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CAC PHU'aNG PHAP GIAI TOAN QUA C A C KY THI OLYMPIC NHA XUAT BAN BAI HQC Q U G C GIA T H A N H P H O H C H I MINH - cAc PHi/dNG PHAP GIAI T O A N Q U A C A C Ki T H I O L Y M P I C Trail Nam Dung (Chu bien) V6 Quoc Ba Can, LePhiic Lff ,0 NHAXUATBAN D A I H O C Q U O C GIA T H A N H P H O H O C H I M I N H Khu 6, Phatag Linh Trung, Quan Thu DCfc, TPHCM So 3, Cong trucrng Quoc te, Quan 3, TP Ho Chi Minh DT: 38239171 - 38225227 - 38239172 Fax: 38239172 - E-mail: vnuhp@vnuhcm.edu.vn PHONG PHAT HANH NHA XUAT B A N D A I H Q C Q U O C GIA T H A N H P H O H O C H I M I N H So Cong trucmg Quoc te - Quan - TPHCM DT: 38239170 - 0982920509 - 0913943466 Fax: 38239172 - Website: www.nxbdhqghcm.edu.vn I I A'! track nhiem xudt ban: NGUYEN H O A N G DUNG Chiu track nhiem noi dung: HUYNHBALAN To ckiic ban thdo va chiu track nhiem vd tdc quyen N H A XUAT BAN DAI HOC QUOC GIA TPHCM Bien tap: CAO NGHI THUC Siia ban in: NGUYEN HUYNH Trinh bay bia: Ma s6 ISBN: 978-604-73-1976-3 So lugng 1.000 cuon; khd 16 x 24 cm So dang ky He hoach xuat ban: 1447-2013/CXB/06-82/DHQGTPHCM Quyet dinh xuat ban so: 194 30/10/2013 ciia NXB DHQGTPHCM Ill lai C ong ty TNHH In va liao bi Hung Phii \ o p luu chieu quy IV nam 2013 LQI NOI D A U Thang 7/2013, Doan hoc sinh Viet Nam dir thi Toan qu6c t^ 2013 tai Colombia da dat tich vang doi voi huy chuang v ^ g va huy chuang bac, c6 02 huy chuang vang cua dai dien cac tinh phia Nam la Pham TuSn Huy (hoc sinh lap 11) va CSn frrSn Thanh Trung (hoc sinh lap 12) cua Truang Ph6 thong Nang khilu - Dai hoc Qu6c gia TP H6 Chi Minh (DHQG-HCM) Thanh tich chac chan se lam nuc long cac em hoc sinh chuyen toan a TP Ho Chi Minh noi rieng va cac tinh phia Nam noi chung De giup cac em dang yeu thich mon toan tiep can vai cac de thi va cac chuyen de Olympic toan hoc, cac giang vien Khoa Toan Tin h9c, Truang Dai hoc Khoa hoc Tu nhien, DHQG-HCM cung cac cong sir da thirc hien viec bien soan cu6n "CAC PHl/ONG PHAP GlAl T O A N Q U A C A C K Y T H I OLYMPIC" Noi dung cuon sach bao gom cac chuyen d8 thuoc tdt ca cac ITnh vuc toan Olympic: Dai s6, Giai tich, Hinh hoc, S6 hoc va T6 hop vai nhung muc dp chuyen sau khac nhau, vi thi, phii hop cho t^t ca cac hoc sinh chuyen toan Ben canh do, cuon sach ciing giai thieu cac dl thi va lai giai ciing nhung binh luan chi tilt cac ky thi quan nhdt ve T o ^ cua Viet Nam nam qua De CO dugc cu6n sach day dan vai nhung noi dung phong phu va bo ich nay, nhom tac gia chan cam an sir dong gop nhiet tinh tir cac thay c6 giao, cac ban hoc sinh, sinh vien, cac vien cua diln dan mathscope Chung toi cung cam an cac d6ng nghiep cua Chuang ttinh dilm quoc gia phat trien Toan hoc nam 2010-2020; cac d6ng nghiep Khoa Toan - Tin hoc Truing DHKHTN, DHQG-HCM; dac biet cam an GS Ngo Bao Chau, GS Le T u k Hoa va TS Nguyen Thj Le Huong da luon dong vien, c6 vu va djnh huang cho chung toi Cam an GS Dam Thanh San (Pai hoc Chicago, Hoa Ky) da gu"i cho chung toi file video ve bai toan MUC LUC cua I M O 2013 va ggi y tuang hinh bia cuon sach Chung toi cam an Cong ty c6 phkn giao due Titan da khich le • • ehiing toi hop tac vai Nha xuSt ban Dai hoe Qu6c gia TP H6 Chi Minh d l xuk ban cuon sach ,4 , Cu6i cimg, chung toi cam an sir yeu thieh va say me toan hoc ciia cac em hoc sinh chuyen toan la nguon dong lire giup chung toi bien soan cuon sach Chue cac em cong fit' Lcfi n o i d a u Nhu-ng d i c u t h u vj v l t6'ng P^.(n) = l * - + 2''+ Trcin Nam ' * ill 'i • • + 77^ Dung ii'i' ' ' Nhom tac gia B o (Ic n a n g l u y thufa va liTng d u n g Trail Minh Hien, Hoang Dang Thien, Vu Xudn Anh 19 rvu B a t b i e n v a nil'a b a t b i e n t r o n g cac b a i t o a n t r o chcTi Nguyen Thanh Khang 47 C a p so' n g u y e n , c a n n g u y e n t h u y va lifng d u n g Nvuxen Dux Lien 62 N g u y e n l y ci/c t r i rofi r a c v a m o t so' i^ng d u n g DiMng Di'fc Lam 79 X u n g q u a n h so' F i b o n a c c i ! LUi Hoang Minh Qudn 98 g i a i m o t b a i t o a n mnf Vd Quae BdCdn 137 I'• • • DifoTng d o t t r u n g va m o t so' b a i t o a n a p di^ng Trdn Ngoc Thdng 144 D i n h l y C a s e y v a tfng d u n g Nguyen Van Linh M p t so phuTcyng p h a p g i a i b a i t o a n t o n t a i t r o n g t o hgTp Nguyen Tat Thu 157 M O t e a c h d6'i b i n v a uTng d^ing t r o n g chuTng m i n h ba't d a n g N H O N G O I E U T H U VI V E thufc Nguyen D a y s6 Van Qui va hoc sink Nguyen Qudn Bd Hong 204 i^(n) = i*^ + 2*^ TONG H — ± p f ' : ' : : = ttn + ( u n ) " , : , j v 232 KieuDinhMinh fi • ' P h a n hoach nguydn v a phSn hoach tSp h^p TranNamDungt i6 ; • mi} 250 LePMcLa M o t hoc s i n h c h u y e n t o a n chac c h a n p h a i b i e t i t n h i e u v e t o n g LoTi g i a i de t h i c h p n H S G t o a n quoc gia n a m 2013 ,,,,,,,.,, ,„.„,, Sk{n) 285 = \'' + 2^+ + N g a y tOr i d p 4, I d p 5, c h a c c h a n c h u n g ta da tiJfng n g h e c a u c h u y e n ve c a u be Gauss da t i n h t o n g cac so n g u y e n tuf d e n 100 c h i t r o n g n h a y m a t : Lflfi giai d l t h i c h p n dpi t u y e n O l y m p i c t o a n quoc gia nam 2013 + + + • • • + 100 = 50 • (1 + 100) = 5050 / » p y j^ CD Chiing minh rang AD Id dudng phan gidc goc A Chu'ng minh Do D chia canh BC theo ti so vang nen ta c6 , Xung quanh so Fibonacci 129 Bai toan 40 Chiing minh rang neu tam gidc ABC Id tam gidc vdng thi goc d dlnh bang 36° ^'ru i^Lt to ftem rtniyj •tL:, \'a/t Chu'ng minh Gia su" ABC la tam giac vang vdi AB = BC = aAC Goi D la diem chia BC theo ti so vang nhu" hinh ve bai 39 Dya vao nhan xet d tren, ta c6 CAD cung la mot tam giac vang va dong dang vdi tam giac ABC, suy ACB = ADC = 2DAC Dat goc ACD = 2x thi tuf AACD, ta c6 BD 2x + 2x + x= 180° =^x = 36° B Vay ^ - C = 72°, = 36° • Sau day la mot so bai toan khac, mdi ban doc ciing thu: siJc Bai toan 41 Chiing minh rang neu tam gidc ABC can, canh ddy AC, CO goc d dinh bang 36° thi tam gidc ABC Id tam gidc vdng Bai toan 42 Cho tam gidc ABC can dlnh B D Id mot diem tren canh BC cho AABC dong dang ACAD Chiing minh rang ABC Id tam gidc vdng vd chi = a Mat khac, ABC la tam giac vang nen ta cung c6 AB = a AC Tvi hai ket qua tren, ta thu difdc : ' ' BD_AB CD ~ AC Hinh chif nhat vang Hinh chU nhdt c6 tl so chieu ddi chia cho chieu rong bang tl so vdng goi Id hinh chQ nhdt vdng A B Dieu chi^ng to AD la diTcJng phan giac goc A ciia tam giac ABC • Nhan xet Ta cung chu'ng minh du'dc rang: Neu ABC Id tam gidc vdng vdi canh ddy AC vd D Id chdn dUdng phan gidc goc A thi CAD cUng Id mgt tam gidc vdng •I h D c •' 130 Cdc phuang phdp gidi todn qua cdc ky thi Olympic Mot so hinh chu* nhat v^ng thiTc tien la cac bufc hoa noi tieng, hay cdc cong trinh c6 dai tren the gidi c6 ti le vang Chang han nhif hinh tufcJng chiia Jesus, biJc tranh nang Mona Lisa cua dai danh hoa Leonardo Da Vinci: Xung quanh so Fibonacci 131 5.4 Day so' Fibonacci trf nhien va kinh te xa hQi Day so' Fibonacci trf nhi^n 5,, Day Fibonacci xuat hien d khap ndi thien nhien Nhffng chiec \i tren mot nhanh cay moc each nhifng khoang tu'dng iJng vdi day so Fibonacci Cdc so Fibonacci xuat hien nhifng bong hoa Ngoai ra, so Fibonacci difdc tim thay su" sap xe'p choi la tren than cay, stog dong vat, gia pha loai ong difc, dang vong xoay qud thong, trai khom (thdm) |: , Hinh chi? nhat vang cung xuat hien d dong ho, hay cong trinh kien true CO Hau het cac b6ng hoa c6 so' cdnh hoa la mot cdc so' 3, 5, 8, 13, 21, 34, 55 hoac 89 Hoa loa ken c6 ba canh, hoa mao lUdng vang c6 canh, hoa phi ye'n thtfdng cd canh, hoa cue van tho cd 13 canh, hoa cue tay cd 21 canh, hoa cue thu'dng cd 34 hoac 55 hoac 89 cdnh Cdc so Fibonacci cung xuat Men cdc bong hoa hiTdng du-dng Nhu-ng nu nho se ket hat d dau bong hoa hu-dng 132 Cdc phuang phdp gidi todn qua cdc ky thi Olympic difcJng diTdc xep th^nh hai tap cac difdng xoan 6'c: mot tap cuon theo chieu kim dong ho, tap cuon ngu'cJc theo chieu kim dong ho So cac diTdng xoan oc hu'dng thuan chieu kim dong ho thudng la 34, c6n ngufdc chieu kim dong ho la 55 Doi cac sS' la 55 va 89, va tham chi la 89 va 144 Tat ca cac so deu la cic so Fibonacci ke tiep (ti so cua chiing tien tdi ti so' vang) Pln*appl* tnd it'i atmn Kung quanh so Fibonacci 133 Ban vanxd Fibonacci thu difdc bang viec chuyen day so Fibonacci day cac no't nhac theo qui tac chuyen mot so nguyen du'dng not nhac sau day: ^ , , , m So I tifdng iJng vdi not Do (C) - t ^0 • So tu'dng iJng vdi not Re (D) • So tufdng iJng vdi not Mi (E) • So tifdng iJng vdi not Fa (F) • So tu'dng iJng vdi not Sol (G) • So ttfdng iJng vdi not La (A) V'' Trai thong ciing vay Vong xoay tuf trung tarn c6 va nhdnh Tr^i khdm c6 ba xoay la 5, va 13 Lai mot bang chiJng nhiJng so' khong phai ngau nhien Thien nhien chdi tro Toan hoc vdi Chung ta? Khong biet nhtfng cac khoa hoc gia suy doan rang cac loai thUc vat moc theo hinh the xoay oc theo nhifng so' Fibonacci vi no tie't kiem nhieu be mat hdn Sap xe'p nhu" the', chung gia tang dieu kien tang tn^dng va do, nhieu dieu kien sinh ton hdn Day s6' Fibonacci Sm nhac Ban vanxd Fibonacci la mot ban nhac ma giai dieu cua no bat nguon tiif mot nhifng day so' noi tieng nha't Ly thuyet so - day so Fibonacci Hai so' dau tien cua day la so va so' 2, cac so tiep theo difdc xac dinh bang tong cua hai so lien tiep tru'dc no day • So tifdng iJng vdi not Si (B) • So tu-dng u-ng vdi not Do (C) • So tu-dng u-ng vdi not Re (D) • Va cu" tiep tuc nhu" vay Vi du, mot day gom so Fibonacci dau tien 1, 2, 3, 5, va 13 tu-dng iJng vdi day cdc not nhac C, D, E, G, C va A De xay difng nhip dieu vanxd ngu'di ta di tim cac doan nhac c6 tinh chu ky ban vanxd Fibonacci Doan nhac difdc gpi la c6 tinh chu ky neu nhif c6 the chia no doan giong het Vi du, doan nhac GCAGCA la doan c6 tinh chu ky, vi no gom hai doan giong GCA Day so* Fibonacci thi trifcTng chu'ng ichoan Fibonacci Retracements (ti le hoan lai theo Fibonacci) la mot cong cu rat bien phan tich ky thuat, no difa tren day so Fibonacci nha Toan hoc Leonardo Fibonacci xay difng vao the ky XIII Fibonacci Retracements dtfdc tao bang each ve du'dng thang ndi ket giffa hai diem gia cao nha't va thap nha't cua thi gia giai doan phan tich va phan chia khoang each theo chieu doc theo cdc ty le Fibonacci quan nhif 23.6%, 38.2%, 50.0%, 61.8% 134 Cdc phuang phdp gidi todn qua cdc ky thi Olympic vh 100.0% ( M o t so'phan m e m giao dich hien nhif Metatrader ttr dong sur dung cdc mu'c Fibonacci 0.0%, 23.6%, 38.2%, 50%, 61.8%, Xung quanh so Fibonacci 135 ^JTSNDM - Mtr 9f 10O012 OtMn 395.72 HI 398.17 Lo « 18 ClOM 36*88 (-3 JHl M o t k h i nhi?ng mu'c dUdc nhan dien thi cac dUcfng nam ngang dUcJc ve va du'cJc suf dung de nhan dien cac muTc h6 trd va cac mu'c khdng cif c6 the c6 Day so Fibonacci nhu" sau: 0, , , 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, • T i le Fibonacci 61.8% du'cJc tao bang cdch chia mot so bat k y day so Fibonacci v d i so h e n sau no V i du, , ^ = 0.6153, = 0.6179 • T i le 38.2% dUdc tao bang each chia mot so bat ky day so' Fibonacci v d i so d v i t r i thu" hai ve ben phai cua no V i du„ ^ = 0.3819 • T i le 23.6% du'cJc tao bang each chia mot so bat ky day so' Fibonacci v d i s o ' d v i t r i thiJ ba ve ben phai cua no V i d u , ^ = 0.2352 Cac nha dau tiT thiTdng SIJ: dung cac mu'c Fibonacci thoai l u i nhU cac mu'c khang ciT va ho trd hieu qua H o sijf dung mu'c de vao lenh mua/ban hoac xac dinh mu'c chan l o s-NWDEx Dôr^laQ012Opôl^B^.72.Hl^MJ7.Le^ô.1ằ,cl + a'^'^abc + b^'c + c'^a M a t k h a c , t h e o b a t d a n g thiJc A M - G M , h{a) = {2-k)a!'^Zko!'-'-k{2k-l){a-\)-~2{k + l) L a n lirdt t i n h d a o h a m c a p m o t v a c a p h a i c u a h{a), t a du'dc I h\a) = k{2-k)o!'-'+ /, b{a + cf = k' = k{2 - k){k - l)a'=-2 + 3kik - l){k - 2)0!^'^ ^ m a x < 3, — ( / c +1)^+1 j • Ke't h d p h a i b a t d a n g thiJc n a y l a i v d i n h a u , ta d e d a n g suy r a 3/0+1;,/= =k{k-l){2-k)a''-^{a-Z) a% + b'^c + d'a ^ b{a + c)^ ^ m a x { 3, D o < /c < v a ^ a ^ n e n h"{a) ^ D i e u n a y chiTng t o h'{a) la h a m n g h i c h b i e n t r e n [0, 3], D o h'{l) k + l ?,k{k-l)a''-^-k{2k-l), (A: + l)*=+i h"{a) • ^ /c'= \ / a+ c = n e n t a c h'{a) > B a i t o a n difdc chiJug m i n h x o n g {k + l) k+l 142 Cdc phuang phdp gidi todn qua cdc ky thi Bai toan mof mdi '^ Bai tap Cho xi, X2, • •, Xn {n ^ 2) Id cdc so thuc khong dm thda man X2 Chvlng minh rang xi + Ta thay rang b^i todn md cua Vasile Cirtoaje thifc cha't chinh la mot trirdng hop rieng cua bai toan sau (da dUdc Pham Kim Hung de xuat [2]): Bai toan Cho cdc so khong dm a, b, c co tong bang Dat: , TA- thuc khong dm co tong bang Chiing minh a%c + b'^cd + c^da + d'^ab a\a2 Bai tap (Anh, 1986) z = vd x'^ + y'^ + cHa bieu Bai ' ^" a^b + b'^c + c'^a + P(Xi, X2, , a'^b + b'^c + c'^a^2 + abc Cho x, y, z la cdc so thUc thoa mdnx + y+ = Tim gid tri Idn nhdt vd gid tri nho nhat Xn) 0:2, , x„ (n ^ 3) Id cdc so thuc khong dm 1- x„ = Tim gid tri Idn nhat cua bieu thiic: = xlxl + xlxl + ••• + X^X? + U^^''-'^X^xl • • • X^ [1] Vasile Cirtoaje, Algebraic Inequalities: Old and New Methods, GIL, 2006 [2] Pham Kim Hung, Sang tao bat dang thiic, Nha xuat ban Tri Thufc, 2006 [3] Vasile Cirtoaje, Vo Quoc Ba Can, Tran Quoc Anh, Inequalities with Beautiful Solutions, GIL, 2009 = + 6^ 0, s = (va r = 1, > 0) Ngoai ra, chung ta cung co lc(i giai cua no mot so trUdng hdp dac biet khac nhiT r = s; r = - \k s = 1 2 (va ngUc(c lai); r = - va s = - - (va ngifdc lai); Nhiftig lieu Bai tap Cho a b, c la cdc sothuc H Hay tim gid tri Idn nhat vd gid tri nho nha't (neu co) cua P{r, s) theo r, ,s M o 143 iJUcfng doi trung vd m^t so bdi todn dp dung Olympic + i 145 Dadng dot trung vd mgt so bai todn dp dung DUdNGDOl TRUNG VA MOT SO BAI TOAN AP DUNG T r d n Ngpc Thang^ That vay, theo dinh ly Ceva va dinh ly Thales, ta c6 TB B a i toan Cho tarn gidc ABC, cdc diem M vd N Ian lam nam tren cdc canh AB vd AC cho MN \\ Goi P la giao diem cua BN vd CM Dadng trdn ngoai tiep cdc tarn gidc BMP vd CNP cat tai hai diem phdn biet P vd Q Chiang minh rang BAQ = CAP Chij-ng minh.^ Gia suT L la giao cua AP va BC, ta se chu-ng minh L Ik trung diem cua BC JTA' WA _ _^ WB" ^ ^ LB r n A-'' ' Ma = nen ta suy = = = - , hay L la trung diem cua MB NB LC ^ BC suy AL la difcJng trung tuyen Do yeu bai toan tu'Ong dufdng v d i viec chiJng minh AQ la diTdng doi trung cua tam giac ABC ke tuf dinh A • MA Trong ky thi Balkan Mathematical Olympiad Ian thtf 26 nam 2009 c6 b a i todn hinh hoc Moldova de nghi c6 noi dung nhu" sau: W NC NA TriTdc he't ta nhac l a i mot so ke't qua lien quan den difdng doi trung cua mot tam giac Dtfcfng ddl trung DJnh nghia Trong mot tam gidc, dadng thang doi xiing vdi trung tuyen qua dadng phdn gidc ke ti( ciing mot dinh daac goi Id dadng doi trung 1.1 Cac t i n h chaft coT b a n '+ Djnh ly Cho tam gidc ABC noi tiep dadng trdn (O) M Id diem thuoc canh BC KM do, cdc menh de sau Id taang daang: (1) AM Id dadng doi trung ke tU dinh A cua tam gidc ABC (2) MB [ABV MC \AC) ' • (3) (BCMN) = - , N la giao diem cua BC vdi tiep tuyen cua dadng trdn ( ) tai A Dadng thang AN daac goi Id dadng doi trung ngodi ke tCc dinh A cua AABC i c r) djM, AB) ^ ^ d{M,AC) - ^ AB AC (5) A, M, S thang hdng, S Id giao diem cua cdc tiep tuyen ke tit cdc dinh B vd C cua dadng trdn (O) tTrUdng T H P T Chuyen Vinh Phuc, tinh VTnh Phuc 144 Djnh ly Cdc dadng doi trung cua mot tam gidc thi dong quy tai mot diem, diem goi Id diem Lemoine 147 Dudng doi trung vd mgt so bai todn dp dung Cdc phucfng phdp gidi todn qua cdc ky thi Olympic 146 C a c bai toan ap dung B a i toan Cho Jam gidc ABC, cdc diem M vd N idn luat nam tren cdc cqnh AB vd AC cho MN \\ Goi P Id giao diem cua BN vd CM Dudng tron ngoai tiep cdc tam gidc BMP vd CNP cat tqi hai diem phdn bi$t P vd Q ChvCng minh rang BAQ = CAP ChuTng minh Do cac tu" giac BMPQ, BMQ vk , ^ BQM suy tam giac QMB CNPQ = BPQ = BPM n o i tiep nen ta c6 = CPQ = t NPC, dong dang v d i tam giac QCN Do do: d(Q, AB) _MB AB_ d ( Q , AC) ~ NC ~ AC TO day, theo tinh chat cua du'dng doi trung ta suy Q nam tren Goi / va J theo thuf tU la hinh chieu cua Q tren cac dUdng thang BM va CN K h i do, (1) nen: dufdng d o i trung ke ixi dinh A hay difcJng thang AQ la difdng doi trung cua tam giac ABC Do vay BAQ = CAP • K h i M, N Ian Imt di chuyen tren cac during thang AB, AC va v d i each chiJng minh tiTdng t\i thi Q nam tren diTdng doi trung ke tijf dinh A cua tam giac ABC Do ta c6 bai toan sau: 9L _ ^ _ ^ 'QJ~'NC~AC MN II BC TO d6, theo tinh chat ciia diTdng doi trung, ta suy Q nam tren dtfdng doi trung ke tiJf A cua tam giac ABC • B a i toan Cho tam gidc ABC, cdc diem M vd N idn luat di chuyen tren cdc dUdng thang AB vd AC cho MN \\ Goi P la giao diem cua BN vd CM Dudng tron ngoai tiep cdc tam gidc BMP vd CNP cat tqi hai diem phdn biet P vd Q Chang minh rang diem Q di chuyen tren mot dudng thing co dinh Nhan xet Dtfdng t h i n g IJ AABC BM) = (PQ, PM) = {PQ, PC) = {NQ, NC) (mod MB) = (PQ, PB) (mod TT) LAJ TT) va (MQ, TO d6 suy ABQM = {PQ, PN) = {CQ, CN) ~ ANQC AIQJ noi tiep nen v d i AQ la dtfdng d o i trung nen QAI = LAJ ChuTng minh Do bon diem B, Q, P, M cung nam tren mot diTdng tron va bdn diem C, Q, P, N ciing nam tren mot dUcfng tron nen ta C O {BQ, That vay, tuf giac vuong goc v d i trung tuyen AL cua , (1) > + AJI suy IJ AL N e u ta Q qua cac diTdng thang tam giac QB'C nen AL nen AL la diTcfng trung = AQI+ QAI AQI = ATI, ket hdp Do do: = TT goi B', C Ian liTcft la cac diem doi xuTng cua AC, AB thi IJ la difdng trung binh cua vuong goc v d i B'C M a t khac, AB' = AC trifc cua B'C TO nhan xet tren ta thu difcJc cac bai toan sau: 148 Cdc phuong phdp gidi todn qua cdc ky thi Olympic Bai toan Cho tam gidc ABC, cdc diem M vd N Idn luat di chuyen tren cdc dudng thing AB vd AC cho MN \\ Goi P Id giao diem cua BN vd CM Dudng trdn ngoai tiep cdc tam gidc BMP vd CNP cat tai hai diem phdn biet P vd Q Goi I, J Idn lucet Id hinh chieu vuong gdc cua Q ten cdc dudng thing AB, AC Chvtng minh rdng dudng thing I J vuong gdc vdi dudng trung tuyen ke tit dinh A cua tam gidc ABC Bai toan Cho tam gidc ABC, cdc diem M vd N Ian luat di chuyen tren cdc dUdng thing AB vd AC cho MN \\ Goi P la giao diem cua BN vd CM Dudng trdn ngoai tiep cdc tam gidc BMP vd CNP cat tai hai diem phdn biet P vd Q Goi A', B ' , C Ian luat Id cdc diem doi xvtng cua Q qua cdc dudng thing BC, CA, AB Chiing minh rang tam dudng trdn ngoai tiep tam gidc A'B'C ludn nam tren mot dudng thing co dinh CMng ly C6\a minh Goi L la giao diem cua AP vdi BC Ap dung dinh cho AABC, ta c6 MA LB NC _ MB' LC' WI~ 149 Dudng doi trung vd mgt so bai todn dp dung Do MN II MA NA BC nen = = = = MB I\ ^ ^ LB W va (1) suy = = = - hay LAy ^ ^ L la trung diem BC Do AQ la du6ng doi trungnen BAQ^ = CAP va ket hcfp vdi tuT giac AJQI noi tiep nen AQI = AJI, suy CAP + AJI = AQI + BAQ = 90° va nhu the, ta c6 AP ± IJ (2) Do cdch xac dinh cac diem B', C nen AB' = AC = AQ hay tam giac AB'C can tai A, ket hdp vdi I J la dUdng trung binh cua tam giac QB'C, ta thu difdc I J \\ va AB' = AC (3) TO (2) va (3) suy AP la dUdng trung trUc cua doan BC suy tam dUdng tron ngoai tiep tam giac ABC nam tren dUdng thang AP hay nSm tren trung tuyen AL cua tam giac ABC • '"' Bang each chufng minh tu'dng tuf B^i todn ta thu difdc bai toan sau: Bai toan Cho tam gidc ABC vd trung tuyen AL {L la trung diem cua doan thing BC) Goi P Id mot diem di chuyen tren trung tuyen AL vd goi A', B', C Ian luat la cdc diem doi xting vdi P qua cdc dudng thing BC, CA, AB Chiing minh rang tam dudng trdn ngogi tiep tam gidc A'B'C ludn nam tren mot dudng thing co dinh ' i Bang each chuTng minh tUdng tu nhu" Bai toan 5, ta dUdc tam dudng tron ngoai tiep tam giac A'B'C nam tren diTdng doi trung ke tuf dinh A cua tam giac ABC Do tir ket qua va Dinh ly 2, ta thu dUdc cac bai toan sau: Bai toan Cho tam gidc A1A2A3 co tam G Goi Pi, P2, PA Idn luat la trung diem cua GAi, GA2, G^la- Goi A'^, B'^, C,' idn luat Id diem doi xvCng cua P, qua cdc dudng thing Ai+iAi+2- Goi d Id tam dudng trdn ngogi tiep tam gidc A'^B'^Ci Chiing minh rang cdc dudng thing A^Oi, A2O2 vd A3O3 dong quy Bai toan Cho tam gidc ^1^12^3 c6 tam G Goi P i , P2, P'i Idn luat la cdc diem doi xiing cua Ai, A2, ^ - qua G Goi A'^, C, Idn luat Id diem doi xiing cua Pi qua cdc dudng thing Ai+iAi+2Goi Oi Id tam dudng trdn ngogi tiep tam gidc A'^B'^Ci ChvCng minh rang cdc dudng thing AiOi, A2O2 vd A3O3 dong quy ' Cdc phuang phdp gidi todn qua cdc ky thi 150 Olympic B a i toan Cho tam gidc A B C vd diem D nam tren canh B C Dudng trdn ngoai tiep tam gidc A B D cat Igi A C tgi E vd dudng trdn ngogi tiep tam gidc A C D cat Igi A B tgi F CMng minh rdng D E = D F vd chi A D la dUdng doi trung cua tam gidc A B C Chufng minh Do tu* giac noi tiep nen AFDC 'FDB = BAG suy Dudng doi trung vd mgt so bdi todn dp dung trung cua tam gidc ABC xong EC_DE tam gidc B D F dong dang v d i tam giac B A C Tuf 66 ta c6 AC suy AB' DB = DF-AB kd tir dinh A B a i toan dUOc chiJng minh ,, , ,, NhSn xet N e u AD la dtfdng doi trung hoSc DE = DF thi E F I I BC That vay, tuf tam gidc BDF dong dang v d i tam giac BAC va tam gidc CED dong dang v d i tam giac CBA, ta dufdc: DF DB 151 (1) • BC ~ AB BF _ DF BC ~ AC' Suy EC BF AC EC BF = AB^AC-AB^^^^^^^- „^ 4';; ' Do ta CO cdc bai todn sau: B a i toan 10 Cho tam gidc ABC vd AD Id dudng doi trung cua tam gidc ABC [D ndm tren cgnh BC) Dudng trdn ngogi tiep tam gidc ABD cat Igi AC tgi E vd dudng trdn ngogi tiep tam gidc ACD cat Igi AB tgi F Chiing minh rang dudng thang E F song song vdi dudng thang BC B a i toan 11 Cho tam gidc ABC vd AD Id dudng doi trung cua tam gidc ABC {D ndm tren cgnh BC) Dudng trdn ngogi tiep tam gidc ABD cat Igi AC tgi E vd dudng trdn ngogi tiep tam gidc ACD cat Igi AB tgi F Ggi P la giao diem cua dudng thang BE vd dudng thang CF ChvCng minh rdng AP di qua trung diem cua BC c B D Tifdng tir tiir giac ABDE noi tiep nen 'DEC = BAC suy tam DE DC giac CED dong dang v d i tam giac CBA Ta thu diTdc AB va do: DE-AC DC = AB • Tur (1) v^ (2), ta c6 DB _ DC ~ [ABy \ACy DE = DF k h i va chi k h i - — DC DF DE fABV AC AC (2) , tuf dang thilc suy , tu-c AD la dirdng doi B a i toan 12 Cho tam gidc ABC vd diem D nam tren cgnh BC Dudng trdn ngogi tiep tam gidc ABD cat Igi AC tgi E vd dudng trdn ngogi tiep tam gidc ACD cat Igi AB tgi F ChvCng minh rang neu E F song song vdi BC thi AD Id dudng doi trung cua tam gidc ABC B a i toan 13 Cho tam gidc ABC vd diem D nam tren cgnh BC Dudng trdn ngogi tiep tam gidc ABD cat Igi AC tgi E vd dudng trdn ngogi tiep tam gidc ACD cat Igi AB tgi F ChvCng minh rang neu E F song song vdi BC thi DAB = PAC P Id giao diem cua dudng thang BE vd dudng thang CF Ccic 152 phiMng phdp gidi todn qua cdc ky thi Olympic Bai toan 14 Cho tarn gidc ABC {AB ^ AC) noi tiep dudng trdn tarn O vd ngoai tiep dudng trdn tdm I Dudng trdn (/) tiep xuc vdi B C tai D Dudng trdn ngogi tiep tam gidc A B D cat lai dudng thdng A C tai E vd dudng trdn ngogi tiep tam gidc ACD cat Igi dudng thdng A B tai F Goi M , N idn luat la trung diem ciia D E , DF Chvcng minh rang 01 ± A D vd chl AD, BN, C M dong quy tgi mot diem I, Chufng minh Bo de tdm O vd B C tgi D doi trung Dudng doi trung vd mot so bdi todn dp dung Ket hdp v d i ' B D C = BCK, tam giac CBK, ta difdc tam gidc D B N dong dang v d i suy ra: , BDN = ^ , CBK Do dd, theo D i n h ly 1, ta dtfdc BN la dUdng doi trung cua tam giac ABC Tifdng tiT, ta difdc CM tam giac la difdng doi trung ke ttt dinh C cua ABC Trifdc he't ta chiJng minh hai bd de sau: Cho tam gidc A B C { A B ^ AC) noi tiep dudng trdn ngogi tiep dUdng trdn tdm I Dudng trdn (/) tiep xuc vdi Khi 01 vuong gdc vdi A D vd chl A D Id dudng ciia tam gidc ABC That vay, difdng tron (/) tiep xuc v d i cac canh BC, C A , A B Ian ItfcJt tai D , E ' , F ' G o i P la giao diem cua diTdng thang E F va du-dng thang BC De thay I P vuong goc v d i AD, mat khac theo gia thiet ta c6 01 vuong goc vdi A D nen ba diem O, /, P thang hang hay OP vuong goc v d i AD Goi D ' la giao diem thiJ hai cua A D v d i (/) va A ' la giao diem thuf hai ciia A D va (O) A p dung dinh ly Menelaus cho tam giac A B C v d i cat tuyen E F P ta dUdc: 'WC_ Tc'Wl' 'PA _ ^ ^ P B _F'B _DB ~^^W~^~~DC~ => {BCPD) = _ 'DB ~W -1 A' Do OP vuong gdc v d i A D nen A D la difdng thang doi cufc cua diem P doi v d i difdng tron (O) Do do, P A , P A ' la tiep tuyen cua difdng tron (O) M a t khac theo chiJng minh tren ta cd (BCPD) = - , suy A D la du'dng doi trung cua tam giac ABC Bo de B N vd C M Id cdc dudng doi trung cua tam gidc ABC That vay, goi K la trung diem cua A C suy B K la diTdng trung tuyen ke tijf dinh B cua tam giac ABC Do tu" giac AFDC noi tiep nen tam giac B D F dong dang tam giac BAC, suy ra: a £ DM ^ _ ^ ~ CA"^ _ 2DN CB ~~ 2CK DB ^ DiV ~ CB 'CK' T r d l a i bai toan, theo Bo de ta cd BN, CM la diTdng doi trung cua tam giac ABC nen AD, BN, CM dong quy va chi AD la diTdng d o i trung cua tam giac ABC hay 01 vuong gdc v d i AD (theo Bo de 1) Tiif dd suy dieu phai chu-ng minh • Bai tap de nghi Bai tap 2001) ,, ; ( V i e t Nam, Cho hai dudng trdn ( O i ) , (O2) cat tai hai diem phdn Met A, B Goi PQ Id tiep tuyen chung cua ( O i ) , (O2) ( P € ( O i ) , Q e (O2)) Cdc tiep tuyen tgi P, Q cua dudng trdn ngogi tiep tam gidc APQ cat tgi diem S, goi H Id diem 154 Cdc phucfng phdp gidi todn qua cdc ky thi Olympic Dudng doi trung vd mgt so bai todn dp dung 155 doi xvCng vdi B qua dUdng thang PQ Chvlng minh rang S, A, H thang hang '••f^-'.'Vrv i : B a i tSp ( M y , 2005) Cho tam gidc ABC nhon noi tie'p dudng B a i tSp Cho dudng trdn co dinh (O) vd mot diem A nam ngodi (O) Xet hai diem B, C e (O) sac cho tam gidc ABC khong can tai A Gid sit AO la dudng doi trung cua tam gidc ABC Chiing minh rang tam dudng trdn ngogi tie'p tam gidc ABC ludn nam tren mot dudng thang c6 dinh PCA B a i tSp Cho hai dudng trdn (O) vd (O') cat tai hai diem phdn biet A, B Cdc tie'p tuye'n tai A, B cua (O) cat tai C M la diem tily thuoc (O) khdc A, B vd nam ngodi dUdng trdn (O') Cdc dudng thang MA, MB cdt Igi dudng trdn {O') tai N vd P theo thvC tu Chvcng minh rang dUdng thang MC ludn di qua trung diem cua NP Bai BC, AD cho tie'p tap Dudng trdn noi tie'p (I) cua tam gidc ABC tie'p xuc vdi CA vd AB tai D, E vd F theo thvt tu Cdc dudng thang vd E F cat tai M Lay M thuoc DF vd P thuoc DE MN II DE vd MP \\ Chiing minh rang tvc gidc EFNP noi mot dudng trdn B a i tap Cho tam gidc ABC vd diem P khong nam tren cdc dudng thang chiia canh cua tam gidc Cdc dudng thang AP, BP vd CP tuang ling cat BC, CA vd AB tai A', B' vd C Goi Q Id giao diem thii hai cua dudng trdn ngogi tie'p cdc tam gidc PBC vd PB'C Chiing minh rdng PQ la dudng doi trung cua tam gidc PBC vd chl A' Id trung diem cua BC B a i tap Chiing minh rang tii gidc noi tie'p ABCD Id tii gidc dieu hda vd chi AC la dudng doi trung cua cdc tam gidc ABD, CBD B a i tap (DiT tuyen I M O , 2003) Cho ba diem A, B vd C thang hang theo thii tu Xet dudng trdn uj di qua A, C vd co tam khong nam tren dudng thang AC Cdc tie'p tuye'n cua u tgi A vd C cdt tgi P, dogn BP cdt oj tgi Q Chiing minh rdng phdn gidc cua goc AQC ludn di qua mot diem co' dinh B a i tap (Ba L a n , 2000).^Oio torn gidc ABC can tgi C Lay diem D tam gidc cho DAB = DBC Goi E Id trung diem AB Chang minh rdng ADE + BDC = n trdn ( O ) P Id diem nam tam gidc cho PAB = PBC vd = PCB Laydiem Q trendudng Chang minh rdng QAP = thang BC cho QA^QP 20QB B a i tap 10 Cho ta gidc loi ABCD noi tie'p dudng trdn (O) Goi M , N theo tha tu Id trung diem AC, BD Chang minh rdng AMB = BMD vd chi ANB - BNC B a i tap 11 ( M y , 2008) Cho tam gidc ABC nhon Goi M, N, P theo tha tu la trung diem cua cdc cgnh BC, CA, AB Trung true cdc cgnh AB, AC cdt tia AM theo tha tu tgi D, E, cdc dudng thang BD, CE cat tgi diem F nam tam gidc Chang minh rang bo'n diem A, N, F, P cung nam tren mot dUdng trdn B a i t§p 12 Cho tam gidc ABC nhon, khong can vd noi tie'p ( O ) Dudng phdn gidc cua BAC cdt (O) tgi D ^ A Goi M la trung diem cua AD vd E Id diem doi xang vdi D qua O Dudng trdn {ABM) cat AC tgi diem tha hai F Dudng thang FM cat dudng thang BC tgi diem N Chang minh rdng fMBV _ NB yMAJ ~ NC B a i tSp 13 Cho tam gidc ABC noi tie'p dudng trdn (O) Tie'p tuye'n tgi B cua (O) cat dudng thang BC tgi S Dudng trdn tam S, bdn kinh SB cdt Igi (O) tgi D Chang minh BD Id dudng doi trung cua tam gidc ABC B a i tSp 14 Cho tam gidc ABC noi tie'p dudng trdn {O) Phdn gidc vd ngodi gdc A Idn luat cdt dudng thang BC tgi D, E Dudng trdn dudng kinh DE cdt Igi dudng trdn (O) tgi F Dudng thang AF cdt dudng thang BC tgi diem M Dudng trdn ngogi tie'p tam gidc ABM cdt Igi dudng thang AC tgi diem N vd dUdng trdn ngogi tie'p tam gidc ACM cdt Igi dudng thang AB tgi diem P Goi I , J lan luat Id trung diem cua cdc dogn thang MN, MP Chang minh rang cdc dudng thang AM, B J , CI dong quy tgi mot diem