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B a i 287 (Cfiu hoi trftc nghi^m) b) , FhiTdng trinh (1) » Phi^cfng trinh nao sau day c6 hai nghi$m X i , X2 thoa m a n bit't d^n thii-c xi < < X2 < a) 3x^ - 5x + = b) 3x - X t = x^ - 2x + (x' - 2x + 2)^ - 4{x'' - 2x + 2) + (2m + 4) = (dieu k i ^ n : t > 1) - t + ( m + 2) = c) Sx"* - 2x - = (*) D a t fit) = t^ - 4t + 2(m + 2) d) 3x - 5x - = +4 =0 PhUOng t r i n h da cho c6 n g h i e m x c=> Phifcfng t r i n h (*) c6 n g h i e m t > • HUdng dan D a t f(x) = ve t r a i phuong t r i n h bac h a i Ta p h a i c6 : B i e n d6'i phuong trinh (1) f3.f(0) < ff(0) < 3.f(2) > f(2) > Trifdng h S • - ^ V, , -2 t2 Dap so : 3x^ - 2x - = (Cau c) B a i 8 (Cau hoi trie nghi^m) Vdi nhffng gia tri nao ciia a thi phi^c^ng trinh 3x^ - ax + a = c6 hai nghi^m X i , X2 thoa man - < X i < < X hoSc a ) a < - V a > c) d) A' = - ( m + 2) = - m S l.f(l) > l.f(1) = 2m + > I Xi Sau d6 g i a i gio'ng b a i 289 Dap so : b) max(y) = v i y = - (x" + 4x^) < (Dfi'u = xdy r a o R , GlAl a) T a c6 y = x^ - 2x + = (x - 1)^ + > 1, Vx £ R Dau = xay r a o x = Dap so : m i n ( y ) = k h i x = R 166 ) c) Phucfng t r i n h da cho o> x = 0) • t = - x"* - x ^ + (t < 5) t^ + 2t + m - 10 - (*) * T i e p tuc g i o n g b ^ i 289, t a c6 : - 25 < m < 11 V m < - 25 m < 11 167 E KI^M TRA CAC KIEN THLfC VE TAM THUTC BAC HAI a i 294 (Cau hoi trac nghiem) Gia silf tam thtfc bgc hai f(x) = (1 - m)x^ + 2mx + c6 bang xet dau : B a i 291 (Cau h6i trSc nghiem) (xi, X2 la hai nghiem ciia f(x)) Tam thtfc bglc hai f(x) = (m^ - 3)x^ + 2mx c6 bang xet dau + +00 f(x) - + c) m = d) m = - b)ml d)|m| k h i x e ( x i ; X2) n e n a = l - m < c:>m>l Dap so': m > (cau a) TU bang xet dau t a c6 f(l) = - > B a i 295 (Cau hoi t r i e nghiem) Tam thiJc f(x) = 2x^ - ax - c6 hai nghiem X i , Xg thoa man dieu Dap so : m = - (cau d) ki^n — + — B a i 292 (Cau hoi trfic nghi^m) Xac dinh cac gia tri cua m de bat phifoTng trinh x^ - 4x + 2m - < (I CO t^p nghi^m S = ? a)ml b) m = - + 00 X2 - f(x) Hay tinh m ? a) m = — 00 X - 00 X b)m>— c)m — a) a = 15 = Tinh a ? b) l a I = 15 c) a = - 15 d) MOt gia tri khac * HUdng ddn 1 X-^ X2 S P Xj •X2 Dap so : a = - 15 (cau c) * HUdng dan Bat phifcfng t r i n h da cho v6 n g h i e m X., - 4x + m - > d n g h i e m diing vdi m o i x E R A' < B a i 296 (Cau hoi trac nghiem) Tam thii-c f(x) = x^ - 2mx + c6 gia tri nho nhat bSng T i n h m ? Dap so': m > — (cau b) a) m = c) I m b)m = - l I = d) I m I >1 * Hudng ddn B a i 293 (Cau hoi tr&c nghiem) • T|lp nghiem cua b a t phufoTng t r i n h _ 2x - >0 1a : (x + l)(-x^ + 4x - 6) a) S = (- ; ^ ) b) S = [- ; 1] c) [- ; ^ ) D a p s o : S = (- 1[ ^] (cau d) > - Vay minff(x)] = - = Dap so : I m d) (- ; ^ ] • HUdng ddn Vi {- x^ + 4x - 6) < 0, Vx e R n e n bat phiTcfng t r i n h da cho tiTOng 2x — 1 duong vdri O v a a + b > c , b + c > a , GlAl • f(x) = (b + c)x^ - 2(a + b + c)x + 2(a + b + c) A' = (a + b + c)^ - 2(b + c)(a + b + c) = (a + b + c)[(a + b + c) - 2(b + c)] = (a + b + c)(a - b - c) , V i a, b, c la dai ba canh ciia m o t t a m gidc nen a, b, c > A' < a < b + ( f(x) > 0, Vx € R B a i 300 T u y theo m.'tim t§p xac dinh cua ham s6' y = ^x'^ + 4x - m + (1) • HUdng d&n • H a m so (1) xac d i n h o x^ + > • D a t f(x) = x^ + 4x - m + 2, t i n h xet dau A' GlAl • H a m so (1) xac d i n h o x^ + 4x - m + > (G9i D la t a p xac dinh cua ham so) • D a t f(x) = x^ + 4x - m + 2, t a "5, CO m + -2 - 00 170 : A' = - (- m + 2) + 171 VECTO Chu^'oii d o I Kicn I) t h i f c coT c) * -> Do dai cua vecta AB la dp dai ciia doan thang AB, ki hieu |AB| CO dai bang ( = ) 3) Hai vector bang Dinh nghla Hai vecta bang nnau chiing ciing hudng va c6 dp dai b&ng Chii y : • a va b bang nhau, ki hieu a = b ban - > C a c djnh nghTa 1) Vector Dinh nghla Vecta la doan thSng da dinh hirdng, nghla la da chon mot diem mut lam diem dau, diem mut lai la diem cuoi • Vdi hai diem phan biet A va B, ta c6 hai vecta khac : AB va BA • Vecta CO diem dau va di§m cuoi trung nhau, chang han: AA , MM , goi la "vector - khong", ki hieu , ' 2) Phifofng, hi^oTng, dp dai ciia vectof a) Dinh nghla Hai vecta goi la ciing phuang hai vecta Ian liicrt nhm tren hai dUcJng thang song song hoac trung H$ qua : Hai vecta cung phuang vdi mot vecta thuT ba thi hai vecta cung phi/ang b) Hai vectcf a va b cung phifcfng thi a va b c6 the cung hxidng hoac ngUdc hi^drng Chii y : • cung hudrng vdi moi vecta • hirdrng Hai vecta cung hudng vdri mot vecta thiJ ba thi hai vecta cung —> 172 a = b va b = c doan th^ng vecta Vecta CO diem dau la A, diem cuoi la B, ki hieu : AB • Dinh nghla > a = c • Cho san a va mot diem O, ta c6 nhat mot diĐm A : OA = a ã Mpi vecta deu bang II) Phep cpng c a c vectd 1) Dinh nghla tong cac vectof Cho hai vecta a va b Tii diem A y, ve AB = a va BC = b Vector AC dUdc goi la tong hai vector a va b , ki hi^u : AC = a + b Chii y : • Tong a + b khong phu thuoc vao vi tri diem A • Quy tac ba diem : AC = AB + BC (ba diem A, B, C y) (hinh 1) • Quy tac dudng cheo hinh binh hanh ABCD la hinh binh hanh =5 AB + AD = AC (hinh 2) A D B^ (Hinh 1) 2) Tinh chat -> ^ a) V6i moi a t a c d : a + = + a= b) V d i m o i a , b , ta c6: a + b = b + a c) Vdi moi a , b, c,tac6:(a (giao hoan) V e c t d d o i c i i a mpt vectof a) Dinhli a + b) X = b) k a = neu k = hoac a = + b) + c = a + (b + c) (ket hcfp) c) Do dai ciia k a la |k a | = i k | | a | 2) Tinh chat cho trUdc, lu6n c6 m^ot vecto x n h a t cho Dinh nghia a) k ( l ) = (k.l) a b) (k + 1) a = k a + a c) k( a + b ) = k a + k b d) a = a ; a = ; k = 3) Dinhli -> Neu vecto a + b = —» t h i vectof b dirge goi la vector doi cua so' thiTc k cho b = k a d a + (- a ) = M i vector c6 m p t vector doi n h a t • -» -> k > • k < Dinh nghia Cho h a i d i e m j h a n biet A va B b) , M chia doan t h ^ n g A B theo t i so' o c) = a+(-b) 1-k OA - ^ k.OB K h i k = - t h i M la t r u n g d i e m ciia doan t h i n g A B Vay : - OM = (k / 1) H§ qua Phep t i m hieu a - b goi l a phep trir hai vectcr > cx> M A = k M B Dinh li Dinh nghia - ' D i e m M chia doan t h ^ n g A B theo t i so k Cho d i e m O tCiy y va k ?i , ta c6 : > > , ' H i ? u c i i a h a i vectof - o a v a b cung hddng -* —> o a va b ngupc h d d n g a) Chiiy: 1) • D i e m c h i a d o a n t h a n g theo ti so' cho trifdfe N e u b l a vector doi cua a t h i a la vector d o i ciia b n e n : va vectd doi cua vector b nghia l a : a - b IV) -> t h i t o n t a i n h a t 4) -» Hi#u cua vector a va vectcf b , k i hi^u a - b , la tong cua vectcf a • M la t r u n g d i e m ciia doan thSng A B o > OM = If OA + OB ( la d i e m y ) Cho ba d i e m bat k i : A, B, C, ta c6 : AC = BC - B A Phep nhan vectd v6l mpt so Dinh nghia T i c h ciia vectcr a v d i so thuc k la m o t vectcr, k i hieu k a , dxxac xac d i n h nhir sau : , -> a va b la h a i vector doi • -> —> _> 2) ^Z", -> Chii y : • ' ;• - Neu h a i vecta a va b ciing phddng va a vector a va k i hieu l a - a • u V d i m o i vectd a , b va m o i so thdc k, 1, ta c6 : 1) a k a cung hudrng v d i vectcr a neu k > ngugc hirdng v d i vectd a neu k < a III) Phep trii hai vectd Vdi m6i a) 5) T r o n g t a m c i i a tarn g i a c • Dinh li • G la t r o n g t a m ciia t a m giac A B C o GA + GB + GC = Q • G la t r o n g t a m ciia t a m gidc A B C o OG = OA + OB + OC Toan K la t r o n g t a m t a m giac Q S U nen K Q + K S + K U = B a i Cho tam giac A B C , goi A' la diem doi xii'ng \6i B qua A, B' la diem doi xxjfng vdri C qua B, C la diem doi xii'ng vofi A qua C G K + ( Q P + S R va phu y OA = OA' + A"A • TiTcfng t u doi v d i OB • AB + BC + CA = va A'A = AB Ta l a i c6 : = OB' (3) Q C UT = - A E SR Tuongtu- OB d = -CA QP : OA = OA' + A'A = OA' + AB CO + UT ) = OC GIAI Ta (2) G K + ( Q K + K P ) + ( sk + K R ) + ( U K + K T ) = (D OA + OB + O C = OA' + O B ' + O C ' • = Cong (1) va (2) ta c6 : Chu'ng minh rang vdti mpt diem O bat ki ta c6 : * HUdng QK + B'B = OB' (1) (vi A"A = AB ) QP (2) + BC (3) OC = OC' + C'C = OC' + CA + UT + SR = (3) va (4) Cong (1), (2) va (3) ta CO : = -EC =:> G K - CA + A E + EC = o GK = = (4) ci> G = K (dpcm) Bai Cho ti? giac A B C D Goi M, N Ian lUgft la trung diem cac canh A B , OA + 013 + OC = OA' + OB' + OC' + AB + BI: + CA C D Chu'ng minh MN = A C + BD = AD + B C GIAI ChiJCng minh = OA' + OB' + O C (dpcni) • B a i Cho luc giac A B C D E F Goi P, Q, R, S, T, U Ian li^grt la trung diem cac canh AB, B C , C D , D E , E F , FA Chu'ng minh rdng hai tam giac P R T va Q S U c6 cung tam • HUdng dan Goi G va K Ian lucft la t r o n g t a m A P R T va A Q S U , ta chu'ng m i n h n - ' u - - G K = bang each chu y • GP + GR + GT = |KQ + KS + KU = M N = AC + B D Ta CO : • AC = A M + M N + N C • BD = BM + M N + N D ' ' • ' \ Cong ve, ta difcfc: AC + B1) = A M + M N + N C + B M + NT) = A M + B M + M N + N C + N D (1) GIAI • G la t r o n g t a m t a m giac P R T nen G P + G R + G T = GK G K + KP + (KP + GK + KR + KR + GK + KT) = + KT (1) = 177 GIAI M l a t r u n g d i e m ciia A B n e n A M + B M = Mk N • l a t r u n g d i e m cua CD n e n N C + N D = L a y d i e m O t u y y , t a c6 : A^i Do d6, (1) t r d t h a n h : AC + BD = M N • + + + A \ (dpcm) - O A j ) + ( OB2 - OA2 ) + = ( OBj = ( O B i + OB2 + + (0B„ - 0A„ ) , T L r o n g t u t a c : AX) + BC = M N T m l a i , t a c6 : M N = AC + B a i Mpt gia d9 dUpc gSn v a c B1) tvCdng = A D + BC nhvC • hinh l a Tam giac A B C vuong V i n d i e m B i , B2, B„ cung Ih n d i e m A i , A2, hieu m p t each khac, cho n e n t a c6 : -> OBj c a n cf diem C Ngifori ta treo vao diem A mpt v|it nang 5N Hoi C O nhang Itfc nao tac dpng vao buTc tvTdng tai hai diem B va C ? + B „ ) - ( O A i + OA2 + -> + OB2 + Tir (1) va (2) AjBi , An nhitog ducfc k i ^ -> + A2B2 + + A^B,, = + B „ = O A j + OA2 + + A „ ) (1) -> + 0A„ (dpcm) B a i Cho ba diem phan bipt A, B, C a) Chiing minh rSng neu c6 mpt diem I nao va mpt so thiic t > > > • cho l A = t I B + (1 - t ) l C thi vofi mpi diem I ' ta deu c6 : • / I-A = t.I-B + ( - t ) l ' C b) diem A, B, C thSng hang Hinh l b Hinh l a Chu-ng to rfing l A = t I B + ( l - t ) l C la dieu k i ^ n o^n va dii de ba GIAI GIAI a) T a i d i e m A , liTc keo F hi/dng t h i n g dijfng xuong difcJi c6 ci^cfng dp Theo gia t h i e t : l A = t I B + ( - t ) I C , t h i v d i m o i die"m I ' , t a c6 : IF + I'A = t N , t a c6 t h e x e m F l a t d n g cua h a i vector Fj va F2 I a n lufcft n&m t r e n h a i dir6ng t h i n g A C v a A B D i thay : f- ir + I'B +( i - t ) i r + rc = t i ' B + ( i - t ) r c + ir | Fj I = | F | va | F2 I = I F | x/2 (do t a m gidc A B C b) vuong can t a i C) I'A = t i ' B + ( i - t ) r c N e u t a chon I ' t r u n g vdi A t h i c6 = t A B + (1 - t ) A C , l a dieu k i e n can va du de ba d i e m A , B , C t h i n g hang Vay : C6 m p t luc 6p vuong goc vdi biJc tirdng t a i diem C v d i ciTdng N , v ^ m p t life keo biJc tifcrng t a i d i e m B theo hi/dng B A v d i cudng dp "5 V2 N (Xem h i n h l b ) —> Chiing minh rfing : A j B j + AjBg + vi tri ciia diem G cho GA + G B + G C + G D = B a i Cho n diem tren mSt ph^ng B a n Minh k i h i ^ u chiing l a A i , A2, An- B a n Mai k i hipu chiing la B i , B2, a i Cho tii giac A B C D a) H a y xac d}nh , B„ + A„'B„ b) Chiing minh rSng vdri mpi diem O, vectof O G l a trung binh cpng ciia bon vectcf O A , O B , OC , OD , ttfc la = OG = - OA + OB + OC + OD (Diem G nhii the' gpi la tam ciia tii giac A B C D ) 17« 179 • Hudng a i dan Siif d u n g c o i i g thufc M A + MB = 2MO (Ola trung diem AB) 3) GA Tiiih + GB GC = +^ G D = OA e) HA + H E + H C = H O f) Du-ofng thflng H O d i q u a t r o n g t a m G c u a t a m g i a c A B C Vi tri cua G OH + GB = 2GI (I la t r u n g d i e m • GG + GD = (J la t r u n g d i e m 2GJ a) AB) ChuTng m i n h t i r a n g t\i t a c u n g co C H // B ' A GA + GB + GC + G D = GI + GA + GB + GC + G D = (gt) ( i ) v a (j) GJ nen GI + GJ = Tir GA + G B = Vay GA + GB + GC + G D b) + GC + G D Ta CO + OA OG = + GO + OB + GO + OC + GO + OD = c) OA + O B + OC + O D - A B ' C H la h i n h b i n h h a n h AH =2 00 OA = OH + HA = OH = OH - 200 = •• =:> A H = B'C B'C =2 00 (dpcm) - OH AH - (OB + OC) = , t a co : « GO ^ (i) (j) O D l a d U d n g t r u n g b i n h cua t a m g i a c B B ' C n e n V a y , G l a t r u n g d i e m ciia I J Chii-ng m i n h G O G o i B ' l a d i e m do'i x i J n g v d i B qua O, t a co B ' C B C V i H l a t r i r c t a m t a m g i a c A B C n e n A H B C V a y A H // B ' C CD) C o n g ve ciia (1) v a (2), t a co : Ma (difcfng GIAI CO : GA o + OC = t h d n g goi l a di^dng t h i i n g - Ic c i i a t a m g i a c A B C ) • b) + OB —> d) (2) GIAI Ta AH = 0D (1) C o n g (1) v a (2) r i sU d u n g g i a t h i e t ) a) C h o t a m g i a c A B C n p i t i e p t r o n g d t T o T n g t r o n (O), H l a trii'c t a m t a m g i a c v a D l a t r u n g d i e m c a n h B C Chiifng m i n h r S n g : OA + OB + OC = OH (dpcm) G l a t r o n g t a m t a m g i a c A B C , t a co : HA + HB + HC = HG = 3( H O = HO B a i C h o d i e m O co d i n h v a difofng t h S n g d d i q u a h a i d i e m A , B co d i n h + OG ) = H O + OA + OB + OG + OC K e t h o p v d i k e t qua ciia cau b t a co : Chiirng m i n h r S n g d i e m M thuQC dUcfng t h ^ n g d k h i v a c h i k h i co s*> HA a s a o c h o O M = a O A + (1 - a) O B d) Vdri d i e u k i ^ n n a o c u a a t h i M t h u Q c d o a n t h a n g A B ? GIAI Ta OM -> ' -> = uOA + (1 - a ) O B < o O M - O B = a { O A - O B ) • 180 = 3Hb + OH = H0 - HO = HO Vi G la t r o n g t a m t a m giac A B C nen 0G = O H , do ba d i e m H , , G t h a n g tCr k e t qua (dpcm) cau b t a co hang CO : -> • + HB + HC Vi B M = a BA -> OM - > - > • - > = u( O A o B M = a B A a i 10 C h o t a m g i a c A B C v a d i e m O t u y y ~ OB ) + OB _ a) H a y x a c d i n h v i t r i d i e m M c h o O M = — B b) Vdri d i e m M d a dtfgfc x a c d i n h d c a u a , t i n h A M theo A B v a + OC o M e d n e n M thuoc d o a n t h S n g A B k h i va chi k h i < a AC ^ 181 : • HUdng a) b) V$y F \k dinh thu- tiT cua hinh binh hknh C B (xem hinh ve) d&n T i n h B M theo B C (can ciJ gia t h i e t ) Cho O = A t a CO k e t qua Ta da c6 : CD = A B ; A E GlAl a) = ^ 30B + OC o OB + B M cj B + B M = B + BC = SOB + b) M So sanh MA + MB + MC va MD + M E + M F = M D + M E + M F = ( M A + AD ) + ( M B + BE ) + ( M C OB + BC = (MA B M Vay d i e m M a t r e n doan t h i n g BC b) Chon d i e m = A t a di/oc : A M • - = BC B M = AE -BC • = - SAB + AC 4( = MC + ' Tifcfng tir : B la t r u n g d i e m D F • 2F'C AB; (1) ' : ' vi CBFA la hinh binh hanh • =>AE = - A F = : > A l a t r u n g d i e m EF C la t r u n g d i e m D E n e n : ; = DE + DF = FD + F"E E B = ED + M E = MA + B C ; M F = MB + C A E"F Cong ve theo vf, t a ducfc : Chu'ng minh rfing cac diem D, E , F khong phu thupc vao vi tri ciia diem M, b) = C"B A E + A F = B1: + C B = - > - > - > MD = BC chufng m i n h t r e n => 2DA Hay xac dinh cac diem D, E , F cho + M B + M C ) + ( A D + B E + CF ) >; AF B M = - B C ( h i n h ve) -> + C¥ ) Ta l a i CO : < B a i 11 Cho tam giac A B C va diem M y, a) va B F = CA n e n D, E, F k h o n g phu thuoc vao v i t r i cua M V i t r i cua M Ta CO : Q M = BI: ve tren hai canh C A v^ 2(D'A + ¥C + E'B) So sanh hai tong vectcf MA + MB + MC va MD + M E + M F o D A + F C = (D'E + ED) + (DF + FD) + (FE + EF) + EB = ( ) o A b + B E = ' + CF = ( ) v GIAI a) • T i r ( l ) va (2) Xac dinh diem D Ta CO: M D = M C + A B o M D - MC = AB o a i 12 • Xac dinh diem E TLforng tir, t a c6 : M E = M A + BC o AE = « 182 dan Ta CO : CH AB C H // A B ' AB' ± AB Tuong tir A H / / C B ' (2) TCr (1) va (2) t a co A H C B ' la h i n h b i n h h a n h Xac diuh diem F Tucfng t y , t a c6 : M F = M B + CA Cho tam giac A B C npi tiep difoTng tron (O) HiCdng B"C Vay : E la d i n h thu" t\i ciia h i n h b i n h h a n h ve t r e n h a i can'' BA va B C (xem h i n h ve) (dpcm) Gpi H la trUc tam tam giac A B C va B' la diem doi xrfng vdfi B qua tam O -> -> -> Hay so sanh cac vector A H va B ' C , A B ' va H C CD = A B Vay : D la d i n h thu" tir ciia h i n h b i n h h a n h ve t r e n h a i caul' A B va AC (xem h i n h ve) => M D + M E + M F = M A + M B + M C o Bli' = CA A H = B'C va A B ' = H C • ' • ' ' 183 Bai 13 C h o h a i h i n h b i n h h a n h A B C D v a A B ' C ' D ' c6 c h u n g d i n h A K h i u = - O M va do Chufng m i n h r g n g : a) BB' + C'C + DD' = b) H a i tarn g i a c B C D v a B ' C D ' c6 c u n g t r p n g tarn Chu y : Cucli chgn dicin O cho v = ' V = ( A B ' - A B ) + ( AC - A C ' ) + ( A D ' - A D ) A B ' f A D ' = A C ( V i A B ' C ' D ' la h i n h b i n h h a n h ) B B ' + C'C + D D ' = A C ' - A C ' - A C + AC = AM = ^ • •• • , • ^ GIAI , Goi , O' I a n luat la t r u n g diem ciia A D va BC, ta c6 : = ( G B ' + B'B ) + ( G C + C C ' ) + ( G D ' + D ' D ) = ( G B ' + GC + G D ' ) + ( B'B + CC' + D'b ) ^ ( G B ' + GC + G D ' ) - ( B B ' + C'C + D D ' ) 00' = OA + A B + BO' 00' = ob Neu G la t r o n g t a m t a m giac B C D t h i GIB + G C + G D til (*) ta cung c6 G B ' + GC + G D ' = 0, i u c OA Ma G cung la t r o n g t a m tam giac B ' C D ' (dpcm) + D C + CO' 0 ' = ( OA + = G B ' + GC + G D ' - = G B ' + GC + G D ' ) ( ) Nen + ob 0"b ) + ( A'B + D C ) + ( B O ' + C O ' ) = (vi O la t r u n g di§m A D ) BO' + CO' = (vi O' la t r u n g d i e m BC) 0 ' = A B + DC o 00' = Vay t r o n g t a m hai t a m giac B C D va B ' C D ' t r i i n g 14 C h o t a m g i a c A B C v a di^ofng t h S n g d T i m d i e m M t r e n dufong > ^ t h A n g d cho vectof u + GC k A B va D N = k D C T i m t a p hdp c a c t r u n g d i e m c i i a d o a n thfing M N (dpcm)' Vdi diem G bat k i ta c6 : Bai • v = ta chon diem O cho GO = —GC H a i t a m g i a c B C D v a B ' C D ' c6 c u n g t r o n g t a m G"B + GC' + 0 ' i 15 C h o til" g i a c A B C D Vdri so k y, l a y c a c d i e m M v a N c h o AIB + A D = AC ( V i A B C D la h i n h b i n h h a n h ) b) '' = (OA + OB + OC) + OC = 3OG + OG + GC = 40G Vay dc = ( A B ' + A D ' ) - A C ' - ( A B + A D ) + AC Nen , G la trpng tam tam giac ABC, ta c6 : B B ' + CC' + D D ' Ma u | = 40M Do d l i vectcf u nho n h a t k h i va chi k h i M nho n h a t hay M la h i n h chieu vuong goc ciia O t r e n d GIAI a) I - > = - > -> M A + M B + M C c6 dp d a i n h o n h a t GIAI Vdi m o i d i e m O ta c6 : u = M A + M B + M C = OA - O M + 013 ~ O M + ( C - O M ) - A B + DC (1) Tuang t\i : va I la t r u n g diem ciia A D va M N nen ta cung c6 : 01 = f I > J AM + DN k.AB Tir (1) va (2) => = k 0 ' + k.DC = O f k i A B + DC (2) => I e dir6ng thSng ( 0 ' ) Vay k h i k thay doi, tap hop cac d i e m I la dUcfng t h i n g M 00' B = OA + OB + C - M Ta chon d i e m cho v = OA + O B + OC = 184 185 GlAl =f Hit&ng d&n T i n h cosA = b^+c^-a^ 2bc Theo cong thufc t r u n g tuyen : b^ng cdch suf d u a g h§ thuTc da cho b^ + c^ GIAI Ta CO : a^ = b^+c^-a^ b +c- a « a^(b + c - a) = b-* + c - a-" o a^b + c) = b^ + « a^ = b^ + c^ - be « b^ + c^ - a^ = be b^ + c ^ - a ^ • 2bc Vay A = 60 = (b + c)(b^ + c - be) , COSA = — 2 2c^ + 2a^ - b^ = m^ = 4(4)^ = 64 (2) • 2a^ + 2b^ - c^ = m,^ = 4(3)^ = 36 (3) 100 b^ 208 b = —9 4%/l3 b= — , , 292 c = - — 2V73 c= B a i 189 Tinh dp dai cac canh a, b, c cua tam giac A B C theo cac goo A, B, C va chu v i 2p ciia tam giac A B C (2) • HUdng dan • HUctng dan * Theo d i n h l i s i n , t a c6 : • Tir (1), chvJng m i n h B = ° (giong b a i 186) • TCr (2) va B = 60°, churng m i n h b = c (dung d i n h h' cosin : c^ = cosC = ' 2b T h a y cosC d (4) v^o (3) t a eo : c^ = a^ + b^ - 2ab a ) (3) (4) sin A Vay K2h) c^ = b^ b = cc::> T a m giac A B C can t a i A Dap so : T a m giac A B C 1^ t a m giac deu B a i 188 D a i hpc ngoai ngff H a Npi - 2000 bdng 5m, 4m, 3m b _ sinB c _ a +b+c sinC sin A + s i n B + s i n C a 2p sin A sin A + sin B + s i n C b 2p sinB sin A + s i n B + s i n C c 2p sinC sin A + sin B + s i n C a = o b = e= 2p.sin A sin A + sin B + sin C 2p.sinB sin A + sin B + sin C 2p.sinC sin A + sin B + s i n C ChuTng minh rSng b -f- c = 2a o sinB -f sinC = 2sinA • HUdng dan T i n h dp dai cac canh cua tam giac A B C Theo d i n h l i s i n t a c6 : i D u n g cong thuTc trung tuyen b^ + c^ = m f + — _ B a i 190 Cho tam giac A B C (BC = a, C A = b, A B = c) C a c di^dng trung tuyen A M , B E , C F c u a tam giac A B C tifcfng vLng hay a = 2R.sinA , Vay b + c = 2a 2b^ + 2c^ - a^ = m^ de lap he phtfong trinh ba an la a^, b^, c^ 280 , (1) Chti'ng minh A B C la tam giac deu • HUdng d&n , t a c6 (1) , y - b a = 2b.cosC o • ^ 10 D a p s6 : a = a + c - b B a i 187 Cho tam giac A B C thoa m a n Theo gia t h i e t t a c6 : a = 2b.cosC 2b^ + 2c^ -a^ = 4(5)^ = 100 + c 2b^ + 2c^ - a^ = ã a = a ô G i a i he (1), (2) va (3) ta duoc : = — m^ + ^ b = 2R.sinB , c = 2R.sinC 2R.sinB + 2R.sinC = 2.2R.sinA o s i n B + sinC = 2sinA (dpcm) 281 B a i C h o t a m g i a c A B C v u o n g t a i A ( B C = a , C A = b, A B = c ) Chtfng m i n h rfing c Bai C h o di^dng t r o n ( C ) c tfim O v a b a n k i n h R = T i ^ d i e m M cy n g o a i ( C ) , v e t i e p tuyd'n M T v a c a t tuye'n M A B vdfi ( C ) ( T , A , B e ( C ) ) = a ~ b + cotgC B i e t M T = 12 v a A B = ( M B > M A ) T i n h O M v a M A , M B sinC • HvCctng ddn • Hiic/ng ddn D u n g d i n h l i s i n va chu y : A = 90° n e n sinA = 1, B + C = 90° n e n sinB = cosC v a sinC = cosB ã ^ , ^Tô/(C) = MT^ =ã M A M B = OM^ - R^ GIAI GIAI b Theo d i n h l i s i n , t a c6 : a = 2R.sinA = = sin A sinB b = 2R.sinB , , c Vay a-b Tinh OM c = 2R hay sinC c = 2R.sinC Ta Vay • 2R.sinC sinC 2R.sin A - R s i n B sin A - s i n B a-b sinC sinC s i n C ( l + cosC) 1-sinB 1-cosC - cos^ C s i n C ( l + cosC) _ + cosC _ sin^C a-b sinC cosC sinC sinC hay > M A ( M A + A B ) = 144 o M A ( M A + 7) = 144 o MA^ + M A - 144 = B a i 194 Cho tam giac A B C c6 trtfc tam H Chu'ng minh rfing cac tam giac H B C , HCA, H A B , A B C c6 b a n kinh dvfdng tron ngoai tiep hkng • HU&ng ddn D u n g cong thiJc O^/^c} = - • L a y d i e m M(xo; yo) e (d) : y = 2x - t a c6 : yo = 2xo - ^ / ( C ) = IM^ - = (xo + D u n g d i n h l i s i n doi v d i t a m giac ABC va t a m giac HBC va chii y BITC = 180° - BXC GIAI ' O M = 13 T i n h M A va M B c=> =f Hii&ng ddn • + (5f = 169 o Vay M A M B = 144 R = >/2 Tim nhijfng diem M tren diicfng th^ng (d) : y = 2x - M/(C) OM^ = (Uf Dap so': M A = va M B = M A + A B = 16 B a i 192 Trong mp(Oxy), cho diiofng tron (C) c6 tam I ( - 5; 2) va b a n kinh cho £P = MT^ = OM^ - R^ c> OM^ - MT^ + R^ M A = - 16 (loai) V M A = (nhan) + cotgC (dpcm) sin C ^I/(C) T a CO : £7^/,c, = M T ^ = M A M B V i t a m giac A B C vuong t a i A n e n sinA = 90° = va sinB = cosC c CO : + (yo - f - 50 • TiTcfng tir doi v d i t a m giac HBC va ABC A = (xo + 5)^ + (2x0 - - ) ^ - = (xo + 5f + (2x0 - 5)^ - 50 = x^ - lOxo ' • ^lliAC) >0 » X o - lOxo > 'M(Xo;yo) e ( d ) : y = 2x - Ket luan o XQ < V XQ > ,^^/(C 282 xo2 > 283 GIAI Goi R , Ra, Rb, Rc Ifin luot la bdn kinh dudng tron ngoai tiep ciia tarn gidc ABC, HBC, HCA, HAB = 2R va sin BAG Ta CO : sin BHC = 2R B a i 196 tam giac ABC c6 di^n tich S, ban kinh di^dng tron ngoai tiep la thoa man h^ thu-c 3S = 2R^(sin^A + sin^B + sin^C) ChuTng minh ABC la tam giac deu • HUdng ddn Dung cong thiJc S = abc vk dinh li sin de tim mgt h$ thuTc giQa BITC = B l i C ' = 180° - BAG (tiJ giac AB'HC' noi tiep) nen a, b, c sin BAG = sin BHC Vay • Siif dung bat d&ng thiJc Cosi (dau = xay ra) R = Ra v^' GIAI Tirmig ty ta cung c6 R = Rb (tarn giac ABC va tam giac HCA) va 9, b , sinA = — , sinB = — , 4R 2R 2R Vay 3S = 2R^(sin^A + sin^B + sin^C) 3.1DC Ta CO S = R = Rc (tam giac ABC va tam giac HAB) Vay Ra = Rb = Rc = R (dpcm) ^ 4R 3ai 195 Cho tam giac ABC vuong tai A : CA = b, AB = c Ve phan giac AD(D e BC) Tinh AD theo b va c Ơ HU&ng dan ã Dung Dung cong thvJc SAABC giac ACD = - bc.sinA doi vdi tam giac ABD va tam = 2R^ ^ „3 a 8R^ b K 33 8R^ c + —- + ,3 ^ 8R^ • Theo bat d^ng thufc Cosi : a^ + b^ + • D i n g thufc ( D o o sinC = c — 2R 3abc = a^ + b^ + c^ (1) > 3^^aVc^ = 3abc (2) Dau "=" of (2) xay 3 o a = b = c o a = b = c (dpcm) B a i 197 (Cau hoi trSc nghi^m) Trong mp(Oxy), cho A(- 2; 0) va B(0; 6) Tinh phi/cmg tich cua M(- 1; 5) doi vdfi dtfofng tron tSm A va qua B B' a) 14 D GIAI Ta CO SAABc = DLr6ng SAABC = - AB.AC.sinA = - be (vi sinA = sin90° = 1) 2 • SAABD = - AB.AD.sinBAD = -c.AD.sin45° = - c A D — 2 2 • SAACD= B a i 198 (Cau hoi trdc nghi^m) Cho goc xOy = 60" va diem M of goc xOy Ve MH Ox va MK Oy (hinh ve) Diet OM = 12 N/S Tinh HK ? - A C A D - s i n C M ) = i b.AD.sin45''= - b A D — 2 2 2bc a) 13N/3 -7= V2(b + c) b) , hay tron (C) tam A va qua B c6 ban kinh R = AB Dap so : iJ^i/io = - 14 (cau b) Vay - b c = i c A D ^ + - b A D ^ 2 2 J84 d) - 26 Vay ^ / , c ) = MA^ - R^ = MA^ - AB^ • _, AD = c) 26 • HU&ng dan SAABD + S^ACD An^K ^ >/2 AD(b + c) — = be b) - 14 AD = J2.hc b+c c) ^ 18 d) 12 285 HUcfng • HU&ng d&n dan sin AIB TJTZ B'inh li 4n,doi v(Ji tam giac OHK : : AIB = 180° - ( A i + B i ) = 180° - (45° + 15°) = 180° - 60° AB => Dudng trdn ngoai tiep tam gide OHK c6 dufdng kinh = OM • CO Ap dung dinh l i sin cho tam gidc AIB, ta c6 : tr6n difdng kinh OM ^ = OM = 2R (R la ban kinh dudng trbn ngoai tiep tam gidc lAB) AB = 2R.sinAIB = x 12 x sin(180° - 60°) sinO = 24sin60° = 24 x ^ Bai 198 (Cau hoi trSc nghi^m) Tinh ban kinh dufcfng tron ngoai tiep tam giac ABC biet BC = a, B a i 201 (Cau hoi trfic nghi^m) Cho tam giac ABC c6 chu vi bSng 12 va sinA + sinB + sinC = — B = 75° va C = 60" ằ)R = a72 Ơ HU&ng b)R=-^ c)R=Đ 72 Tinh ban kinh difoTng kinh di^dng tron ngoai tiep tam giac ABC ? d) R = 2a a ) R = ^ ddn * Hiidng Dap s6': R = - = 72 sin A = 2R o R = sin A sin 45" (cau b) c npi tiep tam giac ABC (hinh ve) Biet ban kinh dtfoTng tron ngoai tiep tam giac lAB b^ng Tinh AB ? 12 d)R=5 ^ = ^ = c ^ a + b+c _ 12 _ g sinA sinB sinC sin A + sinB + sin C 12 , R=^ Dap so : Cau c Cho tam giac ABC vuong tai A va ABC = 30" I la tSm dtfdng tron b) AB = c)R=2 d&n 2R = 3ai 200 (Cau hoi t r i e nghifm) a) AB = 12 73 b)R=— • 12 Dimg dinh l i sin : A = 180° - (B + C ) = 45° Dinh l i sin : = 12 73 Dap so : Cau a H K = OM.sinO = 12 73 sin60° = 12N/3 —- = 18 3ap so : Cau c i86 Ta Vi ( H M = K M - 90° nen tuT giac OHMK npi tiep dudng c) AB = 24 d) AB = 73 B a i 202 (Cau hoi trfic nghi^m) Trong dtfotng tron (C) tam O, ban kinh R, ve goc n$i tiS'p 12cm BAC = 30° Tinh ban kinh AvCiSng tron ngoai tiep tam giac OBC ? a) ^ b)R7S ^ c ) ^ d) ô 7i 24 7i ã HUdng ã d&n BOG = BAG (BOG v^ BAG Ian lugt la g6c tSm goc noi tiep cung ch^n cung BG ) ^ BOC = 60° OBC 1^ tam giac deu BG = R 87 • A p duiiR d i n h l i s i n cho tarn g i a c O B C t a c6 : b d n kin"h d u n g t r b n ngoai tiep tarn g i a c - R „ = 2R' « R ' s i n 60° Vay = R s i n 60',0 BC =^ sin H O C = R ' (R' la OBC) TinhP= R 73 D a p s ' : C a u d B a i 203 (Cau h o i t r i e nghi^m) T a r n g i a c A B C v u o n g t a i A G p i O l a t r u n g d i e m A B , diiitng C O di^cfng k i n h A B c S t dUdng tron (V) t h S n g (CO) t ^ i I v a J a) P = 100 b) P = 50 c) P = 40 d) P = 20 * HU&ng ddn (C,) • L/^,c,= BH.BC c) a)