Bồi dưỡng kiến thức học sinh giỏi lượng giác: Phần 2

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Nối tiếp nội dung phần 1 tài liệu Bồi dưỡng học sinh giỏi lượng giác, phần 2 cung cấp cho người đọc các kiến thức cơ bản và các dạng bài tập bất đẳng thức trong tam giác, vài ứng dụng của lượng giác trong việc giải các bài toán sơ cấp. Mời các bạn cùng tham khảo.

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Nội dung Text: Bồi dưỡng kiến thức học sinh giỏi lượng giác: Phần 2

Dau bang xay ra trong cac ba't dang thtfc tren khi va chi khi ABC la tam giac GAt^V3 (1) 2 2 2 2 o ABC la tam giac deu. . A B C B C 2.3.4 ChiJng minh tu-dng tif 1/. tan — + tan — + tan — >3 do tan — > 0, tan — > 0, tan — > 0 2 2 2 2 2 2 . Bai 2. Trong moi tam giac ABC, ta co: 2 A , B 9 C ^. A B B C C A 3^3 ^ . A . B . C ^ 3 tan — + tan" — + lan" — + 2 tan — tan — + tan — tan — + tan — tan — >3 2 2 2 2 2 2 2 2 2/ 1. sinA + sinB + sinC < . 2. sm — + sm — + sm—< - . 2 2 - 2 2 2 ^. 7 A , B ,C A B B C C A ^ , . A B C ^ 3^3 I o tan" — + tan" — + tan" tan —tan —+ tan —tan —+ tan —tan— > 0 2 2 2 2 2 2 2 2 2 3. cos hcos—l-cos—< . 2 2 2 ~ 2 275
Cty Timn ivrv P VVn Khang vlft A B B C C A ^ do lan — tan — + tan — tan — + tan — tan — = 1 2 2 2 2 2 2 jsjh^it xet: Ta hay hieu the nao la bat dang thufc khong the l a m " t o t hrin". 2 , Ba'l dang thufc R A , B , C) > a doi v d i m o i tam giac A B C g o i la khong the l a m f A B^ ( B tan — - t a n — + tan — - t a n — 2 f c + tan — - t a n — >() (2) 2; 2 " l o t h d n " nc'u nhU' Ve > 0, ton l a i tam giac AnBdC,, sao cho I 2 2J 2 I I 2 RAo, B,„ C„) < a + s (1) V i (2) dung =>{\ dung => dpcm. ^ Bat dang thiJc f(A, B , C) > P doi v d i m o i lam giac A B C g o i 1^ khong the ISm 2. Trifdc hct ta chi?ng minh trong m o i tam giac i h i "tot h d n " neu nhu" Ve > 0, ton lai lam giac A||B„C(| sao cho cot A + cotB + cotC > 0 (3) f r(A„, B„, C„) > p - e (2) That vay neu A B C la tam giac nhon thi (3) hicn nhien dilng V i the chang han de chiJng minh ba'l d^ng IhuTc f(A, B, C) > a d o i vdti m o i Neu A B C khong phai la tam giac nhon khi do gia siir A > 90", i h i B, C la tam giac A B C g o i la khong the l a m " t o t h d n " la lien hanh theo hai biTdc sau: nhon va 0 < B + C < 90". 1. Chu'ng minh vdi m o i tam giac A B C , la c6: f(A, B, C) > a Ta c6: cotA = - c o t ( B + C), ncMi /, i 2. Chi'rng minh c6 (1). De liim dieu nay c6 nhieu cdch, luy nhien each ma ta col A + colB + cotC = colB + cotC - coKB + C) hay siir dung la: ChuTng minh rang ton l a i m o t ho tam giac A , B , C , (phu Ta c6: colB > 0 va do 90" > B + C > C =^ col{B + C) < cotC Ihuoc vao e, \6'\ > 0 dii be)sao cho v d i ho nay la co: ^ colC - cot(B + C) > 0 => cotB + cotC - col(B + C) > 0, vay (3) dung lim r(A,,B,,C,,) = a V i Ic do colA + cotB + colC > S 0 du be. . . A . B , C Khi do: cosAcosBcosC = cos( 180" - 2a)cos^a = - c o s 2 a c o s ' a 3. sm — s m — s m — > 0 . 4. sinA + sinB + sinC > 0. 2 2 2 Tirdo l i m ( c o s A c o s B c o s C ) = - l i m (cos2a)(cos^a) = - 1 c A . B . C , A B C - 5. sm — + sm — + s i n — > 1 . 6. cos — + C O S — + C O S — > 2 . 2 2 2 Vay ba'l dang ihurc (1) khong the lam " l o t h d n " => dpcm. 2 2 2 va cac ba'l dang thuTc trcn khong the lam " l o t h d n " 3.4. Chiang minh tu'rJng liT Vi. Giai 0, B ' > 0, O 0 va A ' + B ' + C = A + B + C = 180" nen co the A. B . C . A coi A', B', C la ba goc cua mot tam giac A ' B ' C khac. Ta c o : 0 < sm — s m — s m — < sm — 2 2 2 2 Theo phan 1/ suy ra cosA' + cosB' + c o s C > 1 ^' ' ^ A') . A . B . C M a t khac: l i m sm — = 0 : • l i m sm—sin—sm — = 0 . . A . B . C , , 2y 2 2 2 =!> sin — + sin — + sin — > 1 => dpcm. A->() 2 2 2 * Vay bat dang Ihufc tren kh6ng the l a m lot hdn. Xet ho lam giac A B C A = 180" - 2 a ; B = C = a v6i a > 0 du be
B6I duang h ^ c sinh g l o l Lupng glAc - rhan Huy Khal Cty Timn nrv P VVH Khang VIft A . B . C lim sin — + sin—I- sin — = lim cos a + 2sin — = 1 A,B = 2R,sin 2 Oh a->{)'^ 2 27 2 A B C , " Vay ba't dang thuTc s i n y + s i n y + sin y > 1 khong the lam "tot hdn" Thay (3) vao (2) va c6: AA, = — 2 R s i n — (4) a 2 6. Do 0 < cos— < 1, 0 < cos— < 1, 0 < cos— < 1 Lap luan lu'dng tiTcd: BB, =~-2RCOS h COS + COS — CC| = 2Rsin —. (6) 2 2 2 2 2 2 , 2 2 A B C 3 + (cosA + cosB + cosC) . ^, ,,. cos —+ cos—+ C O S — > > 2 (Iheophan 1/) Nhan tiTng ve (4) (5) (6) suy ra: 2 2 2 2 • A . B . C Cho ho tam giac ABC vdi A ^ 180" - 2a, B = C = a, vdi a > 0 du be suy ra + a)(a + b ) ^ ^ 2 4Rsin—sin—sin — AA,.BB,.CC, = (h + ' Bat dang thiJc c o s y + cosy + '^osy > 2 khong the lam "tot hdn". Tir (7) (8) di den AA,.BB,.CC, = (b + c)(c + a)(a + b)^^^, , abc i (9) Thco ba't dang thuTc Cosi, thi B. Bat dang thi'tc hidng giac trong cac bai todn hinh hoc phdng b + c > 2^/bc , c + a > 2^[ca., a + b > 2^yab (10) Trong muc nay chung la xct cac bai toan bat dang thiJc liTdng giac, hoac bat TCf (9) (10) suy ra AA|.BB|.CC| > 16R-r =:>dpcm. dang Ihtfc khong co dang lu'dng giac nhu'ng chiJng minh no thi hoan toan di/a Dau bang xay r a o b = c = a o ABC la tam giac deu. vao phcp bien ddi cac he ihiJc lu'dng giac trong tam giac. Cac ba't dang tMc NHn xet: nay gan chat vdi mot bai toan hinh hoc phang lu'dng iJng. Nhif vay de giai 1. Ke't qua noi tren van dung ne'u thay ba du'dng phan giac bdi ba difdng Irung chiing Ihi CO siT phoi hdp giffa lu'dng giac va hinh hoc phang (no phan bict trifc, tiJc thay tam difdng Iron noi tiep I bang tam O diTdng tron ngoai tiep. vdi Ccic bai loan thuan luy ve bat dcing thufc lu'dng giac ma vc mat hinh thifc Khi do AA| = BB| = C C , =2R. CO the thay ngay trong cac bai loan ay khong he c6 mot hinh ve nao ca). Vay AA|.BB|.CC| = 8 R \ B a i 1. Cho lam giac ABC. Ba diTdng phan giac trong cua cac goc A, B, C laa D o R > 2 r = ^ 8 R ' > 16R'r. lu'dt c5t diTdng Iron ngoai ticp tam giac ABC tai A | , B | , C|. Chijrng minh => AA|.BB,.CC| > 16RV AA,.BB|.CC| > 16RV Da'u bang xay ra R = 2r. Giai "J^ABC la tam giac deu. i Ap dung dinh li Plolcme vdi ti? giac noi ticp ^- Ne'u thay I bcli triTc tam H ciia tam giac nhon ABC. A B A i C taco: Ap dung dinh li ham so sin trong A A B A i , ta c6: AA|.BC = AB.A|C + A C . A i B (1); AA, =2RsinABAJ = 2Rsin(B + A ^ ) jf Ta c6: A | B = A.C, nen tif (1) co: = 2Rsin(B + H A c ) " ^~ _ A | B ( A B + AC) (2) • k = 2Rsin(B + 90" - C) BC flp =2Rcos(B-C). Ap dung dinh l i ham so sin trong AABA, (vdi chu y ban kinh cua diTdng iron J t f d n g tir c6: BB, = 2Rcos(C - A ) ngoai tiep tam giac nay cung chinh la R), ta c6:
B6I duang ln>c sinh gioi I n^ing gUic - fhan Huy Khal Cty Timn nrv P VVn Kbang Vift Tir d6 A A , . B B | . C C , = 8 R W ( B - C)cos(C - A)cos(A - B) A . B . C A . B . B . C C . A sm — + s m - + s m - - > 3 sm — sm—h sm —sm hsin — sm — (3) Ta C O the chuTng m i n h diTdc ba't dang ihufc saii day irong m o i tam gi^c nhpn 2 2 z/ , . A . B . C 2 2 2 2 2 2 j cos(A - B)cos(B - C)cos(C - A ) < 8sinysin-sin-. A . B . C^, . A . B . B . C . C A TCr (2) (3) suy ra s m ^ + sm —+ s m Y > 2 s m — s m — + s m — s m — h s m — s i n — 2 2 2 2 2 2 A B C Vay (1) dung => dpcm. T i r d o s u y r a A A , . B B | . C C , < 8 R \ 8 s i n y s i n - s i n - = 16R^r. Da'u bang xay ra A B C la tam giac dcu. V a y neu thay I bdi trifc tam H va gia thiet A B C la tam giac nhon la c6 bai dang thuTc nhi/ng vc)i chieu ngiTrtc l a i . 1. Ncu A B C la tam giac nhon va thay 1 bang triTc tam H , thi ta co ba't dang thtj-c B a i 2. G o i I la tam diTcfng tron npi tiep tam giac A B C . AT, B I , C I keo dai cat vdi dau ngifdc l a i , ti?c la H A + HB + HC > H A , + H B , + H C , . di/5ng tron ngoai tiep A A B C Ian lirm tai A , , B,, C,. That vay do H A C , H A B , HBC va A B C c6 cac vong tron ngoai tiep bang Chu-ng m i n h : l A , + I B , + IC, > l A + IB + IC. nhau, ncn ap dung dinh l i ham so sin trong A H A C . thi A Giai HA = 2R sin H C A = 2R cos A . X7 De tha'y I A = ; IB = ;IC = • l A , = A,B TiTdng tir HB = 2RcosB; HC = 2RcosC. C, . C B sm sm sm Lai C O H A , = 2 H M = 2 H C s i n H C M 2 A A = 2 . 2 R s i n H A C . c o s B =4RcosCcosB. M / Laithay A,B = 2RsinY =>IA, = 2Rsiny TiTdng tir HB 1 = 4RcosCcosA; HC, = 4RcosAcosB. B C Tifdng l i r l B , = 2 R s i n — ; I C | = 2 R s i n y . V a y H A + H B + H C > H A , + H B , + HC, 2R(cosA + cosB + cosC) > 4R(cosAcosB + cosBcosC + cosCcosA) V a y l A , + I B , + I C , > l A + IB + IC o cosA + cosB + cosC > 2cosAcosB + 2cosBcosC + 2cosCcosA (4) . A . B . C 1 1 o 2R sm—hsm—1-sm — > r 2 2 2j . A -+ • ' . B + (4) diTdc chi?ng minh tiTdng liT nhiT tren v d i chii y: sm- sm~ 0 < cosA + cosB + cosC < -. 2 . A . B . Cl „ . A . B . C o 2R s m — h s m — h s m — > 4 R s m — s m — s m — 2. Hoan toan tiTrtng tiT, ta c6 cac ba't dang thiJc sau 2 2 2j 2 2 2 1 1 IA,.IB,.IC, >IA.IB.IC; — + +— — .+ I I 1 1 o A B C A B B . C ^ . C . A , , , s i n y + s i n - + s i n y > 2 s i n y s i n - + 2 s m - s m - + 2 . s m - s m y . (1) IA| IB, < — + —+ — IC| ~ l A IB IC Da'u bang xay ra trong hai ba't dang thuTc Ircn A B C la tam giac deu. Theo bat d i n g thufc cd ban trong tam giac, ta c6: 3. Cho tam giac A B C noi tiep discing tron tam O. G o i I la giao d i e m ciia ba . A , . B^ . C^3 difctng phan giac A M , B N , CP. Gia suf A M , B N , CP keo dai cat di/c^ng tron sm—hsm—+ sm—1. 2 2 2 3 2 . 2 2 2 lA' ^ IB' ^ IC' R . A , . B , . C Giai (2) ~3 sm hsm—hsm — Ta c6: B I A ' = I B A ' l A ' = A'B. . 2 2 2J H i e n nhien ta l a i c6: Lap luan nhu'cac bai tren c6: l A ' = 2Rsin
B6I duding hgc ainb gtdl Lugng glic - Phan Huy Khal B C pgfu bang xay ra A B C la tam giac deu. TiTdng t i r l B ' = 2 R s i n — ; I C = 2 R s i n Y . 2 fjeu thay cac phan giac bang cac trung tuyen va I thay bang trong tam cua 1 1 1 3 tam gi^'-- ' ^ h ' '^'^ dang thiJc .sau: 1 1 1 1 1 3 OA' GB' G C - R . A . B• + : >— 2R " R Xa CO theo he thiTc lu'dng trong du'cfng tron sin sin sin 2 2 2 2 1 1 1 (1) MA.MB' = M B . M C = ^ sin - sin - sin— 2 2 2 4MA 4m., Theo bat dang thiJc Cosi c6: =>MA' = 1 »^ f . A , . B , . C ,GA' = G M + M A ' + - m . , + • sin—hsin—hsin — > 9 . (2) 4m„ . A . B . C 2 2 2 sin sin sin 2 2 2 GA m.. 8mf 2(2b^+2c^-a^) 2b^+2c^-a^ Theo bai toan crtban trong muc A , thi sin — + sin — + s i n — < - . (3) GA' 1 a' 4m,^ + 3a' 2b^ + 2c^ - a^ + 3a^ a^+b^+c^ ^ • 2 2 2 2 3 ' 4m., Tu" (2) (3) suy ra (1) dung dpcm. Tiif do suy ra: Da'u bang xay ra o A B C la tam giac dcu. GA ^ GB GC ^ ( 2 b ^ + 2 c ^ - a - ) + ( 2 a ^ + 2 c ^ - b ^ ) + ( 2 a ^ + 2 b ^ - c ^ ) Nhan xet: GA' GB' GC' a^+b^+c^ 1. N c u A B C la tam giac nhon va thay cac di/dng phan giac b^ng cac du^cfng cao (va goi H la trifc tam). K h i do ta co: = 3 (6) A 1 , . , AA' BB' CC , GA , GB , GC , Laico: ,+ _ _ + _ _ = i+ _ _ + i+ +i + .= 6 (7) (do (6)) HA' HB' HC' " R GA' GB' GC' " ' GA' GB' GC' /p/ That vay theo cac bai Iren ta co: Do A A ' < 2R; B B ' < 2R; C C < 2R H A ' = 4Rco.sBco.sC; 1 1 H B ' = 4RcosCcosA; =>6=M:+M:+^,2R^ ' + • M / GA' GB' GC' IGA' GB' GC' H C =4RcosAcosB. 1 „ , , , , — - > — => dpcm. Tu" do theo bat dang thifc Cosi, ta co: A' GA' GB' GC'~R 1 1 1 1 1 1 1 Dau bang xay ra o A B C la tam giac deu. HA' HB' HC' 4R cosBcosC cosCcosA cosAcosB ' ^ ^ i 4. Cho tam giac A B C . Ba du-dng phan giac trong A M , B N , CP k c o dai cSt 1 diTcfng tron ngoai tiep tam giac tiTdng tfng tai A ' , B ' , C . (41 2 • 4R V ( c o s A c o s B c o s C ) /' /' /' &at A A ' = BB'= /|;,CC= . ChiJng minh ^ + - ^ + ^ >4. T h c o ba't dang thuTc cd ban trong tam giac (va do A B C la tam giac nhon), i ' /„ L /,. co: 0 < cosAcosBcosC < - . Giai (5) 8 AB AM c /., be 'a CO A A B M ^ AAA'C 1 1- H ^ j>— ^ =>dpcm. . AA' AC Tir (4) (5) suy ra b I, HA' HB' HC' ~ R
B6I duOng hpc sinh gkil Lugng gUc - rhan Huy Khal Khi do + + —^>4 m„ mu m,. (1) m„ + t a , mu + i u , m, + 1 ^>4 A A •+ m, 2bccos 2bccos Do I, = < I,, . Ih I,. b+c 2Vbc + ^ > 1 . (8) nr, mu m, /„ < Vbc cos — . (2) Ta co: M A . M A ' = MB.MC hay t„.m, = — , /' 1 4 T i l f ( l ) ( 2 ) d i den ^ > - (3) COS m, 4m.; 2b-+2c^-a- /' 1 TiTdng lirco: - ^ > (4) TiTitng tir b-^ L" ^B' I,- 2 C h cos ^ cos - 2a"+2c^-b" m,, 2a2+2b--c-' I IL I 1 1 1 Vithc(X)c=> Tir(3)(4)suyra ^ + -!i + X > . (5) I" Lt) I' t i;os-^ A eos 2 " cos 2^ 2 2(Y + Z ) - X ci .— Theo bat d^ng thuTc Co si, la c6: 2 2 2 2b2+2c--a'=X>() 2 A A ,B jC 1 I Dat ,^2 _ 2(X + Z ) - Y cos — I - COS" — h cos — >9. (6) 2a-+2c"-b^ =Y > 0 - . 2 2 2j 1A + ' . B+ I, C COS" - COS" COS" 2a-+2b--c"=Z>() 2 2 2i 2 2(X + Y)-Z c = 2A 2 B 2 C 3 + (cosA + cosB + cosC) ^ 9 Do cos — I - cos — h cos — = < —, (7) 2 2 2 2 ' 4 Vlthc-(9)«I±^ +^ + 2 ^ > , /.' /' /' Tif (5) (6) (7) suy ra + + > 4 => dpcm. Y z z 4 — + — + —+ + —+ >6. (10) Dc thay tiTcac lap luan Iren thi da'u bang xay ra ABC la lam giac dcii. I Y xJ i z XJ iz Y Nhgn xet: Khi lhay ba diTclng phan giac lhanh ba difdng Irung luyen, ta c6 Ni' C>o X X ) , Y > 0, Z > 0, ncn ihco bat d;1ng thifc Co si Ihi (10) dung, vay (9) loan sau: dung dpcm. Cho lam giac ABC co dp dai ba trung luyen m,„ mn, m,. Goi M,„ Mb, M , la J'' bang ,xay ra dc thay o X = Y = Z dai cua ba dU'dng trung tuycn ay kco dai cho den khi gap du'(4 o ABC la lam giiic deu. m., m^ m^. Cho lam giac ABC noi tiep irong diTcJng iron lam O. ChiJng minh Giai OA- ^ O B ^ ^ O C ^ Goi cac trung tuycn la A M , BN, CP va giii silr A M , BN, CP keo dai tifdng be ca ab cat diTiIng iron ngoai tiep AABC lao A', B', C . Khi do ta co: AA' = M „ BB' = Mh, CC = M, Giai '-'o dinh li ham .so sin, la co: Dat t, = M„ - m,; tb = M^ - m^; I , = M , - m,.
OA^ ^ OB^ OC' 1 1 1 Giai -+ . (1) be ca ab 4R^ sin B sin C sin C sin A sin A sin B J Neu A|, B | , C| deu nam trong tam giac ABC. Khi do ta c6: Theo baft dang IhuTc Cosi, thi MB|+MC, MB, MC, . ^ 3 ^-^77 ^ = 77r + ^ = smMAB, + s i n M A C 1 • (1) 1 I 1 (2) MA MA MA sinBsinC ^ sinCsinA ^ sinAsinB ^(sin Asin BsinC)^ Hien nhien ta c6 (xem bai 2 § I chu'dng 4) sin M A B | + sin MAC, . M A B | + M A CI • A Lai theo bat dang ihufc Cosi, ihi < sin = sin —. (2) V • . • D • ^ sinA + sinB + sinC 2 ^ 2
Thco ba't dang ihiyc liTdng giac cd ban ihi sin^ —sin-^ —sin^ — < - (6) 4,i(p - a)' + 4b(p - b)- + 4c(p - c)" < 6 7 3 R ' ( 2 R - r) 2 2 2 8 c> a(b + c - a)' + b(a + c - b)' + c(a + b - c)' < 6N/3 R ^ ( 2 R - r) Tur (4) (5) (6) suy ra dpcm irong iriTcfng hdp nay. c> 4abc - (b + c - a)(a + c - b)(a + b - c) < 6 ^ 3 R ' ( 2 R - r). ' Da'u bang xay ra o ABC la lam giac dcu va M la lam cua lam giac a'y. (1) Ap dung cac cong ihiJc: ,;• 2. Ncu A | , B | , C| CO diem nam ngoai lam giac ABC (chang han A, nam ngo^j , abc lam giac ABC) S = p r : S = 4R —; — = (p-a)(p-b)(p-c) Khi do la c6 cac danh gia sau: _ (h + c - a)(a + c - b)(a + b - c) MB| + M C ' 0) + MC MC MC /ID A B C . A . B . C 4 4R cos — cos — cos — 4Rsm —sin —sm —, < 3V3R^ = sinMCA,+sinMCB| . 2 2 2 2 2 2j Do MCA, + M C B = 180 sin MCA, = s i n M C B . o XsinAsinBsinC < 3^3 . (2) Tir do Ihco (*) la c6; Thco bai dang ihiJc Cosi, ta c6: MA, + M B I . MCB + MCB| ^ . C .sin A sin B sin C =_ sinMCB + sinMCB, dpcm. Thco I 11 dang IhiTc lu'dng giac crt ban la co- (Tuy nhicn Irong Iru'cfng hdp nay dii'u biing khong xiiy ra) sinA + sinB + sinC < (4) Bai loan difdc giai hoan loan. Bai 7. Cho lam giac ABC. DiTclng Iron noi licp lam giac licp xiic vdi ba canh Thay (3) (4) vao (2) la lhay (2) dung => (1) dung dpcm. BC, AC, AB [lidng i^ng lai M , N , P. ChuTng minh: Dau bang xay ra ABC lii lam giac dcu. Ba« 8. Goi I la lam dift^ng iron noi tiep lam giac ABC. A I , BI, CI ca'l du^dng iron BA.AP- + C A . B M ' + AB.CN' < ^ R ' ( 2 R - r). ngoai licp lam giac ABC lai D, E, F Wdng u-ng. Giai ChiJngminh: H + I ^ + Z > 3 . Ta CO: lA IB I C ~ b + c- a Giai AN = AP = p - a = Lap luan nhu'cac bai iren, la co: 2 a + c-b lD = 2 R s m - ; I A 4R sin - s m - BM = BP = p - b = 2 A a + b-c CM = C N = p - c = Vi ihc ba'l dang IhiJc can chu^ng minh c6 dang tiTdng diTdng sau:
«.»^ limn mwyvTT nnang vifT BOl tiuuim iitn. snui yiui i-uym^ B . C Giai sin- sin IF 2 A A 4b'c'cos' — Ti/0ngt^c6:|= . c . A ' ^ . A . B• 2sin sin Giainhi/bai4,lac6: ^ = -^ = ^ ^ ^ b c _ l + cosA 2sin- sin - 2 2 2 2 AA2 be (b + c r b c (b + c ) ' 2 V a y theo bat dang thtfc Co si, la c6: - 4bc Thco bat diing ihilc Cosi, i h i < 1. ID IE , IF ^ , J „ . A . B . C (b + c ) ' (1) 1 1 > ji* 8sin—sin—sin— . _ AA| ^ 1 + cosA , lA IB IC~ V 2 2 2 Tir do suy ra — ! - < . 1 Thco bal d i n g ihiJc cd ban Irong lam giac, ihl AAT 2 . A . B . C ^ 1 sin-sin-sin-
Cty TNHH MTV D VVH Khang Vl§t ^ A, = 2 C | A | A = 2C|CA BC BC d o B A ^ + BOC = 18()") ^' R, = = 2(90"- A) = 180"- 2A 2sinBA'C 2sinBOC => sinA, = sit>2A. BC 2RsinA R (1) Ti/dng tiT sinBi - sin2B; sinCi = sin2C. 2 sin 2 A 2 sin 2 A 2 cos A Ta co: A ' O C = 18()" - A O C = 18()" - 2 B Tirdo: S A , B , C | =2R^sin2Asin2B.sin2C. A ^ = A S + C)BC ( d o A ^ = A ^ ) Mat khac SABC = 2R^sinAsinBsinC Tirdo S A | B , C | < S A B C = 180"- 2B + 9 0 " - A o sin2Asin2Bsin2C < sinAsinBsinC (do O B C = O C B va O B = O C (2) va O B C + O C B = 180" - B O C = 180" - 2A =^ O B C = 90" - A ) SsinAsinBsinCcosAcosBcosC < sinAsinBsinC Do sinA > 0, sinB > 0, sinC > 0, ncn =^ Vib = A + B + C - 2 B + 90" - A = 90" + c - B . (3) (2) o cosAcosBcosC < - Do vay O A ' = 2R, sin A ' B O = 2- R -sm 90" + (C - B) Wi (3) dung (theo bat dang thiJc lifdng giac cd ban trong tarn giac) nen suy ra 2 cos A Rcos(B-C) SA|B|C| < S A B C A cos A (2) 2. Neu ABC la tarn giac vuong (chang han t;u A) Li luiin lu-cing iiT co: OB' = Rcos(C-A) (3) Khi d 6 B | = C , ^ A = > S A | B , C | cosB Luc nay hien nhien ta c6: S A | B | C | < ^ A B C • OC' = Rcos(A-B) (4) Vay khi ABC khong phiii la tarn giac lij, cosC Tir (2) (3) (4) suy ra ta luon co: S A | B | C | < S ^ B C OA'.OB'.OC >8R' Dau bang xay ra -t=> ABC la lam giac deu. o cos(A - B)cos(B - C)cos(C - A) > 8cosAcosBcosC Nhan xet: Neu lhay du-dng cao bang diTdng phan giac o cos(A - B)cos(B - C)co.s(C - A) > -8cos(B + C)cos(C + A)cos(A + B). Irong. Khi do la c6 bat ding thiJc vdi dau ngifdc lai. Vi the ket hdp vdi sin-Asin'Bsin-C > 0 suy ra SABC ^ ^A|B|C| • ChiJng minh Wdng lif nhiT Iren va ta co: OA'.OB'.OC > 8R' . A . B . C^1 ^ cos(A - B)cos(B - C)cos(C - A) cos(A + B)cos(B + C)cos(C + A) SABC < S A , B | C , ^ - 8 .sin A,sin B,sin B sin C sin C sin A ~ sin A,sin B.sin B sin C sin C sin A Theo bat dang thtfc lu-dng giac cd ban suy ra dpcm. A, (1 + tanAtanB)(l + tanBtanC)(l + lanClanA) Bai 11. Cho lam giac ABC khong vuong noi tiep trong du-dng Iron tam O ban > -8( 1 - lanAtanB)( 1 - tanBtanC)( 1 - lanCtanA). (5) E)at X = lanA, y = lanB, z = lanC va chii y rang trong moi tam giac khong kinh R. DiTdng thang AO, BO, CO tiTdng rfng cat cac diTdng Iron ngoai ticp vuong, ta co: cac tam giac OBC, COA, AOB lai A', B', C . ChiJng minh: lanA + lanB + lanC = tanAlanBlanC ' ' ' ' OA'.OB'.OC > m\ =^x + y + z = xya (6) Giai Khi do Gpi RI la ban kinh diTdng Iron ngoai tiep tam gidc BCA', thi theo dinh ly han' ( 3 ) 0 ( 1 +xy)(l +yz)(l + zx) > 8(xy - l)(yz- l)(zx- 1) ^ . . j ' ^ : so' sin, ta co: o (z + xyz)(x + xyz)(y + xyz) > 8(xyz - z)(xyz - x)(xyz - y) (7) :,
s S i duong h^c sinh gioi Lu^ng giac - I'h.m Hiiy HhaT Tiif (6) (7) suy ra j j a i 13. Cho A B C la tam giac nhon vii H lii triTc tam. Gia sur H A , H B , H C lar (3) c=> (2x + y + z)(2y + z + x)(2z + x + y) > 8(x + y)(y + z)(z + x). (8) li/cn cat cac du-cfng iron ngoai ticp cac lam giac H B C , H A C , H A B l a i A ' , B' Theo bat d a n g thuTc Co si, Ihi C . Chu-ng m i n h H A ' . H B ' . H C > 8R'COSACOSBCOSC. 2x + y + z = (X + y)(x + z) > 27(x + y)(x + z) Giai 2y + y + x = (y + x)(y + z) > + y)(y + z) Taco: C H A ' = ABC = B (goc CO canh tifdng iJng vuong goc) 2z + X + y = (z + x)(z + y) > 2^(x + y)(x + z ) . M a i khac nhU'da bict du'ilng Iron ngoai Tir do suy ra (8) dung => d p c m . tiep tam giac H B C c6 ban kinh ciing Dc thay da'u bang xay ra A B C la tam giac d e u . chinh la R (ban k i n h dudng iron ngoai B a i 12. Cho tam giac A B C . G o i I la tam di/dng tron noi t i c p tam giac. Gia sir tiep A A B C ) . A I , B I , C I ti/dng iJng ca't cac diTdng tron ngoai tiep cac tam giac I B C , l A C , l A B tai A ' , B', C . ChiJng m i n h : l A ' . l B M C < 8 R ' . ^ A p dung djnh l i ham so sin trong tam giac H B A ' , Giai taco: H A ' = 2 R s i n H B A ' G o i A ' la tam du-dng tron bang = 2Rsin(HBC + A ^ ) tiep goc A ciia tam giac A B C = 2 R s i n ( 9 0 " - C + B) = 2Rcos(B - C). => B A ' va C A ' tufdng uTng la cac Lap luan tu-dng tir c6: H B ' = 2Rcos(C - A ) , dirdng phan giac ngoai cua goc B, C HC' = 2 R c o s ( A - B ) . => I B 1 A ' B va IC 1 A ' C Tir do H A ' . H B ' . H C = 8 R ' C O S ( A - B)cos(B - C)cos(C - A ) . ( 1 ) => du^dng tron du'5ng kinh l A ' chinh la diTdng tron ngoai ticp M a t khac theo chiJng m i n h trong bai 11, ta c6: tam giac I B C . V i the trong l a m cos(A - B ) C O S ( B - C)COS(C - A ) > cosAcosBcosC. (2) giac v u o n g I B A ' , ta c6: Tir ( I ) (2) suy ra dpcm. Dau bhng xay ra A B C la lam giac deu. IB IB IB Bai 14. Cho tarn giac A B C va G la trong tam l a m giac. D a t BGC = a, IA' = cosBIA A + B . C cos sm — C G A = 3, A G B = Churng minh rang: • 2 a b c . a .3 . "v ^„ . A . B . C -f sin — + s i n - - f - s i n — 4R s i n - s i n - - sin - +mh mt,+m, m,+m, ~ 3 2J 2 _ 4Rsin — . Do I B = =>IA' = . B . B . C . B . C sin- sin - Giai sin - sm sin - 2 2 2 2 2 B C Ta CO 2sin^ cosa, v i the ap dung TiTdngtirco I B ' = 4 R s i n - ; I C ' = 4 R s i n y . •; c. dinh ly ham so cosin, ta c6: A B C V I the l A ' . I B ' . I C < 8R^ o 64R'sin—sin-sin-< 8 GB'+GC^-BC^ cosa=: 2 2 2 2GB.GC . A . B . C^ 1 ,,. sin — s i n — s i n — < - . (I) 9/ , , N 2 2 2 8 Theo bat d^ng thiJc lifdng giac cd ban trong tam gi^c t h l ( 1 ) ddng => dpcm. D a u bSng xay ra o A B C la tam giac d c u .
^ i d i m n g hpc slnh glol Lupag glAc - Phan nuy nnai Cty TNltn MTV DVVn Khang Vl^t , a Aiml+ml)-9'j} 4(m|, + m , f - 9 a ^ - 8 m t , m , A B C =>l-2sin" —= - ^ (5) 2 8m^m,^^ 8m|,m^. cos h cos—h cos — /\ dung bat diing thiJc Cosi, la c6: . 2CV , 4 ( m h + m J - 9 a 2 2 2 2 c 2 => sm — = 1 (1) >9. (6) 2 /c+/a /a+/h a b c Theo bat dang ihtfc Cosi, thi 4(mh + m,)' > 16mHm, Tlif (3) (6) suy ra 4(m|, + m , )^ - 9a- ^ A{m^ + m , ) ' - 9a^ 1-- > (7) / K + / C /c+/a /a+'b A B C (vdi chii y la do GB + GC > BC => - (nib + m j > a => 4(mh + m,)^ - 9a^ > 0) cos + cos + COS Thco bat diing thiJc lU'Png giac cO ban trong lam giac thl 2 2 2.j ^ ^ 4(mh+m,)^-9a^ 4(m,+m,)'-9a' A B C 3N/3 4(mh+m,)' " 16mi,m,, (8) 0 < cos h c o s — h c o s — < 3^ r>2sin - T i r ( 7 ) ( 82) c 6 : 2a 2 2 •>S. (9) /,+/, k+l, 4(m|,+mj' 2 (2) Do a + P-i-Y = 271 = ^ ^ - 1 - ^ + ^ = 7 1 . 2 3 2 + m, . (Y -sin — < DT nhicn ^ X ) , ^ X ) , - > ( ) , ncn c6 the coi —, - , - la ba goe cua mot tarn Da'u bang trong (2) xay ra mn = m, 2 2 2 2 2 2 ^ , 2 . 3 ^ b giac niio do. Vi the lai thco ba'l dang IhiJe lu"dng giac ed ban la co: c: Lap luan tiTdng tiT, la co: - s i n - < (3) 3 2 m^. + . (\ , 3N/3 2.-1^ c (4) sin —-I-sin-+ s i n - < -sin-< 2 sin — + sin--I-sin — Dau biing trong (3), (4) tifdng iJng xay ra ABC la lam giac deu. 22 2 2 22 (10) 3 2 m^, + mt, Nhdn xet: Ta c6 bai toan tiTdng tiT sau: Cho tarn giac ABC vdi I la tarn diTdn^ Tir(9)(10).suyradpcm iron noi licp. Dal BIC = (v, CIA = 3, AIB = -i. ChiJrng minh: '^ai 15. Cho lam giac ABC. Goi I la lam du'dng iron noi licp tam giac ABC. Goi a . > 2 sin—h sin — + sin — 0 | , O., O, irin lu-dt la tarn du^dng Iron ngoai licp eac lam giac IBC, lAC, lAB. 2 2 2 Chu"ng minh: S,'()|()2()3 ^ S. /b+'c K+k k+k 3 Giai Giai Ta c6: Goi M, N, P Ian lirm la trung diem ciia lA, IB, IC => S,v,Ni> = ;|;S (do AMNP ^ AABC vdi ti so dong dang k = ^ ) a 1 p f 1 1 1 1 ^ani Oi la giao ciia hai diTdng trung trifc cac ^ I) - + - - +- ^•anh IB, IC. -b + c; - Ic di, Ui bj A B A Tam O; lii giao ciia hai du'dng trung iri/c cac 2bccos . , 2accos . . 2abcos , 2_ ^ + ^ _| 2_ a + c ^ 2_ + " '^^nh IC, lA. b + c ' be a-he ac a-fb ab
B6t duSng tipc slnh glol Lupng glAc - Fhan Huy Kbal Cty TNnnnrvDVVHKhang V/ft Tam la g i a o c i i a h a i dir(:>ng I r u n g trifc c a c c a n h lA, IB Lap l u a n ti/dng t y c o : Rj = R ; R3 = R Khi do dc lha'y O,, O 2 , O., liTOng ilng c u n g la t a m c a c diTcfng tr6n bang lio'p 2cosB 2cosC tiep cac goc M, N, P cua AMNP. Goi R ' la ban kinh di/dng tron ngoai ticp AMNP, thi Ta co: 0 , 0 , = O 3 K + 0 , K = ^R.i f JR? - — SMNP = 2(R')'sinMsinNsinP Theo phan dang thiJc trong tam giac ta c6: N 2VcWc 2VCWA O1O2 = 4R'COSY; = yCtanC + lanA). P Tu-dng liTco O2O1 = y d a n B + lanA). Vi the 0,03 = 4 R ' c o s - So,0203 = sinO^O;02 180"-NIP = N + P ^0,03.0,02 R5 0 3 0rang = c6: , 0 2 la R ^ sin(C + A) .sin(B + A) / „ \ smO^OiOj = s m. N— ^+—P = cos— M = 7; f. ' ^in(l80"-BOc) 1 M N » cos C. cos A cos B cos A Vi the' So,0203 = - 0 , 0 2 . 0 , 0 3 . s i n 0 3 0 | 0 2 = 8(R'r cosy cosycos- 8 cosBcosCcos^ A 4 cosBcosCcosA _ R ^ >sinBsinCsin2A V^'the'So,0203 S R^ sinBsinCsinA Taco: S„|0203 > S " 03>S it;-; M N P R ^ sinBsinCsinA > 2R\sinAsinBsinC 8 ( R ' ) " c o s — C O S Y C O S — > 4 S M N P 4 cosBcosCcos A 8(R')- c o s — c o s — c o s - > 4.2(R')'sinMsinNsinP o cosAcosBcosC < - . (1) 2 2 2 8 . M . N . P^1 Theo bat dang thiJc lu-rfng giac cd ban trong tam giac thi (1) dung => dpcm. sm — sm — s m — < - . (1) Dau bang xay ra ABC la lam giac deu. 2 2 2 ~8 Thco bat dang thufc liTdng giac cdban trong tam giac thi (1) dung => dpcm. ^ O H xet: Dau bang xay ra MNP la tam giac deu ABC la tam giac deu. Neu ihay O bang trong tam G cua lam giac ABC, thi neu goi O,, O2, O3 Ian Bai 16. Cho tam giac nhon ABC. Goi O la lam diTcJng tron ngoai tiep A A B C liTdt la tam du-^ng Iron ngoai tiep cac tam giac GBC, GAC, GAB thi la van Goi 0|, O 2 , O3 Ian liTcU la tam diTcJng tron ngoai tiep cac tam giac O B C c6 ba'l d^ng thu-c 80,0203 > 8 . OAC, OAB. Chu-ng minh: 80,0203 > S Chiang minh xin danh cho ban doc. ^ Giai 17. Cho tam giac ABC, M la trung diem cua AC. Dal CBM = (p. Goi R|, R 2 , R3 Ian liTdt la ban kinh cac ChiJngminh: colA > ^ ^ - ^ c o s ^ ^ dirdng iron ngoai tiep cac tam giac OBC, OAC, OAB. sinip Ta co: , Giai R , = BC a 2RsinA Taco: ZsinBOC 2 sin 2 A 4 sin A cos A 2 cos A cn,^^ 2V2-3cos^
hdl du6na hpc sinh glot lufng giic - fhan tluy hJial pgfu bang xay ra a, = aj; P, = P2; y, = y: cot A + 3cot(p > •. (1) O la tam diTc^Jng tron noi ticp AABC. sirup P^i 19. Cho tam giac nhon ABC va goi O la tam diTcfng tron ngoai ticp. Goi r,, Thco clang thu-c trong tam giac, ta c6: ^ - r^, r, Ian lifcn la ban kinh diTdng tron ngoai tiep cac tam giac OBC, OAC, b^+c'-a^ RM^+MC^-MC^ OAB. Ch^ng minh: r, + rj + r^ > 3R. , cotA + 3cot(p = h3 4S 4S Giai 2a^+2c^-b2 ^ ^2 Ap dung dinh ly ham so sin trong cac AABC, BOC ta c6: a b c 4S " 2S R = 2 sin A 2sinB 2sinC _ 4a- + 2c^ - b^ _ (2a^ + 2c^ - b^) + 2a^ a a ' , 2S 2S 2,sin BOC 2sin2A _ 4BM^+2a^ _ 2 B M ^ + a ^ , b TiTdng liTco: =• 2S ~ S 2sin2B 2sin2C Thco bat dang thi?c Cosi, ta thay 2BM- + a^ > 2%/2 BM.a. (3) Khi do r, + r: + r, > 3R 2V2BM.a 2^/2 a b c TCr (2) (3) suy ra cotA + 3col(p > — - — ; — = . o >3R 2sin2A 2sin2B 2sin2C BM.asm^? smif Vay (1) dung ^ dpcm. Dau bang xay ra o 2 B M ' = a' o a = m, N/2 . 2R,sinA 2RsinB ^ 2RsinC o >3R Bai 18. O la diem bat ki nilm trong tam giac ABC. Chi?ng minh: 4 sin A cos A 4 sin B cos B 4 sin C cos C BAC ABC ACB^ o >6 (1) OAcos + OBCOS + OCcos >p. . cos A cosB cosC 2 2 2 A Thco ba'l dang thifc Co si, thi Giai p 1 1 1 (cos A + cos B + cosC) > 9 Ki hicu tti, a2; Pi, P:; 71, 72 (xcm hinh vc). UosA cosB cosC Ke OM 1 BC, ON 1 AC, OP 1 A B . 3 Do 0 < cosA + cosB + cosC < ncn tiT (2) suy ra (1) dung =^ dpcm. Taco: ^ 2 j Dau bang xay ra ABC la lam giac deu. p = ^ (BM + MC + CN + NA + AP + P B ) Chu y: Neu ABC lii lam gitic c6 g()c tu, ihi kel luan ciia bai loan chu'a cha'c dung. = ^(OBcosp: + OCcosy: + OCcosy, + OAcosa: + OAcosa, + OBcosP,) That vay xet tam giac ABC c6: A = 1 2 0 " ; B = C = 30". => p = i o A ( c o s a , + costt:) + ^OBCcosp, + cosp.) + ^OC(cosy, + cosy:) C>c iha'y luc nay r, = R BAC tt|- p = OA cos cos —! + - cos cos + - cos BC A cos — - ' 2 2 2 2 2 2 ^'0 canh bang R, ncn BAC ABC ACB => p < OAcos + OB cos + OCcos—-— 2 RV3 RV3 r, dpcm. ^Sy r, + r. + r 3 = R + R>/3 < 3R £•1 V f,-' :.' 2 2 2
lidJ duany / 7 ^ f Siinn yiUI Lupng 2. Tir R > 2r, ncn hicn nhien la c6: 21. Cho lam giac A B C . Goi 1 la tarn du'cJng Iron noi licp. r, + r: + r, > 6r ChiJngminh: IA.IB.IC dpcni Dau bang xiiy ra ABC la lam giac dcu Ciai Thco ba'l dang ihuTc Co si, la c6: Ap dung dinh li ham so sin irong tam giac H A B , la c6: A . B . C HA AB HA c r, + r 2 + 1 , : - 2 R s i n - + 2 R s i n - + 2 R s i n - (do HAB + C = 180") sinABli sinAHB cos A sinC _^ „ ^ 2RsinC.cosA . > 3,4R\sin-sin-sin-. (3) => HA = = 2RcosA. V 2 2 2 •sinC Vi r = 4Rsin —sin —sin—=>HR''sin—.sin^sin^ = 2R^-. TiTdng tiT HB = 2RCOSB; H C = 2RcosC. 2 2 2 2 2 2 (. Vay H A . H B . H C < R ' ( Thay vao (3) va c6 r, 4 r, + r-, > 3 V 2 R ^ (4) 8R'(COSACOSBCOSC) < R ' Ma t khiic la lai c6: R > 2r -t> 2R-r > Sr" (5) (1) Thay (5) vao (4) va c6: r, f r, + r, > 6R cosAcosBcosC < - . ^'K-O bal diing thi^c lu'(;ng giac ciJ ban ihi (1) dung => dpcm. Do la dpcm. '^iiu bang Xiiy ra ABC la lam giac dcu. Dau bang xay ra « ABC la lam giac dcu.
Nhdn xet: I C = 4Rco.s—co.s- 1. K h i A B C la tarn giac viiong (chan han vuong lal A ) . K h i do 2 2 H =A HA = 0 H A . H B . H C = 0. N h i r v a y I , A + I^B + I,C < 9R H i c n nhicn bat dang thiJc H A . H B . H C < R ' la dijng (d day dau bKng khong A B , B C C A o 4R cos — COS h COS —COS h cos —COS — R = a • i ? Do thay Ta CO H A = a; HB = HC = A B B B C, C A 2A 2B ,C ;os—COS — + cos—cos — + cos—cos — < cos — + cos — + cos — (2) cos HA.HB.HC > 3a'> a ' = ^r^, 2 2 =^ H A . H B . H C > R' V, cos^ A + , o s 2 ^ + cos^ ^ = 3 + cosA + cosB + cosC V a y ket luan cua bai loan Irong trifclng hdp nay chu'a chac dting. 2 22 2 ' ; B a i 23. Cho lam giac A B C . G o i I.„ 1^, 1 Ian liTdt la tarn du-cfng Iron bang ticp cae 3 , -' cosA + cosB + cosC < - (bat dang thifc lu-ctng giac cct ban Irong tam giac), goc A , B, C cua tam giac. 1. ChtJng minh: I,A.IhB.I,C < 2 7 R \ ncn tij-(2) suy ra {1) dung => dpcm. 2. Chii-ng minh: I.,A + I,,B + I,C < 9R. Dau bang .xay ra o A B C lii lam giac deu. " ' Giai Hai 24. ChiJng minh rang Irong moi lam giac A B C , la c6: Nhqn xet: Ta thay rang 21 suy ra 1/ h.. 27 tl That vay thco ba't diing thiJc Cosi, ta c6: 1. ^ > / ^ . 2. + /h /.. ~ V R /,. " V R I , A + I , B + I,C > 3;^r,A.IhB.I,C . Giai 1. Ta luon c6 the gia su^ B > C (ncu khong ihi chi vice Inio ddi vai tro giffa hai dinh B va C) va c6 hinh vc ben. Vc du-iJng cao A H va phan giac A D (khi do A D d ben phai A H - xem hinh ve) Ta c6: -fit = — = sin A D H = sin /.. AD c+- • Tirdo ili!-> j f i ^ s i n ^ C + - ^ R cos- C - 2 ~ R B 2} . A . B . C >8s!n—sin—sm — 2 2 2 B cos 2 C - B-C B+C > 4 s i n — cos cos 2 ^ . . A B~C , . 2A COS" >4sm—cos 4sm — 2 ~ 2 2 2
p a i 25. Cho tam giac A B C noi tiep trong dU'&ng tron tam O. Dtfdng phan giac C-B - . A cos— ^^'^^ >0. (1) trong cua cac g6c A , B, C c^t du'dng tron ngoai tiep tam giac tai A|, B , , C|. V i (1) dung = > d p c m ChiJng minh r i n g A A , + B B , + CC, > A B + BC + C A . ^ g Giai D a u b i n g xay ra c o s — ^ - = 28111^ H D Ta c6: A B + A C = c + b = 2RsinC + 2RsinB C-B . C + B „ . A A ^„ . B + C B-C 2cos s i n — — = 2sin—cos— = 4Rsin cos . (1) 2 2 L L 2 2 o sinC + sinB = 2sinA b + c = 2a Theo dinh ly ham so sin trong cac tam giac < » b, a, c lap thanh mot cap so cpng. 2. Theo tren ta c6: A B A : , A C A i ta c6: h h u h , B-C C - A , A-B AA, =2RsinABA| =2Rsin 2 i ^ + ^ + - i - = COS + C0S + COS—T— AA| = 2 R s i n A C A | = 2 R s i n h,. h, h,. 2r Vay ^ + ^ + - T ^ > 3 3 | - - I . •ai /b - AIR ( . A . B . C Tur do suy ra 2 A A | = 2R sin + sin C + - B-C C-A 2] C O S ^ - + COS- + COS > 6 sin—sin—Sin — (1) 2) 2 2 ' ' 2 2 2 2 . B + C + A B-C B-C Theo ba't dang thufc Cosi, ta c6: = 4Rsin cos = 4Rcos , (2) . . 2 2 B_c e o s - ^ C-A A-B + c o s ^ ^ + cos^— > 1 B-C C-A A - B 3 ^ c o s - ^ c o s - ^ c o s - ^ (2) Tir(l)(2)diden 2AA| > AB + AC. (2) Lap luan tu-dng tir 2 B B | > B A + BC, (3) Ta se chtfng minh rkng trong m o i tam giac thi 2CC| > C A + CB. (4) c„.iz£c„,s^cosA^>8si„As,„S.„^. (3) Cong iCrng vc (2) (3) (4) suy ra dpcm. 2 2 2 2 2 2 Nh^it xet: Ta c6 the chiJng minh (2) bang each " p h i lu^dng g i a c " nhiTsau: That vay Theo dinh l i Plolcme, ta CO: .'-'t^i}} 8cos-cos-cos-cos^^cos^^cos^^>8sinAsinBsinC 2 2 2 2 2 2 AA|.BC = A B . A | C +AC.AiB ^ . B + C B-Cl , . C+ A C - A l ^ . A + B A - B = A|B(AB + AC) (doA,B = A,C) 2 sin cos—— 2sin cos—— 2 sin cos 2 ' 2 2A|B(AB + AC) =>2AA| = >8sinAsinBsinC BC o (sinB + sinCXsinC + sinA)(sinA + sinB) > > 8sin A s i n B s i n C (4) Do 2A|B = A , B + A|C > BC nen suy ra 2 A A | > A B + A C , vay (2) dung. . Theo ba't dang thiJc Cosi, thi ^'^'^ 26. Cho A B C la tam giac nhpn. Cac di/dng cao ke tir A , B, C cat diTdng tron sinB + sinC > 2 V s i n B s i n C , ngoai tiep tam giac A B C tai A , , B,, C,. Chiang m i n h : ..^ .,, , sinC + s i n A > 2 V s i n C s i n A , A A , + B B , + CC, > A B + BC + C A . sin A + sin B > 2 Vsin A sin B . Tir 66 suy ra (4) diing => dpcm. '^P dung dinh ly ham s6' sin trong A A B A , , ta c6: , fj + ,^ . D a u bang xay ra o A B C la tam giac deu. '^A, = 2RsinABAT
Cty TrainnrvownHhang Vi^t Vi ABA, = ABC + CBA| = B + CAA| »! a + b 2ah _ ( a - h ) ^ H i c n nhien la c6: " —'- X) = B + 90" - C 2 a+ b 2{a + b) b + e 2bc e+ a 2ea => sin A B A , = sin(9()" + B - c) = cos(B - C) ~Y~" bT^ ~ 2 ~ " ^ - ' ^ ^ ' ' y =>AAi- 2Rcos(B - C). => A A , + B B , + C C | > a + b + e =5 dpcm. , . TifiJng tir t o B B , = 2Rcos(C - A ) ; CC, = 2Rcos(B - C ) . Hal 27. Cho lam giac A B C . M la m o l d i e m bal k i Irong lam giae. G o i x, y, / Ian Nhirthc lu-dl la Ciic khoang each liT M xuong BC, A C , A B . A A | + B B | + C C | = 2R(cos(A - B) + ats(C - A ) + cos(B - C)) (1) r-. V Chirng minh: N/X + ^ + < J a" - + b" + c" 'tn ' . i , Co the chufng minh rhng 2R A B C Giai COS(A - B) + COS{B - C) + COS(C ~ A ) > 4 c O S y c o s — C O S y (2) Ta c6: SM,JC + S M A C + Svi,\|j = S V i Ihc lhay (2) vao ( I ) va c6: A A | + B B | + C C | > 2p = A B + BC + C A =^ dpcm. -.i ^ ^MBC I SMAC , S.MAB, _ | s s s Nhaii xet: -^ + ^ =1 Ta CO k c l qua lifdng tiT khi lhay du'dng cao biing cac di/fJng Irung luycn (d h.. 1 c day dung v(?i m o i l a m giac) 2 X y / h., + hh + h, = (h, + hh + h,) —+ —+ — (I)! Ta c6: A M . M A , = B M . M C = — K hh h . la eo TiY (1) va Iheo bal dang ihiJc Bunhiaeopski, MA, = 4 m.. K + hh + h,. > "> •> , 2 . 2 , 2 => A A , = m , M a" 4m,, + a b +c 4m., 4m., 2m, (2) c^ + a^ TiTdng liTco: B B , ; cc, H 2m,. Da'u bilng Irong (2) xay ra o — = — = . 2mK X y /. R6 rang Irong m o i A A B C , ta c6: Trong m o i l a m giac A B C , Ihi h,, = bsinC, h^ = csinA, h,. = asinB, nen 2m., < b + c; 2mh < c + a; 2m, < a + b. /bc + ae + ab 7h;, + h^ + h,^^ = V b s i n C + c s i n A + asinB = ^ ^ ^ ^ Tif do suy ra ^R A A D D b-+c c~+a- a'+b Thco bal dang ihufe Cos], la eo: A A | + B B | + CC| > 1 + b+ c c+ a a+ b a^+bVe^ 2 be 2ca 2ab (3) A A , + B B , + CC, > ( b + c) - + (c + a) - ^ — + (a + b) - — — - 2R b + c c+ a a+ b Dau bang irong (3) xay ra a = b = e. AA, + BB, + C C , a- + b ' + e ' ' a+ b 2ab f b +_c _ 2be e+ a 2ea 'riV(2)(3)e6: sf^ + ^ + ^f^. < J - • dpcm. >a + b +c + + b+ ej + , 2 e + a, , 2 a+ bj f^3u bang xay ra o A B C la lam giac dcu va M la l a m eiia lam giac.
7 — rnarM MMMMJT MmMMmmm cty mnn mv p vvn Kbmng vift B a i 28. Dirdng tr6n noi tiS'p tarn giic tiep xuc BC, CA, A B tai A,, B , , C,. G o i p. Giai la nii'a chu v i tarn giac A|B|C|. ChuTng minh: p, 90" - hinh 2 ) Chu y rSng du'cfng tron ngoai tiep ABC|A| chinh la dufcfng tron ngoai tiep ti? . A ll-cosA giac BC|HA| \
B 6 i duang hpc sinh gioi Luring giac - nan nuy nrtST Vay P I = - ( A | B , + B , C , + C , A , ) A K p - a ) s i n y + B K p - b ) s i n y + CI(p - e ) s i n ^ = i(AA| + AA|) ( d d a y B|C| = 0) 1 •+. B 1 + -. C > ( A I + BI + C I ) l (1) A A B C sin cos sm cos sm cos = AA| 2 2 2 2 2 2 Taco p = i ( A B + AC + B C ) ^ ^ p = ^ ( A B + AC + BC). b-c a, = ^^AKp - a ) s i n y ; a^ = ^BI.(p - b ) s i n y ; a, = ^ C I ( p - e ) s i n - D a l A B = c, A C = h, BC = a ==> A A , = 1 u 1 u ' K h i d o pi < ^ p ^ < i - ( a + b + c) b| = • f . A . A= • " 2 = — .r B= = \, = ' B . C C sm cos sm cos , sm cos 2 2 V 2 2 V 2 2 4bc < a" + a{b + c). (4) Ta CO a- = b ' + c' > 2bc. (5) (khidoaibi = J A I - J ^ — ^ = V A I . A I = A I : a.b, = B I ; a^b, = C I ) , L a i C O b + c > 2Nybc , cos 2 •vtJ'dw,: a = V b 2 + c - >x/2x/b^ Suy ra (1) dung => dpem. . =::>a(b + c) > 2 V 2 b c > 2 b c . (6) r>.A' 1 V •> ai a-> • a-1 Dau bang xay ra o - ^ = — = -±- Tijf (5) (6) suy ra a ' + a(b + c) > 4bc ^ (4) dung b| b, b, T o m lai la da chiJug minh du^Oc Pi < ^a da'u bang xay ra A B C la lam A B C la lam giac dcu. 2. Thco djnh ly hiim so sin Irong lam giac B I C , ta c6; giac d c u . . C ^ . A . B jg g(- asm _ bsm csm Bai 30. G o i I la tarn dicing Iron noi tiep lam giac A B C . C . B+C ^'B=—^.Ti/.ngurIC = — ^ : IA = —A 1. Chufng minh; ab + be + ca > ( A I + B I + C I ) ' . sin sm cos eos cos 2 2 2 2 2 lA.IB.IC 2. ChiJng minh: alA^ + b I B - + c I C - Nhu-vay l A . l B . I C = abetan — l a n - l a n - . (2) 2 2 2 Thco bat dang thiJc Cosi, ta eo; I . K e I M 1 BC, I N 1 AC, IP 1 AB. alA' + b I B - + e I C - >3\/lA^IB'IC^ . (3) Taco: A N = AP = p - a ; BP = B M = p - b; Tir(2) (3) suy ra C M = C N = p - a. lA.IB.IC 1J A B C . il'i' T , - 3 ; ' ' / t a n ^ — l a n ^ — l a n ^ — 2 2 2 2 2 2 ~ V 2 2 2 ., • Tif do la c6: ab + be + ca > ( A I + B I + CI)" ^ } A B C 1 ( tan — t a n — t a n - < - ; = . (5) o 2S 1 > ( A I + Bi + C I ) ' V 2 2 2 "73 sinA sinB sinC, ^hay (5) vao (4) va c6 dpem. Dau bang xiiy ra A B C la lam giac deu.