Thực hành luyện giải đề trước kỳ thi Đại học 3 miền Bắc - Trung - Nam Toán học: Phần 2

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Nối tiếp nội dung phần 1 tài liệu Luyện giải đề trước kỳ thi Đại học 3 miền Bắc - Trung - Nam Toán học, phần 2 giới thiệu tới người đọc một số chuyên đề ôn thi và các bộ đề thi các khối A B, D của trường chuyên Lê Quý Đôn - Đà Nẵng có hướng dẫn. Mời các bạn cùng tham khảo.

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Nội dung Text: Thực hành luyện giải đề trước kỳ thi Đại học 3 miền Bắc - Trung - Nam Toán học: Phần 2

Luy^n J/' !f :i Ihi tJH 3 miETl Dm:, I rung, num Iuun iiyi. - iigujcn run it X A C D|NH C A C Y E U TO C U A T A M G I A C Qoi A ' la d i e m d o i xifng cua A qua G thi de dang tinh diTpc A ' 10 . Khi TRONG MAT P H A N G TQA DQ •3'3j ^6 A'B //// C id C , nen PT A CC, A''BB la 14x - 13y + 51 = 0. B a i 1. Trong mat p h i n g tpa dp Oxy cho tam gidc A B C c6 A ( 4 ; - 1 ) , B ( l ; 5^ [8x-y = 3 rx = l C(-4; -5). Gii suf B(x; y ) ta c6 he PT ^ ^1 _s ^B(l;5). V i e t phiTdng trinh (PT) c^c diTcfng t h i n g sau: ^ 14x-13y = -51 y ~ 1) Dircfng cao A D ( D e BC). ' /• TiTf^ng tir tCf A ' C // B B , tinh diTdc C ( - 4 ; - 5 ) . 2) Cac di/cJng trung tuyen B B , . CC, ( B , e A C , C, e A B ) . *' ' Cdch 2: D a t B(x; y ) va C,(x,; y , ) i h l 8x - y - 3 = 0 va 14x, - 13y, - 9 = 0. 3) Cac dirfJng phan gi^c BB2, CC2 (B2 e A C , C2 e A B ) . .3 V I C , trung d i e m canh A B nen x + 4 = 2x,, y - 1 = 2y,. i ^ rx=i Toa dp B la n g h i p m cua h? PT T ^ ^ •B(l;5). 1) V I A D d i qua A (4; - 1 ) , vu6ng g6c v d i BC = (-5; - 1 0 ) ^ [14x-13y = -51 y-5 nen c6 vectd phap tuyen n = ( 1 ; 2). o ^ Tinh toan ti/dng tif ta thu di/dc C ( - 4 ; - 5 ) . Rill' Do d6 PT A D : X + 2y - 2 = 0. C Biki 3. T r e n mat p h i n g tpa dp Oxy cho lam gidc A B C c6 A ( 4 ; - 1 ) va phi/dng 2) Do B, la trung d i e m canh A C nen B,(0; - 3 ) . Suy trinh hai di/dng phan gidc BB2: x - 1 = 0; CC2; x - y - 1 = 0. T i n h tpa do cac ra PT dircJng B B , m 8x - y - 3 = 0. Ti/dng l\i dInh B, C. B D 8« C, H i f d n g dSn g i a i PT dirdng CC, la 14x - 13y - 9 = 0. mnh7.1 Theo tinh chat cua difcfng phan gidc, cic d i e m doi xtfng cua A qua BB2 va 3) Ta v i e t PT cac difcJng phan gidc b^ng each tinh tpa dO c^c d i e m B2, C2. CC2 deu thupc BC. G p i D , la d i e m d o i xtfng cua A qua CC2 thi difcfng t h i n g Taluonco ^ = ^-l=>AB,=^AC. 3 8 A D , qua A ( 4 ; - 1 ) va vuong g6c v d i CC2 nen c6 vectd phap tuyen n = ( l ; l ) . B,C BC 5 Vay PT A D , la X + y - 3 = 0. N e u A D , c i t CC2 tai H , thi tpa dp H , la nghiem x - 4 = -3 x = l x-y =1 ! Gia sur BjCx; y), luc do ta c6 h^ PT: •B- cua h? PT •H,(2;l) y.l=-- x+ y= 3 V I H , la trung d i e m A D , nen D|(0; 3). V a y PT di/6ng BB2 m x - 1 = 0. ^8 5^ fl - y m Tifdng tif neu D2(x; y ) la d i e m d o i xtfng cua A qua BB2 thi D2(-2; - 1 ) . Tifdng tif V I A C 2 = - A B => C 2 .3-3; SuyraPTdifdngBCiaPTdifdngD,D2: ^ ^ = ^ ^ o 2 x - y + 3 = 0. Suy ra PT di/cJng C C , x - y - 1 = 0. x=l _ B a i 2. T r e n mat p h i n g tpa dp O x y cho tam gidc A B C c6 A ( 4 ; - 1 ) va phifdng n dp cua B la nghiem cua h? PT: T ^ ° o •B(l;5) trinh hai di/5ng trung tuyen B B , : 8x - y - 3 = 0, CC,: 14x - 13y - 9 = 0. Tinh [ 2 x - y = -3 y = 5 ' tpa dp cac dinh B , C. [x-y = 1 Jx = ^ oa dp ciia C la nghi?m cua h$ PT: • C M ; -5). Hi^dng dSn g i a i [ 2 x - y = -3 y = -5 De chuan b i cho cic thi du diTdc phat trien d phan sau, chung t o i tnnh 4. T r 6 n mat p h i n g tpa dp Oxy cho tam gidc A B C c6 C(-4; -5) va phifdng hai cdch g i a i cua thi du nhy. Wnh difdng cao A D : x + 2y - 2 = 0, difdng trung tuyen B B , : 8x - y - 3 = 0. Cdch 1: (C6 sit dung trpng l a m G cua A A B C ) . Dat G(x; y) thi Tinh tpa dp cdc dinh A , B . 1 H i r d n g d i n giili X = — r8x-y = 3 3 . E>tfcfng t h i n g C B qua C ( - 4 ; - 5 ) va vuong g6c v d i A D nen c6 vec t d phap il4x-13y = 9 l3''3j 1 tuyen n = ( 2 ; - 1 ) , do d
CdngtyTNHHM'n in-,Khum; Tpa dp B la nghi^m cfia h$ PT: P'' ^ •'ol* ^=>B(1;5). - fheo thi du 3, D2(-2; -1) la diem doi xjJng cua A qua B B 2 ndn PT dtfdng BC la 2x - y + 3 = 0. Gpi C(x; y) va B,(x,; y,) thi 2x - y + 3 = 0 va 8x, - y, - 3 = 0. [8x-y = 3 [y = 5 Iviat khac x + 4 = 2xi, y - 1 = 2yi nen 8x - y + 27 = 0. , Gii sil A(x; y) vk B,(x,; y,) thl x + 2y - 2 = 0 8x, - y, - 3 = 0. Mat khic x Tpa dp C la nghipm ciia hp: 2 x - y = -3 fx = -4 - 4 = 2x,, y - 5 = 2yi => 8x - y - 33 = 0. 8 x - y = - 2 r ' ^ 1 y = -5^ >CM;-5). fx + 2y = 2 fx=4 pal 8- TrS" '"^t P^^"g ^9 Oxy cho tam giic ABC c6 B(l; 5) va phifdng trinh Toa do Aia nghi^m cua hp PT ^ =>A(4;-1). drf£»ng cao AD: x + 2y - 2 = 0, difdng trung tuyen AA,: 2x + 1 ly + 3 = 0. Tinh [8x-y = 33 [y = - l toa d6 cic dinh A, C. Bai 5, Tr&n mat phing tpa dO Oxy cho tam gidc ABC c6 B(l; 5) va phifdng tr,nh Hifthig din giai 1 ! dir^ng cao AD: x + 2y - 2 = 0, difdng phan gidc CC2: x - y - 1 = 0. Tinh tpa Toa dO cua A la nghi$m cua h§ PT i * ^ ~ ^ => A(4- -1) dO cdc dinh A, C. [2x + lly = - 3 i Hi/dng din giai Drfdng thing BC theo bai 5 c6 phifdng trinh 2x - y + 3 = 0. Tpa dp diem A, Vi BC qua B(l; 5) vh vuong g6c vdi AD n6n c6 vectd phdp tuyen n = (2; ~i) (trung diem doan BC) la nghipm cua h$ PT. suy ra PT BC la 2x - y + 3 = 0. Vay tpa dO C la nghi^m cua hp PT 2 x - y = -3 [2x-y = - 3 2x + lly=^-3 >C(-4;-5). : Gpi B' la diem d^i xtfng ciia B qua CC2 thi BB' c6 vectd phdp tuyen II VAN DyNG TfNH CHAT CUA CAC H I N H i^; m = (1;1) nen PT BB' la''-^y - 6 = 0. Neu BB" dt CC2 tai K(x; y) thi X +=y 6_^.rZ;5UB'(6;0). VA CAC TINH CHAT D^C Bl|T CUA H I N H K U 2) Bai 1. Cho tam gidc ABC vu6ng cSn tai A, M(l; -1) la trung di^m cua canh BC, Do 66, PT di/dng AC la PT B'C: x - 2y - 6 = 0. i (1 \ x + 2y = 2 x-2y = 6 trpng tam tam gidc ABC la G —;0 . Tim tpa dO c^c dinh ciia tam giic d6. Suy ra A(4; -1). Htfdng din giai Bki 6. Tren mat phJIng tpa dp Oxy cho tam giic ABC c6 A(4; -1) va phiTdng Nhfin xet: TCf tinh chat trpng lam GA = -2GM , ta tim dtfdc tpa dp diem A. trlnh dirdng trung tuyen BB|: 8x - y - 3 = 0, phiTdng trinh di/dng phln gi^c Do lam gidc ABC vuong can nen trung tuyen AM cung la diTdng cao, do d6 CC2: X - y - 1 = 0. Tinh tpa dO cdc dlnh B. C. ta Viet diTdc PT di/dng thing BC. Mat khdc BM = CM = AM, ta tim diTqfc tpa Hi/dng dSn giai ^0 ciia cdc diem B va C. Theo bai 2. nd'u cho A va trung tuyen BB, ta tinh dtfdc C(-4; -5). Theo bai 3, Hoac Viet PT canh BC, va difOng thing AC, AB (di qua A tao vdi diT&ng cho A va phSn giic CC2 U'nh dir(?c B(l; 5). thing AM g6c 45"). A B^i 7. Tren mat ph^ng tpa dp Oxy cho tam gidc ABC c6 A(4; -1) va ph\i
LMV^/ truOc Ihi DH 3 mijn Bdc, Trung, Nam Todn hoc - Ngiiyln Van Thdng l(x - 1) + 2(y + 1) = 0 X + 2y + 1 = 0. f^hi d6 N c d c h d e u h a i diTcJng t h i n g AB. AD. Mat k h a c N. M ciing phia d o i Gia sic diem B(-2a - 1; a), ta c6: MB = AM y,ji dff^ng thing AB va cilng phia doi vdi diTdng t h i n g AD. Ttfc la x + 2 y - 2 2x + y + l (2x + y + l)(2 + 2 - 2 ) > 0 I 2J l 2 ; l 2; I 2j Bai 2. Viet phiTdng trinh canh AB cija hlnh chOr nhat ABCD biet canh AB, Be CD. DA Ian liTdt di qua cac diem M(4; 5); N(6; 5); P(5; 2); Q(2; 1) va diCn vay PT diTdng t h i n g chffa c a n h AC la x - y + 3 = 0. Uch hlnh chur nh$t bkng 16. PT diTdng c h e o BD la l(x - 1) + l(y - 2) = 0 o x + y - 3 = 0 ^ Hi^dng d i n giai ' - , , • TiTd6 B(4;-l). D ( ^ ; 7). Nh§n xet: Ta can xac djnh kich thirdc cua hlnh chiy nhat. Kich thirdc hint, TSm c u a hlnh thoi la d i e m 1(0; 3). tiT tpa dO ,/ chff nhat b^ng khoang each tff mot diem thuoc mot canh cua hlnh chO nhai r4 den canh do'i di$n. dilm A, suy ra tpa dp d i e m C - ; — . (h.7.5) Gpi phiTdng trinh canh AB (di qua diem w 3y M(4; 5)) a(x - 4) + b(y - 5) = 0(a^ + b^ ^ 0). Khi d6 phiTdng trlnh diTdng th^ng BC (di qua N(6; Chu y : C6 the v i e t phiTdng trlnh c a n h BD b ^ n g cdeh: ^ Wi^t PT dffdng t h i n g qua M tao v d i h a i dffdng t h i n g 5) vuong g6c vdi AB) la b(x - 6) - a(y - 5) = 0. AB. AD hai g d c b ^ n g nhau. Di?n tich hlnh chff nhat la ABCD ^ Ta cung c6 the tlm tpa dp d i e m B, D b l n g c d c h suf d u n g c o n g thffc tinh d i e n a(5 - 4 ) + b(2 - 5)| [b(2 - 6 ) - a(l -5)| ^ 4[(a -3b)(a - h) SABCD = 2SABD = 2(SAMB + SAMD) = a(d(M; AB) + d(M; AD)) = d(P; AB).d(Q; BC) = U'ch hlnh thoi. Gpi c a n h hlnh thoi la a. Khi dd d i p n tich hlnh thoi la M4t khac SABCD =2SABD =2AB.ADsinBAD = 2a^sina , trong8ad6 a \h gdc Theo bai ra ta c6 1.2+2.1 4 3 hai diTdng thing AB, AD diTdc xic dinh bdi cosa= r- r- =—=>sina=- 4 ( a - 3 b ) ( a - b ) = 1 6 o ( a - 3 b ) ( a - b ) =4(a^+b^): b = l,a = - l giffa 5 5 S.S b = l;a = - j Vly ta c6 = a 2. -6 => a = 475 Vay PTcanh AB m -x + y - 1 = 0hoSc -x + 3y - II = 0 75 5 3 B&i 3. Cho hlnh thoi ABCD c6 phiTdng trlnh hai canh AB. AD theo thuT i\i 1^' TiT dd ta cung tlm diTdc tpa dp cAc diem B. D (thupc hai diTdng thing da cho X + 2y - 2 = 0, 2x + y + 1 = 0. Canh BD chffa diem M( 1; 2). Tim tpa dp cdc dinh each A mot khoang bkng a). HiTdng d i n giai 4. Cho tam gidc ABC can tai B. phffdng trlnh canh AB cd dang Nh§n xet: Tim tpa dp diem A de dang. Theo tinh chS't hlnh thoi, diTdng cht > ' ^ x - y - 2 > / 3 - 0 . tarn diTdng tron ngoai ticp tam gidc la 1(0; 2), B e Ox. cung 1^ diTdng phan giic, nen ta c6 the vie't phiTdng trlnh duTdng ch6o kc Tim tpa dp cdc dinh tam giac. dinh A. Mat khdc, dffdng ch6o BD qua M vuong g6c vdi AC nen cung >> HiTdng dSn giai dinh diTdc phffdng trlnh. x^t: Tpa dp diem B xac djnh de dang. Diem A cung xac djnh de dang, ™ ta sit dung tinh cha't lA = IB = R. Tpa dO diem A nghi^m h$ PT [x + 2 y - 2 = 0 4 5 tim C, ta sur dung tinh cha't doi xffng cua C va A qua dffdng phan giac 12x + y +1 = 0 ' •3*3 tam giac ABC can tai B) hoac viel phffdng trlnh dffdng thing chffa canh Gpi N(x; y) thuOc tia phSn giic AC (cua BAD). b^ng cdch xac djnh g6c giffa dffdng thing BC va true hoanh.
Ll^n gidi di^ ln/// < iniin Bdc, Trun^. Sam IK::: < r , Nguyln VOn TMng po dinh A thupc dffdng thing x + y - 2 = 0 nen A(t; 2 - t). ^ Tpa dp B(2; 0). Goi A(&;S&-2y[3) jChi ^ I= vdia?t2. Ta c6: IA = IB po A cd ho^nh dp dffdng nen A ( > ^ ; 2 - > ^ ) . •y I r /- \ Fa ==2 (loai) o a ^ + ( N / 3 a - 2 V 3 - l ) =8 o ^ ^ ' [a = l + 73 pifa v^o hinh ve, ta de suy ra tpa dp dinh: V$y tpa do diem A(I + >/3;3-N/S). B =B 3 ' Do g6c giCfa IB vdi true hoinh \k 45", g6c giffa \ o \ >/6 AB vdi true ho^nh b^ng 60", suy ra i B C = 30" ^ c =c 4i ;2 V tiviiiw:; Hinh7.7 ' hay gdc giffa B C vdi true ho^nh \l 30". Vay hp ; i- pal 6. Cho lam giac A B C vdi dffdng cao AH cd phffdng trlnh x = 3V3 , phffdng so g6e eua difcJng thAng B C k = tan30" = -|=. ;( , trinh hai dffdng phan giac Irong gdc ABC v^ ACB Ian Iffdt 1^ y = — x ; \/3 . Vay phffdng trinh dffdng thing chffa eanh B C y = -U(x - 2 ) . v3 y = —|=x + 6. Bdn kinh dffdng trdn npi tiep tam gidc b^ng 3. Viet phffdng v3 Gpi tpa dp diem C (c-2) vdi c ?t 2. trinh cdc canh cua tam gidc, bie't dinh A cd lung dp dffdng. Hffdng dSn giai 1 \ e=2 (loai) . Nhqn xet: Ta thay hai dffdng phan gidc v i dffdng cao dong quy tai mot diem, Ta c6 IC = IB o + (c-2)-2 = 8: c = ^/3-I canh BC song song hoSc trilng vdi true hoanh. Hai dffdng phan giac lao vdi true ho^nh hai gdc b^ng nhau nen tam gidc n^y can tai A. Vay c(V3-l;I->/3). (h.7.8) Do dffdng cao AH cd Bai 5. Cho tarn giic A B C c6 phiTdng trlnh canh BC la y = 2. dinh A thupc di/tlnJ phffdng trinh \ 3\f3 nen 1 dffdng thing B C song song y = - thing x + y - 2 = 0 v a dipn tich tam giac la . Tim tpa dp edc dinh cu| hoac tr&ng vdi true ho^nh. tam gidc ABC, biet A c6 ho^nh dp dffdng. ' Hai dffdng phan giac tao vdi Htfdng d i n giai true hoanh hai gdc b^ng nhau Nh§n xet: Tff gid thiet tam gidc ABC deu va dipn tich da biet, ta xac djnl b k g 30" k = ± , nen difdc dp d^i canh cua tam gidc, tff dd tinh dffdc dp d^i dffdng cao AH. khdc, AH bSng khoang cdch tff dinh A den dffdng thing B C , tff dd ta tfn^ tam giac A B C deu. Hinh 7.8 dffdc tpa dp dinh A. Do B C song song vdi true ho^nh va AH vuong go*- Tam dffdng iron npi tiep la l(3>/3;3). Khoang each tff I den B C bing 3, nen true ho^nh nen dffa v^o dp dii canh da biet ta de d^ng xac dinh dffdc phffdng trinh dffdng thing B C la y = 0 hoSic y = 6. Neu phffdng trlnh dffdng cuaB,C. * l n g B C m y = 6, thi lung dp cua diem A Ik -3 (loai). (h.7.7) Gpi a la dp d^i canh tam gidc deu ABC. ^ay phffdng trinh canh B C 1^ y = 0. 2 41 2 =>a = , - 8. Toa dp cic diem B va C IJl B(0; 0), C(6N^;0) . '• '3 \ Ta CO dipn tich tam giac 1^ S^BC ^ * "4" dffdng thing A B cd hp so g6ck = yf3,vh dffdng thing C A cd h^ so' gdc P ' = - 7 3 , vay phffdng trlnh cua chiing Ian lff(?tm y = V3x; y = -V3x + i8. AH = a. : ^ = x/2.
Luy?n gi&i 3S IfUOt Jg> IM UH J mten-BiB^TTimgrnam lomnpc- lyguym van i nong %huyht M IS: piSu ki$n di dudng thdng tigp xiic vdi elip. pi/^jng t h i n g A x + By + C = 0 la t i ^ p tiiyen da elip n d i trdn (1) k h i vk chi BA DtfdNG C 6 N I C ( E U P , H Y P E R B O L , P A R A B O L ) ' ^hi:AV + B V = C^ w , I (Gidi tfafch: t f a e o chiftfng tdnh mdi) 1. T 6 M T A T Li T H U Y E T ^ ' ^ f)inhnghia hypebol A.EUP ' M / ' ^ -f^:' V. . ^a..,!:,.; Tr6n mat p h i n g cho hai d i e m co dinh F, va F 2 , v d i F 1 F 2 = 2 c > 0. Tap hdp /. Dfnh nghta Elip: . v c^c diem M cua mat p h i n g sao cho | M F , - M F J ] = 2a (trong do a la mot so - Trong mat p h i n g cho hai d i e m co dinh F i , F 2 vdi F 1 F 2 = 2 c > 0. Tap hpp difdng khong d d i nho hdn c) gpi la mpt hypebol. d i e m M cua m5t p h i n g sao cho ta luon c6 M F i + M F 2 = 2a (a > c) (a la h^n so) goi Ik mpt elip. F I va F 2 gpi la cac tieu d i e m cua hypebol. ^ - Hai d i e m F,, F 2 g p i Ik tieu d i e m cua elip. ' '^^^ " ' ' ^ ' Khoang each F 1 F 2 = 2c gpi la tieu ciT cua hypebol. - 2c g p i 1^ tieu c\i cua elip. N^u M n^m tren hypebol, thi MF|, M F 2 gpi la ckc ban kinh qua tieu diem - N e u M n^m tren elip thi M F i , M F 2 gpi 1^ cdc bdn kinh qua tieu d i l m cua M. cuaM. . ^.j 2. Phuang trinh chtnh tdc cua elip. 2. Phuang tiinh chtnh tdc vdcdcyiutdcim hypebol. - Chpn he tpa dp Oxy sao cho Fi vh F 2 c6 c^c tpa dp F i ( - c ; 0), F 2 ( c ; 0). Khi d( Chpn he tpa dp Oxy sao cho F|(c; 0) k h i do phi/dng trinh cua hypebol la 2 ..2 elip c6 phi/dng trinh ^ + ^ = 1 (b^ = a^ - c^) (1) - ^ - ^ =1 (2) vdib' = c'-a' a^ b^ a^ b^ ( 1 ) g p i \k phiTcfng trinh chinh t^c cua elip. \ phi/dng trinh tren gpi la phiTdng trinh chinh t i c cua hypebol. i j uv, Elip c6 bon dinh A , ( - a ; 0 ) . A 2 ( a ; 0), B , ( 0 ; - b ) , 6 2 ( 0 ; b). - Cic diem A i ( - a ; 0) va A 2 ( a ; 0) gpi la cac dinh ciia hypebol t^ 2a gpi Ik dp d^i ciia true Idn A 1 A 2 ; 2b gpi Ik dp d^i cfla true nh6 B 1 B 2 cua elip Ox gpi la true thiTc, Oy gpi la true ao cua hypebol (2) (do no khong c i t true Oy); F|(-c; 0), F 2 ( c ; 0) la hai tieu d i e m cua elip. Fi(-c; 0) va F 2 ( c ; 0) gpi la hai tieu diem cua hypebol, con 2c gpi la tieu ciT Ta CO cong thufc sau de t i m cic bdn kinh qua tieu: Ne'u M ( X o ; yo) n^m tren hypebol; elip t h i M F , = a + ^ ; MFj = a - ^0 •2a gpi la dp dai true thiTc, c6n 2b la dp dai cua true a o ; ' a a y = — x , y = - — X la hai dirdng t i p m can cua hypebol; - D a i liTdng e = — g o i la tam sai cua elip. NhiT vay 0 < e < 1. a a '^nf'' •"' ' ' a Hinh chi? nhat tao bdi cac diTcfng t h i n g x = ±a, y = ±b g p i la hinh chff nhat cd - H i n h chff nhat g i d i han b d i ckc difdng t h i n g x = ±a, y = ±b g p i 1^ hinh chff sdciia hypebol. c j i i , .vc , nhat cd sd cua elip. Cac cong thtfc ban kinh qua tieu - Elip c6 hai di/dng chuan: Neu x„ > 0 thi M F , = a + ; MF2 = - a + ^ + DiTcJng chuan x = - la difcfng chuan tfng v d i tieu d i e m F 2 ( c ; 0) a a , e a Neu x„ < 0 thi M F , = - a ; MF2 = a - - ^ + DiTcfng chuan x = - - la di/dng chuan uTng v d i tieu d i e m F i ( - c ; 0) a a • •• e 3. Dinh It sai e ciia hypebol diTdc djnh nghia e = - . NhiT vay v d i hypebol ta c6 e > I . N e u M(x,); y„) thupc elip. K i hi?u M H : , M H 2 tiTdng tfng Ik ckc khodng
Luyen gfM di truOc thi DH 3 miin Bdc, Trung, Nam Todn hoc - Nyuvl^n Van ThOng + Di/dng c h u a n x = — g o i 1^ diTdng c h u a n c u a h y p e b o l ttfdng uTng v d i ij^^ :-2py diemF|(-c;0) jgu diem F ; dirdng c h u a n y = — ^ piSu ki$n tiip xuc ciia dudng thdng vdi parabol + Goi H,, H2 tiTdng lirng 1^ hlnh c h i c u tCf M d e n h a i diftlng c h u a n x = , x - ^ c ' Ax + By + C = 0, (A^ + B^ > 0) Ik tiep tuyen cua parabol y^ = 2px dieu Ici^ncan vadu ia:pB^ = 2AC Khidotaco: MH, llll-L^ MH,L = e . , rfn!' ••--t: "nhfin Chti y: Vdi parabol c6 phiTdng trlnh dang y^ = -2px, (p > 0) thi dieu kipn ticp 3. Diiu ki^n di dudng thdng tiSp xOc vdi hypebol xdc la-pB^ = 2AC. 1 - Dc dirdng thdng Ax + By + C = 0 (A^ + > 0) la ticp tuycn vdi hypcrbol (i) dieu kiOn can va dij la A V - B V = PHl/CfNG TRiNH C A C D l / d N G C d N I C C.PARABOL im^^-^U'ifi'^i^^^ Bai 1- Trong mat phdng vdi h? toa dp Oxy, cho Elip (E) co phiTdng trlnh 1. DinhnghJa. ^' „2 y2 + — = 1. X a c d j n h tpa d p c d c tieu d i e m , tinh d o dai c a c true vk t a m sai - Parabol la tap hcfp cdc diem cua mat phdng each deu mot dirc:>ng thang A (.6 25 16 djnh va mot diem F co djnh khong thupc A . ciJa elip (E). + Diem F di/dc goi la tieu diem cua parabol. Hifdng din giai + DiTclng thdng A goi li di/dng chuan cua parabol. 2. Phuang trlnh chtnh tdc cm parabol x^ Chpn he true toa do: True Ox 1^ difcJng thdng di qua tieu diem F va vuong Phifdng trlnh chinh tac cua (E) c6 dang: —a^ +b^ i— = 1 ( a > b > 0 ) goc vdi drftfng chuan A, hi/dng difdng tif P den F (P \h giao diem cua Ox vdi Theo de ra, ta c6: a = 5, b = 4 ^ c = Va^ - b^ = 3 ''^ A). True Oy la trung trifc cua PF. Tpa dp cde tieu diem: F,(-3; 0); F2(3; 0). - Trong hp true niy F ,P Dp dai true Idn: 2a = 10. Dp dai true bd: 2b = 8. TSm sai: e = - = - . \2 ) \) a 5 - Difdng c h u a n A c6 phifOng trlnh * = " ' ^ Bii 2. Trong mat phdng vdi hp tpa dp Oxy, lap phiTdng trlnh chinh tic cua Elip - Phirpng trinh c u a parabol \k: y^ =2px v^ g o i Ik phiTdng trinh c h i n h tac cua (E) c6 dp dai true Idn b^ng 4>/2 , cic dinh nkm tren tryc nho vk cac tieu parabol diem cua (E) ciing n^m tren mpt difdng lr6n. - p> 0 goi la tham s6 tieu (chu y ring p b^ng k h o a n g cdch tif tieu diem F den I HUdng din giai diTiJng c h u a n A). x^ Elip (E): ^ + i - = l ( a > b > 0). Theo gi5 thi^t a = 2V2 , cdc dinh trdn Oy la - Neu M(x; y) nkm tren parabol y^ = 2px. thi MF g o i Ik hkn k i n h q u a tieu CD^ a b diem M. Ta c o c o n g thiJc de tinh b a n kinh qua tieu: MF = x + B,(0; -b); B2(0; b); F,(-c; 0). Ttf gikc F,B,F2B2 la hlnh thoi. theo gid ihict 4 3. Vdi dQng phuang trinh khdc cua parabol dinh Cling nkm tren duOng tr6n. nen F,B|F2B2 trd thanh hlnh vu6ng =>h = c • y^ = -2px ma a^ = b^ +c^ 8 = 2b^ => b = c = 2. Tieu diem F ; dirCJng c h u a n " = ^ Suy ra phifPng trlnh ciSa EUp (E) la: — + — = 1. 8 4 • x^ = 2py Tieu diem F 0;il ; diftlng c h u a n y = —j 3. Trong mat phdng vdi hp tpa dp Oxy, cho Elip (E) c6 phi/dng trlnh , 2) Ox. ^ +Oy Ian = 1.lirpt VicttaiphiTdng A, B saotrlnh chotiep = 2B0.d cua (E) biet d d l hai true toa dp AO tuy^n
Hi/dng din giai * Do tinh cha't doi xiJng ciia clip (E), ta chi can xet triTcJng hdp x > 0, y > 0. '2 ^ 2 =>S2 = 4 7 1 - — + 7t Goi A (2m; 0), B(0; m) la giao diem cua tiep tuye'n cua (E) vdi true tao (j^ l3 J 3 7t + 37t +2 (m>0). . ' •i 0ai 6- '^''""S '"^^ phing vdi he tpa dp Decac vuong g6c Oxy, cho Elip (E) c6 ii X V ^ PhiTdng trinh diTcJng th^ng AB : +—=1 2m m > ' phi/dng trinh + =^• ' ^ i ^ ' " M chuyen dpng tren tia Ox va diem N o X + 2y - 2m = 0, AB tiep xiic vdi (E) o 64 + 4.9 = 4m' o 4m^ = 100 chuyen dpng tren tia Oy sao cho diTcfng thing M N luon tiep xuc vdi (E). Xac j o m ^ = 2 5 o m = 5(dom>0). 'MS H v dinh tpa dp cua M , N de doan M N c6 dp dai nho nhat. Tinh gia tri nho nha't do. Vay phi/tfng trinh tiep tuye'n la: x + 2y - 10 = 0. ' Hi^dng dSn giai V i tinh chat doi xtfng nen ta c6 4 tiep tuyen la: ^ ^ Cdch I : Gia suT M ( m ; 0) va N ( 0 ; n) \k hai diem chuyen dpng tren hai tia Ox X + 2y - 10 = 0; X + 2y + 10 = 0; X - 2y - 10 = 0; x-"2y + 10 = 0. Oy. s Bai 4. Trong mat phing vdi h§ tpa dp Oxy, hay viet phiTOng trinh chinh ta'c cila 2 Di/dng thing nay tiep xuc vdi (E) khi va chi khi: 16 fr +9 = 1. Elip (E) biet (E) c6 tam sai blng ^ va hinh ciia chOr nhat cd sd cua (E) co ,m; .Mil Theo bat d i n g thtfc Cosi ta c6: chu vi b i n g 20. 2 2 r iA^_9_^ Hifdng dSn giai MN^ = m^ + n^ = (m^ + n^) = 25 + 1 6 + 9 ^ > 2 5 + 2 x/r6:9 m Goi phi/dng trinh chinh t^c cua Elip (E)lk: — + ^ = 1, a > b > 0. = 49 => M N > 7 a b 16n^ ^ 9m^ i grcux ioJj « ' ^5 ,2 m n a 3 .2 Ding thiJc xay ra m'+n^ = 4 9 o m = 2V7,n = N/2T TCr gia thiet, ta c6 he phiTdng trinh: 2(2a + 2b) = 20. Suy ra a = 3; b = 2. m >0,n > 0 ^« A ' qs 11-. t Ke't luan: Vdi M(2N/7;0) , N(0;V'2T) thi M N dat gid tri nho nhat vk gia tri X y nho nhat la ( M N ) = 7 . Vay phiTdng trinh chinh tic cua (E) IS: — + ^ = 1. Cdch 2: Gia sijT M (m; 0 ) va N(0; n) la hai diem chuyen dpng tren hai tia Ox Bai 5. Parabol y^ = 2x chia diOn tich hinh tron x^ + y^ = 8 theo ti so nao? v4 Oy. Dirdng thing M N c6 phiTdng trinh — + ^ -1 = 0 Hifdng dSn giai 2 Hinh tron x^ + y^ = 8 c6 R = 2>/2 , E^irdng thing nay tiep xuc vdi (E) khi \k chi khi: 16|( -— +99 ( -Ar + ^ ] =1 do CO dien tich la nR^ = 871. "Theo ba't ding thtfc Bunhia-copxki, ta c6: S Ta can tinh tl so — (hinh ve) trong do: r 16 9 1 ( 4 3^ MN^ = m^ + n^ = (m^ + n^) > m.— + n — = 4 9 o M N > 7 . 2 \ Im' n^J ^ m 4 3 •I: S| = | ( > / 2 x - x ) d x + S 4 „ , „ r t n O A B . iit?i'^f \ m: — = m: — a;nfjilq I lit!.: ., m n ^^ng thij-c xay ra 2 _2 + n ^ = 4 9 o m = 2 V 7 . n = V2T nni'i -xV2x -X Q+SquallronOAB ~ " J " m >0,n >0 .
U^^n^^^jru^^ T O . / , hoc - N ^ W r T T ^ K6i luan: V d i M(2>/7;0) . N(0;V2r) thl M N dat gid trj nh6 nhaft gia trj H i r i n g dSn giai nh6nha'tla(MN) = 7. Elip (E): ^ +^ = 1 c6 hai tieu diem la F,(-VrO;0), F2(Vl0;0). Hypebol CdcT. 3; Phircing trinh tiep tuyd'n tai diem (x„; y.,) thuoc — 16 + — -1• (H) c6 cung tieu diem vdi elip, suy ra phiTdng trinh cua Hypebol (H) c6 dang 9l va N f n0;— fffis'l' ,: a b Suy ra toa dp cua M va N la M s , Hypebol (H) c6 hai tipm can la y = ±2x = ± - x o - = 2 o b = 2a (2). 2 Q2 2 2 \ fl6^ 9'] Si H .o' =>MN^ = 4 ^ y9oJ V ll6 ^0 yl) T i r ( l ) va (2)suyraa^ = 2;b^ = 8. , ^ x^ ' SuT dung baft d^ng thtfc Co-si hoSc Bunhia-c6pxki (nhiT cdch 1 \k 2) ta c6 Vay phiTdng trinh Hypebol (H). ^ = 1. >M : 2 8 :)ni;an£l rin, ' MN^ ^ 7 ^ ng thurc xSy ra o x„ = -j-; yo = —j- Bai 9. Trong mat ph^ng vdi hp tpa dp Decac vu6ng gdc Oxy cho Elip (E): 2 y2 Bki T.Trong mat phing vdi true toa dp Oxy, cho d i l m C(2; 0) Ik Elip (E): - - - — = 1, va cdc diem M ( - 2 ; 3), N(5, n). Viet phi/dng trinh cdc di/dng 4 1 = 1. T i m toa dp cdc diem A, B thupc (E) biet r^ng hai d i l m A, B doi thing di, dj qua M va tiep xiic vdi (E). Tim n de trong so cdc tiep tuyen cua 1 E di qua N c6 mpt tiep tuyen song song vdi d| hodc d 2 . xiJng nhau qua true hoanh v^ tarn gidc ABC Ik tarn gidc deu. Hvtdng dSn giai Hifdng dSn giai x^ v^ Gia suf A(x„; y„). Do A, B doi xtfng nhau qua Ox nen B(x„, -y,)). (E): ^ +^ = l,M(-2;3),N(5;n). . Mif Ta CO A B ' = 4 yl vk AC^ = (x„ - 2)' + yl. Nhan xet (E) cd hai tiep tuyen thing duTng x = ±a = ±2 va dp x = - 2 la tiep 2 2 ViA6(E)ncn ^ + y?,=l=>y?,=l--ii- (1). luyc'n cua (E) di qua M . 4 4 Gpi dj la difilng thing qua M cd hp s6 goc k ,j V i A B = ACncn(xo-2)'+y?,=4y?, (2). dj: y = k(x + 2) + 3 o kx - y + 2k + 3 = 0 x„=2 dj: tiep xuc (E) « a ^ A ^ + b^B^ = C^ /3 Vdi X o = - thay vao (1). la CO y „ = ± — • ==> A / / d 2 = > k A = k d 2 = - | 7 ^ / i - \ (2 4^1 (2 AS' (2 AS] • 2 2 (2 4S] hoac A , B A qua N (5; n) cd hp so goc k a = — j , A : y = - j (x - 5) + n Vay A V 7 ; . B [r 7 J =^ -2x - 3y + 10 + 3n = 0; A tiep xuc (E) o 4(-2)^ + l ( - 3 ) ' = (10 + 3n)^ x^ y_=i 3n^ + 20n + 25 = 0 o n = -5 V n= - | Bai 8. Trong mat phJng vdi h? true Oxy, eho ehp (E) • j ' phUdng trinh hypebol (H) e6 hai diT^ng ti^m can la y = ±2x va c6 hai n= loai vl khi do A = dz. Do dd N(5; - 5 ) . ^'^ *' 3 , ' .'i diem la tieu diem cua elip (E). ,
Luyen giii dS trade thi DH 3 man Bdc. Trung, Nam Todn hoc - Nguyln Van Thdng COng(y TNIIIIMr\Klu^n, Vie, la chu v i thie't d i $ n th^ng 1 la dp dai canh ben r K H 6 I D A DI$N, THfi TfC H K H 6 I CH6P, K H 6 I L A N G TRV. Sxq = Ph Lang t r u diirng: ,.. T H E TfC H C A CK H 6 I T R 6 N X O A Y piachuviddy ' 1. T 6 M T A T L i T H U Y E T . v h la chieu cao A . H i N H L A N G TRg Hinh hpp chiynhSt: S,p = 2(ab + be + ca) V; 7. Dinh nghla: a, b, c la k i c h thirdc cua hinh hpp chff nh§t: H i n h lang tru la hinh da di?n c6 2 mat song song g p i 1^ day, cac canh 6. Thi ttch • „„„^,, . . khong thuoc 2 day song song v d i nhau. . T h ^ tich cua hinh hop chff nhat: I v ^ a b c l a. b. c la kich thffdc - 2. Ttnhchdt Trong hinh 13ng t r u : . The tich cua hinh laip phiTdng: V = a3 A la canh Cdc canh ben song song va b i n g nhau ""^^ . T h ^ tich lang t r u : |V = B.h| B la d i ^ n tich d^y, h la chieu cao; - Cac mat ben, mat ch
Luyen gUli dS trudc k9 thi DH 3 mijn Bdc. Trung. Nam Todn hpc - Nguyen van T/tang Di0n lich loan phan: p. HlNHTRy y £)fnhnghia ;••n'f' ••. : • O J i ' | . ! ; : M ; f b :v; 3. TMttch ' ; ' i t ? ti Hinh try 1^ hinh sinh ra bdi hinh chff nh3t O ' O M M ' quay xung quanh canh OO' The tich hinh ch6p V^-i-B.h Canh OM sinh ra hinh tr6n 6Ay. 3 Canh M M ' sinh ra mat tru tron xoay ' The tich tiir di^n: V = —abdsina : J | ? ! f s t J r i o qu.^ IViM' goi la dff5ng sinh, OO' la true cua hinh tru ' 6 O ji = OO' Ih chieu cao M a, b: do dai hai canh doi R = OM ban kinh day d: do d^i doan vuong goc chung 2, Di$n tich hinh tru a: g6c giffa hai canh doi. Dien tich xung quanh: S^^j = 27tRh Ti so the tich cua hai hinh chop tarn gidc c6 chung dinh 3 canh ben V s A B T ' SA'.SB'.SC R: ban kinh day SA.SB.SC h: chieu cao ^SABC C. HINH CHOP C g i S,p=2jtRh + 2;iR^ 7. Dinhnghia * 3. ThS t(ch hinh tru V - TiR^h ' Hinh chop cut la phan hinh chop nam giffa day va thiet dien song song vdi R: Ih ban kinh ddy day. •:' ' Hinh ch6p cut c6 tff hinh chop deu goi la hinh ch6p cut deu h: chieu cao A'B'C'D'*ABCD E. HiNHNON / , SH SA' A ' B I. Dinhnghia SH' SA AB Hinh non la hinh sinh ra bdi tarn giac vuong OMS quay xung quanh canh goc S , p = S , , + B + B' vuong OS. 2. Dmttch Canh OM sinh ra hinh tron ddy Di?n tich xung quanh cua hinh chop cut deu: S , , = i ( n a + na').d Canh SM sinh ra mSt n6n tron xoay. SM goi Ih dirdng sinh, SO Ik true, dffdng cao. n:socanhddy R = OM ban kinh ddy > a, a': canh day H = SO chieu cao d: do d^i trung doan, chicu cao cua mat ben 2- Di^ntkh V = V,-V2 I^i?n tich xung quanh hinh non: 3. Thitkh V: The tich hinh chop cut R: ban kinh day V , : T h e tich hinh chop 1: do dai dffcJng sinh V2: The tich hinh chop tren Di?n tich toan phan: S,p = 7 i R l + nR = n R ( l + R) ^- Thitkh B, B'm di^n tich day The tich hinh n6n: V = -7tR^h h \k chieu cao 3
R: M n kinh day f icp tuyen cua mat cau la du^dng thing c6 mpt diem chung vdi mat cau h: chieu cao v&^p. picu kien de du-dng thing A tie'p xiic vdi mat cau la: d(0, A ) = R. F. H i N H N O N C y T i1 dnfrN 2 piSn t(ch mat cdu: |s = 47iR^ 1. Dinhnghia ' , > • • -t^^-i^-r.uMf^ ^uim, ^^ J fhi ttch mat cdu: Hinh n6n la phin hlnh n6n giffa ddy vJk mftt thi^t didn vu6ng g6c vdi true. 3 Hinh n6n cut sinh bcli mpt hinh thang vuong OMM'O' quay quanh OO' . „. B A I T A P M I N H H Q A h = 0 0 ' : chieu cao pji 1. Cho hlnh chop S.ABCD c6 day ABCD la hlnh chCT nhat vdi AB = 1, MM'= 1 : dirdng sinh /^D = a>y^, SA = a va SA vuong goc vdi mat phing (ABCD). Gia sii" I la 2. D^ntich giao diem cua BM va AC. Gpi M, N Ian liTdt la U-ung diem cua AD va SC. Dipn tich xung quanh: S,4=7t(R + R')l n m the tich tur dien ANIB. Xiib flriht • R. R':Mnkinh ddy Hirdng dSn giai ^ o£3 ir- 1: difcJng sinh Gpi O la tam cua day ABCD. Di$n tich tokn phan: S,p=7c(R + R ' ) l + JtR^ + 7iR'^ Trong tam giac SAC, ta c6 NO la dirdng trung 3. Thittch blnh nen NO//SA, ttfc N O l (ABCD) va NO = - . 2 The tich hlnh n6n cut: V = i7i(R^+R'^ + RR')h Tacd VAN,B=VN.A,B=^SA,B.NO = iSA,B (1) J o R,R': bdnkinhddy Ta tinh dipn U'ch tam gidc AIB. h: chieu cao X6t hlnh Chi? chat ABCD. Do MA = MD G.HlNHCAU =>MA = - B C = > A I = - I C 2 2 /. Dinhnghia AC^ 2a2+a2 Hlnh cau tam O, ban kinh R la tap hdp nhffng diem M trong khong gian thoa =>AI = i-AC=>Al2 = 3 dieu ki^n OM < R. „2> Mat cau tam O bdn kinh R la t§p hdp nhufng diem M trong khong gian thoa 2a^ L?i c6: B I = - B M => B I ^ = - B M ^ = - a 2 + . dieu ki?n OM = R. 3 9 9 Thiet dipn qua tam m hlnh trdn Idn tam O ban kinh R. Oo dd A I ^ + BI^ = a^ = A B ^ nen A I B la tam gidc vuong dinh I . Thiet di?n ciia hlnh cau vdi mpt mat phing la hlnh tr6n c6 tam H la hinh 1 a^/3 &S 3^72 chieu cua O tren mat phing va bdn kinh: SAIB=-IA.IB = - . (2) AIB 2 2 3 6 r, = V R ^ a^^ Thay(2)vao(I)tac6: V^NIB (dvtt) 36 R: ban kinh hlnh cau chia khoa: d: khoang each tir tam tdi mat ph^ng Cho hlnh chop S.ABC. Lay A', B', C tiTdng xtng trgn canh SA, SB, SC. Khi dd: d = OH a' V s A B C ^ SA' SB- SC Tiep di$n cua mat cau la mat phiing c6 1 diem chung vdi mat ca" Vs.ABC SA • SB • SC Dieu ki$n de mat phing a tiep xiic vdi mat cau la: d(0, a) = R
Luytn gUU di truOc thi DH 3 miin Bdc, Trung, Nam Todn i imVi Van ThiOng Chtfng m i n h : A B C = 60". H i n h chieu vuong goc B ' l e n ( A B C ) trOng v d i trong t a m tam K e A ' H ' va A H cilng vuong g6c mat p h i n g (SBC). gi^c A B C . T i n h the tich ti? dien A ' A B C . K h i d6 A ' H V / A H va S, H ' , H th^ng hang. Hi/dng dSn g i a i r;ri, ' 1 Goi G la trong t a m tam giac A B C ta c6 B ' G J. ( A B C ) . ^S.A'B'C _ '^A'.SB'C _ 3 Tit d6 g6c B ^ = 6O" la goc ma B B ' toa v d i mat p h I n g ( A B C ) . Vs.ABC VA.SBC ISABC-AH ; Trong tam giac vuong B B ' G ta c6 ngay: B G = ^ ; B ' G = J S B ' . S C ' .sina.^ ^ Ua.^^ .sina.A'H n' _ SB^ S C S A ' p a t A B = 2x, trong tam giac vuong A B C , ta c6: ;aH »'•)!) => dpcm -SB.SC.sina.AH A C = x; B C = X 7 3 (do goc A B C = 6 0 " ) 2 GiasufBGnAC = Nthi BN =^ B G =^ f^^Vf'i.f/'' '.^ 6 d a y a = B ^ ' = BSC C h u y : K e t qua tren v a n dung neu nhiT trong cac d i e m A ' , B ' , C c6 the c6 ^ A p dung dinh l i Pitago trong tam giac vuong B N C ta c6: ' cdc trircfng hdp A s A ' ; B = B ' ; C = C . BN^ = NC^ + B C ^ ^ ^ ' % ^ + 3 x ^ = . x 2 = ^ (1) . Bai 2. Cho hinh chop S . A B C D c6 ddy A B C D la hinh thoi canh b ^ n g VScm, 16 4 52 ^ dirdng ch6o A C = 4 c m . Doan t h i n g SO = 2%/2 c m va vuong goc v d i day, j day O la giao d i e m cua hai diTcJng ch6o AC va B D . G o i M la trung diem cua Tac6: V , , , 3 e = l s , , , . B ' G ' = l . l A C . B C . ^ = i ^ x . x 7 3 = ^ (2) canh SC. G i a suT m a t p h i n g ( A B M ) c i t SD t a i N. T i m the tich hinh chop J c ,4^,3 2 2 12 4 * S.ABMN. y(l)vao(2)tac6: V ^ . ^ e c - f ^ (dvtt) H i f d n g din g i a i B^i 4. Cho hinh lang tru dtfug A ' B ' C ' A B C c6 day la tam gidc vu6ng A B C tai B , Ta c6 A B / / D C => A B / / (SDC) gia sijf A B = a, A A ' = 2a, A C = 3a. G o i M la trung d i e m cija A ' C va I la ^ ( S A B ) n (SDC) = M N // A B ( N e SD) giao d i e m cua A M va A ' C . T i n h the tich tiJ dien l A B C V i M la trung d i e m cua SC nen N la trung d i e m cua SD Hif(7ng dSn g i a i Ta c6: Vj^BMN = "^SABN + ^S.BMU (1) Trong tam giac vuong A * A C , ta c6: A C - V9a^-4a^ = a^fs ^''^' Theo b a i todn c d ban ta c6: Tit 66 trong tam giac vuong A B C , thi: B C = Vsa^ - a ^ = 2a J- i f I VSABN _ S N _ 1 V _1 _ i Tr ~ T ^S.ABN - r VS.ABD - 7 ^S.ABCD Do ( A A ' C ' C ) 1 ( A B C ) nen trong ( A A ' C ' C ) k c I H 1 A C ( H € A C ) 'S.ABD ^ Z t => I H 1 ( A B C ) . \WL_SNSM_I _1 - 4 ^"^^-^^^ - gVsABCD Theo dinh l i Talet ta c6: AA' CA' Tir (1) suy ra: VS.ABMN = g ^5 ABCD (2) lA' A'M lA'+CI 3 it:,:.) \. DS thSfy: VS.ABCD = ^SABCDSO = U . A C . B D . S O = 1.4.2.272 = ^ =^ J i . _ C L = 2 ^ ^ . ^ 4 a M A C . AA' CA' 3 3 Tilf (2) va (3) suy ra: VS.ABMN = ^ (^^tt) B a i 3. Cho hinh lang tru x i e n A B C . A ' B ' C c6 canh b e n BB* = a va B B ' tai ^ T a c 6 : V.^BC = l s A B c I H = i . l A B . B C . I H = l a . 2 a . ^ = i ^ (dvtt). mat phJlng A B C g6c 6O". Gia sijf A B C la t a m gidc vuong t a i C va fi L
Luy(n giOi dS truOc kp thi DH 3 miSn Bdc, Trung, Nam Todn hoc - Nguyin Van ThOng Ta c6: Bai 5. Cho hinh chop S.ABCD day la hinh vuong ABCD canh a, mat ben S la tam gidc dcu va nSm Irong mat phing vuong g6c vdi day ABCD. Goi (vj VA.BDM ="VA.BDE - ' V M . B D E = - S B D E A A ' - - S B D E . M C = - S B D E ( A A ' - M C ) N, P Ian lirot la trung diem cua SB, BC, CD. Tinh the tich tuT dien CMNP. ' Hxt&n^ dSn giai - i . i B D . E O b - ^ l = i.a^^^a^/I.-^ = ^ ( d v t t ) 3 2 2. 6 2 2 4 ui'.>-,'> Goi H la trung diem cua AD. ihi SH 1 AD. Do (SAD) 1 (ABCD) nen suy ra- pai 8* ^ ' " ^ '^^"^ ^^^^ S.ABCD c6 canh day AB = a, canh ben SA = nyjl . SH 1 (ABCD) va SH = (vi SAD la tam giic deu canh a). Goi M , N , P Ian liTOt la trung diem cua SA, SB, CD. Tim the tich tiJ dien SH aV3 AMNP. Ke M K // SH (K e HB) => M K 1 (ABCD) va M K - Hir8 .. Do MS = M A ==> d(A, (MNP)) = d(S, (MNP)) (•) J = '^•aoM 1 1 a^ a>/3 d?S (dvtt). Vay: V ^ , c N P = - S c N p M K = - . y . — = => V A . M N P = Vs.MNP SM SN 1 mXUzd) :i;v • Bai 6. Cho hinh chop tam giac S.ABC c6 day la tam giac deu canh bang Theo bai todn ccl ban, ta co: VS.MNP Vs^BP SA SB 4 SA = 2a va SA vuong goc vdi mat phing (ABC). Goi M , N tiTdng uTng la hinh chie'u vuong goc cija A tren SB, SC. Tim the tich khoi ch6p A.BMNC. => Vs.MNP 7-^SABP.SO - ^ . f ^ AB.HP.SO = Hifdng d§n giai 4 3 4 3 2 »( J .J t ( O va H tiTdng tfng la tam cua day A B C D va trung diem ciia A B ) Ta c6: V^BMNC = ^S.ABC ~ ^S.AMN (1) VSAMN _ SM SN 1 ^2_a^^aV2 Theo bai t o i n chia khoa thi: = —.a.a. (2) 'S.ABC SB SC 24 V ^ 2 48 VI AB = AC => SB = SC aV2 •^iiSHJiiA f i t ; T i r ( l ) v a (2)suyra: V^ ^^^p = (dvtt) a Ta c6: SA^ = SM.SB = SN.SC => SM = SN. 48 V a y : : ^ ^ = ^ (2) SUf DyNG PHLfdNG PHAP THE TICH DE TJM KHOANG CACH ^S.ABC SB Khoang each tiJf mot diem den mot mat phing, khoang each giuTa hai difcing SA^ th^ng, trong nhicu triTcJng hdp co the quy ve bai todn the tich khoi da dien. Ta c6: SA^ = SM.SB SM = SB Vi$c tinh cdc khoang cdch nay difa vao cong thiJc hien nhien sau: h = — Vay tCr (2) c6 s 6 day V , S, h Ian liTcft la the tich, dien tich day va chieu cao cua mot hinh Vs.AMN _ SA^ SA^^ 4a^ 16 Y = — Vs.AMN - ^S.ABC (3) ''S.ABN SB^ AB^ 4a^+a^ 25 25 chop nao d6 (hoac h = — doi vdi hinh lang tru). s 9 1 a^v/S ^ 3a^>/3 Tiir (1) va (3) co: V^BMNC = ^ VS.^BC = ^ • ^ ' ^ ^ • 2 ^ =" (dvtt) 9. Cho hinh chop S.ABCD day la hinh thoi ABCD c6 SO vuong goc vdi day 50 va O la giao diem ciia AC va BD. Gia sijf SO = 2 V 2 , difdng ch6o AC = 4, Bai 7. Cho hinh hop chff nhat A B C D . A ' B ' C ' D ' ddy la hinh vuong canh b^ni: ' canh day AB = %/5 . Goi M la trung diem cua SC. Tinh khoang cdch giii:a hai chieu cao A A ' = b. Goi M la trung diem cua canh C C . T i m the tich tu" d'^' BDA'M. JS Trong (ACC'A'): A ' M n AC = E CJpi M la trung diem ciia SC. Khi do ta co OM//SA => SA//(OMB) Goi O la tam cua day ABCD vi M la trung diem cua C C , nen ta ^ => d(SA, M B ) = d(SA, (MOB)) = d(S, (MOB)) = d(C, (MOB)) (1) CE = AC = a>^. (doMS = MC) 341
Ta c6: OB^ = AB^ - OA^ = 1 OB = 1 ^jj^ng minh S C D la tam giac vuong. finh khoang each tiif H tdi mat phSng ( S C D ) . Ke M H ± (ABCD) => M H = - S O = y/l. ^' HuTdng dSn giai 2 72 po A B 1 B C => S B 1 B C (djnh li ba difSng vuong goc). ' ' ' ' Vay V ^ o B C = | S o B c M H = i loB.OC.MH = i . 1 . 2 . ^ = (2) ' pi/a vao djnh l i Pitago trong cdc tam giac vuong S A B , A B C , S A D de dang 1_. 1 tinh di/cfc: S B ' = 3a^ S C ' = 4a^ S D ' = 6 a l Taco: OM = -SA = -sf8 + 4 = S Ttfdd c6: D C ' = 2a^ vay suy ra: S D ' = S C ' + C D ' => S C D la tam gidc vuong tai C => dpcm Gpi h Ik khoing cich tH C tdi (MOB), ta c6: 2 Ke HK va B F cilng vuong goc vdi mat phing ( S C D ) => HK // B F va S, K, F . VcMOB = ^SMOB-h = i . ^ O B . O M . h = ^ (3) « , HK. SH thing hang. Ta c6: (1) Ttf (2) va (3) suy ra ^ ^ ^ h^ ^ . (4) Trong tam giac vuong S A B c6: ,^ y, ^ , 2^/6 SA' = S H . S B = > S H - ^ ^ Tfif (1) (4) di den: d(SA. MB) = nfnl 2a^/3 BJki 10. Cho Mnh iSng tru di?ng A B C . A ' B ' C day 1^ tam giac ABC vuong tai B Gia sijr AS = a, A A ' = 2a; AC = 3a. Goi M la trung diem cua A ' C va I giao Vaytir(l)c6: — = — ^ = - r > H K = -BF (2) i l ,V BF aV3 3 3 diem cua A M vk A ' C . 1. Tim the tich tif div*n I ABC. Tac6: Vs^CD = - S B C D . S A = - . - B C . C D s i n l 3 5 " . S A =-a.a>/2.—aV2 2. Tim khoJng cdch tif A tdi mat phdng IBC. 3 2 Hi/dng d§n giai 3^72 4a^ 1. T a c 6 V,ABc = Mat khdc: Vs gen = -SscD-BF = -SC.CD.BF ... • 3 • 6 2. Ke I H 1 AC I H 1 (ABC) Ki HE 1 B C => IE 1 BC (djnh ly ba diTdng vuong goc). Tiifdotaco: ^ ^ = -2a.aN/2.BFr:>BF = - . (3) „^ . HE CH CI 2 „c 2^^ 2a ' 6 6 2 uJ^^''-^=^=^X^=J=^"^=3^^=T- TCr(2) va (3) suyra: HK = d(H, (SCD)) = | D o v ^ y IE = VnFT]i^ = Jl^^i^ = 2ax^ •^"i 12. Cho hinh lap phiTdng ABCD.A'B'C'D' c6 canh b^ng 1. Goi M , N Ian V 9 9 3 •wt la trung diem cua AB va CD. Tinh khoang each giflra hai dirdng thing A'CvaMN. „;0: M N // (A' BC) ^ , 9 3 2 3 5 => d(MN.A'C) = d(MN. (A'BC)) = d(M, (A'BC)) , ' Bai 11. Cho hlnh chop ttf gidc S.ABCD c6 day ABCD la hinh thang vuong tron- ^^thay V ^ . ^ , B c 4 w A A ' = ^.i.i.l.l = :l (D do ABC = BAD = 90", BA = BC = a; AD = 2a. Gia suf SA vuong g6c vdi da^ ABCD va SA = a>/2 . Gpi H la hinh chieu cua A tren SB. ^ CB 1 (BAA'B) z:> CB 1 BA' => A'BC la tam giac vuong tai B.
•"6 '7 Tiirdd: V ^ . M B C = V M . A B C = ^ S A . B C - h (2) Trong tarn giac S N M ve difdng phan gidc N K cua goc S N M ( K e S M ) . ' V i D C // A B => D C // (SAB) => ( K D C ) n (SAB) = QP // A B . T i r ( l ) va ( 2 ) t a c 6J:: — i - ^=i - i. -AA' B ' B. B . BCC. .hh== - iV- V 2 2. h . h==>^ h = —^ (3) . (QeSA;PeSB). 12 3 2 6 4 Khi 66 C D Q P la thiet dien phai difng. V i S M N la tarn gidc deu otn KS = K M => Q va P ti/dng ufng la trung d i e m cua SA va SB. ^ T i r ( l ) . (2), (3) suy ra: d ( M N , A ' C ) = ^ ; ' Gpi V| la the tich phan hinh chop n^m tren thiet d i e n , ta c6: ., Bhi 13. Cho hinh ch6p S.ABC. G p i M la trung d i e m cua SB. D i f n g thict dicn vrjj V| = Vs.QPCD = Vs.QPD + Vs.PDC (1) hinh ch6p qua M , song song v d i SA va song song v d i B C . Chufng minh thit}', Tacd: ^ ^ -^'-Hi; dien chia k h o i ch6p thanh hai phan c6 the lich bkng nhau. SQ S VsQPD^^SQ SPP ^_J ^1 1 _ 1 _ 1„ _ 1 , / 1 Htfrfng d§n glai fiT.sn Vs.ABD SA • SB SA SB ~ 2 • 2 ~ 4 2'2 ^•^^'^ ~ " g ^S.ABCD = g V (2) (Hoc sinh tif ve hinh) d day V = Vs.ABCD Ke M N // SA ( N e A B ) , M Q // BC (Q e SC) '- TMng tiT, ta c6: >^S.PDC - ^ V __1_' v - ^S.PDC - - V (3) K e NP//BC (P e A C ) =^ QP//SA. V,S B DC 4 Thiet dien la hinh bmh hanh M N P Q . TO(1). ( 2 ) . ( 3 ) t a c6: V , = V s . o D c p = - V + - ! - V = - V (4) De thay M , N , P. Q Ian liTdt la trung d i e m cua SB, A B , A C va SC. 8 4 8 G p i V| la the tich phan hinh chop n^m ben trai thict dien M N Q . Tir M kc V, 3 thtfc (4) dan den: M R // A B (R e SA va c6 RS = R A ) . 5 Ta CO R M Q . A N P la hinh lang tru va c6: ' CAC B A I T O A N T H E T I C H KH6| D A D l | N C O BAN VA NANG C A O V | = Vs.RMQ + VRMQ.ANP (1) Bai 1 . Cho hinh chop S . A B C D , day A B C D la hinh thang vuong t a i A va D : Gia sur S = SABC. h la chieu cao cija hinh chop S.ABC k c tir S. K h i do do lha\ A B = A D = 2a, C D = a. Goc giCTa hai mat ph^ng (SBC) va ( A B C D ) b^ng 60". chieu cao cua hinh chop S . R M Q la hinh lang tru R M Q . A N P d e u b5ng ^ Gpi I la trung d i e m cua A D . B i e t hai mat ph^ng (SBI) va (SCI) ciing vuong g6c v d i mat p h i n g ( A B C D ) . T i n h the tich hinh chop S . A B C D . C u n g thay ngay SRMO = S A N P = T ^ . nen tilf (1) c6: HiTiJng dSn giai 4 ^ . . 1 . h S h l I V Gpi H la hinh chieu cua I tren B C (h.6.31). ^' - ~ * R M Q T + ^ANP ~ - ~ - —•-'3" - — • Tif gia thict suy ra SI vuong g6c v d i mat ph^ng ddy. .yik .1. 2 3 Tinh diTctc I C = a>/2 , I B = B C = aN/s . TH do: = 1 ti?c la thiet d i y n chia hinh chop thanh hai phan b^ng nhau. ft Vj S A B C D = ^ A D . ( A B + CD) = 3a^ Obvh) 1 B a i 14. Cho hinh chop ti? giac deu S . A B C D , goc giiTa mat ben v^ d^y b^ng 60" DiTng thie't dien qua D C va mat ph^ng phan gidc cua. goc tao bcli hai m3| T a c 6 ^ I H . B C = S , B c = S ABCD ~ S^B, - ScDi ph^ng ( S D C ) va ( A B C D ) . Thiet dien chia khoi ch6p thanh hai phan c6 th*^' 3a^ = 3a^-a^- ^ tich V , va Vz.Timtiso 2 2 Hifdng d i n giai Nen I H = ^ =^ = ^ S I = IH>/3=^a, BC ^/5 v/5 Gpi M , N Ian liTctt la trung d i e m cua A B va C D . K h i 66 S N M = 60" la g''^^ 3%/3 3 giffa mat ben (SCD) va day ( A B C D ) . Ttr do V,s . A B C D = - ^ a - (dvtt).
i/uvcn gun ae iruuc Ky im i Bai 2. Cho hinh lang tru lam giac ABC.A'B'C c6 BB' = a. G6c giffa difdng th^og BB' vk mat phang (ABC) bling 60": tam giac ABC vuong tai C rrtf (1) va (2) suy ra — ^ - 4SA2 V.suyraSA = ^ BAC = 60". Hinh chieu vuong g6c cua diem B' len mat phing (ABC) triing 2a vdi trong tam cua tam gidc ABC. Tim the tich hinh chdp VAABC- a^^/6 v a y Vs.ABc =^^]y^ (^v")- Hifdng dSn giai p^i 4. Cho lang tru deu ABC.A'B'C cd canh day b^ng a, khoang each tam O Gpi M la trung diem cua AC, H la trpng tam tam gidc ABC (h.6.32). TinhdiTdc B H = - : B M . A a ; B - H = ^ ciia tam giac deu ABC den mat phing (A'BC) bkng - . Tinh the tich hinh 2 4 2 6 DatBC = xthi C M = - A C = - . BC= ^ - — ^ . lang tru deu dd. . , _ 2 2 tan BAC 2V3 Ht/dng dSn giai Tac6: BM^ - B C ^ +CM^ Gpi M la trung diem cua BC, H la hinh chieu cua O len A'M (h.6.34). 9 2 2 X ' •=:>X 2 = —27a 2 , Cd AM 1 BC; AA' 1 BC => BC 1 (AA'M) 16 a - \ 12 52 . => BC J. OH. Do dd OH l(A'BC) o — 2 Tird6 S A B C = ^ B C . A C = ^ ^ d(0; (A'BC)) = OH = J , dat AA' = x, ta cd 104 6 3 a Dod6 VA.ABC =VB.ABC = 9a (dvtt). AMOH OH _ MOdong dang MA'A, ndn 208 => X = — a . Bai 3. Cho hinh chop S.ABC cd tam giac ABC vuong tai C, AC = a. AB = 2a.SA AA'~MA' aV3 x^ + ^a^ .. vuong goc vdi day. Goc giffa mat ph^ng (SAB) va mat phing (SBC) bang 4 60". Gpi H, K Ian lifPt la hinh chieu cua A len SB va SC. Churng minh r^ng 3/2 AK 1 HK va tinh the tich hinh chop S.ABC. Suy ra V A B C . A B C = A A ' . S ^ B C = - r ^ a ^ (dvtt). Hif(}ng dSn giai u lo -, (h.6.33) Bai 5. Cho hinh lang tru diJng ABC.A'B'C cd ddy ABC la tam gidc vuong tai B. • Ta cd SA 1 BC va AC 1 BC ^ BC 1 (SAC) AB = a, AA' = 2a, A'C = 3a. Gpi M la trung diem cua doan thing A'C, I la B C l AK. giao diem cua AM va A'C. Tinh the tich khoi tif dipn lABC theo a. Ma AK 1 SC nen AK 1 (SBC) => AK 1 HK. Hi/dng dSn giai (h.6.37) De dang tinh dUpc S^g^ - (dvdt). Tinh dirpc AC = a^fs ; BC = 2a. Trong tam giac vuong AKH cd Vi I la trpng tam tam gidc AA'C, n6n I A - - A M . AK = AH.sin60" = — A H . 3 2 ^M.ABC 3 ^ Cac tam1 giac SAB, 1 SAC 1 vuong 1 tai 1A' 2 1 nen (1) ^^
COng ty TNHH MTVD WH Khang Hi^dng d i n g i a i ' H i f d n g dSn g i a i L a y M G A C ; N G A D sao cho A M = A N = a. Gpi O va O' theo thdr tif la lam cua hinh vuong Tac6 BM = - A C = a , B N = aN/2 . ^ A B C D va A'B'C'D", M la giao d i e m cua A K va OO". 2 Qua M ke dU'dng th^ng song song vdi B D c i t M N ^ = AM^ + AN^ - 2 A M . A N . c o s M A N = 3 a ^ BB', D D ' Ian li/Pt tai E, F. B' 1 A' Suy ra M N = a>/3 . Khi d6, thiet di?n tao bdi (a) va hinh lap K / / D o 66 tarn gi^c B M N vuong tai B. Hinh 6.38 phi/dng chinh la hinh binh hanh AEKF / / (h.6.40). / / V I A B = A M = A N nen hmh chieu cua A tren ( B M N ) la tarn H cua dircJng V i O M la dirdng trung binh tam gidc A C K nen ' —"" tr6n ngoai tiep A B M N , H cung chinh la trung d i e m ciia M N . -/ -" ^^•^^
Hvtdng d i n giai p a t B C = x , t a c 6 BJ = ^ ^ va A K = ^ Gia suf AB = a,-mat ph^ng qua AB tSm O cua 2sin^ ^ hlnh cau noi tiep c^t CD tai diem E (h.6.42). Dat S = SABE 1^ dien tich thiet di^n c^n tlm. Xhay BJ, A K vao (4) ta dffdc ^smp ^ x ^ ^ . ^ ^ ^ 2cos^ - >/3sina (5) Theo cong thurc (1) ta c6 2sin- ^ 2 2 pqsina 2 ,. ^ABCD 3a 3 a V$y he thffc (5) chinh la moi quan he giffa a v^ p. Ket qua nay gon hdn ket Do thiet di?n di qua tarn hinh cau npi tiep nen qua trong chffdng trinh dan gidi neu tren. mat phing OAB la mat phing phan gidc cua Bdi 11- Cho tff dien DABC c6 DA vuong goc vdi mat phing (ABC), nhi dien g6c nhi di$n canh AB, do do canh DB la nhj dien vuong, DB = ayfz , BDC = 45", DAB = a (0 < a < -) 2S ABC .S.sin — 2 'ABCD = y C A B E + ^ D A B E Xic djnh a de nhi dien canh DC b^ng 60". 3a Nhan xet: Bai nay da dffdc giai bllng cdch 2pqcos— difng mat phIng di qua A va vuong g6c vdi DC Tiif (2) va (3) suy ra S = tai F, c i t BD tai E, tff do xac dinh goc nhi di?n p+ q canh DC, roi tinh tana, ket qua tana = — . Bai 10. Cho hinh ch6p tarn gidc deu D.ABC. Goi a \k g6c giffa mat ben DBG 2 mat My ABC, P la g6c giffa mat ben ke nhau ADB va ADC. Xic dinh quan Neu ta sff dung cong thffc (1) cua mpnh hp giffa a va p. de thi Idi giai thu^n Idi hdn nhieu. Nh0n xet: Bai n^y da difdc giai trong hffdng Hifdng dSn giai dan giai de thi tuyen sinh b^ng phffdng phdp V ^ 1 - L D A . A B . B C ^ - ^ ^ ' " " ^ " "^'"•^•^•'4 sau: Dffng mat phing (DAK) vuong goc vdi BC 3 2 6 ' tairK: Dffng mat ph^ng (BCJ) vuong g6c vdi vdi DC = 2a, DA = a x ^ c o s a , AB = a N ^ s i n a , AC = a72(l + sin^ a). AD tai J tW A K D = a v^ BJC = P (h.6.43). Do ABC Ih tam giac deu nen BK = KC vk Do SQBC i SDAC =a^cosaVl + sin^a nen dp dung cong thffc (1) cua m$nh de ta dffdc B J K = C J K = ^ . Chieu tam giic DAB len mat 2 72a-'sin2a 2a^(a^ cosa J l + sin^ a)sin60" , i ^— = ^ o 2V2sina = J3(l + sin^ a) phing (DAK) ta c6 S^AK = SDAB-COS^, suy ra: 6 6a ^ O sin^ a = - . Suy ra sina = , / - (do 0 < a < —). DK.AKsina = DA.BK. c o t ^ roi dan den tan^ a 3 t a n 2 P - l = 4 . 5 V5 2 2 ^ ^ i 12. Chffng minh rkng trong mot tff di$n thi tich cua cac cap canh do'i chia mdng din giai cho tich cAc sinh ciia cdc nhi dipn cua tffng cap d6 la b^ng nhau (Dinh l i sin _ 2SDAB-SDAC sinp ^SpBc-SABC-sintx chotffdipn). , , Ta c6: VABCD 3DA 3BC Hi^dng din giai SoAB-sinP _ S ^ B c s i n a BJ.DAsinp _ AK.BCsina X6t tff dipn ABCD. Gia sff AB = a, CD = b. va a, p theo thff tff la gdc nhi ' DA BC DA BC dien canh a. b cua tff dien do (h.6.45). Hay BJsinP = AKsina (4) Gpi S| = SABC; S2 = SABD; S3 = SDAC; S4 = SDBC; V = VABCD-
CdngtyTNHHMTVDWli Kh.,n, 1,,/ Theo cong thtfc (1) cua m$nh dc ta c6 _ 2 S 1 S 2 sing _ 2S3S4 sinp ,2 u2 VABCD - ; V^BCD - — . 2 • • 2^ + ^ +2abcotg.cotP) = sm a sin p ^ 9V2 2S,S2 sing ^28384 sinp 3U.. ra S.S; sing ^ S3S4 sinp ^ 3V 3a 3b ' ^ a b 2 ' Mhirtig theo cong thtfc (1) thi V = ^^1^2 s ' " " nen. ,_4Sfs2 3a sin^g 9V^ ab 4S1S2S3S4 Do do b^ 452^2 J, sing sinp 9V^ ; i/png tir cd -sin^P —- =—9 V1'^ . Thay vao ding thufc d tren ta thu dirPc Ap dung tiTdng tif cho cac cSp canh doi con lai ta c6 dieu phai chtfng minh. Bai 13. Gia suf a, b la do dai cSp canh doi cua mot tiJ dien; a, p \k goc nhi djei^ a^ + b^ + 2abcotg . c o t p = ^ . i ^ (s^S^S.^S^f tifdng uTng vdi cic canh nay. Chufng minh r^ng dai li/dng sau khong pf^, 9V^ 9V2 ^ thupc vao c^ch chpn cdc cSp canh doi cua tuf dien: m = + b^ + 2abcotg.cotp. 4S?S^4S3^S^•(S^S^S2-Sn' Hay a^ + b^ + 2abcotg.cotp = • ^ IX Htfdngdingiai la mot 9V^ X6t tur di0n ABCD. Gia suf AB = a; CD = b. dai li/Png khong doi. DSt S ] = SABCJ S2 = SABD; S3 = SDBC; S4 = Vay M = a^ + b^ + 2abcotg.cotp khong phu thupc v^o c^ch chpn cdc cap SDAC^= VABC^^C = m ^ C = n; Dat A ^ C canh doi nhau cua li? dipn ABCD (dpcm). i a = DPH ; y = D M H ; 9 = DNH la cdc g6c , C A C BAI TAP KHOI C A U , KHOI TRy, KHOI IRON XOAY nhi di^n tiTdng tfng vdi cdc canh a, m, n. ^ nki 1. Cho hinh chdp ti? gidc dcu S.ABCD cd canh ddy AB = a canh ben SA = a. Trong dd H \k chan diTdng cao ciia tur di?n ha a) Chufng minh r^ng nam diem S, A, B, C, D cung nkm tren mot mat cau (gpi la tir D xuong mat ABC va DH = h, (h.6.45). mat cau ngoai ticp hinh chdp). Hay xac dinh tam va tim bdn kinh cua mat Neu H n^m trong tam gidc ABC thi Hinh 6.45 du dd. F.T V ^ A H B + ^ B H C "•" '^CHA = S] o ah, cottt + mh, cot Y + nh| cot9 = 2S, (6) b) Chiang minh r^ng cd mot mat cau tiep xuc vdi nam mat ciia hinh chdp (gpi la Do a, y, 9 bien thien tilf O" den I8O" nen (6) van diing cho tnrdng h^p H nam miit cau npi tiep). Hay xdc djnh lam va tim ban kinh cua mat cau dd. n g o i i tam giiic ABC. Hi/dng dan giai 3V 2S a) Vi S.ABCD la hinh chdp ttf gidc dcu nen SA = SB = SC = SD = a va tuf giac Thay h, = — vao (6) ta diTqlc: acota + mcoty + ncot9 = — p (7) Si 3V1 ABCD la hinh vuong canh a, nen AC = BD = asfz . Viet cac ding thiJc ti/dng tiT (7) cho cic mat cd di$n tich S2, S3, S4, sau do Tam giac SAC cd SA^ + SC^ = AC^ (vi cdng b^ng 2a^) nen la tam giac vuong cpng cac d i n g thiJc cd S, vdi ding thufc cd S2 roi triT di ding thtfc cd S3, S4 t»i S. Suy ra: OS = OA = OC (1) nhandiTc^c Tufdng tir, tam giac SBD vuong can tai S nen OS = OB = OD (2) u o s?+s^-s?-s5 Tilf (1) va (2) cd OS = OA = OB = OC = OD = — . acota - bcotp = - J ^^^-^2 1 2 ^§y nam diem S, A, B, C, D cilng n^m _(s^s^s.^-s^)' Suy ra a^cot g + b^cot P - 2ab.cotg.cotp = 9V^ ^^n mat cau tam O, ban kinh . 2 Thay cot^ g = ^ - l ; c o t 2 p = ^ - 1 vao ddng thuTc tren ta cd ' ^Pi M la trung diem cua BC. sin^g sin^p ^^ong lam gidc SOM vc phan gidc M I ( I € SO). 5 55
Luyengiiii dc'iru>k ky tin Dll i mu\i lliU . T-un; jV«" ^ '••in hQC - NguySn Van ThOng Ceng fy TNHH MTVDWH Khang m TO I ke I H 1 S M , suy ra l O = I H . AP 'iMng B D T Cauchy cho ba so' di/Ong ta c6 Nhan thay SO L ( A B C D ) . O M 1 EC nen S M L BC (dinh li ba diTcJng vuo,^ 2 8sin2 ^" + 3 - 4 s i n 2 - + 3 - 4 s i n 2 - goc). Suy ra BC 1 ( S O M ) . ma I H e ( S O M ) , suy ra I H 1 B C , do do IH f „ • 2 « 2 2 2 8sin — 3 - 4 s i n 2 ^ = 8. (SBC). V a y d i e m I each deu hai mat p h l n g ( A B C D ) va (SBC). 2 I 2; Tu'cJng tif ta tha'y I each deu cac mat con l a i eua hinh chop. X e t tarn gj^^, S O M c6 M I la phan giac nen. p o do Vs A B C ~ " 2 ^ " ^ ^ • ^ ^ " S '^'^'^ '"^ ^^"^ diyPi a I . 'i.nU '•lAi iff, 10 MO 10 MO SO.MO a>/2 8sin^- = 3 - 4 s i n ^ - o s i n ^ - = - « a = 6O" , } w ' _ 2 2 _ 29 ' 10 = ^2 2 4 IS SM ^ S O ~ M O + SM M O + SM a a ^ 2(l + ^/3) S . A B C la tur d i c n deu. ' 2"^ 2 Vay khi a = 6 O " t h i the tich tuT dien S . A B C Idn nha't. , .72 B^i 3. Cho hinh chop tam giac dcu S . A B C c6 canh ddy la a va canh ben hpp v d i V a y mat cau tam I , ban kinh la mat cau npi tiep hinh chon 2(1 + N/3) mat day mot goc 45". M o t mat cau tiep xuc v d i mat day A B C tai A va tiep S.ABCD. xuc vdi canh BS kco dai tai H . Qua tam I cua hinh cau va diTcJng cao B D ci4a B a i 2. Cho hinh cau ban kinh R. tif d i e m S tren mat cau di/ng ba cat tuyen bbg day ta diTng mat p h i n g (a). ^ n h a u c ^ t m a t c a u t a i A, B . C s a o c h o ASB = ASC = BSC = a . a) Hay tinh d i p n tich cua mat cau va the tich cua hinh cau do. a) T i n h the tich cua tuT d i c n S A B C theo R va a . b) Tinh goc hdp b d i mat phang (a) vdi mat ddy (chinh xac den dp). b) Xac dinh a dc the tich cua tuT dien S A B C Idn nhat. Hif(}ng dSn giai HUdng dSn g i a i a) Mat cau tam I ban kinh R tic'p xiic v d i mat p h i n g ( A B C ) tai A va tig'p xuc vdi BS tai H nen l A 1 ( A B C ) , I H 1 BS va l A = I H = R. a) V i SA = SB = SC va A S B = ASC = B S C . 1 \ Goi O la U-png tam tam giac A B C , ha I K 1 SO. : nen A A S B = AASC = ABSC (c.g.c). I t \ I 1 \ Ta CO SBO = 4 5 " , A B = B C = C A = a Suy ra A B = BC = A C . ; K e SH 1 ( A B C ) thi H A = H B = HC. I nenSO = O B = ^ va SB = ^ . Vay S.ABC la hinh chop tam gidc dcu. 3 3 G o i SH c^t mat cau tai P, A H ca't BC tai M Ta CO ti? giac l A O K la hinh chi? nhat nen tW M B = M C va SP = 2R, SAP = 9 0 " . l A = O K = R; IK = AO = — . Dat SA = SB = X , ta c6: x^ = SA^ = SH.SP = SH.2R SH = 2R Ta CO I H ' + HS' = IK' + SK' (= IS') a . a BCVJ /r . a M C = x . s i n — nen B C = 2xsin in—; AM= = xV3sin—. nen R^ + a - 2 2 3 2 2 (2 Ma SC^ = SH^ + HC' = SH' + -AM rimdirdcR= |(2>/2-N/3). 3 4 2 • 2a 2 4 2 Do do dien tich mat cau: ( I ; R) la S.iu = 47tR^ = Tta' {2^/2 - N/J)^ . Suy ra = -i--x^sin — o x ^ =-R^ 3-4sin — 4R^ 3 2J The tich hinh cau ( I ; R) la V^^„ = ^nR^ =^na^ (2V2 - sf3f ^ . 1 BC.AM . a [z . a Taco: VCARC = - S H . S A D r S.ABC 3 =-.—. 3 2R 2 = 31 x^ —.xsin—.xV3.sin— 2R 2 2