Tài liệu Chuyên đề bồi dưỡng học sinh giỏi-giá trị lớn nhất, giá trị nhỏ nhất phan huy khải (phần 7)

Chuy6n dg BDHSG To^n gia tr| lan nha't g\& tr| nhd nhSt - Phan Huy Kh5i ,atBB| = x ; B , C 3 = S + S, De lhay A 3 B B , M , CCjMA,, 1+ •iB = A 3 B 1 ; A M S2 = ZSj + — ,- (3) cung chinh la AA3B1C2. MB1C3 la ^ cac tam giac deu vdi canh lan liTdt la x, y, z, cua chung thi: (5) >— S (6) 3 O S1+S2+S3 ^S; (do a = (difa vao nhan xet hicn nhien sau: Ne'u a + b + c = 1, thi a^ +b^ \di ) S ^ ) (z^ x ^ + y ^ + z ^ V ^ y (x + y + z)^ x^+v^+z^ ^ S. (1) s/ +S^\MB,C2 +SAMA,B, ro up SAA,B|C| ^ S A ^ A J C J om .c ok bo ce fa w. ww la dien tich tam giac ay. Tim gia trj Idn nha't ciaa - ( x + y + z ) ^ nen tir (1) suy ra: >is (3) . T C f ( 2 ) ( 3 ) s u y r a : ^ < | . (4) Da'u b^ng trong (4) xay ra o c6 da'u bang trong (3) • o x = y = z o M l a tam cua A deu ABC. Hudng ddn gidi Goi S la dien tich tam giac dcu ABC canh a, thi S = ( S M A , A A , +SMC2CC3 +SMB2BB1 \ a Da'u b^ng trong (3) xay ra x = y = z = - . Bai 2. Cho tarn giac dcu ABC canh a. M la diem tily y n^m ben trong AABC. Xet tam giac c6 3 canh la MA, MB, MC. Goi ^ 2 s'l +S2 + S3 CO dau b^ng trong (6) o O la trong tarn AABC. B2C2, A3C3 ^ fy^ + y + z) (7) is O la trong tarn AABC. S /g O la trong tarn AABC. --S. 3 Qua M ve ba doan thang A|B,, CA (xcm hinh ve) X Tac6:<^= S S S 1 Da'u bang trong (6) xay ra ^S = S^ - ^S = 3- o Tilf (6) (7) suy ra: P > 2S + 3 7S S • . . Tirddtacd: S,+S2+S3= >3' Dau bang xay ra <=> a = b = c = ^ ) Vict lai (5) du'di dang: P = 2S + ^S2 ^S3 Ta + ^ = l,ncnlac6: 7S Vay min P = — o „ nen neu goi S,, Sj, S3 tiTdng ufng la dien tich P = S , + S 2 + S 3 = 2 ( S , + S 2 + S 3 ) + l(sf+S^+S^) P= — o = C2B, iL ie uO nT hi Da iH oc 01 / (4) Tir (2) (3) (4) suy ra: O = A 3 C 2 ; CM Do cac tam giac M B . A , . M A . C . va S3 = 2S3 + O ; Vay lam giac co ba canh M A , M B , MC s i . , tfdoco: (2) s Lap luan tiTdng tiT, ta c6: AA3MC2 cac hinh thang can. c2 \, Sp •S + S, =s 1 + — - + -^ = S + 2 S , + ^ = > S , - 2 S , + ^ S s s^ Vi: ^ + ^ =z y;C3C (1) DoS = a'V3 4 minS = ^ ^ 12 o M la tam cua AABC. tifdng iJng song song vdi AB, BC. 357 j^.. ^ ^ . . w ^ i w u M t j i d I I I luii MiiciL v d y i d u | iiMu i i r i a i - r n a r i nuy ^nai Bai 3. Cho lam giac ABC c6 AB = c,• AC" = b va A=n . Xet lap hdp cac du, "^«ni. thang A qua A va khong U-ung vc'ti hai canh AB, AC. Goi P la tich khoang each if B va C tdiyi^A. Tim nhat ciia \, gia tri IdnHii(fng danP.gidi DiTcJng thang A qua A chia thiinh hai loai: Nhom I: Dirftng A cii doan BC. Dal BAH = 3 . Kc B H l A, C K I A . Khido: '' ' P = BH.CK = bcsinpsin(a - p) '; Tir(l)tac6: P <-bc{l-co.sa) = b c s i n ^ - . 2 , 2 Dau bilng irong (2) xay ra bcsin^ ^ , neu A la goc iD (90" < a < 180") ^bc, ne'u A la goc vuong ( a = 90") NhiT vay ne'u A la goc nhon, ihi maxP = bccos^ y khi A la phan giac ngoai cua goc A va ne'u A la goc tu, thi maxP = bcsin^ y, khi A la phan giac trong cua g6c A va ne'u A 1^ goc vuong thi maxP = - b e khi A la phan giac trong I -•••^'A- (2) ' /g ro up cua gc)c A la dai li^ctng P dcU gia tri \&n nhat va gia tri do bSng: P| = be sin^ y. ww w. fa ce bo ok .c om Nlwm 2: Difilng A qua A va citt doan B C , & day C la diem dol xuTng cua C qua A. Kc BH va CK cimg vuong goc A. Kc C'K" 1 A =^ CK = C'K' =:> P = BH.CK = BH.CK' Ap dung li luan phan 1 vao lam giac ABC", la lhay tich BH.CK'(cung la tich BH.CK) li'm nha'l khi A la phan giac , trong ciia goc BAC' (luTc A la phiin giac ngoai ciia goc A) va gia trj Idn nhat do = bccos — . 2 2 Theo nguycn li phan ra, thi: maxP = max{Pi; P:} = max|bcsin^y; 358 ubccos — 2 « s/ Vay trong cac di/ctng thdng thuoc nhom 1, thi du'clng phan giac trong hang p. = be sin hoSc la phan giac ngoai cua goc A. „ Nhqn xet: Trong bSi tren da stf dung nguyen li phan ra cua bai loan tim gia tri Idn nhat va nho nhat, di nhien ke't help vdi cac lap luan ve hinh hoc phang c6 iJng dung lifdng giac! Bai 4. (Bai toanToricelli) Cho tarn giac ABC c6 max {A, B, C} < 120" (tiJc la mpi goc cija tarn giac deube hdn 120"). Hay tim trong tam giac ABC mot diem M sao cho MA + MB + MC dat gia tri be nhat. HUdng ddn gidi Cdch 1: (Lc(i giai cua Toricelli) , , Durng ba tam giac dcu ABC,, BCA,, ACBi ra phia ngoai tam giac ABC. De tha'y ba dUcJng iron ngoai ttep cua ba tam giac deu ay dong quy tai diem T, va T la 3 diem nhin 3 canh cua tam giac BAC dUdi ba goc b^ng nhau va hlng 120". Do CJTA = ABCJ = 60", nen Afc = 120"=^ C,, T, C th^ng hang. Li luan tiTdng tir c6 B, T, B, thing hang va A, T, A| cung thang hang. La'y diem I tren TB|, sao cho TA = TI i=> AAB,I = AACT (c.g.c) 359 Ta <=> cos(a - 2P) = 1 <=> P = Y o A lii phan giac Irong cua goc A. iL ie uO nT hi Da iH oc 01 / 2 . cos(a-2P)-cosa bccos^ —, ne'u A la goc nhon (o < a < 90") Civ => B|I = C T => T B + T A + T C = B T + T I + I B , = BB| (1) V a y M ( i la d i e m nhin ba canh tam giac durdi ba goc bling nhau = 120", tuTc Mo L a y d i e m M baft k i thuoc m i e n tam gidc A B C . Ta se chiJng m i n h r k n g : M A + M B + M C > T A + TB + TC , „: !NMM M IV r i W I I Kh.il"! Viet_ la d i e m T o r i c e l l i phai t i m . (2) C^ch3: That vay diCng lam giac A M E sao cho E cf nufa m a t phang hci A M khong chuTa B. ^ ^ :r m- y Ro rang A A M C = A A R B , => M C = E B , => M A + M B + M C > T A + T B + TC R6 rang ta c6 b d de h i c n nhien sau day ( v i the chung t o i bo qua chtfng minh). :=> (2) diing. D a u bang xay ra N c u QNP la tam giac deu, I h i v d i m o i vj t r i cua d i e m M n a m trong m i e n tam M = T giac QNP, ta l u o n c6 long khoang each tif M t d i ba canh cua tam giac Q N P "Toricelli". la hang so' (c6 the tinh di/dc ngay hang so nay bSng chinh c h i c u cao cua tam J^'i . , 2: L a y M la d i e m tijy y trong m i e n lam giac A B C . ^ giac d c u QNP). R A p dung bo de ay ta g i a i bai toin T o r i c e l l i nhifsau: Thirc h i c n phep quay R" ( B , 6 0 " ) Khido: R-(B,60") C-yA, B c u a tam giac dvtdi ba goc bang nhau, tuTc la: A T B = B T C = A T C = 120". * • Q u a A, B, C dirng ba M ^ M ' Hdi/dng T h e n tinh cha'l cua phep quay suy ra. M B M ' la tam giac d c u => M B = M M ' . Ngoairado: * ^tai MC = M'A,. (1) up jjnj^ s/ I V a y M A + M B + MC = M A + M M ' + M'A,. '' 7jy (*) vuong TB, TC. goc Ba Hdirdng nay c^t nhau diTcfng gap khiic, ta c6: (2) Q, N , P. De thay QNP la tam gi^c deu. L a y M la d i e m tiiy y trong A A B C . G o i h , , h2, hi la khoang each N h i r vay tir (1) (2) ta da chi^ng m i n h difcfc rhng v d i m o i vj t r i d i e m M thuoc tijr M t d i ba canh NP, m i e n A A B C , ta luon c6: M A + M B + M C < A A , . ok (3) A, M , A2 thang hang. bo Da'u bang xay ra o cua phep quay ta c6 goc tao b d i M,)C va bkng 6 0 " =j^MoMoC = 6 0 " B M o C = 120". M„A| fa chat w. tinh A M „ B = I20". ww Theo ce Gia sur Mo la vj t r i cua M ma A, M„, A, thang hang Do BMoMo = 60" .c om /g M A + M M ' + M'A, .'1'VV,,,., Ta Cdch • ' iL ie uO nT hi Da iH oc 01 / V a y d i e m M can t i m chinh la d i e m T n 6 i tren. D i e m 66 thi^dng g o i 1^ diem PQ, Q N cua AQNP. m. M Ro rang ta c6: M A + M B + M C > h| + h2 + h j H • • T h e o bo de neu tren (ap dung vao tam giac deu QNP), ta c6: h , + h2 + h j = T A + T B + T C Tir do suy ra: M A + M B + M C > T A + T B + T C (*) • Da'u b^ng xay ra trong (*)<=> M s T . Do chinh la dpcm. m:(ich4: Sis dung b6 &G h i e n nhien sau day (ChuTng m i n h ra't ddn gian va x i n danh cho cac ban doc) B6 de: Gia si^ A M B = B M C = C M A = 120" va M A = M B = M C t h i : M A + M B + M C = 0 - : Do gia thie't m a x ( A . B , C) < 120", nen t6n tai duy nhaft d i l m T trong tam giac sao cho A T B = B T C = C T A = 120" . 160 361 Cty TIMHH MTV DVVH Khang Vi^t ChuySn de BDHSG Toan g i i t r i Ion nhat va g i i Iri nh6 nha't - Phan Huy KhAi ' . , TA TB TC An dung bo de suy ra: 1 1 =0 ' • ^ Gia sur M e BC. Goi H, I , J ti/(tng i?ng la hinh chic'u cua M xuong AB, BC, CA. (*) . b > c). * G o i L = M K n BC Su" dung cac bat dang thtfc hien nhien sau: 1. a|+a2+... + a„ . < . + «2 + ...+ Delhay: VF(2) = (2)' MC.TC TC > MA.TA TA + MB.TB TB MC.TCi, +• TC (MT +TA)TA (Mf +TB)TB TA TB = MT TA TB TC TA TB TC r^ - o a b c a a 2a Do vay S = — + — + - = — + — = — . (2) X TC MeBC +TA+TB+TC. . T i i f ( * ) v a ( 3 ) s u y r a : VF(2) = TA + TB + TC. z X (3) 'i (4) /g .c ok trj Idn nha't, nho nha't trong cac biii toan khac noi chung (trong hinh hoc bo phang noi rieng) cflng c6 rat nhicu each gitii khac nhau. Bai toan tren la mot ce vi du dien hinh minh chufng cho tinh da dang ciia cac phiTdng phap tim gia tri fa w. BC, CA, AB ti/dng uTng. Gia suT M la mot diem di dpng tren diTctng tron ngoai tie'p tam giac ABC. Goi x, y, z Ian liTdt la khoang each tif M den cac canh max M l M,)Io MeBC Ta c6: M,,!,, = BI„tan MoBI,, = - t a n ^ Tiif do suy ra: min S = — MeBC , 2a A =:4C0t—. a tan A 2 2 Cung gio'ng nhu" trong cac bai loan dai so', giai tich, v6'\c bai toan tim giii B a i 5. Cho tam giac ABC vdi a > b > e, d day qua a, b, c ki hicu do dai cac canh (3) 2 Lap luan ttfdng tu", co: i B C ' min S = 4cot— va min S - 4 c o t — . MeAC 2 MeAB (4) 2 Theo nguyen l i phan rii, ta co: min S = min min S, min S, min S MeBC M6AC MeAB (5) Tif (3) (4) (5) suy ra: BC.CA, AB. A B C min S = min 4col—,4cot—,4cot— 2 2 2 Tim gia tri be nha't cua dai ii/dng S = — + — + - . X y z Do a > b > c =^ 180" > A > B > C > 0 HUdngddngidi 2a 2a MeBC om Nfuln xet: Da'u bang xay ra <=> M s T. Do chinh la dpcm. ww X d day Mo la Irung diem ciia BC ro Ttr (1) (2) (3) (4) suy ra: M A + MB + MC > TA + TB + TC. Idn nha't va nho nha't ciia mot dai lufdng cho tri/dc. X ^2a^ Nhuf vay min S = min {MT + TC)TC ^ y Ta TB + p a cung CO ACLM - AABM => - = — z X s/ MB.TB up TA •+ (1) Tir(l)(2)=^^ +^ =- ^ l i l ^ =i . y z X X 2. a.p <|a|. p ; ta c6; MA.TA De thay A B L M - AACM ^ - = — . y X iL ie uO nT hi Da iH oc 01 / K^ro y^Ar^ MA.TA MB.TB MC.TC M A + MB + MC + + TC TA TB „ B 9 0 0 >A^ >B- ^ C >|>0^cot- M giao diem cua cac diTdng phan gi^c trong cua cac goc A, B, C, D cua ti? giac. Ve di/dng tron ngoai tiep tam giac ABM va gpi ABN la tam giac can npi tiep c6 dinh la N sao cho ANB = AMB. Gpi h la khoang each tif N xuong AB, con h, la khoang each tir M xuong AB. Khi do ta c6: h > hi va hi = r. T Ta CO' AABD = 2htan ANB = 2htan AMB (2) up ro /g om ww w. fa ce bo ok .c Dau h\ng trong (2) xay ra o M each deu C va D. AMB htan CMD Tir (1) (2) c6: P = AB + CD > 2r tan (3) 2 2 Dau bkng trong (3) xay ra o dong thcfi c6 dau bkng trong (1) (2). Taco: AMB-f CMD = 1 8 0 " - A l l + 180" - C + D = 180^' tan AMB = cot CMD tan 1 - . Vi theo bat dang thiJc Cosi, suy ra: CMD 1 H-tan ^ AMB CMD tan 1-tan tan CMD 364 iL ie uO nT hi Da iH oc 01 / Ti/Png ixi ta c6: CD > 2r t a n ^ ^ . Ta (1) s/ AB>2rtan AMB Dau bang trong (1) xay ra o h = hi o M each deu A va B. AMB Dau bang trong (4) xay ra <=> - ^ — = 4 3 . Tir (3) (4) di den: P > 4r (5) Dau bang trong (5) xay ra <=> dong thcti c6 dau bang trong (3), (4) M each deu A, B; M each deu C, D va AMB = CMD = 90" o ABCD la hinh vuong ngoai tiep dufdng tion ban kinh r da cho. Tom lai min P = 4r. Gia tri nho nhat dat du-dc khi v^ ehi khi ABCD la hinh vuong ngoai tiep diTdng tron ban kinh r cho trifdc. Bai 7. Xet tat ea cae ti? giac ABCD ehi c6 duy nhat mot canh Idn hdn 1. Tim gia tri Idn nhat cua S, d day S la dien tich tu* giac ABCD. Hiidng ddn giai ^ Gia siJ AD > 1. Khi do ta c6: AB < 1, BC < 1, CD < 1. Dat AC = X va gpi M la trung diem cua AC Ta eo: A B ' + B C ' = 2BM^ + AC^ >2. (4) .2 =^2BM^+ — < 2 = > B M < - V 4 - x ^ 2 ~ 2 (1) Do AC < AB + BC < 2 0 2 Dau b^ng trong (1) xay ra o AB = BC = 1. Ke BH 1 AC. Ta c6: (2) -1A C . B H < -1x . B M . < X < SARr= 2"" "~2 Dau b^ng trong (2) xay ra o H s M o AB = BC Tir (1) (2) c6: SABC < - x V 4 - x^ . (3) 4 Dau bang trong (3) xay ra o dong thcfi c6 dau bang trong (I) (2) o A B = BC=I. (4) Tac6:SACD= -CA.CDsinACD<-x. Dau b^ng trong (4) xay ra o CD = 1 va AC 1 CD. Tir (3) (4) ta c6: S = SASBCD = SABC + SACD < ixVTI x^+ix, 2 2x + x V 4 - x ^ (5) hay S < Dau b^ng trong (5) xay ra o dong thdi c6 dau b^ng trong (3), (4) o AB = BC = CD = 1 va ACD = 90". 365 Cty TNIiH MTV DV'VH Khang Vigt ChuySn gg BDHSG Toan g\i trj Idn nhS't va g\i tr| nh6 nha't - Phan Huy KhSi _ , Ta co: 2x + x V 4 - x ^ 4 = H i c n nhicn ta c6: A B < CE = aV2 . x/, , , r f\ + 1 + V4 - x"^ j . 4 A p dung ba't dang thiJc Bunhiascopski, ta c6: ( l + l + V 4 ^ ) < 3 ( 2 + 4 - x ^ ) hay 2 + V 4 ^ < 7 3 . 7 6 ^ + x\/4-x^ fZ 73 T i r ( 6 ) (7) s u y r a : c6 dau bang trong (7) Da'u bang trong (2) xay ra o CD=:1 N/3 "(2)' A B la diTcJng chco cua hinh vuong. M Huding ddn gidi ro /g cac bo hinh vuong sao cho A va B nam tren cac canh cua hinh vuong ay. G o i / la tong ce cac khoang each tiT A den cac dinh cua hinh vuong. T i m gii tri nho nha't ciia /. fa Gia sijf C D E F la m o t hinh vuong tuy y canh a va gia suT A e C D . Dat A B M = (v, N B C = 3. D o do canh \dn nha't trong cac doan A C , A D , A E , ,;l Do M B N = 45" =i> a + 3 = 45" tan((v + 3) = 1 :A * tan (x + t a n 3 , x+ y = 1=^ 1 - x y^ = 1 l - t a n a + tan3 >x Ta + y = 1 - xy. c6: _ ^ X y ~ 2 2 D a t t = xy. K h i do TacotCfd): (1) ••.yy.A,. S = SBMN = SABCB - ww w. Dat A M = X, C N = y, nhiT vay x, y e [0; 1 ]. om ok B a i 8. T r e n mSt ph^ng cho hai d i e m A , B v6i A B = d. X e t tap hdp tS't ca AE>a, AF>a. ' gia tri Idn nhat va nho nha't ciia S. A B C D la nii-a luc giac deu canh 1. Taco: AC + A D = a 72) = d ( l + 72). A B va C D sao cho M B N -=45*'. Gia sijf S la di?n tich tarn giac B M N . T i m (10) HUdng ddn gidi + i 9. Cho hinh vuong A B C D canh bang 1. D i e m M va N Ian liTdt di dpng tren .c o + AB = A B ( I 72 Gici trj nho nha't dat diTdc khi A B la diTdng cheo cua hinh vuong. A B C D la nufa luc giac deu canh 1. T 6 m l a i max S = — + ^ T o m h i i : m i n / = ( l + 72)d. A C I C D o >4^ AF Nhir vay ta d i den: / > ( l + 7 2 ) d . s/ AC = x = N/3 . /3 I AB do TiJf (1) ta c6: A F > A B , nen ket hdp l a i suy ra (9) Tir do ket hcJp v d i (5) co: S < — . 7 3 = 4 4 4i (7) =lox-V3. Lai iheo ba't dang thiJc Cosi, ta c6; cVri-x^ < = 3. Da'u bang trong (9) xay ra o , >a> AE (8) T o74-x^ 72 iL ie uO nT hi Da iH oc 01 / T v / - / ; \ / ' 7 \x AB TiS do suy ra: A C + A D = a > - (6) 2 ta c6: [x + y SACM - SBCN - (l-x)(l-y) ^ S = 1-t A x M SMDN 1-xy 2 (2) 1-t [xy = t A F la A E hoSc A F (thi du la AF) nen theo dinh l i V i e t x, y la hai nghi^m thoa man dieu k i e n 0 < x, < X2 < 1 Ro rang trong cac d i e m tren bien cua hinh vuong cua phiTdng trinh: thi d i e m each xa A nha't phai la mot trong cac dinh. x2_(i _t)x + t = 0 (3) De (3) CO nghiem nhiT vay, ta can c6: Tir do suy ra: AB 0 -6t + l>() (X,-1)(X2-1)>0 <=> X| +X2 <2 (*) >0 X| +X2 >0. l-t<2 a - p = 45" t <=>00 . , a - p = -45" (3) , „ <=> a = 45";P = 0 a = 0;p = 45" 42 : 2 "M = D ; N = C M = A; N = D :S M Nhdn xet: Mot Ian ni^a ta thafy du'dc tinh da dang cua cic phtfcJng phap d6ng de giai bai toan lim gia tri Idn nha't va nho nha't trong cac linh vifc khac nhau: Tir(2) (3) suyra: V 2 - 1 < S < (4) i ' 2 X = 0; y = 1 M = A;N = D S = - o t = 0<=> 2 M = D;NsC x = l;y = 0 dai so, giai tich, so" hoc, hinh hoc, liTdng giac... ai 10. Cho nufa di^cfng Iron bdn kinh R, di/dng kinh AOB. C la mot diem tily y up s/ Ta tren nufa di/dng tron khong trung vdi A va B. Trong hai hinh quat BOC va ro o M , N tiTdng tfng la chan diTdng phan giac cua ABD, DEC . om /g 1 M = A,N = D Vsiy maxS = - <=> .c minS= V ^ - I o M = M„;NsNo, bo ok cl day BM,, va BN,, Ian liTdt la phan giac cua cdc goc ABD va DBC . ce Nhgn xet: Ta c6 each giai khac bai toan tren nhiTsau: (diTa vao li/cfng giac) 1 2 w. 42 ww \ 2 cosa cosp 2 fa 1 Ta c6: S = SBMN = - B M . BNsinMBN 2 cos(a + 3 ) - c o s ( a - P ) 4i >/^ + 2cos(a-(3) (5) (do a + p = 45") V i a, p e [0; 45"], nen - 4 5 " < a - p < 45" V2 => — < cos(a - p) < I 368 (7> Ta thu lai ket qua tren! l-t>0 1 1 «cos(a-p) = iL ie uO nT hi Da iH oc 01 / S=V2-1 t^ - 6 l + l > 0 (*)o ' Lai c6: S = - o cos(a - p ) = l < = > a = P = 22"30' 2 ' Ap dung dinh l i Viet vdi (3) ta c6: x, + X2 = I - t; x,X2 = t nen t-(l-t) + l>0 242 2 2 + V2 hay V 2 - 1 < S < ^ . ', X|X2>0 >0 X, +X2 Tir (5) (6) suy ra: X | X 2 - ( X | + X 2 ) + 1>0 <2 X| +X2 X|X2 i.ii ihry ^' ^ AOC ve hai di/dng tron npi tiep. Gpi M v^ N 1^ hai tiep diem cua hai diTdng tron ay vdi diTcJng kinh AB cua nuTa diTcJng tron da cho. DSt 1 = M N . Tim gia tri nho nha't cua 1. • > HUdngddngiai Gpi Oi, O21^ tarn ciaa hai di/cJng tron. Dat CON = 2a (nhir vay 0 < a < 90") D 3 t M 0 , = R; NO2 = R2. De thay O ^ = ^CON - a , O.OM = - C O M = 90" - a. 2 A Ta CO / = M N = O M + ON :p R,cot(90" - a) + R2Cota = Rjtana + R2Cota Trong tarn gidc vuong 0 | M 0 , ta c6 R) = OO|Sin(90" - a) = (R - Ri)cosa => R i ( l + cosa) = Rcosa =:> R| = Rcosa (2) 1 + cosa Hoan to^n ti/dng tiT, ta c6 R2 = (6) (1) T i r ( l ) ( 2 ) (3) suy r a / = Rsina (3) 1 + sina Rcosa sina Rsina cosa 1 + cosa cosa 1 + sina sina 369 l + cosa •H Rcosa s i n a + coscx + l = K1 + sina (1 + s i n a ) ( l + c o s a ) a 2 cos - sin 2 = R 2cos2^ . a sin 2 Cty TNHH MTV DVVH Khang Vigt Dau bSng trong (2) xay ra <=> A = B = C a + COS ~ 2 +COS 2R -. s i n a + cosa + 1 L a i Iheo ba't dang thuTc Cosi, thi R A B B C , C, A ^ . J 2A , 2B. 2C 1 = t a n — t a n — + tan—tan — + tan —tan — > 3 : V t a n — t a n —tan — a ( . a cos sin + cos 2 I 2 2 a 2 2 2 (3) <^a = 4 5 " 2 la ba goc cua m o t tarn giac A B C . ro s/ up E. Diem quarngtso bai todn tim gid tri Idn nhd't, nho nhd't trong luang gidc Chu y rhng P > 0 vdi m o i tarn giac A B C . T i f do ta c6 maxP = VmaxP^ , ^2 Taco 2A B C I , . 2A P = cos —cos—cos—= — 1 - sin — 2 2 2 2) /g B a i 1. Clio tarn giac A B C . T i m gia trj nho nhat cua d a i liTdng sau: om \ 2 A V . AA 1-sin — 2 ok .c P = tan^ A + tan^ ^ + tan^ ^ - tan^ ^ t a n ^ ^ t a n ^ ^ 2 2 2 2 2 2 bo HuHng ddn giai ce tan^ — + tan^ — > 2 t a n — t a n — 2 2 2 2 . 2B 2C , B C tan^ — + t a n ' ' — > 2 t a n — t a n — 2 2 2 Docos^—^<1, 2 sin — + cos 2 nentir(2)c6 Da'u bang trong (3) xsiy ra o cos w. ww tan ^ 2 .2A C A +tan^ — > 2 t a n I —tan —. — 2 2 2 -2 A . 2B B B tan ^ + tan^ ^ + tan'' ^ > tan—tan — + tan — t a n — + tan — t a n — . 2 , 2 2 2 2 2 2 2 2 Trong m o i tarn giac A B C , thi V P ( 1 ) = 1 , nhiT vay ta c6 : 1. (2) Ta B-C 2 , + sin • — Al 2 j 1-sin^-^ 2 2 CO — 1^ 1 Tijr do theo ba't dang thuTc Cosi, c6 ^ 1-sin^A 2) A; 1-sin^A^ 2) , 1 1 . A) + sin — 2) + sin — 2 Da'u b^ng trong ( 4 ) x a y ra o cos- 1 B +C B-C^ • + cos - B-C^l 2 P^<-|l-sin^2 1 2 fa Thco bat dang ihu-c Cosi, ta c6 Hitfing ddn giai Ta V a y min/ = 2 R ( N/2 - 1), k h i C la di em chinh giffa cua A B . Tir do suy ra: B C cos —cos— d day A , B, C 2 2 B a i 2 . T i m gia trj Kin nha'l cua bicu thuTc P = '^"'^ C\A triing d i e m ciia A B 2 2 26 Nhi/vay minP = — o A B C la tarn giac deu. / = 2 R ( N ^ - 1 ) 2 26 Da'u bang trong (3) x a y ra o A = B = C. T i f (2) (3) c6 P > — . => 0 < sina + cosa < yfl ,\\e tiT (3) c6 : * 2 2A 2B 2C 1 tan'' — tan —tan — < — 2 2 2 27 Do sina + cosa = yfl cos(a — 4 5 " ) , ma 0 < a < 90". 2R 2 2 iL ie uO nT hi Da iH oc 01 / Rsina = 1 , 1 . A^ + sin — o B =C A\ , • A 2-2sin1 + sin — 2 A 2 . • A' 2-2sn I 2> • A— 1 + sin I (. • A ^ (, • A^ + I 1 + sin 2 ; + I 1 + sin 2) n3 4 ,116 '27 A A A 1 2 - 2 s i n — = 1 + s i n — <=> s i u y = - . (4) Cty TNHH MIV DVVH Khang Vijt Chuyen dg BDHSG Join gii trj I6n nha't vh gia tri nhd nhat - Phan Huy KhAi A B C Bai 4. T i m gia tri nho nhat cua dai lifdng S = tan'' — + tan^ — + tan'' —, d day A, 27 B, C la ba goc ciaa mot tarn giac. P^=-^<:>B = C v a s i n ^ = i 27 2 3 V a y tiTCl) ta c6 m a x P = 16 27 I <=> A B C la tarn giac can dinh A v d i A = 2arcsin ^ . \, 'ii B a i 3. X e t cac tarn giac A B C thoa man he thtfc tanA + tanC = 2tanB. gidi I cosB = 2cosAcosC (do sinB = sin(A + C)^Q) => => cosB = cos(A + C) + cos(A - C) 2C0SB = C O S ( A - C). , (1) s/ Tur (3) suy ra S > 0, nen m a x S = V m a x S ^ . . . CO 2 + z ' > (x^ + y^ + 2 ce (X .»1, (2) 7})'. 2 + y + z) 1 =^ = -. (3) (4) ViU. = B = C. /g o A 9 T i r ( l ) ( 4 ) CO S =i o A =B =C 9 1 V a y minS = ^ fa ; D a u bang trong (4) xay ra o dong thcfi c6 dau b i n g trong (2) (3) la tarn giac deu. w. B a i 5. X c t cac tarn giac A B C v d i A la goc Idn nha't. ww T i m gia t r i Idn nhat cua dai lifdng S = sin2B + sin2C + sin A Hiidng • r ddn gidi 2 ]_ Ta CO S = 2sin(B + C)cos(B - C) + — — sin A = 2sinAcos(B - C) + sin A =^c:>2-2cosB = l +2cosBocosB = - . 8 4 c o s ( A - C ) = 2cosB max S = - r - . TCf do suy ra x ' + y^ + z^ > ^ . ok bo ^?::ii^^i?Ki±l£^ 2 x-" + y + z > (4) V i the theo bat dang thiJc Co si di den CO om .c (3) 2 Vx, ^ y . + y + z)(x' + y ' + z') > (x^ + y^ + v?)^ ro (2) Tir(l)(2)suyraS=2sin|^i±^. TiT do ta up A +C A - C - . B l + cos(A-C) = 2cos cos = 2 sin—J 2 2 2\ 8 CO ( X Lai L a i CO S = cosA + cosC n2 ta x^ + ,. (1) Da'u b^ng trong (1) xay ra <:5> A = B = C. A B B C C A D a t X = t a n y t a n - j , y = tan — t a u y , z = t a n y t a n y thi x + y + z = 1 ..^^ => • ( 2 - 2 c o s B ) + (l + 2cosB) ddn gidi Apdungba'tdangthuTcBunhiacopskichohaiday sin(A + C) 2sinB — = cos A cos C cosB Taco S ^ = 4 s i n 2 - i i ^ ^ = 2 2 ' Ta . A ,. \ r. tanA + tanC = 2tanB ' (chiiy x > 0 , y > ( ) , 7 . > 0 ) T i m gia t r i Idn nhat cua dai liTdng S = cosA + cosC. T„ ' Ta CO ' *' , TA ^B ,B iC TC. ^A De lhay S > tan — t a n — + t a n — t a n ' — + tan- — t a n " — . 2 2 2 2 2 2 4N/3 HUdng ddn '* Hiidng iL ie uO nT hi Da iH oc 01 / Tfif (3) (4) di den <=> cosB = — 4 <=> 3 B = arccos— 4 • , , j f Do sinA > 0 va cos(B - C) < 1, nen lif (1) cd: S < 2sinA + 2 sin A (2) 373 Chuyfin 66 BDHSG Toan gii trj Idn nha't vi giA trj nh6 nhat - Phan Huy Kh&\ Cty TIMHH MTV DWH Khang ViSt Da'u bang trong (2) xay ra <=> cos(B - C) = 1 <=> B = C. Dau bang trong (4) xay ra Do A la goc l(^n nhii't trong lam giac => A > ^ => ~ Xct ham so f(x) = x ' + x - — vdi 0 < x < 1 < sin A < 1 ci> B=C ' >3 ; i i b i f <^ X (do A > 90" ^ — > 45" X c l ham so i\x) = 2\ - wYi — 0 Vx e (0; 1], ncn c6 bang bicn thicn sau iL ie uO nT hi Da iH oc 01 / 1 0 X r(x) X , ,Vay ^ l a x i(\) = { yf3] v 2 . 90" =o B + C < 90" Ta CO cosBcosC = ^[cos(B + C) + cos(B -C)J . > -[cos(B + C)cos(B-C) + cos(B-C)] P =>cosBcosC > - c o s ( B - C ) l l + cos(B + C)] = c o s ( B - C ) c o s ^ - ^ i ^ 2 .. 2 A => cosBcosC > sin^ —cos(B - C ) 2 => cosBcosC > sin^—(cosBcosC + sinBsinC) 2 =>(1 - sin" — )cosBcosC > sin^ — sinBsinC. 2 2 Do B, C nhon => cosBcosC > 0. „2A cos L . A col 2 9 ., sin B sin C „ 2A ^ > =>tanBtanC cos(B - C) <=> B = C => (1) di/cJc chiJng minh! ' Tir do ta CO Thay (3) vao (2) ta di den P < c o l ' - + c o l 2 2 ' cos(B + C) > 0. ok Bai 6. X e l tarn giac ABC vdi A > 90". Tr\idc hct ta c6 tanBtanC < cot" A ' u ; om <=> ABC la tarn giac dcu. P - l o B - C v a c o t —= 1 2 « ABC la tarn giac vuong can tai A .c 7V3 1 P<1 773 Vay maxS = + 0 - < — < — =>cos — < 0 3 2 2 2 4 2 Tur do CO f'(x) < 0 V 0 < X < 1, nen CO bang bien thien sau: f'(x) "3 (1) B-C =1 cos2 A= ^ 2 o ABC la tam giac vuong can dinh A. 3A . • • i- • • 1 : Ta 3A - Dau bkng trong (1) xay ra <=> s/ 7t I V2 _ V2 , , x= l sinA + 2xcos— 2 cos Xethamsof(x)= 2- v d i O < x < 1 z:>f'(x) = — 2 A • A cos A + 2x sm — (cosA + 2 x s i n y ) ^ 2 ^ 1 Tiif do suy ra f(x) > — + 1 . 3A 7t ^ min g(A) = g 2" C Datx=cos I + 2 iL ie uO nT hi Da iH oc 01 / => - < A < - . sin A + 2 sin i i 1 g'(x) 71 sinA + sinB + sinC n"'" 71 \/2 Ttf d6 ta C O ket qua sau: minP = ^ + o ABC la tam giac vuong can dinh A. Bai 8. Xet ABC la tap hdp cac tam giac nhpn. Tim gia tri nho nha't cua bieu thuTc: A B C P = tanA + tanB + tanC + tan —tan—tan —. 2 2 2 Hiidng ddn gidi R5 rang trong mot tam giac nhon, ta c6: A B C tanA + tanB + tanC > cot — + cot — + cot —. Z Z Zr (1) . Dau bang trong (1) xay ra <=> A = B = C. A B C 1 NhiTvay P > cot — + cot — + cot — + A B C 2 cot — cot—cot — 2 2 2 A B C A B C cot—cot—cot — = cot — h cot — + cot —, Trong moi tam giac ta c6 2 2 2 2 2 2 cot— + cot — + cot — > 3 VJ. 2 2 2 A B C Tiy(2) (3) (4) s a u k h i d a t x = cot — + cot—+ cot—, ta c6: 2 2 2 377 ChuySn dg BDHSG 7oAn gia Irj Idn nha't gia trj nh6 nhU't - Phan Huy Khii !S/hdn xet: Xet cac each giai khac sau day: 1. Xet cac vectd ddn vi e , , C 2 , C 3 nhtfhinhve. Ta Xethams6'f(x) = x + 1 CO (e, + 62 f'(x) = l - ± , nen c6 bang bien thien sauc, 3v^ f'(x) f(x) +00 1 1 1 28N/3 9 B ^ cos(62,33) > 0 . - »- ••; • = 1 => 3 - 2(cosA + cosB + cosC) > 0 =>S = cosA + cosB + cosC < —. 2 3 Lai CO S = — o C | + 62 + 63 = 0 28^3 o A = B = C. o e , + e 2 = -e3 Ta s/ up Bai 9. Tim gia tri Idn nhat cua dai lifdng S = cosA + cosB + cosC, ro d day A, B, C la ba goc ciia tarn giac ABC. /g Hitdng ddn gidi om S = 2cos^^-tlcos^—^ + l - 2 s i n 2 2 2 2 ^ . C A-B , ^ . 2C = 2sm—cos t-l-2sin — .c ok w. •2A-B -sir-—-+3 Tir(l)suyra 3 Tom lai maxS = 2 '^"^ Siac deu. Ta thu lai ket qua tren. IT IT •2'2 thi cox+cosy ^ < cos Da'u bKng xay ra o x = y. Tdrdotaco: , ww 2sin--cos— 2 2 B = 60". Tifdng ty A = C = 6O" => ABC la tam giac deu. Neu x, y e fa 4sin^cos^^—^ + 2 - 4 s i n 2 2„, ,,,,,2„.,::„,.,,.,„f 2 MBP = 120" o Ta CO nhan x6t hien nhien sau: bo 2 o Xet each giai khdc nffa nhU'sau: ce 2 ^v, o BMNP la hinh thoi c6 B M = BN = BP = 1 28>y3 Nhir the minP = ^ o ABC la lam giac deu. (1) ^, IT IT . , „ cosC4tosA,n C+ cosA + cosB , A+ B , 3 A+B+C+ --H ^ cos l-cos - - - 2 2 < 2 2 — <- ^.Qj. (2) 2 • A-B „ sm = 0 T . C A-B 2 . . 2 2sin — = cos Vay maxS = - ,|cos(e,,e3) + 2|t Do X = 3yf3 2 , cos(e|,e2) Do cos (c|, 62) = cos (18()" - B j = - cos B . - A= B=C Taco + 2 Ti/dng tiT cos(e|,e3) = - c o s A va cos(e2,Cj) = - c o s C . Tirdo m i n = f ( 3 ^ / 3 ) = x>>/3 , Ket hcJp lai suy r a P > + + 62 + 2 + >0 iL ie uO nT hi Da iH oc 01 / X +63)^ => o A = B = C. cosA + cosB + cosC + IT 1 ^ —2- < cos— = — => cosA + cosB + cosC < —. 4 ~ 3 2 2 Dafu bkng xay r a < » A = B = C = ^ . ABC la tarn giac deu. 379 -- V a y maxS = - « ABC o — .., I nqii i i u y iMim r\ndl la tarn giac deu. Ta cung thu lai ket qua tren. M U C L U C 3. L a i x6t each g i a i khac nffa nhu'sau: Ta c6 S = cosA + cosB + cosC A + B = 2cos- C O S - — — 2,. D a u b^ng trong (*) hMng A-B xay ra o . A-B = 1 o A 3 §1. V a i bai loan m 6 dau + cosC < 2sin — + cosC 2 cos 1. M('/ d i u v6 gia tri l«tn nha't va nho nhS't cua ham s6' 3 §2. N h i n l a i cac bai loan ve gia t r i Idn nha't va nho nha't cua h a m so trong cac k l t h i t u y e n sinh vao dai hoc, cao dS^ng = B 13 • '^i;' iL ie uO nT hi Da iH oc 01 / §3. C d sd l i thuye't cua b a i toan t i m gia trj Idn nha't va nho nha't cua h a m so X e t h a m s c r f ( x ) = 2 s i n - + cosx v d i O < x < 7 t cua h a m s6' T a c 6 f'(x) IT X = J = cos--sinx= 2 c o s - 1 - 2 s i n —, 2 2j § 1. Phifdng phap sit dung bat d^ng thufc Cosi (do 0 < X < j i ) V d i chu y k h i 0 < X < j i t h i cos J > 0, ta c6 bang b i e n thien sau: 3 om suy ra S < - , va S = 2 2 la tam giac deu. <^ 1.3. PhiTdng phap t h e m bdt hang so 63 1.4. PhiTdng phap t h e m bdt hang tuT 73 1.5. PhiTdng phap n h d m cac so hang 90 1.6. PhiTdng phap siJ d u n g k i thuat ngiTdc dau trong bat dSng thiJc C o s i 111 §2. PhiTdng phap stir dung bat dang thiirc Bunhiacopski t i m gia t r i Idn nha't va nho nha't h a m so 119 §3. Cac phifdng phap thong dung khac siir dung ba't dSng thiJc de t i m gia t r i Idn nha't va nho nha't ciia ham so 133 1. PhUdng phap xuat phap tiif mot ba't ddhg thiJc da biet tClT tri/dc 133 va nho nhS't cua h a m s5' ww Nhi/ the maxS = ^ 50 141 191 ChiTcTng 4. Phi/dng phap c h i ^ u bien thidn h a m s6' tim gia trj Idn nhd't w. ~ 3 1.2. Phirdng phap suT diing trifc tiep bat dang thiJc Cosi Fhifrfng phap Ivhfng giac h6a tim gia trj Idn nhS't va nho nhfl't c u a h a m s6' fa ABC 42 Chrfdng 3. ce A = B 42 1.1. Sijrdung ba't ding thuTc Cosi c d ban v^ nho nha't cua h a m so ok Tir do k e t hdp v d i (*) 33 2. Cac bai toan khac su-dung ba't ding thiJcde t i m gia t r i I d n nha't .c 2 bo 13J ^ _ TV o s/ 1 - 3 IT Vay maxf(x) = f 0 ro 1 f(x) + •' /g 1 f'(x) n up 0 Ta IT X 30 hiidng 2. Phrfefng phap suT dyng bS't dang thrfc de tim gia trj Idn nhSft va nho nhS't ' ''^^^ 206 §1. SuT diing triTc tiep chieu bien thien ham so de t i m gia trj Idn nha't, nho nha't ... 206 la lam gidc deu. §2. SU diing chieu bien thien h a m so c6 k e t hdp t h e m cac phiTdng phap khac B a i t o a n mof d ^ u c u 6 n s a c h n a y d a chtfng to srf p h o n g p h i i v a d a d a n g c u a c a c phif(/ng p h a p d u n g de t i m g i a t r j I d n nhfi't v a n h o nhS't c u a n i g t b i 6 u thuTc h a y m o t h a m so' c h o trxidc. phiTcTng p h a p t i m g i a t r j Idn nM't b a y k l lufdng t r o n g c u d n s a c h n a y mdi 216 ChiYdng 5. Phrfdng phap mien gia trj ham so'tim gia trj Idn nhS't, nho nhS't ciia h a m s6' ....227 B a i t o a n n e u t r f i n k h e p l a i c u 6 n s a c h n a y c u n g vdi m p t t h d n g d i $ p Cac de t i m gia t r i Idn nha't, nho nha't r^ng: va n h o nhS't m a c h u n g t d i d a t r i n h that phong phii l a m sao! Chifdng 6. Phrfdng phap siJ dung 66 thj hoSc hinh hqc d4 tim gia trj idn nhS't va nho nhS't cua h a m so' 243 C h i f r f n g 7 . L f n g d u n g c u a g i a t r f l«?n nha't v a g i a t r j n h o n h a t t r o n g b a i t o a n g i a i phi/(fng t r i n h v a bS't phi/«/ng t r i n h c o t h a m .s(Y 263 § 1 . M o ' i l i e n h e giiJa g i a t r i i d n n l i a l , n h o nha't c u a h a m so • v a sU C O n g h i p m ci5a p h i f i t n g t r i n h v a b a t phU(•! t ^ v a b a t p h U d n g t r i n h k h d n g c 6 t h a m so 296 § 2 . G i a t r i I d n nha't v a n h o nha't c u a h a m so p h u t h u p c t h a m so 308 A . B i e n l u a n t h c o t h a m s o ' g i a t r i I d n nha't v a n h o n h a t c u a h o h a m so 314 up c u a c a c h a m so p h u t h u o c t h a m so' s/ B . C a c iJng d u n g c u a v i § c k h a o s d t g i a t r i I d n nha't v a n h o nha'l ro § 3 . G i d i t h i O u m O t s o b a i t o a n g i a t r j I d n nha't, n h o nha't t r o n g so h o c , 330 /g h i n h h o c , li/dng giac .c B . D i e m q u a b a i t o a n g i a t r i I d n nha't, n h o n h a t t r o n g h i n h h o c t o h d p om A . V a i b a i t o a n v e g i a t r i I d n nha't, n h o nha't t r o n g so h o c ok C . V a i b a i t o a n v e g i a t r j I d n nhaft v a n h o nha't t r o n g h i n h h o c k h o n g g i a n bo D . D i e m q u a v a i b a i t o a n v e g i a t r i I d n nha't, n h o n h a t t r o n g h i n h h o c p h d n g 330 342 348 355 w. fa ce E . D i e m q u a m o t so' b a i t o a n t i m g i a t r j I d n nha't, n h o nha't t r o n g l i f d n g g i a c . . . . 3 7 0 ww ' 308 Ta p h u t h u o c t h a m so'