PHAT TRIEN NANG LUI) TIT DUY SANG TAO CHO HOG SINH KHA GIOI THONG QDA DAY HOC DAI TAP HlNH HOC KHONG GIAN 6 TRUCNG TRUNG HOC P H 6 T H O N G
TS. C A O THI HA-
Mi
I uc tieu day hoc (DH) mdn Toin d tardng phothdngkhong chl nhiim cung cap trithuc I toan hpc cho hpc smh (HS) macdn ren luyen cho cac em eac kJ nang toan hpc va phat trien cac nang It/c tu duy, dae biet la nang It/c tu duy sang tao (NLTDST). Da cd rat nhieu nghien eiiu ve phat trien NLTDST cho ngudi hpe trong DH toan dtardng pho thdng, trong docac nghien ciiu t^p trung chii yeu vao viec ren luyen cho hpc sinh (HS) biet nhin van de dudi nhieu gde dd khac nhau (nham ren luyen tinh mem deo ctia tuduy); ren luyen kha nang tim kiem nhieu ldi giaiehomdtbaitoan(BT)(tInh mem deo vatinh nhuan nhuyen ciia tuduy); ren luyen kha nang phan tieh BT de tim ldi giai dpc dao (tin h dpc dao cua tu duy). Trong bai viet nay chiing tdi de cap den viee phat trien NLTDST choHS kha, gioi thong qua DH bai tap hinh hoc khdng gian (HHKG) dtnidng trung hoe phothong (THPT).1. Ren luyen cho HS biet nhin van de dirdi nhieu goc dp khac nhau
Trong qua trinh DH mdn Toan, Cling mdt ndi dung cd
&ie ducc the hien dudi nhieu hinh thtic khac nhau. Vi vay, de gop phan phat trien NLTDST choHS.GVcan ren luyenchocacembietnhinBT dgt ra dudi nhieu gdc dp khac nhau. Boi neu lam dupc dieu nay, HS seconang hjc chuyen
tiJrhoatdpngtrituenay sang hoatdpng tri tuetrituekhac;
dong thdi, dua ra each giai quyet van de mot each sang tao. Tlf dd, ren luyen dupc tinh mem deocua tuduy.
BT 1: Cho tam dien vudng OABC, vudng tai 0 . Gpi G la trpng tam cua tam giac ABC va 1 la tam mat cau ngpai tiep tam dien OABC. Chiing minh r^ng 6 . G,l thang hang (/i/h/j /).
Phan ftb/i; Thong thudng, HS tiep can BT biing each sau: De chtJmg minh ba diem thing hang, ta
48 Tap chi Gido due so 305
chiing minh ba diem ciing nam tren hai mat phang phan biet c§t nhau. Tuy nhien, doivdi BT nay, viec chi ra 0 , G, 1 cLing thupc hai mat phing la khd khan nhung b i n g stf linh hoat trong giai toan, GV cd the hudng d i n HS each tiep can BT theo mdt khia canh khac. Gpi G' la giao eua 01 va matphing (ABC), ta se ehung minh G' la trong tam ciia tam giac ABC.
That vay, goi M,N lan lupt la tnjng diem ciia ABva OC, Tu M ke Mx//OC; tir N ke Ny//OM, gpi 1 la giao ciia Mx vdl Ny. Khi dd, tam mat cau ngoai tiep tam dien OABC la 1. Do 01 va CM ciing thupe matphing (P) nen 01 giao vdi MC tai G', Suy ra G'chinh la giao diem ciia 01 vdi mat phang (ABC). Lai cd; M1//0C, suy r a ^ ^ ^ ' l - y ^ i . Do G'thupe trung tuyen MC nen G' la trpng tam eua tam giac ABC. Tiiday tacd dieu phai chiing minh.
Co the ndi, viec ren luyen cho HS each nhin mot van de dudi nhieu gdc dp khac nhau cdy nghla rat Idn trong viec phal trien NLTDST, gitip HS thay dudc nhimg khd khan va thuan Idl trong tiing each nhin de tiidd dua ra each giai mdi
2. Ren luyen cho HS kha nang tim nhieu Idi giai khac nhau cho mpt BT
De lam dupc dieu nay, GV can tao chp HS thdi quen khdng chap nhan mpt each giai quen thupc hoae duy nha't, ma ludn suy nghT, tim tdi va de xuat nhieu each giai khac nhau eho mpt BT. Mud'n vay, GV ean cdst/dinh hudng each phan tieh BT cho HS.
Ben canh dd, HS can phai ed sunhuan nhuyin, mem deolinh hoat trong tuduy khisu dijng eac kien thtic da cd, tap hpp nhieu each gial khac nhau va tim ra duoc phudng an toi uu. Thi/c hien dupc bien phap nay se gdp phan ren luyen tinh nhuan nhuyen - mpt trong nhiing dae tmng co ban ctia TDST.
S r ^ ; C h o ttidien ABCD cd AB = CD = a; AD = BC = b; AC = DB = c. Hay tinh thetich cua tiidien theo a, b, c.
' Tnroni Dai bpc ur pbaa, Sai bQC Thai Neuytn
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Trong chuong Irinh Hinh hoc 1 1 , HS da quen thupc voi cong Ihiic linh thelich cua mot hinh chop la v - j M (•). Neu HS ap dung cong thi?c (•), cac em se g3p I<h6 I<h5n trong viec tinh di/ong cao cija tudien.
V|y, diing truoc I<h6 l<iian do ta se giai quyet BT nhu thenao?
Phan tich:Va\ BT nay, GV cothe huAig dan HS xet BT theo cac phuong dien: - Nhin tiidlen gan deu la mptbp phan ciia tudien vuong: - Nhin tudien gan diu la mpt bp phan cua hinh hop: - Nhin tiJ dien gan diu lam6t tli'dien.
Nhuvay, ta se co cac each sau de giai BT;
Cach 1. Van de dat ra la ta xetxem tudien da cho CO lien quan gi den tudien deu hay tudien vuong hay khong.
That vay,trang m a t p h i n g (BCD), dung tam giac MNP sao cho B, C, D lan lupt la trung diem cua IVIN, NP, MP {hinh 2). Khi do: AB = CD = ™ => AMAN vuong t?i A. Tuang tu, ANAP,
AMAP vuong tai A va '',.,. = ji',.,„.. T l i dien AMNP vuong nen ta co '.,.. = T^^r Af^Af. Gpi dp dai 3 canh A M , AN, AP lan lupt la X. y, z.
Khi dp: x^ + y* = 4a^: x^ + z^ = 4b^; y^' + z^ = 4c^
Wall 2
G lai hp tren ta dupc: .,S?
• TV Vay, r . „ . i J2„,-+/.•-,'I Va,.'*.-'-/.'. ^ -
Do do: 1'.. • j j j * ' ' + ' ' ' - ' ' i A ' ' * ' ' - * ' ! v ^
Cach2:Km\\ tiidipn va hinh hop co quan he mat thiet vdl nhau; n§'u cho trudc tii dien, ta luon co hinh hOp ngoai Hep tiJdlen do. VcS tiidien npi tiep trong hinh h0p, ta cd: tli dien dat d m pt goc cua hinh hop cd the tich b4ng i the tich cua hinh hop va tiidlen tao thanh ti>6 ducfng cheo cua 6 mat hinh hop co the tich bang
^ the tich hinh hop. Trong tardfng hpp hinh da cho la hinh hft) chunhatthi tudien tao boi6 dudng cheo cua 6 mdt cOa hinh hop la tii dien cd cac cap canh dol bdng nhau. Do do, ta tinh thetich cua tiidien ABCD thong qua thetich hinh hop chu nhat.
Trudc het, ta tinh dp dai ba canh cija hinh hop chUnhat khi biet dp dai cac canh cua tli dien ABCD.
Gpi do dai ba canh cua hinh hop chi? nhat la X, y, z.
Khi do:
x'+y'=a' • x'+z' =h' • r + j -
{hinh 3). Gial he 3 phuang trinh nay ta dupc:
|(»-'+/>-'-<-)" ^ h<i'+c'-h') tu'+li'-a=) V 2 •••"V i -"°V i •
•.V(i('+/i'-<-') 7("'+ ("•'-/>') •jlr+lj'-u'\
Mat khac: f',„
Vay:''•*» = —77>/i"'+/>'-i-').7i'i'+' -/»')•Ji,-*h--a'j, Cach 3. Do khi sudyng cdng thiic, HS gap khd khan trong viec tinh toan thetich tiidien ABCD [hinh 4). Vi vay, HS ed the thiet lap mpt cdng thii'c khac thuan loi hdn trong viec tinh toan. Ta cd;
vcri IJ la doan vudng gdc Chung ciia AB va CD;
a=(ABycB).
Vi tli dien ABCD la tii- dien gan deu nen' AB = a
= CD;DA = CB = b:AC = BD = c va doan vudng gdc chunglJnenl,Jlanluptla
tojng diem ctia AB va CD. Xet tam giac vudng AU ed:
Ji-=JA'-IA'- Vdi:
Hinh 4
JA' = AC+Aiy ciy_h'+(^
2 4 Ir+r-er
IA- = -
2 4 ' 4 ' jr-=" "^' " .GoiM la trung diem eiia B C , t i ) M k e
MN //AB; MP//CD => MNP =a.TUdngti/,tam giac IZ+tZ-i'
Suy ra: ixyaa |M/VN
U'-'-^L,
Jtr +lr -<•• v<r
Vay, V.,: "6V2 ^..lir+iT-i^Y'jtr+1^-if .jr^tj-d
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Tap chl Giae due so 305 49
3. Ren luyen c h o HS biet each phan tich BT de tim ra each giai dpc dao
Mpttrong nhung khdkhan ldn nhatcua HS khi giai cae bai tap HHKG la sukhae nhau giua kien thiic ve hinh hpc phang da hpc d ldp dudi vdi kien thuc ve HHKG. Degiai quyet dupe khd khan nay, GVed the hudng dan HS phan tieh BT,tim sulien he VCT nhimg kien thiic trong hinh hpc phang da biet, tiidd dua viec giai quyet BT HH KG ve BT hinh hpe phang. Viec lam nay khdng chi giiip ngudi hpc tim ra phudng phap giai BT ma cdn ren luyen tinh ddc dio cua tu duy.
67"5.'Cho hinh tiidien ABCD cd dien tich ba tam giac ABC, ACD, ABD bang nhau va tong ciia ba gdc phang goc tam dien dinh A b&ng 180° (hinh 5) Chiing minh rang tii dien da cho cd cac
cap canh doidien bing nhau tung ddi mpt
^ Chdng minh:)(enrong mat phang (ABD), quay cac tam giac ABC, ADC lan lupt quanh cac taic AB, AD den n^m tren mat phang (ABD) sao eho D,C, n i m khac phia ciia dudng thing AB;
cac diem B, C^ nSm khac phia eua dudng thingAD(/)/n/7 6).
K h i d d : A C , = A C ,
= AC; BAC = BAC,; . . . ^ ^ . - ^ DAC = DACj.Theo gia thiet: ba gdc phang gdc tam dien dinh A b i n g 180°, nen A, C,, C^ thang hang.
G o i H , K lan lutJtlahinh chieu vudng gdc eiia B, D tren dudng thang C,Cj. Ta cd: S^„,.^ -5^,„-. suy ra BH = DK; vi AC, = AC^, dong thdi BH // DK nen tii giacBDKH la hinh chl? nhat v a C , C j / / B D .
Ke dudng eao AE eua tam giac ABD. Khi dd: AE - DK = B H . M a t k h a c 5 , „ „ = 5 , „ , n e n A E . B D = DK.AC3, suy ra BD = AC^. D§n den tiigiac ABDC^ va ADBC, la nhiJng hinh binh hanh (cd mot cap canh doi song song va bang nhau). Vay, AB = DC^ (hay AB = CD);
AD = BC,(hayAD = BC).
Deren luyen eho HS giai m d t s o B T HHKG mot each d'octfa'o trong nhung tn/dng hpp cdthe, GV nen khuyen khich eac em xem xet, phan tich BT de tim ra nhung moilien hetrong nhutig sukieri ben ngoai tudng nhu khdng cd lien he vdi nhau. Ching han, nhin be ngoai thi tu dien va hinh hop hien nhien khdng cd st/
lien he, cac tinh ehatse khac nhau nhung trong thucte, nhieu BT ve tii dien chiing ta cd the giai quyet thdng
50 Tap chi Giao due so 305
qua hinh hop bdivi bat ciimpt tiidipn nao eung cd mdt hinh hop ngoai ti^,dupc gpi laphuong phap longghep.
S r ^ . C h o tudien ABCD vad la khoang each giiia hai dudng thing AB va CD. a = ( A B ^ ) . Chiing minh rang: v^„ = UiAB.CD-!i[t\a.
Dung hinh hop AC'BD'A'CB'D ngoai tiep tudien ABCD [hinh 7}.
la Do
v..
Mat Kv,>, Lai
co: /T^
do: \ \
= V„,,„+4V»j;;„. \ '•
khac: \ , 0 3 c6: 1'.=,'-^,,., =^.fMflcijs
* • • ' \
''-•--••^<X
Hint} 7 na, s u y
Trong DH toan dTHPT, HHKG la mpt npi dung toan hpc mang tinh tniu tupng cao. De hpc \6\ npi dung nay ddi hdi HS phai ed su tu duy linh hoat, dc phan doan nhanh khi xuli cae hinh ve, Vi vay, GV can cd nhdng phuong phap su pham phti hpp, giiip HS phat tnen NLTDST thdng qua vipc giai cac bai tap toan HHKG. •
(1) Nguyen Ba Kim. Phmmg ph^p day hoc m6n Toiin.
NXB Dgi hoc sic phgm, H. 2004.
(2) Tran Luan. " D a y hgc sang tao mOn Toan a trirdng ph6 th6ng". Tap chi NghiSn ciiu gido due, sO' thang 3/1995.
(3) Ton Than. Xdy dung hi thd'ng cdu hdi vd bdi tdp nhdm boi duang mot s(^yiu td'cua tuduy sdng tgo cho hgc sinh khd vd gidi & trudng Irung hgc ca s&
Viil Nam. Luan an li^n sT khoa hpc Giao due, Vi$n Khoa hoc giao duc, 1995.
SUMMARY
The goal of teaching math in high school not only provides the mathemaUcal knowledge but also to prac- tice the skills and development thinking capacity, particularly the capacity of creative thinking for stu- dents. There have been many studies on creative thinking capacity development for the teaching of mathemaUcs in school. In which the researchers fo- cused mainly on: Students train to look at problems in different comers (to train ffexlble of thinking): train- ing capable to find many solutions to a problem (to exercise flexibility and clever way of thinking): train- ing ability to analyze problems to lind a unique solu- tion (to train the unique way ofthinkmg). In this arti- cle we refer to the development of creative thinking capacity for leaming through teaching assignments spatial geometry inhigh school.
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