th gi l trng tm ca 4 im A, B, C, D. Tuy nhin, ngi ta vn quen gi G l trng tm ca t gic ABCD. b)TrngtmGltrungimcamionthngnicctrungimhai cnh i ca t gic, n cng l trung im ca on thng ni trung im hai ng cho ca t gic. www.VNMATH.com S ha bi Trung tm Hc liu i hc Thi Nguynhttp://www.lrc-tnu.edu.vn 55 c)TrngtmGnmtrncconthngnimtnhcatgicv trng tm ca tam gic to bi 3 nh cn li. Bi8.ChotgicABCD.HaiimM,NthayitrncccnhAB, CDsaocho.CDCNABAM= GiP,Qlnltltrungimcahaingcho AC, BD, I l trung im ca MN. Chng minh 3 im P, I, Q thng hng. Bi9:ChotgicABCDngoitipngtrntmI.GiE,Fln ltltrungimcaccngchoAC,BD.ChngminhrngI,E,F thng hng. Hng dn hoc li gii Bi 1 Ta cOB OB OA OM OB OA OM + = + = ) ( ) 1 ( o o o d M BA BM OB OA OB OM e = = o o ) (VBA BM o =nn M thuc on thng AB khi v ch khi1 0 s soBi 2. T gi thitNB NC 6 = 6 3 85 5 585AC ABAN AP AMPN PM = = + = Suy ra M, N, P thng hng. Bi 3. Cch 1.Ta c: ID IC IBDC DBIC IN IB IANB AN= = = = + =2 30 2 32 4 33 Vy = + 0 ID IA A, I, D thng hng v I l trung im ca AD. A BC N M P A B CD I N www.VNMATH.com S ha bi Trung tm Hc liu i hc Thi Nguynhttp://www.lrc-tnu.edu.vn 56 Cch 2T 433 3CB CACN NB NA NB AN+= = =(*) Mt khc:1 1, 3 2 02 3CN CI DB DC CB CD = = = Thay vo (*), ta c:02= + += ID IACD CACIVy I, A, D thng hng, I l trung im ca AD. Bi 4 a) Ta gi k, l, m l cc s sao choB C m A C A B l C B C A k B A1 1 1 1 1 1; ; = = =BCkkBA C A k B A= =11 1 1 BAmBC B C m A C= =111 1 1 lBA l BCBB A B l C B= =11 1 1 1 1 1) 1 (1111BC mllBAkklBB = A1, B1, C1 thng hng1 . . 1 ) 1 (1111= = m l k mllkkl Ba imA1, B1, C1 ln lt i xng vi 3 imA2, B2, C2 qua trung im on thng BC, CA, AB nn ta c: 2 2 2 2 2 2; ; A C k A B B A l B C C B mC A = = = Nu 3 imA1, B1, C1 thng hng th 3 im A2, B2, C2 cng thng hng v ngc li. b) Gi M, N, E ln lt l trung im ca BC, CA, AB G, G1, G2 ln lt l trng tm ca tamgic ABC, tam gic A1B1C1, tam gic A2B2C2. Ta c 31 1 1 1GC GB GA GG + + =32 2 2 2GC GB GA GG + + =www.VNMATH.com S ha bi Trung tm Hc liu i hc Thi Nguynhttp://www.lrc-tnu.edu.vn 57 ) ( ) ( ) ( ) ( 32 1 2 1 2 1 2 1GC GC GB GB GA GA GG GG + + + + + = + =2. 0 ) ( = + + GE GN GM2 1GG GG = . Vy G, G1, G2 thng hng, G l trung im ca G1G2. Nhn xt: C th dng kt qu nh l Mnelayt ( c chng minh v d 1 (bi 19- tr8-SBT HH10-nng cao) kim tra kt qu ca bi 2, v bi 4a. Bi 5a)Qua M k A1B2 song song AB, A1eBC, B2 eAC k B1C2 song song BC, B1eAC, C2 eAB k C1A2 song song AC, C1eAB, A2 eBC tam gic MB1B2, tam gic MC1C2, tam gic MA1A2 u. | |2 1 2 1 2 121MC MC MB MB MA MA MR MQ MP + + + + + = + +=| | | | | |1 2 1 2 2 1212121MB MA MA MC MB MC + + + + +=| | MC MB MA + +21 =MO23 Vy:| | MO MR MQ MP MK2131= + + = M, O, K thng hngb) K NP // AC (P eAB) NQ // AB (Q eAC ) T gic APNQ l hnh bnh hnhTa c:) (21NF NE NJ + = = 1( )4NB NP NQ NC + + + A BCP Q R A1 A2 B1 B2 C1 C2 M B A Q P E N F J O C www.VNMATH.com S ha bi Trung tm Hc liu i hc Thi Nguynhttp://www.lrc-tnu.edu.vn 58 =) (41NC NB NA + +=NO NO43. 3 .31=Vy N, J, O thng hngBi 6: Qua M k A2B3 // AB; A2eBC; B3 eAC K B2C3 // BC; B2 eAC; C3 eAB K A3C2 // AC; A3 eBC; C2 e ABKhi : A MA2A3 ~A C2C3M ~A B3MB2 ~A ABC VvyMA1,MB1,MC1lnltl ng trung tuynca 3 2 3 2 3 2, , C MC B MB A MA A A A . Ta c: ) (311 1 1 1MC MB MA MG + + =( ) ( ) ( )((

+ + + + + =3 2 3 2 3 221212131MC MC MB MB MA MA MG MGMG MC MB MA2121) (611 = = + + = M, G, G1 thng hng v G1 l trung im ca GM. Nhn xt: cho tam gic ABC u ta c kt qu bi 5a. Bi 8 Theo gi thit ta c) 1 0 ( ; ; s s = = k CD k CN AB k AMGi P v Q ln lt l trung im ca AC v BD. Ta c) (21) (21CD AB k CN AM PI + = + = (1) ) (21CD AB PQ + =(2) B C B1 B2 A C3 C1 C2 G A1 A3 A2 B3 M www.VNMATH.com S ha bi Trung tm Hc liu i hc Thi Nguynhttp://www.lrc-tnu.edu.vn 59 T (1) v (2)PQ k PI = hay P, I, Q thnghng.V1 0 s s k nnIthuc on PQ. Nhn xt: Cho k=21, ta c kt qu bi 7b. Bi 9 Gi M, N, P, Q ln lt l cc tip imca (I) vi cc cnh AB, BC, CD, DA;x, y, z, t l cc khong cch t A, B, C, D n cc tip im tng ng. t IMIMAB a =1; ININBC a =2

IPIPCD a =3; IQIQDA a =4 Ta c ( ) )( ( ) (4 3 2 4 3 2 1CB DC AD a a a AB a a a a + + + + = + + +) ; cos( . . ) ; cos( . .) ; cos( . . ) ; cos( . . ) ; cos( . . ) ; cos( . .) ( ) ( ) (4 4 3 33 3 2 2 4 4 2 24 3 3 2 4 24 4 3 3 2 2DC a DC a AD a AD aCB a CB a DC a DC a CB a CB a AD a AD aDC a AD a CB a DC a CB a AD aCB a DC a CB a AD a DC a AD a+ ++ + + + =+ + + + + =+ + + + + =Theo gi thitDA a CD a BC a AB a = = = =4 3 2 1; ; ;Ngoi ra d thy:) ; cos( ) ; cos() ; cos( ) ; cos() ; cos( ) ; cos(4 33 24 2DC a AD aCB a DC aCB a AD a = = = 0 ). (4 3 2 1= + + + AB a a a aChng minh tng t ta c:0 ). (4 3 2 1= + + + BC a a a aA B C D F E I M N P Q 1a2a3a4aA B C D Q P I M N www.VNMATH.com S ha bi Trung tm Hc liu i hc Thi Nguynhttp://www.lrc-tnu.edu.vn 60 0 ) (4 3 2 1= + + + a a a a0 . . . . = + + + IQ DA IP CD IN BC IM AB (vIM=IN=IP=IQ) 0 ) ( ) ( ) ( ) ( = + + + + + + + IQ x t IP t z IN z y IM y x0 ) )( ( ) )( (0 ) ( ) ( ) ( ) (= + + + + + = + + + + + + + ID IB z x IC IA t yIA t ID x ID z IC t IC y IB z IB x IA y = + + + 0 ) ( ) ( IF z x IE t y I, E, F thng hng. 2.3.4 Chng minh hai ng thng vung gc Vn dng cc kin thc v PPVT gii quyt cc bi ton v quan h vung gc s cho li gii kh r rng, ngn gn. Thng thng vi dng ton trn, ta c th qui v bi ton chng minh hai ng thng song song, hay t nh ngha tch v hng ca hai vct ta c th suy ra: - nub a,l 2 vct khc0th0 .= b a b aVybitonchngminh2ngthngvunggccthquivbi ton chng minh tch v hng ca hai vct bng O. V d 1 Cho tam gic ABC cn ti A; M l trung im ca BC, H l hnh chiu ca M trn AC, E l ttrung imca MH. Chng minh rngBH AE Hng dn gii: Bc 1.Tm hiu ni dung bi ton Trc ht hc sinh phi tm hiu bi ton mt cch tng th: y l dng ton chng minh 2 ng thng vung gc. Tip theo phi phn tch bi ton cho. - Bi ton cho bit g? (cho tam gic ABC cn ti A, M l trung im ca BC, H l hnh chiu ca M trn AC, E l trung im ca MH). - Bi ton hi g? (Chng minh rngBH AE ) - Tm mi lin h gia ci phi tmvi ci cho. www.VNMATH.com S ha bi Trung tm Hc liu i hc Thi Nguynhttp://www.lrc-tnu.edu.vn 61 Bc 2:Xy dng chng trnh gii: chng minh AE vung gc BH, ta phi chng minh nhng g? ( phi chng minh ng thc vct0 . = BH AE ) s dng gi thitAM BC ( hay 0 . = BC AM ), v AC MH (hay0 . = AC MH ), ta phi phntchvctBH AE,theo nhng theo nhngvct no? Khi ? . = BH AE Bc 3: Thc hin chng trnh gii 2 . ( )( ) AE BH AM AH BM MH = + + = BM AH MH AM + BH AEMH MH MH MH MH HMMC MH MH AM BM MH AM MH AM = + = + =+ = + + =0) (2 2

Bc 4:- Kim tra v nghin cu li gii. - Kim tra li cc bc gii ca bi ton. Vd2:ChohnhvungABCD,E,Flccimxcnhbi CD CF BC BE21,31 = = ,ngthngAEctBFtiI.Chngminhrng AIC=900 Hng dn gii: Bc 1. tb AD a AB = = , . Chn2 vct ny lm vct c s. Mivct trong bi ton u phn tch ( biu din) c qua 2 vct ny. Bc 2. Gi thit cho ABCD l hnh vungAD AB hay0 .= b aA B C M H E A B C DF EI www.VNMATH.com S ha bi Trung tm Hc liu i hc Thi Nguynhttp://www.lrc-tnu.edu.vn 62 -GisBF k BI = .VA,E,IthnghngAI AE, cngphng.Biu dinAI AE,theo 2 vct c sb a,sau dng iu kin cng phng, ta s tm c k. -iuphichngminhlAIC=900,tngngvi0 = CI AI ,vi CI AI ,u phn tch c quab a, ( s dng gi thit0 = b a ) Bc 3. ta b a b AD a AB = = = = ; , (a l di cnh hnh vung) Ta cb a AD AB AC AB AE31313132+ = + = + =AD k ABkBF k AB BI AB AI + + = + = + = )21 (b k ak+ + = )21 (VAI AE,l 2 vct cng phng nn: 5232131: 1 : )21 ( = = + = + k kkkk. Vyb a AI5256+ =Mt khc b a AC AI CI5351 = =CI AIa a b a b a CI AI = = + = 0256256)5351)(5256( .2 2 Bc 4- Kt lun AIC=900

- Kim tra li li gii.Nhn xt: Qua 2 v d trn, ta thy rng khng phi lc no vic p dng qui trnh 4 bc gii bi tonHH bng PPVT cng c r rng v thun li. Vvy,trongqutrnhgingdy,giovincnphilinhhottrongvic hng dn hc sinh phng php tm li gii bi ton theo 4 bc ca Plya hay s dng qui trnh 4 bc gii bi ton HH bng PPVT. www.VNMATH.com S ha bi Trung tm Hc liu i hc Thi Nguynhttp://www.lrc-tnu.edu.vn 63 *H thng bi tp. Bi 1. ( Bi 8-tr52-SGK HH10-nng cao) Chng minh rng iu kin cn v tam gicABC vung ti A l 2AB BC BA =Bi 2. (Bi11-tr40-SGK HH10 - nng cao) Tam gic MNP c MN=4, MP=8, M=600.Ly im E trn tia MP v t MP k ME = . Tm k NE vung gc vi trung tuyn MF ca tam gic MNP. Bi 3.Cho tam gic ABC. Gi M l trung im ca on thng BC v H l im nm trn ng thng BC. Chng minh rng AB2- AC2 = 2 MH BCl iu kin cn v AH BC. Bi 4.Cc ng AM, BE, CF l trung tuyn ca tam gic ABCa) Chng minh rng BE2+ CF2 = 5AM2 l iu kin cn v BAC = 900 b)ChngminhrngAB2+AC2=5BC2liukincnv BE CF. Bi 5. Cho tam gic ABC vung cn ti nh A, trn cc cnh AB, BC, CAtalnltlyccimM,N,Esaocho.EACENCBNMBAM= = .Chngminh rng: AN ME. Bi6.ChotamgicuABC.LyccimM,Nthamn AB AN BC BM31;31= = giIlgiaoimcaAMvCN.Chngminhrng gc BIC = 900. Bi 7. Tam gic ABC cn ti A ni tip ng trn (O). D l trung im ca AB, E l trng tm ca tam gic ADC. Chng minh rng OE CD Bi 8. Cho tam gic ABC ni tip ng trn (O), ngoi tip ng trn (I). B l im i xng ca B qua O. (I) tip xc vi cc cnh BA, BC ti P, Q. Trn BA, BC ly cc im K, L sao cho BK = CQ, BL= AP. Chng minh rng BI KL www.VNMATH.com S ha bi Trung tm Hc liu i hc Thi Nguynhttp://www.lrc-tnu.edu.vn 64 Bi 9. Cho tam gic ABC. Gi O, I ln lt l tm ng trn ngoi tip, ni tip tam gic. Trn cc tia BA,CA ly cc im E, F sao cho EB = BC = CF.Chng minh rng OIEF. Bi 10. Cho t gic ABCD. Chng minh rng iu kin cn v t gic c 2 ng cho vung gc l tng bnh phng cc cp cnh i din bng nhau. Bi 11. (Bi 22 - Tr 41- SBT- HH10 - Nng cao )T gic ABCD c hai ng cho AC v BD vung gc vi nhau ti M, gi P l trung im on thng AD. Chng minh rng MP BC khi v ch khi MD MB MC MA . . = Bi 12. Cc ng cho ca t gic ABCD l vung gc v bng nhau. Trn cc cnh AB, BC, CD, DA ta ln lt ly cc im P, Q, R, S sao cho: ) 1 ( ; ; ; = = = = = k SA k DS RD k CR QC k BQ PB k APChng minh SQ PR Bi 13. (Bi 23 - Tr 41- SBT- HH10 - Nng cao )ChohnhvungABCD,imMnmtrnonthngACsaocho 4ACAM = . Gi N l trung im ca on thng DC. Chng minh rng BMN l tam gic vung cn. Bi 14. Cho hnh ch nht ABCD. K BK vung gc vi AC. Gi M, N ln lt l trung im ca AK v CD. Chng minh rng gc BMN vung. Bi 15. Cho hnh vung ABCD. Cc im M, N thuc cc cnh BA, BC sao cho BM=BN. H l hnh chiu ca B trn CM.Chng minh rng DHN=900 Bi 16. Cho t gic ABCD ni tip ng trn (O;R). Chng minh rngAC BD 2 2 24R CD AB = + Bi 17. (Bi 32 - Tr 43- SBT- HH10 - Nng cao )Trongngtrn(O;R)chohaidycungAA,BBvunggcvi nhau im S v gi M l trung im ca AB. Chng minh rng SM AB. www.VNMATH.com S ha bi Trung tm Hc liu i hc Thi Nguynhttp://www.lrc-tnu.edu.vn 65 Bi 18 (Bi 35 - Tr 43- SBT- HH10 - Nng cao )ChoimMnmtrong gc xOyvgiM1,M2lnltl hnhchiu ca M trn Ox, Oy. V ng trn () qua M1, M2, ng trn ny ct 2 cnh Ox, Oy ln lt N1, N2. K ng thng vung gc Ox N1 v ng thng vung gcviOyN2.Gis2ng thng vung gc vinhauN. Chng minh ON M1M2.Hng dn hoc li gii. Bi 2. Ta c MN MP k ME NM NE = + = ) (21MN MP MF + =0 ) )( ( = + MN MP k MN MP MF NE5216 6416 16) .() .(22=++=++=++= MP MP MNMN MP MNMN MP MPMN MP MNkBi 3. Ta c AB2-AC2 = CB AM AC AB AC AB 2 ) )( ( = + = CB HM AH ). ( 2 += CB HM CB AH 2 2 +NuBC MH CB HM AC AB CB AH BC AH 2 2 02 2= = = NuBC AH CB AH MH BC AC AB = = 0 22 2 Bi 4. S dng ng thcAC AB AM CB CA CF BC BA BE + = + = + = 2 , 2 , 2

22 2 2AC BC BABC BA +=Bi 5. tb AC a AB kEACENCBNMBAM= = = = = ; ; . Ta c0 = b a vb a =M N P F E www.VNMATH.com S ha bi Trung tm Hc liu i hc Thi Nguynhttp://www.lrc-tnu.edu.vn 66 CAkkCE BCkkBN ABkkAM1,1,1 +=+=+= 11 1 1k kAN AB BC AC ABk k k = + = ++ + +

11 1ka bk k= ++ + = ME ABkkACkABkkAE1 111 ++=+

11 1ka bk k= ++ + = ME AN1( )1 1ka bk k++ + 1( )1 1ka bk k++ + 0 = . Vy AN ME. Bi 6. tb BA a BC = = ; . V tam gic ABC u nnb a =Gi s ) 1 (= = x IM x IA) 1 ( = = y IC y INTa c xBM x BABI=1 BCxxBAx ) 1 ( 3 11=bxaxx+=11) 1 ( 3 Mt khcbyayyyBC y BNBI) 1 ( 321 1 + == 611 ) 1 ( 3) 1 (3211= = = yyyxxyx Vyb a NC b a BI32;7471 = + == NC BI 0 )21821.21221.7471( )32)(7471(2= + = + a b a b aVy BI CN A BC M N E A BCM N I www.VNMATH.com S ha bi Trung tm Hc liu i hc Thi Nguynhttp://www.lrc-tnu.edu.vn 67 Bi 7. Ta c: ) 2 (21) (21OC OB OA CB CA CD + = + =

1( )3OE OA OD OC = + +

) 2 3 (61) ) (21(31OC OB OAOC OB OA OA+ + =+ + + = Do = OE CD 6 . 2 ) 2 ( OC OB OA + ) 2 3 ( OC OB OA + +

0 4 ) ( 44 4 4 32 2 2= = = + + =CB OA OC OB OAOC OA OB OA OC OB OA Vy0 = OE CD , tc l OE CD. Bi 8. Ta c: '. '.( ) IB KL IB BL BK =

0 . . . .. .. ' . '= = = = =QC PA PA QC PA BK QC BLBK PA BL QCBK IB BL IB Vy IB KL Bi 9. Gi M, N, P ln lt l hnh chiu ca O tren BC, CA, AB. X, Y, Z ln lt l hnh chiu ca I trn BC, CA, AB. Ta c:) ( . CF BC EB OI EF OI + + == CF OI BC OI EB OI . . . + + = CF NY BC MX EB PZ . . . . + + =( ) ( ) ( ). BP BZ BE BX BM BC CY CN CF + + A BC P I Q K L O B B CX M A F N Y E I Z O P A BC D E O www.VNMATH.com S ha bi Trung tm Hc liu i hc Thi Nguynhttp://www.lrc-tnu.edu.vn 68 =CF CN CF CY BC BM BC BX BE BZ BE BP . . . . . . + + =aba c p aaa b p a b p ac2) (2) ( ) (2 + + =O abc pab p b pc= |.|

\| + + + 2 2 2 Suy ra OI EF Bi 10. Trong mi t gic ABCD ta c: DB AC DB DB ACCD CB AD AB ACDA CD BC AB ACCA DA CD AC BC ABDA CD DA CD BC AB BC ABDA CD BC AB2 ) () () () ( ) () )( ( ) )( (2 2 2 2= + = + =+ = + =+ + + == + Nh vy trong mi t gic ABCD, ta c h thc sau: DB AC DA CD BC AB 22 2 2 2= + 0 = BD AC BD AC 2 2 2 2 2 2 2 20 DA BC CD AB DA CD BC AB + = + = + Bi 11. ) ).( ( 2 MB MC MD MA BC MP + =MB MD MC MAMB MA MC MD MB MD MC MA = + = (V0 = = MC MD MB MAdo AC BD) T ta c: MD MB MC MA BC MP BC MP = = 0Bi 12. Gi s AC ct BD ti O doPB k AP =SA k DS RD k CR QC k BQ = = = ; ;(k 1 = ).D A B C M P C A B D www.VNMATH.com S ha bi Trung tm Hc liu i hc Thi Nguynhttp://www.lrc-tnu.edu.vn 69 Nn:

+++=+++=+++=+++=OAkkODkOSOCkkOBkOQOBkkOAkOPODkkOCkOR1 111 111 111 11

+++= =+++= =BDkACkkOQ OS QSBDkkACkOP OR PR1111 11 Ta cO ACkBDkkQS PR =++=2222.) 1 (1) 1 (.QS RP Bi 13tb AB a AD = = ; khi ta c) (4141b a AC AM + = =2ba DN AD AN + = + =T suy raAM AB MB =) 3 (41) (41b a b a b + = + =) (412b aba AM AN MN + + = =) 3 (41b a + =Ta c) 3 ).( 3 (161. b a b a MN MB + + = O b a b a = + + = ) . 8 3 3 (16122

CA B D P Q R S O AB CD N a b M www.VNMATH.com S ha bi Trung tm Hc liu i hc Thi Nguynhttp://www.lrc-tnu.edu.vn 70 2 2 2 2285) . 6 9 (161) 3 (161a b a b a b a MB = + = + =2 2 2 2285) . 6 9 (161) 3 (161a b a b a b a MN = + + = + =Vy MB MN v MB = MN, tam gic BMN vung cn ti nh MBi 14: tc BK b BC a BA = = = ; ; KhitacO c b a O b a = = ). ( ; . Tac) (21) (21c a BK BA BM + = + = . CN BC MB MN + + =1( )2 2 2 2BA a cBM BC a c b b = + + = + ++ = 2.2. . . 22cc bca b a MN BM + = Occc= + =2 2222Vy gc BMN vungBi 15: t a l di cc cnh ca hnh vung,= o MBH= BCH Ta c:) )( ( . CD HC BN HB HD HN + + =HC BN CD HB . . + =(VCD BN HC HB ; ) HC BN BA HB . . + =. . . . HBaCos NBCHCos o o = +2 2. . . . MBaCos NBaCos o o = += O (V MB = NB) A D NC BabM K cA D B C M N H www.VNMATH.com S ha bi Trung tm Hc liu i hc Thi Nguynhttp://www.lrc-tnu.edu.vn 71 Bi 16:AB2 + CD2 = (22) ( ) ( OC OD OA OB + ) =4R2 - 2 ) . . ( OD OC OB OA += 4R2 - 2R2(CosAOB+CosCOD) Do AB2 + CD2= 4 R2 CosAOB+CosCOD=O AOB + COD = 1800 AC BD Bi 17Xt tch v hng) ' ' ).( (21' ' . SA SB SB SA B A SM + =1( . ' . ' . ' . ')2SASB SASA SB SB SB SA = + Ta c:. ' SASB O = do SA SB O SA SB = ' . do SB SA ' . ' . SB SB SA SA =T suy ra:O B A SM = ' ' . Nn SM AB Bi 18:Ta c:2 21 1 . . ON OM ON OM = (*) Xt tch v hng) (. .1 2 2 1OM OM ON M M ON =21. . ON OM ON OM = 1 12 2. . OM ON OM ON = (**) T (*) v (**) suy raO M M ON =2 1. Hay ONM1M2 A D B C O B B A A M O S O M1 N1 M2 N2 NM x z y ( ) www.VNMATH.com S ha bi Trung tm Hc liu i hc Thi Nguynhttp://www.lrc-tnu.edu.vn 72 2.3.5 Chng minh ng thc vct ng thc vct l mt ng thc m c hai v l cc biu thc vct. Mibiuthcchacchngtlvctvchngcnivinhaubi cc du ca cc php ton vct hoc mt trong hai v ca ng thc lO (hoc O). chng minh cc bi tp dng ny, ch yu ta s dng cc quy tc 3 im, quy tc hnh bnh hnh dng cc vct c cho hai v ca ng thc, s dng cng thc trng tm ca tam gic, trung im ca on thng, tnh cht ca cc php ton, cc tnh cht ca tch v hng ca hai vct rt gn hai v... V d 1: Chng minh rng vi 4 im A,B,C,D ta cO BC AB DB AC CD AB = + + . . .(*) Hng dn gii:Bc1:ChnvctAD AC AB , , lmccvctcs.Mivctxut hin trong bi ton u phn tch c qua cc vct ny. Bc 2: Bi ton cho di dng ngn ng vct. Bc 3:. . . ABCD AC DB ADBC + + = O AD AC AC AD AC AB AB AC AB AD AD ABAB AD AC AD AD AC AB AC AC AB AD ABAB AC AD AD AB AC AC AD AB= + + = + + = + + =) . . ( ) . ( ) . . (. . . .) ( ) ( ) ( Bc 4Nhn xt:1. ng thc vct (*) c gi l h thc le. C th dng h thc le chng minh: Trong tam gic 3 ng cao ng quy. Tht vy, gi s cc ng cao k t B v C ca tam gic ABC ct nhau ti H. p dng h thc le cho 4 im H, A, B, C ta c:O AB HC CA HB BC HA = + + . . .www.VNMATH.com S ha bi Trung tm Hc liu i hc Thi Nguynhttp://www.lrc-tnu.edu.vn 73 DoHB CA,HC ABnnO AB HC CA HB = = . . tO BC HA = . tc HABC2. Kt qu va chng minh l s m rng ng thc. . . ABCD AC DB ADBC O + + =khi A,B,C,D nm trn mt ng thng. V d 2: (Bi 28- tr42- SBT- Hnh hc 10 - nng cao ) Cho mt im O bt k nm trong tam gac A1A2A3 Gi B1, B2, B3 ln lt l hnh chiu ca O trn A1A2, A2A3, A3A1.

t 331 3 3223 2 2112 1 1. ; . ; .OBOBA A aOBOBA A aOBOBA A a = = =Chng minh rng:o a a a = + +3 2 1 Hng dn gii:Ta c: ) , cos( ) , cos() )( () ( ) (3 2 3 3 2 3 3 1 2 3 1 23 2 3 3 1 23 2 3 1 3 22 1 3 2 2 1 3 2 1A A a A A a A A a A A aA A a A A aA A A A a aA A a a A A a a a = = + =+ = + + Theo gi thit: 3 1 3 3 2 2; A A a A A a = =Ngoi ra d thy:) , cos( ) , cos(3 2 3 3 1 2A A a A A a =0 ) (2 1 3 2 1= + + A A a a a . Do vct) (3 2 1a a a + +vung gc vi ng thng A1A2

Chng minh hon ton tng t, ta c vct) (3 2 1a a a + + vung gc vi ng thng A2A3. Vy) (3 2 1a a a + + =0 . A1 A2 A3 O B1 B2 B3 1a2a3awww.VNMATH.com S ha bi Trung tm Hc liu i hc Thi Nguynhttp://www.lrc-tnu.edu.vn 74 Nhn xt: 1)ChoimOtrngvitmngtrnnitiptamgicA1A2A3l im I, ta c kt qu sau: 01 3 2= + + IB c IB b IB aVi a=A2A3, b=A3A1, c=A1A2 2)Ktqutrnngviagic A1A2.....An bt k (nh l Con nhm). Vi: . 1 +=k k kA A a(xem An+1A1) . 1 2.......na a a + + + =0 . (Ccvct ka cgilcclng nhm) Vd3.ChotamgicABCv A1A2A3 c cng trng tm G. G1, G2, G3 ln lt l cc trng tm ca cc tam gic BCA1, ABC1, ACB1. Chng minh rng03 2 1= + + GG GG GGHng dn gii: G1 l trng tm tam gic BCA1, ta c:01 1 1 1= + + A G C G B GHay01 1 1 1= + + + + + GA G G GC G G GB G G1 13 GA GC GB GG + + = (1) Tng t, ta c: 1 23 GC GB GA GG + + = (2)

1 33 GB GC GA GG + + =(3) Cng tng v cc ng thc (1), (2), (3) ta c: 1 1 1 3 2 1( ) ( 2 ) ( 3 GC GB GA GC GB GA GG GG GG + + + + + = + + ) Do G ng thi l trng tm ca hai tam gic ABC v A1B1C1 nn suy ra: 0 ) ( ) (1 1 1= + + = + + GC GB GA GC GB GAVy03 2 1= + + GG GG GGAn A1 naA2 1aA3 2aA4 3aA5 4awww.VNMATH.com S ha bi Trung tm Hc liu i hc Thi Nguynhttp://www.lrc-tnu.edu.vn 75 *H thng bi tp. Bi 1. Cho tam gic ABC, G l trng tm. Chng minh rng1.0 . . . = + + AB MC CA MB BC MA2. MA2+ MB2+MC2=3MG2+GA2+GB2+GC2 3. GA2+GB2+GC2=31(a2+b2c2), vi a, b, c l di 3 cnh tam gic ABC. 4. Nu tam gic ABC ni tip (O;R) th: OG2=R2 -91( a2+b2c2) 5.NutrngtmGcatamgicABCthamndiukin 0 = + + GC c GB b GA ath tam gic ABC u. Bi 2. Cho tam gic ABC, gi H l trc tm. I l tm ng trn ni tip. Chng minh: 1.0 = + + IC c IB b IA a(a, b, c l di cc cnh tam gic ABC ) 2.0 . tan . tan . tan = + + HC C HB B HA A3.0 . . . = + + MC S MB S MA Sc b a,trongMlimbtknmtrongtam gic ABC, Sa, Sb, Sc theo th t l din tch ca tam gic MBC, MCA, MAB. 4. a.IA2+b.IB2+c.IC2=abc. Bi 3. Cho tam gic u ABC tm O, M l im bt k tong tam gic. HMD,ME,MFlnltvunggcvicccnhBC,CA,AB.Chng minh rng: MO MF ME MD23= + +Bi4.ChotgicABCD.GiI,JtheothtltrungimcaAC, BD. Chng minh rng: AB2+ BC2+CD2+DA2=AC2+BD2+4IJ2 Bi5.ChotgicABCDvsk= 0;k= 1.TrnccngthngAB, BC, CD, DA ta ly cc im tng ng A, B, C, D sao cho: www.VNMATH.com S ha bi Trung tm Hc liu i hc Thi Nguynhttp://www.lrc-tnu.edu.vn 76 A D k DD D C k CC C B k BB B A k AA ' ' , ' ' , ' ' , ' ' = = = = .GiOltrngtmcat gic ABCD. Chng minh rng:0 ' ' ' ' = + + + OD OC OB OAHng dn hoc li gii Bi 1. 1. y chnh l h thc le chng minh v d 1. 2. Ta c: 2 2 2 2 2 2) ( ) ( ) ( GC MG GB MG GA MG MC MB MA + + + + + = + + 2 2 2 2) ( 2 3 GC GB GA GC GB GA MG MG + + + + + + = MA2+ MB2+MC2=3MG2+GA2+GB2+GC2 (1) Nhn xt: -imctngbnhphngcckhongccht nccnhca tam gic nh nht chnh l trng tm ca tam gic. - Nu tam gic ABC ni tip ng trn (O;R) th: 3(R2-OG2)=GA2+GB2+GC2 (ly M trng vi im O) 3. Cho MA th (1) thnh c2+b2=4GA2+GB2+GC2 Cho MG th (1) thnh a2+c2=GA2+4GB2+GC2 Cho MC th (1) thnh b2+c2=GA2+GB2+4GC2 Cng tng v cc ng thc trn ta c: 2(a2+b2+c2)=6(GA2+GB2+GC2) GA2+GB2+GC2=31(a2+b2c2) 4. Cho MO th (1) thnh: 3R2=3OG2+GA2+GB2+GC2 OG2=R2-91((a2+b2c2) 5.) ( 0 GC GB GA a GC c GB b GA a GC c GB b GA a + + = + + = + +GC c a GB a b ) ( ) ( = . www.VNMATH.com S ha bi Trung tm Hc liu i hc Thi Nguynhttp://www.lrc-tnu.edu.vn 77 - Nu b-a= O, thGCa bc aGB=iu chng tGC GB,ng phng. iu ny v l, vy b= a(c-a) O GC =c- a c= a VyAABC uBi2:Tasdngphngphpphntchmtvcttheohaivct khng cng phng gii v d nya) Phn tchICtheo IA vIBDnghnhbnhhnhIACBtacIA IB IA IB IC | o + = + = ' ' vvct ' IB vIB ngc hng nncbB AC AIBIB = = =11'o(tnh cht phn gic ) Tng t caA BC BIAIB = = =11'|VyIAcaIBcbIC . =O IC c IB b IA a = + + .b) Xt trng hp tam gic ABC c 3 gc nhn- Dng hnh bnh hnh HACB ta cHB HA HB HA HC . . ' ' | o + = + =ta c1 11 1. ' tan. tanBC BB CotC HA AHA B A BB CotA Co = = = = 1 11 1. ' tan. tanAC AA CotC HB BHB AB AA CotB C| = = = = VyHBCBHACAHC .tantan.tantan = HayO HC C HB B HA A = + + . tan . tan . tanChng minh tng t cho trng hp tam gic c 1 gc tB A B A C A1 I B1 C1 A BC A B H A1 B1 C1 www.VNMATH.com S ha bi Trung tm Hc liu i hc Thi Nguynhttp://www.lrc-tnu.edu.vn 78 c) Cch 1:Gi giao im ca cc tia AM, BM, CM vi BC, CA, AB ln lt l A1, B1, C1

Dng hnh bnh hnh MACB, ta c: .MB MA MB MA MC | o + = + = ' '. baMABMBCSSSSAICHA BC BMAMA = = = = =11'oTng t:bcSS| = VyMBSSMASSMCcbca = 0 = + + MC S MB S MA Sc b a Cch 2: Gi A l giao im ca ng thng MA vi BC Ta c:C AC AB AB A '''' =MCBCB AMBBCC AC AB AMCC AB AMBMA' '''1''' + =++= Nhng ''''MA C MAC bMA B MAB cS S S A CA B S S S= = = MCS SSMBS SSMAS SSBCB AS SSBCC Ac bcc bbc bcc bb+++= +=+= '''(*) Mt khc: c baMAC MABC MA B MAMACC MAMABB MAS SSS SS SSSSSMAMA+=++= = =' ' ' '' MAS SSMAc ba+ = ' . Thay vo (*) ta c: 0 = + + + = MC S MB S MA S MC S MB S MA Sc b a c b a A BC M A B A1 B1 I H C1 A BC A M www.VNMATH.com S ha bi Trung tm Hc liu i hc Thi Nguynhttp://www.lrc-tnu.edu.vn 79 Nhn xt: - ChoMtrngvitrngtmtamgicABC,tacktqu: 0 = + + GC GB GA- ChoMtrngvitmngtrnnitiptamgicABC,tackt qu cu a). - Nu tam gic ABC u th vi im M bt k trong tam gic, ta c: 0 = + + MC z MB y MA xTrong x, y, z l khong cch t M n cc cnh BC, CA, AB. - Nu M nm ngoi tam gic ABC, chng hn M thuc gc BAC, chng minh tng t ta c kt qu:0 = + + MC S MB S MA Sc b a d) Cch 1. Theo kt qu cu a) ta c: O IC c IB b IA a = + + . 0 ) . (2= + + IC c IB b IA aabc cIC bIB aIAabc c b a cIC bIB aIA c b abc a c ab abc IC cb ca c IB bc ba b IA ca ab aCA IA IC ca BC IC IB bc AB IB IA ab IC c IB b IA aIB IC bc IC IA ac IB IA ab IC c IB b IA a= + + + + = + + + + = + + + + + + + + + + = + + + + + + + + = + + + + + 2 2 22 2 22 2 2 2 2 2 2 2 22 2 2 2 2 2 2 2 2 2 2 2 2 2 22 2 2 2 2 2) ( ) )( (0 ) ( ) ( ) ( ) (0 ) ( ) ( ) ( . . .0 2 2 2 . . . Cch2.GiH,KlhnhchiucaItrnAB,AC.Ththc O IC c IB b IA a = + + . , ta c: 2 2 2 2 2) ( cIC bIB IA IC c IB b cIC bIB aIA + + = + + ) 1 ( ;) ( ) (AC KC c AB HB bAC IC c AB IB bIA IC IC c IA IB IB b+ =+ = + = VvctHBvvctAB cnghng,vctKC vvctAC cng hng nn: b c p AC KC AC KC c b p AB HB AB HB ) ( . ; ) ( . = = = =(2) www.VNMATH.com S ha bi Trung tm Hc liu i hc Thi Nguynhttp://www.lrc-tnu.edu.vn 80 T (1) v (2) suy ra: abc ab c p b p b c p c c b p b cIC bIB aIA = + = + = + + ) ( ) ( ) (2 2 2 Nhn xt: Bng phng php ca cch 1, ta c th chng minh c kt qu tng qut hn: - Cho tam gic ABC. im I xc nh bi: ) 0 ( ; 0 = + + = + + | o | o IC IB IA ta c:) . . . . . . (1. . .2 2 2 2 2 2b a c IC IB IA o | | o | o | o + ++ += + +Bi 3. Qua M ta k cc ng thng song song: B2C2 song song BC, A2C1 song songAC,A1B1songsongAB.TaccctamgicMA1A2,tamgic MB1B2, tam gic MC1C2 l cc tam gic u. Ta c: MG MF ME MDMGMC MB MAMC MB MB MA MC MAMC MC MB MB MA MAMF ME MD233) ( ) ( ) () ( 21 1 2 2 2 12 1 2 1 2 1= + + =+ + =+ + + + + =+ + + + + == + + Bi 4. Ta c: AB2+BC2+CD2+DA2=AC2+BD2+4IJ2 A BC M D E F A1 A2 B2 B1 C1 C2 O A B C D I J www.VNMATH.com S ha bi Trung tm Hc liu i hc Thi Nguynhttp://www.lrc-tnu.edu.vn 81 CD AB AB CDCD AB BC BD AC AD CDCD AB BC BD CD AC AD