www.vnmath.comCHUYN TH TCHBi 1 Cho hnh chp tam gic S.ABC c y l tam gic vung B,cnh SA (ABC) . T A kAD SB vAE SC . Bit AB = a, BC = b, SA = c.Tnh th tch ca khi chp S.ADE?Phn tch - tm li giiAD,AE l cc ng cao trong tam gic SAB,SAC ABCDESTnh ng cao: ABC vung ti B nnAB BC Gi thit cho :SA (ABC) SA BC BC (ABC) AD BC AD l ng cao trong tam gic SAB

AD SB

AD (SBC)

AD SC Mt khc :AE SC SC (ADE) Hay SE l ng cao ca hnh chp S.ADE di SE:

2 2 2 2AS.AB AS.AB a.cADSBAS AB a c + + 2 22 2 2 2 2AS.AC SA.AC c. a bAESBSA AC a b c+ + + +p dng Pytago trong tam gic SAE c:

2 2 22 2 22 2 2c (a b )SE AS AE ca b c+ + += 22 2 2ca b c + +www.vnmath.comDin tch tam gic ADE:DE = 2 2AE AD += 2 22 2 2 2 2c .b(a b c ).(a c ) + + +S = 1.AD.AE2 = 2 22 2 2 2 2 2 21 c .b ac. .2(a b c ).(a c ) a c + + + +

= 3 32 2 2 2 21 a.c .b.2(a b c ).(a c ) + + +Th tch: V =1 1.SE. .AD.DE3 2 = 3 32 2 2 2 2 2 2 21 c 1 a.c .b. .3 2a b c (a b c ).(a c ) + + + + + 2 42 2 2 2 21 a.b .c.6 (a c )(a b c )+ + + Xt mt cch gii khc nh sau: DE(SAB) BC (SAB) => DE // BC Pytago trong cc tam gic vung: SD2 = AS2 - AD2; SE2 = AS2 - AE2SB2 = SA2+AB2SC2 = SA2+AC2 = SA2 + AB2 + AC2Lp cc t s:

2 22 2 2cSA AEa b c+ + + 2 2 2 22 2 2 2 2SA AD SA AE.SA AB SA SB SC + + +

2 2 2 2 22 22 2 2 2 22 2 2 2 2c .a c (a b )c ca c c a b.a c c a b+ + + ++ + +4 2 22 2 2 2 2 2 2c b .c.(a c ) (c a b )+ + +=> 32 2 2 2 2SA SD SE b.c. .SA SB SC (c a b )(a c )+ + + SADESABCV SA SD SE. .V SA SB SC = 32 2 2 2 2b.c(c a b )(a c ) + + + =>SADEV = 32 2 2 2 2b.c(c a b )(a c ) + + +.SABCV = 32 2 2 2 2b.c(c a b )(a c ) + + +.1 1.SA. .AB.BC3 2 = 2 42 2 2 2 21 a.b .c.6 (c a b )(a c ) + + + (vtt)www.vnmath.comBi 2:Cho hnh chp tam gic S.ABC c mt bn SBC l tam gic u cnh a.Cnh SA (ABC) , gc 0BAC 120 .Tm th tch ca khi chp S.ABC? ABCDS

Trnh by li gii:Xt hai tam gic vung SAB v SAC c: SA chung SB = SC =>SAB =SAC (c.c) => AB = AC =>ABC l tam gic cnGi D l trung im ca BC ta c : tanCAD = CDAD => AD =CD a2. 3 tanCAD Din tch y: 2ABC1 a . 3S AD.BC2 4 SD l ng cao trong tam gic u SBC cnh a nn : SD =a 32p dng nh l Pytago trong tam gic SAD ta c:www.vnmath.comSA2 =SD2 - AD2=2 2 23.a a 2a4 12 3 =>SA = a 23Th tch cn tnh:V = ABC1SA.S3 = 3a . 236 (vtt) Tng qut ha ta c bi ton sau: Cho hnh chp S.ABC c mt bn SBC l tam gic u cnh a, gc 0BAC (0 90 ) . Bit SA vung gc vi mt phng y. Tnh th tch ca khi chp S.ABC theo a v ? Mt cch hon ton tng t ta c li giinh sau:AD = AD a2.tantanCAD Din tch tam gic: 2ABC1 1 a aS AD.BC a2 2 2.tan 4.tan SD l ng cao trong tam gic u SBC nn SD =a 32p dng nh l Pytago trong tam gic SAD ta c: SA2 =SD2 - AD2=2 2 23.a a 2a4 12 3 =>SA = a 23Th tch cn tm:

ABC1V SA.S3=1 1 1.B.h .SA. .AD.BC3 3 2 = 3a 212 3.tan (vtt)Bi 3 Trong khng gianh ta Oxyz cho hnh chp S.ABCD c y l hnh thoi. AC ct BD ti gc ta O . im A(2;0;0), B(0;1;0), S(0;0;22 ). Gi M l trung im ca SC v mt phng (ABCD) ct cnh SD ti N. Tnh th tch ca khi chp S.ABMN?www.vnmath.com ABCSMNDO Li giiTa nhn thy mt phng (SBN) chia khi chp S.ABCD thnh hai khi chp S.ABN v S.MBN Theo nh ngha v th tch ta c: S.ABMNV = S.ABNV + S.MBNV

S.ABNS.ABDV SN 1V SO 2 =>S.ABNV = 12S.ABDV= 14S.ABCDVTng t ta c: S.MBNS.BCDV SM SN 1.V SC SD 4 =>S.MBNV = 14S.BCDV = 18S.ABCDVDo vy: S.ABMNV= 14S.ABCDV + 18S.ABCDV= 83S.ABCDVTh tch khi chp S.ABCDV = ABCD1 1.SO.S .SO.AC.BD3 6= 832 Th tch cn tnh: S.ABMNV = 2 (vtt)Nghin cu li giiGi V1 l th tch khi a din nm di (ABMN): V1 = S.ABMNVKhi : www.vnmath.com

S.ABCDV = S.ABMNV+ ABCMNV hayV= V1+S.ABMNVTa c :V1 = N.ABDV + B.CDMNV

N.ABDV = S.ABD1V2= V4 ABCSMNDOHai hnh chp B.SCD v B.DCMN c chung nh v mt phng cha y nn: DCMN SDC3S .S4Th tch ca chp S.ABCD l: V = ABCD1 1.SO.S .SO.AC.BD3 6 = 832 Th tch cn tnh: Bi 4www.vnmath.com Cho hnh vung ABCD c cnh a, cc na ng thng Ax v Cy vung gc vi mt phng (ABCD) v cng mt pha so vi mt phng y. Ly imM A trn Ax, lyN C trn Cy. t AM = m ; BN = n.Tnh th tch ca khi chp B.AMNC theo a, m, n? Trnh by li gii Theo gi thit ta c:CN (ABCD) CN CB , O l tm y nn OB AC OB (ACMN) hay OB l ng cao di OB = AC2= a 22. Mt khc MA ACNC AC' MA // NC nn t gic ACMN l hnh thang ACMNSMA NC 2AC a(m n)2 2+ +Th tch khi chp:

2ACMN1 aV OB.S (m n)3 6 +( vtt ) Nghin cu li gii Nhn thy doAM AB ,AM AD ,AB AD nn ta a vo h trc taOxyzsaochoA(0;0;0), B(a;0;0), M(0;0;m), D(0;a;0)ttaxc nh c ta nh C(a;a;0) sau p dng cng thc tnh th tch ca khi hp: 1V AB, AC AM61 ]uuur uuur uuuur = 2a(m n)6+ (vtt)Bi 5: Cho hnh chp tam gic S.ABC c y l tam gic u cnh a. Cnh SA (ABC) , SA = 2a. Gi M, N l hnh chiu vung gc ca A ln cc cnh SB, SC. Tnh th tch ca khi chp ABCMN?www.vnmath.comABCSMN Trnh by li giiXtSAB vSAC c AB = AC, SA chung, A = 090

SAB =SAC SB =SC mt bn SBC l tam gic cn.p dng nh l ng cao trong cc tam gic SAB v SAC ta c:

2 2AB.ASAMAB AS+ =2a5 2 2AC.ASANAC AS+ = 2a5p dung nh l Pytago: SM = 2 24aSA AM5

2 24aSN SA AN5 Ta c cc t s: SMSB = SNSC = 45 S.AMNS.ABCVV = 1625

S.AMNV = 1625S.ABCV = 38a 375Th tch :

ABCNMV= S.ABCV- S.AMNV= 3a 36- 38a 375 =33a 350 (vtt) Nghin cu li giiTa c th gii bi ton trn bng phng php ta bng vic a vo htrctaOxyztrongA(0;0;0), B(a;0;0), S(0;0;2a). Taxcnh c ta ca C, M, N, sau s dng cng thc sau: www.vnmath.com 0S.AMN1V AM, AN AS 60 BAC61 ]uuuur uuur uuur S.AMN1V AM, AN AS61 ]uuuur uuur uuur ABCNMV= S.ABCV- S.AMNV =33a 350(vtt)Bi 6: Cho hnh lng tr tam gic ABC.DEF c BE = a, gc gia ng thng BEvi mt phng(ABC)bng 060 . TamgicABCvungti C, gc 060 BAC , hnh chiu vung gc ca E ln (ABc) trng vi trng tm ca tam gic ABC. Tnh th tch ca t din D.ABC?ABCDEFMG Trnh by li giiTa c: EG (ABC) nn EG l ng cao ca chpp dng h thc lng trong tam gic vung EGB ta c: EG = EBsinB = asin060= a 32p dng pytago:

2 2BG BE EG = a2 m BG = 23BM BM = 32BG = 3a4p dung Pytago trong tam gic BMC:www.vnmath.comMC = MBsin015=3a4sin015 , AC = 2MC = 3a2sin015 , BC = ACtan060=3a2sin015 3Th tch ca khi chp: V = 3ABC1 3EG.S a .3 42 0sin 15(vtt)Bi 7: Cho t din ABCD gi d l khong cch gia hai ng thng AB v CD,l gc gia hai ng thng . Tnh th tch ca t din ABCD?ABCDE Trnh by li giiDng hnh bnh hnh ABDE, do AE // (BCD) nn ABCD E.BCD B.ECDV V V = ECD1S .d(B, CDE)3 =1 1 1CE.CD.sinECD AB.CD.d.sin3 2 6 (vtt) Nghin cu li giiTa xt mt cch gii khc nh sau:www.vnmath.comBCDEFMNADng hnh hp ch nht AEBF.MDNCngoi tip t din ABCD, AB (ABEF) , CD (CDMN) V (ABEF) // (CDMN) nn chiu cao ca hp bng dTh tch cn tnh: ABCD hp MDNC1 1V V S d3 3 = 1 1MN.CD.d.sin3 2 = 1AB.CD.d.sin6 Bi 8:Trong khng gian cho hnh hp ch nht ABCD.EFGH c nh A trng vi gcta, imB(a;0;0), D(0;a;0), E(0;0;b), Mltrungimca CG.tnh th tch ca khi t din BDEM theo a v b?www.vnmath.comBCDEFGHM(0,a,0)(a,0,0)(0,0,b)A(0,0,0) Trnh by li gii M l trung im ca CG nn: C GMC GMC GMx xx a2y yy a2z z bz2 2+ + '+ Ta cc vect: BMuuur(0;a;b2), BDuuur(-a;a;0), BEuuur(-a;0;b)Xt tch hu hng:

2b b0 a a 0 ab abMB, BD , , , , a2 2a a 2 2a 0 0 a _ _ 1

] , ,uuur uuurTch v hng: MB, BD BE 1 ]uuur uuur uuur = 23a b2Th tch: V = 1MB, BD BE6 1 ]uuur uuur uuur = 1623a b2 = 2a b4 (vtt) Nghin cu li giiKCO BD , ko di EM ct SO ti N, mt phng (BDM) chia khi chp thnh hai khi chp E.BDM v N.BDM nn E.BDM E.BDN1V V2www.vnmath.comV M l trung im ca SG nn: CN = CADin tch tam gic BDN: S = 1 1 3 2aBD.NO a 2.2 2 2= 23a2ADFGHMNCBEOTh tch: E.BDM E.BDN1V V2= 121623a22b = 2a b4 (vtt)Bi 9: Tnh th tch ca khi t din ABCD c cc cnh i bng nhau AB = CD = a, AC = BD = b, AD = BC = c? Trnh by li giiDng t din APQR, y l t din vung ti nh A, tht vy: AD = BC = PQ2 BC l ng trung bnh ca tam gic PQR BC = QD = DP AD = QD = PD AQ AP Hon ton tng t ta c: AQ AR ,AR AP Ta c:APQR ADBQ ABCD ACDP ACBRV V V V V + + + www.vnmath.com ABCD APQR AQR1 1 1 1V V AP.S4 4 3 24 AP.AQ.ARp dng nh l Pytago trong tam gic APQ, AQR, APR

2 2 2 2 2AP AQ QP 4BC 4a + 2 2 2 2 2AQ AR QR 4CD 4c + AP =2 2 22(a b c ) + , AQ = 2 2 22(b c a ) + , AR = 2 2 22(a c b ) + Th tch: ( )32 2 2 2 2 2 2 2 21V 2 (b c a )(a c b )(a b c )24 + + + = 2 2 2 2 2 2 2 2 22(b c a )(a c b )(a b c )2+ + + (vtt)Bi tp ngh Bi 1 ( khi A - 2007)Cho hnh chp S.ABCD c y l hnh vung cnh a, mt bn SAD l tam gic u, (SAD) (ABCD) , gi M, N, P l trung im ca SB,BC,CD. Tnh th tch ca khi chp CMNP theo a?Bi 2 ( Khi A - 2009) Cho hnh chp S.ABCD c y l hnh thang vung ti A v D, AD = AB = a, gc gia hai mt phng (SCD) v (ABCD) bng 060 , gi I l trung im AD, bit hai mt phng (SBI) v (SDI) cng vung gc vi mt phng (ABCD). Tnh th tch ca khi chp S.ABCD theo a?Bi 3 (Khi B - 2006) Cho hnh chp S.ABCD c y l hnh ch nht AB = a, AD = a2, SA = a, SA vung gc vi mt phng (ABCD). Gi M, N l trung im ca AD, SC, I l giao im ca AC v BM. Tnh th tch ca t din ANIB? Bi 4 (Khi A - 2008) Cho hnh lng tr ng ABC.DEF c di cnh bn bng 2a. y ABC l tam gic vung ti A, AB = 2a, AC = a 3 . Hnh chiu vuong gc ca D ln mt phng (ABC) l trung im G ca cnh BC. Tnh th tch ca khi chp G.ABC?2. Th tch ca khi lng trTrongmc ny ta s dng nh l sau: Th tch ca hnh lng tr bng mt phn ba din tch y nhn vi chiu caoV B.h trong : B l din tch y h l chiu caowww.vnmath.comBi 1 Cho hnh t gic u ABCD.EFGH c khong cch gia hai ng thng AD v ED bng 2. di ng cho mt bn bng 5. Tnh th tch khi lng tr ?ABCEFGHKD Trnh by li giiGi KlhnhchiuvunggccaAlnED AK ED , AB// EF (EFD) do AB // (EFD)nn d(A,EFD) = d(AB,ED)M EF (EFDA) nnEF AK AB AK AK=d(A,EFD) = d(AB,ED) = 2t EK = x ( 0 x 5 ). Trong tam gic vung AED ta c: AK2 = KE.KD 4 = x(5-x) x2 - 5x + 4 = 0 x 1x 4

Vi x = 4 ta c AE = 2 2AK KE 5 + V = AE.ABCDS =5( vtt)Vi x = 4 ta c AE = 2 5 V = 10 5( vtt)Bi 2y ca khi lng tr ng ABC.DEF l tam gic u. Mt phng y to vi mt phng (DBC) mt gc 030 . Tam gic DBC c din tch bng 8. Tnh th tch khi lng tr ?www.vnmath.comABCDEFK Trnh by li gii t CK = x, DK vuong gc vi BC nn DKA = 030Xt tam gic ADK c: cos030= AKDK AK = x 3 , DK = 2xDin tch tam gic BCD: S = CK.Dk = x.2x = 8, do x = 2 AD = AK.tan030= x 3 33 = 2Th tch khi lng tr:V = 13AD.CK.AK = 83(vtt)Bi 3Cho khi lng tr ng ABCD.EFGH c y l hnh bnh hnh v BAD = 045 , cc ng cho EC v DF to vi y cc gc 045v 060 . Chiu cao ca lng tr bng 2. Tnh th tch ca lng tr ? Trnh by li giiT gi thit:GAC = 045 , BDF = 060 , AC = AG = 2, BD = 2.cot060= 23p dng nh l cosin trong tam gic: 2 2 2BD AB AD 2.AB.AD. + cos045

2 2 2 0AC AD CD 2.AD.CD.cos135 +

2 2 0 0BD AC 2.AD.AB.cos45 2.CD.AD.cos135 +www.vnmath.com = -22AB.AD 43 - 4 = -22AB.AD AB.AD = 43 2- Th tch cn tm: V = AB.AD.EA.sin045=43 2 222 (vtt)Bi 4 Cho hnh lng tr tam gic ABC.EGH c y ABC l tam gic vung cn, cnh huyn bng 2. Bit mt phng (AED) vung gc vi mt phng (ABC), AE =3 . Gc AEBl gc nhn, gc gia mt phng (AEC) vi (ABC) bng 060 . Tnh th tch ca lng tr?ABCKEMHG Trnh by li giiwww.vnmath.com H EK AB(K AB) EK (ABC) . V AEB l gc nhn nn K thuc on AB K KM AC EM AC ( theo nh l ba ng vung gc ) EMK = 060 . Gi s EK = x ,2 2EK EA EK + = 23 x MK = AK.sinKAM = 2223 x M MK = EKcot060= x3, do : 2223 x = x3 x =35Vy V = EK.ABCS= 12AC.CB.EK = 3 510 ( vtt)Bi tp nghBi 1Cho lng tr t gic ABCD.EFGH c y l hnh thoi c di cnh bng a. Gc BAD = 060 ,AF BH . Tnh th tch ca khi lng tr ?Bi 2Cho lng tr ABC.DEF c y l tam gic vung cn ti A, BC = 2a. M l trung im ca AD, gc BMC = . Tnh th tch ca lng tr ?3. Th tch ca khi hp ch nhtTrong mc ny ta s s dng nh l sau: th tch ca khi hp bng tch di ba kch thc V = a.b.c = B.h trong : a, b, c l ba kch thc B l din tch y h l chiu caoBi 1Cho khi hp ABCD.EFGH c tt c cc cnh u bng a,BAD = EAD = ( 0 090). Tnh th tch khi hp ?www.vnmath.comABCEFGHKDM Trnh by li giiH EM AC(M AC) (1) tam gic EBD cn ti E ( do EB = ED ) BD EO MBD AC BD (BAO) BD EM (2)T (1) v (2) ta c: EM (ABCD) hay EM l ng caotEAO = , hEK AB MK AK (nh l ba ng vung gc) cos2cos = AM AKAE AM= AKAE = coscos = coscos2EM = a.sin = 22cosa 1cos2= 2 2acos cos2cos2 Th tch cn tnh: V = AB.AD.EM.sin = 2a .sin2 2acos cos2cos2 = 3a .sin22 2cos cos2 (vtt)www.vnmath.comBi 2:Cho khi hp ch nht ABCD.EFGH c y l hnh ch nht c AB =3 , AD =7 , hai mt bn (ABDE) v (ADEH) ln lt to vi y cc gc 045v 060 , di tt c cc cnh bn u bng 1. Tnh th tch ca khi hp ?ABCEFGHMKDN Trnh by li giiK EK (ABCD), (K ABCD) , KM AD(M AD) , KN AB(N AB) Theo nh l ba ng vung gc ta c: AD EM, AB NK Ta c: EMK = 060 , ENK= 045 ,t EK = x khi : EM = 0xsin60 = 2x3AM = 2 2EA EM = 23 4.x3 = KN m KN = x.cot045 Nn x = 23 4.x3 do x = 37Th tch khi hp ch nht: V = AB.AD.x =7 . 3 .37 = 3 (vtt)Bi 3www.vnmath.com Cho hnh hp ch nht c di ng cho bng d, ng cho to vi y gc , to vi mt bn ln gc ,tnh th tch ca khi hp ?BCDEFGHA Trnh by li gii ng cho AG c hnh chiu ln (ABCD) l AC,ln mt phng (BCGF0 l BG nn: GAC , AGB p dng nh l Pytago trong cc tam gic: ACG, GBA, ABC c CG = d.sin, AC = d.cos,AB = d.sin, BC = 2 2AC AB = d.2 2cos sin ta c V = AB.BC.CG = d3.sin.sin.2 2cos sin M: ( )1 cos2 1 cos2 1cos2 cos22 2 2+ + = cos(+).cos( - ) Vy V= d3.sin.sin. cos( ).cos( ) + (vtt)Bi tp nghBi 1Cho hnh hp ch nht ABCD.EFGH c AB = a, AD = b, gc BAD , ng cho AD t vi y gc . Tnh th tch khi hp ch nht ?Bi 2www.vnmath.com Cho t din ABCD c AB = CD, AC = BD, AD = BC, qua mi cnh ca t din k mt phng song song vi cnh i din, cc mt phng nhn c xc nh mt hnh hp:1) Chng minh hnh hp ni trn l hnh hp ch nht?2) CMR hhcn ABCDV 3V 3) Gi IJ, EF, MN l cc dng trung bnh ca t din.CMR: ABCD1V3IJ.MN.EFBi 3Cho hnh hp ch nht ABCD.EFGH, M l trung im ca AD, mt phng (ABM) ct ng cho AG ti I, tnh t s th tch ca hai khi a din c to bi mt phng (EBM) ct hp?4. Bi ton cc tr th tchBi 1 Cho hnh chp S.ABC c SA = x, SB = y, cc cnh cn li bng 1,vi gi tr no ca x, y th th tch ca khi chp l ln nht, tm gi tr ln nht ?ABCSNM Trnh by li giiGi M, N l trung im ca SA, BC, ta c: S.ABCV = 2.S.MBCV, cc tam gic ABS, ACS c: BA = BS, CA = CS ABS = ACS v l cc tam gic cnTa c: BM SA, CM SA SA (MBC) SM (MBC) ,www.vnmath.comSM l ng cao, SM = x2Tnh din tch y: MB = MC = 2x14, MN = 22BCBM4 = 2 2x y14+

MBCS = 12MN.BC = 2 2y x y12 4+Th tch: S.MBCV = 1 x y3 2 2 2 2x y14+ = xy12, S.ABCV = 2 2xy x y16 4+Ta c: ( x-y)2 0 x2 + y2 2xy 2 2x y xy4 2+

S.ABCV = 2 2xy x y16 4+ xy6xy12 1622 xy(xy)2 16xy xy2 (2 xy)2 2p dng BT Cauchy cho ba s xy2, xy2, (2-xy) ta c: xy2xy2(2-xy) 33xy xy(2 xy)2 23 _+ + , = 1627

V 161627 = 2 327, du bng xy ra khi 2 2x y 2xyxy2 xy2 + ' x = y = 23Bi 2 Cho hnh chp t gic u S.ABCD c khong cch t nh A n mt phng (SBC) bng 2a, gi l gc gia mt bn vi mt y, vi gi tr no ca th th tch ca khi chp l ln nht?www.vnmath.comA BCSHMIND Trnh by li gii M, N l trung dim ca BC v AD nn SMN , v AD // BC suy ra AD // (SBC) d(A,SBC) = d(N,SBC) (1)Mt khc: MN BC,SM BC BC (SMN) (SBC) (SMN) Do SM = (SBC) (SMN) , kNH SM NH (SBC) , d(N,SBC)= NH(2)t (1) v (2) ta c NH = 2aH thc lng trong tam gic vung MNH ta c: MN = NHsin = 2asinDin tch y : 22 2ABCD4aS AB MNsin Gi I l tm y th ta c SI = MI.tan = a atansin cos Th tch: th tch V = 13SI.ABCDS = 324a3sin .cos 2minV sin .cos t GTLN cos(1 - 2cos ) t GTLN t x = cos, xt hm s y = x - x3 trn (0,1), xt du hm y ta c www.vnmath.commaxy = 3 2 3y( )3 9khi v ch khi x = 33 cos = 33Vy 3minV 2a 3 ( vtt )Bi 3Cho tam gic u OAB c AB = a, trn ng thng i qua O v vung gc vi mt phng(OAB) ly im M, t OM = x,Gi E, F ln lt l cc hnh chiu vung gc ca A ln MB v OB.ng thng EF ct d ti N. Xc nh x th tch khi chp ABMN l nh nht?AOBMEFN Trnh by li gii Gi V l th tch khi t din ABMN ta c V = M.OAB N.OABV V + = OAB OAB1 1OM.S ON.S3 3 += OAB1(OM ON).S3+Do th tch V nh nht ( OM + ON ) t GTNN Hai tam gic OMB: OFN suy ra: OM.ON = OF.OB = hng s v O, F, B c nh, ta c: OM + ON2 OM.ON du = xy ra OM = ON nn ( OM + ON ) t GTNN OM = ON = xV OM.ON = OF.OB 22ax2 a 2x2( OF = a2, OB = a )www.vnmath.comVy M thuc d sao cho OM = a 2x2th th tch l nh nht , khi :V = OAB1(OM ON).S3+ = 2a 612 ( vtt )Bi 4Cho hnh chp S.ABCD c 7 cnh bng 1, cnh bn SC = x. Tinh sth tch ca khi chp, vi gi tr no ca x th th tch l ln nht?ABCSDOH Trnh by li gii SH l ng cao trong tam gic C nn ta c: SH.AC = SA.SC SH = 2SA.SC xACx 1+Ta c: OB2 = AB2 - AD2 = 2(3 x )4 nn OB = 23 x2 ( 0 x3)Din tch y ABCDS = AC.OB = 213 x2 .2x 1 +Th tch V = ABCD1S .SH3 = 2 221 x(3 x )(x 1)6x 1 ++ = 21x 3 x6Ta c: V2 = 2 21x (3 x )36 v 2 2x (3 x ) + = 3 khng i nnMax2 2x (3 x )1 ] = 94 2 2x (3 x ) 6x2 www.vnmath.comV2 t gi tr ln nht l 94.36 6x2 max1V46x2 Bi 5 Cho hnh chp S.ABCD c y l hnh vung cnh a, hai mt bn (SAB) v (SAD) cng vung gc vi y, mt gc 0xAy 45 chuyn ng trn y quay quanh im A cc cnh Ax, Ay ct CB v CD ti M, N, t BM = x, CN = y, tm x, y th tch caAMCNV t GTLN?BCDSMNA Trnh by li giiTr ht ta chng minh ng thc: x + y =2a xya Ta c 0BAM NAD MAN 90 + + , 0BAM NAD 45 + t 0BAM NAD 45 + tan( ) + = 1 Ta c tan tan11 tan tan + , m x ytan , tan2 2 suy ra 2x y21 x yxy1a+ + = 2a xya, www.vnmath.com ta c 2AMCN ABCD ABM ADNax ayS S S S a2 2

2AMCN AMCN1 a ax ayV SA.S (a )3 3 2 2 = 2a(a xy)6+ ta c xy 2(x y)4+ suy ra Max (xy) = 2(x y)4+

maxV (xy) t GTLN khi (xy) = 2(x y)4+ t GTLN suyra22(x y)a(x y) a4++ + suy ra x y 2a( 2 1) + 2maxaV (2 2)3 khi x = y =a( 2 1) Bi tp nghBi 1 Cho hnh chp S.ABC trong SA (ABC) , ABC l tam gic vung cn ti C. Gi s SC = a. Hy tm gc gia hai mt phng (SCB) v (ABC) th tch khi chp l ln nht, tm gi gi tr ln nht ?Bi 2 ( s 21- Chuyn luyn thi vo H - Trn Vn Ho) Cho ba tia Ox, Oy, Oz vung gc vi nhau tng i mt. Xt tam din Oxyz. im M c nh nm trong gc tam din. Mt mt phng qua M ct Ox, Oy, Oz ln lt ti A, B, C gi khong cch t M n cc mt phng (OBC), (OCA), (OAB) ln lt l a, b, c. Tnh OA, OB,OC theo a, b, c th tch khi t din l nh nht? 5. Chng minh h thc hnh hc chng minh cc h thc trong khi adin ta c th s dng cc kin thc v th tch gii nh sau: Gn bi ton cn chng minh vo mt h thc no v th tch, cc h thc ny thng l: Th tch ca mt khino c th biu din thnh tng hoc hiu cc th tch ca khi a din c bn ( nh khi chp, khi lng tr )

Vi cc h thc v th tich y sau cc php bin i tng ng n gin ta nhn c iu phi chng minh. Bi 1Cho t din ABCD, im O nm trong t din v cch u cc mt ca t din mt khong r. Gi A B C Dh , h , h , h ln lt l khong cch t cc im A, B, C, D n cc mt i din.www.vnmath.com CMRA B C D1 1 1 1 1r h h h h + + +ABCDHKhi t din ABCD c chia thnh 4 khi t din OBCD, OCAD, OABD, OABC. Ta c: OBCDABCD AV rV h OCADABCD BV rV h OABDABCD CV rV h OABCABCD DV rV hCng v vi v ca cc ng thc trn ta c: OBCD OCAD OABD OABCABCD A B C DV V V V 1 1 1 1rV h h h h 1 + + + + + + 1 ]

ABCDABCD A B C DV 1 1 1 1rV h h h h 1 + + + 1 ]

A B C D1 1 1 1 1r h h h h + + + ( pcm ) Bi 2www.vnmath.com Chng minh rng tng khong cch t mt im nm trong t din n cc mt i din ca n khng ph thuc vo v tr ca im nm trong t din ?AB COFGHKGi s M l im ty thuc min trong ca t din u ABCD Gi 1 2 3 4d , d , d , d l khong cch t im M n cc mt (BCD), (ACD), (ABD), (ABC)Gi 1 2 3 4V, V , V , V l th tch ca 4 khi t din chung nh M, V l th tch ca t din ABCD ta c:V = 1 2 3 4V V V V + + + 1 2 3 4V V V V1V V V V + + +V ABCD l t din u nn khong cch t nh xung mt i din bng nhau. Ta gi s khong cch ny l h, hai t din ABCD v MBCD c chung ynn: 1 1d Vh VHon ton tng t ta c kt qu: i iV dV h(i= 2,3,4)Do : 1 2 3 4d d d d1h h h h + + + h = 1 2 3 4d d d d + + + ( pcm)Bi 3 Cho gc tam din vung Oxyz nh O trn Ox, Oy, Ox ln lt ly cc im A, B, C sao cho OA + OB + OC + AB + AC + BC = L, gi V l th tch ca t din ABCD. CMR: 3L ( 2 1)V162www.vnmath.com Hng dn giiAC OB t OA = a, OB = b, OC = c p dng BT Bunhiacpxki:a + b 2 22(a b ) + , a + c 2 22(a c ) + , b + c 2 22(c b ) +cng v vi v cc BT trn: 2 2 2 2 2 22(a b c) (a b ) (a c ) (c b ) + + + + + + +(a + b + c)( 1 + 2) 2 2 2 2 2 2(a b ) (a c ) (c b ) (a b c) + + + + + + + + (a + b + c)( 1 + 2) L (1) du = trong (1) xy ra khi a = b = cp dng BT Cauchy cho a, b, c ta c: a + b + c 3abc (2)Ta c V =abc6, BT (2) a + b + c 33 6V (3)Du = trong (3) xy ra khi a = b = cT (1), (3) ta c3L 3.(1 2).3. 6V +hay3L ( 2 1)V162(4) Du = (4) xy ra khi a = b = c = L( 2 1)3 Bi 4 Cho OABC l t din vung ti nh O, vi OA = a, OB = b, OC = c, gi r l bn knh hnh cu ni tip t din. www.vnmath.com CMR: 1 1 1 1 3 3r a b c a b c + + ++ +Hng dn giiA BCOHK OH (ABCD) v gi s OH = h Do OABC l t din vung nn a, b, c, h l 4 ng cao ca t din ln lt k t A, B, C, O theo kt qu ca bi tp 2 ta c: 1 1 1 1 1r a b c h + + + (1) T (1)suy ra BT cn chng minh c dng: 1 3 3h a b c+ +(2)V OABC l t din vung nn ta c kt qu sau: 2 2 2 21 1 1 1h a b c + + (3)Theo (3) v p dng BT Cauchy ta c: 32 2 2 2 2 2 21 1 1 1 1 1 13h a b c a b c + + (4)Ta li c: ( a + b + c )2 2 2 2 39 a b c (5)T (3), (4), (5) ta c: 2 2 2 21 93h a b c+ +www.vnmath.com 1 3 3h a b c+ +Vy (2) ng suy ra pcm, du = xy ra khi a = b = cT bi ton trn ta c kt qu: h1 3r +Tht vy: Theo (1)1 1 1 1 1r a b c h + + +Do:22 2 2 21 1 1 1 1 1 13 3a b c a b c h _ _+ + + + , , hay 1 1 1 3a b c h+ + 1 1 3 h1 3r h h r + +( pcm ) Bi tp nghBi 1Chng minh khong cch t mt dim nm trong hnh lng tr n cc mt ca n khng ph thc vo v tr ca im nm trong lng tr ?Bi 2Cho hnh chp tam gic c a b csin sin sin , trong a, b, c l ba cnh ca tam gic y. Ccgc, , tng ng l cc gc nh din cankj a, b, c. Chng minh tng khong cch t mt im O trn mt y n cc mt xung quanh ca hnh chp l mt hng s