NGHIEN CtrU & UNG DUNG
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SU DUNG PHUONG PHAP DAY HOC NEU VAN DE THONG HEN LUYEN TU DUY SANG TAO CHO HOC SINH THPT KHI GIAI BAI TAP HINH HOC 10
ThS. Ta Khac Bjaft Trudng THPT Diin Chdu 2, Tinh Nghp An SUMMARY
In this article, a teaching situation is built: solving geometry problems grade 10^-the natural sciences department so as to enable students to act creatively on their own with the teacher's guidance. At the same time, it can train students the ability to self-convey from knowledge arui skills to the new situations, the ability to combine between the old solutions and the new creative problems.
Keywords: Pratice creative thinking, geometry homework.
Ngdy nhpn bdi: 2/7/2013, Ngdy duyPt ddng: 29/6/2014 Rdn luyfn tu duy eho hgc sinh Id mpt trong nhttng myc tiflu ciia qui trinh dgy hpc ndi ehung vi dgy hpc Toin ndi rifng nhim hinh thinh cho hgc sinh khi ndng ddc ldp suy nghT, khi ndng vin dyng Iinh hogt sing tgo tri thUc, khi ndng ty truyen tii cie tri thdc vi ky ndng sing tgo nhihig tinh hudng mdi, kha ning phit hifn ra nhieu ldi giii, nhieu cich nhin nhgn mdi ddi vdi vifc tim kifm Idi giii, khi ning kflt hpp nhttng Idi giii da biit thinh nhttng phuang thdc mdi cd tinh sing tgo.
Chdng ta hdy xudt phdt tir bdi todn ca bdn trong SGK hinh hgc 10 Ban TN
VI dy 1. Dgy hpc bii toin ve trgng tam tam giic:
"G Id trgng tdm M BC khi vd chi khi GA+GB + GC = 0 (1) ". GV hudng hpc sinh vio vifc phit hifn vi ehOmg minh bii toin vi ttt dd rfln luyfn cho hpe sinh kiin tgo dupc eic tri thUc mdi.
Trude khi hpc sinh hpc vf bii toin niy thl eic em da biit vi mdt tinh chdt cua tiung diim li: "M Ii trung diim cua dogn thing AB khi vi chi khi M S + M B = 5 " . Di dua ra bii toin vi tigng tdm Gciia tam giic ABC d trfn thl ta ed thi di ttt eii di biit bdng cich xem dogn thing li mgt tam giic d^c bift cd ba dinh thdng hing ehdng hgn vdi C Ii trung diim eiia AB khi dd diim M sf li tigng tim ciia tam giic d$e bift dd. Nhu v?y khi MA+RIB - 0 nic li bii toin tifln dUng trong trudng hpp djc bift niy. Biy gid ta chung minh cho tam giic bdt ki.
Diiu cdn chung minh: G li trpng tdm tam giic ABC tuang duong vdi ddng thUc GA+GB + GC = 0, biy gid ta xem xdt ding tiiUc edn chiing minh di tim ra cich chUng minh bii toin.
• Niu xem vecla 0 dudi khIa cgnh li tdng ciia
hai vdcto ddi nhau ta ed hudng chimg minh nhu sau:
Ta biin ddi biiu thuc GA+GB+GC thinh tdng ciia hai vfcto ddi nhau bing cich dya vio tinh chdt ciia trpng tam.
+ G 14 trgng tam tam giic ABC khi vi chi khi G thudc trung tuyin AM vi GA = 2GM.
Dyng hinh binh hinh GBDC ta cd M Ii trung diem ciia GD.
V4 suy ra G Ii trung diim ciia AD v4 ta cd GA=-GD (2)
V^y G 14 trgng tim liABC khi vi ehi khi GA=-GD (2), mi theo quy tic
hinh binh hinh ta cd: GD=GB+GC (3). Nfn (2)
«GA = -GB-GCoGA+GB+GC = 6 . Niu ddi tupng hpc sinh li hgc sinh khi gidi, GV cd tiii hudng hpc sinh suy nghT md rdng bii toin tdng quit hon, giup phit triin tu duy sing tgo cho hgc sinh tdt hon ntta. Theo tir duy bifn chUng till cii chung tdn tgi frong edi riflng, thdng qua eii riflng di biiu thi sy tdn tgi eua minh, nfn ta ed thi tim eii chung trong cii riflng. Vl vgy niu ta xem bii toin tien chi li mpt ttudng hpp rieng eiia mpt trudng hpp ting quit, biy gid ta hdy tim xem cii chung, tdng quit han li gi?
Niu ta xem frgng tam G Ii mgt diim d$c bift nam frong tam giic thda man
i s . Khi dd:
3 ABC
GA+GB+GC=0
^3 ABC GA+:^S,_GB+is^^GC = 0 (4) 3 ABC
TAP CHi THlfTBIGlAO DMC-stf 1 0 7 - 7 / 2 0 1 4 • 33
NGHIEN cuu SUNG DUNG
Ta df y ring tdng cdc hf sd cua bidu thUc vf trdi cua (4) bdng S^^^^. Tir dd ta xem xdt mdt kit qua tdng qudt hon nhu sau:
Bdi todn 1: ^M Id diim bdt ki ndmAABC. Bgt
^ l ~ ^ M B C ' ^ 2 ~ ^ M C A ' ^ 3 ~ ^MAB
Chungminh: "S MA+S MB+S MC = 0 "
>MA = - ^ M B
SiMA+SjMB + S^MCsO fMC (5) Dfl chUng minh (5) ta dyng hinh binh hdnh MEAF nh§n MA lim dudng chdo, MEviMF Idn lupt thude cic dudng tiidng BM, CM.
Theo quy tic hinh binh hinh ta cd:
TirddMA = - M M B - | ^ M C MB MC Hay MA = - % B M B - % 1 ^ M C Sj Si Do AE//MC vi AF//MB nfln MA = - ^ M B - - 3 . M C • c S
Sj S^
Tir trudng hpp rieng, ta da md rdng bii toin ra eho trudng hpp M Ii diim bdt ki vi ta di ed tfnh chdt tdng quit ban. Nhung theo phdp bifn chiing duy vgt thl mdi eii riflng dupc chUa dyng trong nhiiu cii ehung, cii bao frUm nd theo mdt sd quan hf nio dd vi ngupc Igi, nhifu cii riflng cd thi chira dyng trong ciing mdt eii mpt chung theo mpt mdi quan hf nio dd gitt^ cie ddi tupng. Nhu v^y ta ed thi md rpng tinh chit ndy ntta hay khdng nflu ta xem S^3p, \cK\K^ chi li mgt trudng hpp rieng cua mgt bd hf so cd tinh chit chung dd la tdng cua cie hf sd dd bdng S^^? Vi diim M ndm trong tam giic chi Ii trudng hpp rifng eiia mdt diim M bit ki?
Nhu vgy nflu diflm M nim ngodi tam giic ta sfl cd kit qui nhu thi nio?
Nflu diflm M ndm ngoii tam giic, chdng hgn ta xdt diim M ndm trong miin gde tgo bdi hai tia CA vi CB khi 06 ta ci S ^ , + S ^ ^ : S , ^ = S ^
Chimg ta sf cd tinh chit tdng quit nhu sau:
Bdi todn 2: "Cho Mid diim ndm ngodi tam gidc vd thu0c miin gde tgo bdi hai tia CA vd CB.
Chdng minh: SjMA+SjSffi-SjSiC = 0 vdi
^ l ~ ^ M B C ' ^ 2 = Sj^CA» ^ 3 = S , ^ j ^
Tacd: S MA+S,MB-S,MC = 0
< = > - ! - ! ^ + ^ M B = S i C
s s
S, s. (6)
Df chUng minh (6) ta dyng hinh binh hinh CPMQ vdi P, Q Idn lupt nim tten cic tia MA vi MB.
Theo quy tdc hinh binh hinh ta cd:
MC = MP+MQ =—MA+-5!:^MB MA MB
Vi CP//MB v i CQ//MA nfn
'^MBC
s.
MA+-2.MB = MC Ta sf cd kit qud nhu thi nio neu ta xem diim M Id tryc tim tam giic? Tim dudng frdn ndi tiip tam giic? Tim dudng frdn ngogi tiip tam giic? Tim dudng trdn bing tiip gde C?GV cd the hudng dan hpc sinh xiy dyng cic bii toin khic khi cho M l i cdc diim d ^ bift cua tam giie.Thye ra day li mpt dgng khic ciia bii toin v^
tdng qudt hern. Khi ta nhin nd dudi gde dd "dipn tich tam gidc". Thi m4 rdt nhiiu hpc sinh khi, gidi khi g^p bii niy hiu nhu khdng lim dupc, khdng biit bit ngudn tU ddu, cdi mdu chdt ciia nd, d ddy ngudi GV cdn hudng din cho hgc sinh di "quy Ig vi quen".
Nh^t xf 11: 6 bii toin I kit qui khdng phy thudc vio diim M, niu ta thay M bdi I Id tdm dudng trdn nfi tiip AABC vd r Id bin kfnh dudng trdn thi hpc sinh cd thi sU dyng bii toin 1 di giai bii toin sau.
: • TAP CHt THI^BIGlAO Dgc-sd 107-7/2014
NGHIEN cuu & UNG DUNG
III
Bii todn 3: Cho AABC, gpi I li tam ducmg tr6nnOitiep MBC.
Dat BC = a,AC = b,AB = c. Chimg minh i«ngaiA + b 5 + ciC = 0 (T)
HD giii: Ap dyng bii toin 1, ta cd:
S.iA + SjiB + SjiC
o - r a i A + i r b i B + - r c I C = 0 2 2 2 o a i A + biB + c i c = 0
Nhin x6t 2: Ngu ta nhin cic cgnh duiM "g6c dp"
g6c, thi ta l^i c6 bii t o ^ sau.
Bdi todn 4: Cho hABC, gpi I li tim dudng tr6n nOi tijp AABC.
D}t BC = a,AC = b,AB = c- Chiing minh ring:
iAsinA + iBsinB + iCsinC = 0
Nh|n x&t 3: Ta thiy vj tri M thu^c miln trong tam giic, v$y AABC nhpn thi tim O du6ng tr6n ngoji tilp AABC thupc miM trong. Do d6, ta c6 bii toin mdi.
Bii todn 5: Cho AABC nhpn, BC=o,AC=b,AB=c.
Gpi O li tim duimg trdn ngo?! tifip AABC. ChOng minh ting: OAsin2A + OBsin2B+OCsin2C = 0 (8) HD giai: Ta c6: S, "-OCjOB sin ZBOC - — sin2A ^
R^ R^
Tuong t|r: S, =—sin2B vi S, =—sin2C.
Theo bii toin 4: S,MA+S,^*8,^ = 0 Doddtacd: sin2AOA+sin2B6B+sin2c6c = 0 Nh$n xit 4:Wdi M = 1 li tim dudng trdn n$i tilp.
Luu J ring a. h, = b. h^ = c. h^ = 2 S^^j^ vi
^ - ^ - ^ - ? p Thay vio ding thiJrc (7) ta sinA sinB sinC
CO bii toin mdi.
Bii loin 6: Cho AABC, gpi I li tim dudng trdnnOitijp MBC.
D51 BC = a,AC = b,AB = c. Chiing minh ring: h X ' I A + h . h , l A + h X I C = 0 (9)
Nhan xa 5; Vdi O li tarn dudng trdn ngo^i ti8p MBC.
Nlu ta biln ddi ding thtrc (8):
sin2A • OA + sin2B • OB+sin2C • OC = 0 o 2 s i n A c o s A O A + 2 s i n B c o s B O B
+ 2 s i n C c o s C O C = 0
^ sin(B+C) g g ^ s i n ( A + 0 g g cosB • cosC cos A • eosC
s-oc=o
^ sin(A+B) ^ cosAcosB
o {tanB+tanC)OA+(tanA+tanC)OB +(tanA+tanB)OC=0
Vdy ta cd bii toin.
Bdi todn 7: Trong MBC nhpn, gpi O li tim dudng frdn ngogi tiep tam giic eua nd. CMR:
{tanB+tanC)OA+(tanA + tanC)OB +(tanA+tanB) OC = 0
Kit lu$n: Qua vi^c giii quyit cic bii toin trfln, ta thdy ring diim M ludn thudc miin trong tam giic.
Ddy cd thi li vdn di ddc trung, cii mdu chdt cho cic bii dgy thudc logi trfln. Khi g$p edc bii toin cd lifln quan dfln diflm M v4 cic vecto, hay egnh. TuJ thudc ddi tupng hgc sinh GV cd thi hudng cho cie em cich lifln tudng eic bii toin da hgc. Cd thi tu bii toin trfln khai thic thflm vgn dyng sing tgo, giai quyit bii toin thudc logi niy, hay sir dyng eic bii toin khie. Vi§c khai thic tiem ndng sich giio khoa Ii diiu cyc ky quan trgng trcmg qua trinh dgy hpc.
Tii li|u tham khdo:
1. Doin Quynh (tdng ehu bifln). 2006. Hinh hgc nang eao 10. NXB Giio dye Vi|tNam.
2. Doin Quynh (tdng chu bifln). 2006. Hinh hpe ning cao 10 sich gido vifln. NXB Giio dye Vi?t Nam.
3. Nguyin Vin Ldc (CB) vi cdng sy - 2006.
Tim tdi cic Idi giii khic nhau cila bii toin hinh hgc 10 nhu thi nio? yXB Dgi hpc Qudc gia TP.HCM
TAP CHl THlfTBIGlAO DUC-sd 107-7/2014 • 35