Bach Phucme Vinh Tap chi KHOA HOC & CONG NGHE 83(07) REN LUYEN MOT SO HOAT DONG TRI TUE CHUNG CUNG VOI CAC HOAT DONG TRI TUE PHO BIEN TRONG TOAN HOC CHO HOC SINH LOfP 9 THONG QUA BAI TAP HINH HOC PHAN[.]
Trang 1REN L U Y E N M O T SO H O A T D O N G TRI TUE C H U N G C U N G VOI
CAC H O A T D O N G TRI T U E P H O BIEN T R O N G T O A N H O C C H O
HOC SINH LOfP 9 T H O N G Q U A BAI TAP HINH H O C P H A N G
Bach Phuong Vinh
Trii-dng Dai hoc Su pham - DH Thdi Nguyen
TOM T A T
Day hgc giai bai tap hinh hgc phang d ldp 9 nham thuc hien mdt trong nhQ-ng nhiem vu ciia mdn hgc la phat trien tri tue cho hgc sinh; dieu nay se cd y nghTa sau sac ban neu ngudi giao vien ludn tao co- bdi cho hgc sinh thuc hien cac boat ddng tri tue chung: phan tich, tdng hgp, tuong tu, khai quat hda, dac biet bda cung vdi cac boat ddng tri tue phd bien trong toan hgc: phan chia trudng hgp, lat ngugc van de, xet tinh giai dugc trong qua trinh hgc sinh di tim ldi giai va suy nghT khai thac bai tap binh hgc
Tir khoa: Hogt dong tri tue, tu duy, hoc sinh, bdi tap hinh hoc, lap 9
Phat trien tri tue cho hgc sinh (HS) la nhiem
vu ciia mgi mdn hgc trong trudng phd thdng,
nhat la ddi vdi mdn toan d trudng trung hgc
ca sd (THCS) cang cd nhieu dieu kien thuan
lgi de thuc hien nhiem vu nay Ddi vdi day
hgc giai bai tap hinh hgc phang d ldp 9, de
thuc hien nhiem vii tren ngudi giao vien (GV)
phai ludn tao cho HS co- hdi thuc hien cac
hoat ddng tri tue (HDTT) chung: phan tich,
tong hgp, so sanh, tuong tir, khai quat hda,
triiu tugng hda, dac biet biet hda cung vdi
cac HDTT phd bien trong toan hgc: phan chia
trudng hgp, lat ngugc van de, xet tinh giai
dugc trong qua trinh HS di tim ldi giai cua
bai toan Dieu nay se cd y nghTa sau sac hon
neu GV ludn tao co- hdi cho HS thuc hien cac
HDTT chung eung vdi cac HDTT phd bien
trong toan hgc khdng chi d viec HS di tim ldi
giai ciia bai toan hinh bgc ma d ca viec HS
nghien ciru khai thac bai toan Dd cung chinh
la muc dich day hgc cua mdn bgc nham phat
trien tu duy sang tao cho HS
Cac dang toan hinh hgc phang ldp 9 rat phong
phii va da dang Mdi dang toan deu cd nhQng
phuang phap (PP) giai ca ban va dac trung,
tuy nhien khdng phai liic nao tuan theo nhung
phirang phap dd deu giai dirge bai toan; ma
con ddi hdi HS phai biet nhin bai toan mdt
each tdng hgp de phan tich bai toan quy la ve
quen biet phan chia trudng hgp so sanh khai
quat hda, dae biet hda, tdng quat hda, biet lat ngugc van de va xet tinb giai dugc ciia bai toan de lira chgn nhQng PP va each thu-c phii hgp, hieu qua nham giai quyet bai toan, dua ra dugc ldi giai, tien tdi cd ldi giai hay, ngan ggn, ddc dao tii- dd de xuat nhung bai toan tuang tu, dac biet va cung cd the la nhQng bai toan "khai quat, tdng quat hon", nhQng bai toan mdi Trong qua trinh dd HS dugc ren luyen cac HDTT chung cimg vdi cac HDTT phd bien trong toan hgc, gdp phan phat trien cho HS kha nang quan sat, nang luc phat hien giai quyet van de va tu duy sang tao Sau day la bai tap binh bgc Idp 9, xuat phat tii-viec di tim ldi giai va khai thac bai toan nham ren luyen cho HS mdt sd HDTT chung cimg vdi cac HDTT phd bien trong toan hgc
Vi du 1 "Cho tam gidc deu ABC ndi Hep
trong dirang trdn (O) Diem M thudc cung
BC
Chimg minh rdng MA = MB + MC "
• Phdn tich bdi todn tim cdch gldl
Muon chirng minh MA = MB -i- MC (phan tich tach ra nhung thudc tinh ciia bai toan (cai toan the)) ggi cho HS lien tudng den viec tao doan thang AD nam tren MA sao cho
AD = MC (hoac AD = MB), hinh (H 1); khi
dd chi cdn phai chimg minh MB = MD (hoac
MC = MD) Dieu nay cd dugc tu cac cap tam giac bang nhau
133
Trang 2Bach Phuang Vinh Tap chi KHOA HOC & CONG NGHE 83(07): 133- 138 Neu nhin bai toan theo quan diem bien hinh
(til- mdi quan he giQ-a hai each giai bai toan
theo PP tdng hgp va PP bien hinh), ggi cho
HS lien tudng den viec ddi MC den MA, d
day siJ- dung phep quay tam B gdc quay 60°
chieu quay ngugc chieu kim ddng hd; dua vao
tinh chat ciia phep quay suy ra dieu phai
chimg minh
• Trinh hdy ldi gidi (HD tdng hop - hi/p Itii
ctic phdn cua hdi todn )
+yi Cdch 1: Phirang phdp tong hap
Lay D G A M s a o c h o M C = DA (1),
cd AABD = A C B M (c.g.c) => MB = DB va
BMD = 60" (gdc ndi tiep chan cung AB)
=> ADBM diu => MB = MD (2)
Til- (1) & (2) => MA=MD + DA=MB -^ MC
dieu phai chirng minh (dpcm), hinh (H I)
+^ Cdch 2: Phucmg phdp bien hinh
Theo gia thilt (gt): (MC, MA) = 60"
=> Q(B, 60°); MC -> MA
C ^ A (vi AABC diu)
M ^ D e MA
(chieu quay ngugc chilu kim dong hd), theo
tinh chat ciia phep quay => MC = DA (1) va
BM = BD, MBD = 60° => A BMD diu =>
MB = MD (2) Til-(1)& (2) t a c o :
MA = MD-^ DA = MB + MC (dpcm)
• Khai thdc bdi todn
> Khai thdc bdi todn theo hudng tim
them nhieu cdch gidi khde nhau
*) HD Phdn tich bdi todn theo PP gidi
4- Chung minh MA = MB -^ MC theo each
1, dat MC tren MA bang each lay D e MA sao
cho MC = DA va chimg minh MB = MD; ma
MA, MB, MC cd vai trd nhu nhau vi chiing
deu la day cung ciia (O);
*) HD tuang tw vd xet tinh gidi duac
Cdch 1.1: Dat MC tren MB bang each
lay D thugc tia doi ciia tia MB sao cho MD
= MC Khi do, chiing minh MA = MB -I- MC
^ MA = DB C= A MAC = ADBC (c.g.c),
hinh (H 1.1);
Cdch L2: hoac dat MB tren MC, bang each
lay D thugc tia ddi cua tia CM sao cho CD =
MB Khi do, chirng minh MA = MB -H MC
<= MA = MD <= A A M D diu <= A M A B
= A DAC (c.g.c), hinh (H 1.2)
i- Cdch 3: Chung minh MA = MB -i- MC,
ggi cho HS lien tudng den PP chung minh dang thuc hinh hgc nhd cac ti sd cd tir hai tam giac ddng dang Xet A MBE ' ^ A MAC
va A M C E ~ A M A B
- ^ N 1 B ^ _ ^ MC_ EC
MA " AC ' MA " BA
^ MB MC BE EC MB-hMC
=> + = -I- c^ = 1
MA MA AC BA MA
=> MA = MB + MC (dpcm), hinh (H 2)
B^
(H1.2)
( H 2 )
Trang 3I Ciich 4: Chung minh MA = MB -H MC,
agi cho HS lien tudng din PP chiing minh
dang thuc hinh hgc dua vao tinh chat cua
dirdng phan giac trong tam giac
Theo gt ta co BMA = AMC = 60", nen MA la
dirdng phan giac cua gdc BMC
MB _ BE _^ _ MC.BE
EC
MC.BE
=>
MC
r> MB + MC
EC
MC.AB
- + MC =
EC MC(BE-HEC)
EC
MA (vi AMCE -^MAB)
CE
(dpcm), hinh (H 2)
i Cdch 5: Chimg minh MA = MB + MC,
ggi cho HS lien tudng den PP van dung djnb
ly Ptoleme vao tii giac ABMC ndi tiep
du-gngtron (O) tacd:
AB.MC + AC.MB = BC.AM vi AB, BC, AC
la cac canh ciia tam giac deu, nen MA =
MB + MC(dpcm), hinh(H2)
Trong day hgc giai bai tap hinh bgc phang
neu ngudi GV ludn tao cho HS thdi quen tim
nhieii ldi giai ctia bai toan, dieu dd khdng chi
la CQ- hoi de HS dugc ren luyen cac HDTT,
ina HS con dugc he thdng hda cac kien thu-c
ki nang da hgc, the hien d cac dang tri thirc:
tri thuc ngi dung, tri thuc chuan, tri thirc gia
trj va dac biet la tri thuc phirong phap
> Khai thiic bdi todn theo huang de
xuat bill todn mai (bai toan tuong tu, khai
quat hoa, tdng quat hda, )
*) Ren luyen HD phdn tich tdng hop, twang
tir, khiii qudt hda cimg v&i HD lat ngirac
van de, phan chia trudng hop vd xet tinh
gidi dirge ciia hdi todn
Nham tra ldi cho cau hdi: M e cung BC tbi
MA = MB + MC, ngu-gc lai, nlu cd MA =
MB + MC thi M cd thudc cung BC khdng?
*) Ren luyen HD phan tich vdi HD phan chia
trudng hop va xet tinh giai dugc: Xet cac vj
tri tu-o'ng dii ciia dilm M vdi AABC va
du-dng tron (0) ngoai tilp tam giac
V M e BC (til- kit qua cua vi du 1) deu cd
tinh chat MA =
MB-^MC-Neu M d trong AABC (M 7^ B, M ^ C) thi
M A < M B + MC, hinh(H3);
N I U M d ngoai AABC ( M g BC ) thi MA < MB-HMC, hinh(H4)
That vay, ta cd gdc hgp bdi dudng thang MC vdi AM khac 60" nen trong phep quay Q(B
60 ), chieu quay ngugc chieu kim ddng hd
M G MC cd anh la D g AM => ABMD la tam
giac diu =^ MB = MD = BD
=> AABD = ACBM (c g c) =^ MC = DA
ma MA < MD + DA (do xet AMAD)
Dodd M A < M B - ^ M C Nhu vay, chi cac diem M e BC thda man MA
= MB + MC, nen ta cd bai toan dao ciia vi du
1 nhu sau:
Bdi todn J.L Cho tam gidc deu ABC, neu MA
= MB + MC thi M ndm Iren cung BC cua dirirng Irdn ngogi Hep tam gidc deu ABC
*) HD tdng hap: Ket hgp vi du 1 va bai toan
1.1 di den bai toan quy tich:
Bdi todn 1.2 Cho tam gidc deu ABC Chu-ng minh rdng quy tich nhung diem M thod mdn
MA = MB ^- MC Id cung BC cua dudng tron tvjoai tien taw vide deu A RC
(H3)
(H4)
135
Trang 4Bach Phuang Vinh Tap chi KHOA HOC & CONG NGHE 83(07): 133- 138
*) Ren luyen HD tong hffp, khdi qudt hda tu
ket qud cua HD phdn tich ciing v&i HD
phdn chia trir&ng hap:
Tu- cac ket qua cua vi du 1 bai toan 1.1; 1.2
di den bai toan khai quat hda:
Bdi todn 1.3 Trong mat phdng cho tam gidc
deu ABC vd mot diem M bdt ki Chiing minh
rdng MA < MB + MC Ddu bdng xay ra khi
vd chi khi M ndm tren cung BC cua dirdvg
trdn ngogi tiep tam gidc deu ABC
*) Nhan xet: Tir bat dSng thu-c MA < MB +
MC ggi cho HS lien tudng den bai toan cue
trj hinh hgc va di den bai toan mdi sau:
Bdi todn 1.4 Cho tam gidc deu ABC ndi Hep
trong dirdng trdn (O) Hay xdc dinh vi tri ciia
diem Mtren cung BC sao cho tdng MA + MB
-I- MC cd gid tri lan nhdt
Theo kit qua tren MA + MB -^ MC = 2MA
<» M G cung BC Tong MA -i- MB + MC se
Idn nhat khi MA Idn nhat, MA la mdt day ciia
(Q) nen Idn nhat khi nd la dudng kinh ciia
(O) Vay M = I (I = AO n BC ) la dilm chinh
giua cua cung BC , hinh (H 5)
*) Quan sdt hinh ve (H 6) va tiep tuc phdn
tich bai toan: TQ- bat dang thuc FMB -i- MC >
MA, ta cd do dai MA ludn thay ddi Neu lay
mdt diem N d ngoai (O) va thudc mien trong
gdc BAC tbi MB -^ MC + MN > AM + MN
> AN
Do dd neu B C, N cd djnb => A cd djnb =>
Tdng MB -I- MC -^ MN cd gia trj nhd nhat la
AN <» A M, N thang hang hay M =
AN n BC , ta di den bai toan mdi:
Bdi todn 1.5 Xdc dinh diem Q thudc mien
trong tam gidc ABC sao cho tdng QA QB
-i-QC cd gid tri nhd nhdt
Dung tam giac deu BCN sao eho A va N nam
ve 2 phia cua BC Dung dudng trdn ngoai tiep
A BCN ^> Q = AN n BC cua dudng trdn
ngoai tilp A BCN, hinh (H 7)
*) Nhan xet: De cd giao diem Q thi AABC
phai cd cac gdc khdng Idn hon 120 Trudng
hgp AABC cd gdc Idn hon 120° thi Q chinh
la dinh ciia gdc Idn nhat Cd the xac dinh
diem Q nhu sau: Q = B P n AC cua dudng
trdn ngoai tiep tam giac deu ACP (hoac Q =
C M n AB ciia dudng trdn ngoai tiep tam
giac diu ABM), hinh (H 8)
(H5)
(H6)
(H7)
*) HD tdng hgrp, tir kit qua ciia bai toan va
cac nhan xet tren, de xuat cac bai toan chung minh sau:
Bdi todn 1.6 Cho tam gidc ABC dimg cdc tam gidc deu MAB, NBC, PAC thudc mien ngodi tam gidc ABC Chimg minh rdng 3 duang trdn ngogi Hep 3 turn gidc deu do cimg
di qua mot diem
Bdi todn 1.7 Cho tam gidc ABC dimg cdc tam gidc deu MAB, NBC, PAC thuoc miin ngodi tam gidc ABC Chimg minh rdng duang thdng MC, NA, PB ddng quy tai mdt
Trang 5diem chinh Id giao diem cua ba duang trdn
ngogi tiep ba tam gidc deu
Bdi todn 1.8 Cho lam gidc ABC dimg cdc
lam gidc deu MAB, NBC, PAC thudc mien
ngodi tam gidc ABC Chimg minh rdng
MC = NA = PB vd gdc tgo bai hai dogn
ihdng bdng nhau dy bdng 60"
*) Nhu vay, qua trinh phan tich bai toan tim
Idi giai va khai thac vi du 1, HS dugc ren
luyen cac HDTT chung ciing vdi eac HDTT
pho bien trong toan hgc:
- Thuc hien HD lat ngugc van de, xet bai toan
1.1 la bai toan dao ciia vi du 1;
- Thirc hien HD tdng hgp: ket hgp vi du 1 va
bai toan 1.1 cd bai toan 1.2 la mdt bai toan
quy tich;
- Thuc hien HD phan chia trudng hgp, xet
tinh giai dugc ddi vdi vj tri tuang ddi ciia cac
hinh da cho ket hgp vdi HD tdng hgp va khai
quat hda tu cac ket qua cua vi du 1; bai toan
1.1; bai toan 1.2 ta cd ket qua khai quat hda la
bai toan 1.3;
- Thirc hien HD tdng hgp tir bat dang thirc cua
bai toan 1.3 ggi cho HS lien tudng va di den
cac bai toan cue trj: bai toan 1.4, bai toan 1.5;
- Xet cac trudng hgp cua bai toan cue tri ta
phai giai quyet bai toan dung hinh va de xuat
dirge bai toan chimg minh cac dudng ddng
qui: chii'ng minh cac dudng trdn ddng qui
-bai toan 1.6; chirng minh cac dudng thang
dong qui - bai toan 1.7;
Tong hgp cac ket qua tren de xuat dugc bai
toan 1.8 rat thii vj, ma tri thuc phuong phap
hinh thanh trong bai toan 1.8 dugc van dung
de giai mdt chudi cac bai toan tuong tu va ind
rgng cua bai toan 1.8
*) Tri thirc phuong phap ap dung de giai bai
toan 1.8 (sii- dung PP tdng hgp hoac PP biln
hinh):
- Chung minh cac cap tam giac bang nhau;
- Tir hai tam giac bang nhau suy ra cac yeu td
tuong irng bang nhau;
- Van dung tinh chat: Trong hai tam giac cd hai gdc tuong irng bang nhau tung ddi mdt thi gdc tuo-ng irng thu- ba cua chiing cQng bang nhau, di den dpcm
*) Tri thuc PP giai bai toan 1.8 cd the sir dung
de giai cac bai toan tuong tu va md rdng tu-bal toan 1.8 dugc de xuat bang each thay ddi dieu kien cua bai toan
Bdi todn 1.9 Cho tir gidc ldi ABCD, dimg
cdc tam gidc deu MAB NCD thudc mien ngodi cita tir gidc vd tam gidc deu PBC thudc mien trong ciia tir gidc Chirng minh rdng MP
= AC, PN = BD vd gdc tcio bdi hai dogn thdng bdng nhau bdng 60'
Bdi todn 1.10 Cho 3 diem thdng hdng A, B,
C theo thir tw do Tren nira mat phdng bd'
AC dwng cdc tam gidc deu MAB, NBC Chirng minh rdng AN = CM vd gdc tgo bdi hai dogn thdng do bdng 60"
• %
\ • " V
"-vA /
* ^ ' „
-/ '^
/ '•;
/ I
/ / /' /
B^
'4'
(H9)
Bdi todn PIP Cho 3 diem thdng hdng A, B, C theo thir tir dd Dimg tam gidc deu M4C', NAB thudc ve hai nira mat phdng ddi nhau hd AC
137
Trang 6Bach Phuonn Vinh Tap chi KHOA HOC & CONG NGHE 83(07): 133- 138
Chimg minh rdng MB =
dogn thdng dd bdng 60
NC vd gdc tgo bai 2
*) Md rdng cac bai toan tren thay viec dung
cac tam giac deu thanh dung cac hinh vudng;
sir dung tri thuc PP hinh thanh d bai 1.8
chung minh dugc hai doan thang bang nhau
va gdc tao bdi giua chimg bang 90
Bdi todn 1.12 Cho tam gidc ABC Dirng cdc
hinh vudng ABDE, ACFG thudc mien ngodi
tam gidc ABC Chimg minh BG = CE vd gdc
tgo bai hai dogn thdng dd bdng 90
Bdi todn 1.13 Cho tam gidc ABC Dimg cdc
hinh vudng ABDE, ACFG thudc mien ngodi
tam gidc ABC Goi M la trung diem cua BC;
Ol, O2 Idn luat Id tdm ciia cdc hinh vudng
tren Chimg minh MOi = MO? vd gdc tgo
bai hai dogn thdng do bdng 90
Sii' dung ket qua bai toan 1.12 cho bai toan
1.13 Ap dung nhanh chdng PP giai bai toan
1.8 cho bai toan 1.14 la sir dung ket qua ciia
bai 1.13
Bdi todn 1.14 Cho lirgidc ldi ABCD Dimg
cdc hinh vuong ABEF, BHCI, CDPQ, DARS
thudc mien ngodi cua tir gidc Goi Oi, O2 O3,
O4 kin lugt Id tdm ciia cdc hinh vudng tren
Chirng minh O1O2 = O3O4 vd gdc tgo hdi 2
dogn thdng do bdng 90'
Bdi todn 1.15 Cho 3 diem thdng hdng A.B, C
theo thir lu do Pren nua mat phdng bd AC
dimg cdc hinh vudng ABDE, BCFG Chimg
minh rdng AG = CD vd gdc tgo bai hai dogn
thdng dd bdng 90
Bdi todn 1.16 Cho 3 diem thdng hdng A,B, C
theo thir dd Dimg cdc hinh vudng ACDE,
ABFG thudc ve hai nira mat phdng ddi nhau
bd AC Chimg minh rdng BE = CG vd gdc
tgo bdi hai dogn thdng dd bdng 90
Xuk phat til- viec di tim ldi giai va khai thac
phat trien bai toan hinh hgc mot each cd he thdng vdi nhQng HD toan hgc, GV tao dieu kien eho HS tu nhin nhan phan tich bai toan tim each giai va de xuat nhung bai toan mdi bai toan tuong tu, bai toan md rdng cua bai toan ban dau, tir bai toan 1.1 den bai toan 1.16 Dd la ca qua trinh HS dugc ren luyen cac HDTT chung: phan tich, tdng hgp, tuong tir, khai quat hda ciing vdi cac HDTT phd bien trong toan hgc: lat ngugc van de, phan chia trudng hgp va xet tinh giai duge thdng qua giai cac dang toan hinh hgc d ldp 9, da gdp phan khdng nhd vao viec phat trien tri tue
va tu duy sang tao cho hgc sinh
TAI LIEU THAM KHAO
[I] Hoang Chung (1997), PPDH Todn hoc a
trudng pho ihdng THCS, Nxb Giao d'uc
[2] Phan Diic Chinh, Ton Than, Nguyen Huy Doan Pham Gia Due, Truang Cong Thanh
Nguyen Duy Thuan (2008), Todn 9 - Tap 1 vd 2,
Nxb Giao due
[3] Vu Huu Binh (2008>, Ndng cao vd phdt trien
todn 9 - tiip 2, Nxb GD
[4] Nguyen Ba Kim, Vuang Duang Minh, Ton
Than (1999), Khuyen khich mot sd hogt ddng tri
tue cita HS qua mdn Todn d Trudng THCS, Nxb
Giao due
[5] Ton Than, Pham Gia Dire, Truong Cong
Thanh, Nguyen Duy Thuan (2008), Bdi tap todn 9
- Tap 2, Nxb Giao due
[6] Vu Duang Thuy, Pham Gia Dire, Hoang Nggc
Hung, Dang Dinh Lang (1998), Thuc hdnh giai
Todn, Nxb Giao due
SUMMARY
PRACTISING SOME OPERATIONS WITH GENERAL INTELLECTUAL ACTIVITIES IN COMMON INTELLECTUAL MATHEMATICAL GRADE 9 STUDENTS THROUGH EXERCISE PLANE GEOMETRY
Bach Phuong Vinh
College of Education - TNU
Teaching award exercises the plane geometry in grade 9 to fulfill one's duties is subject to the intellectual development of students will have a deeper meaning if the teacher is always an opportunity for students to perform general intellectual activities: analysis, synthesis, similar generalizations, especially of with intellectual activity common in mathematics: division cases, reverse the problem, considering the resolution is in the process of students finding the solution and thinking exercises exploit geometry
Key words: intellectual activity, thinking, students, homework geometry, grade 9