PHƯƠNG PHÁP GIẢI TOÁN TỰ LUẬN HÌNH HỌC KHÔNG GIAN - TRẦN THỊ VÂN ANH

8/10/2019 PHNG PHP GII TON T LUN HNH HC KHNG GIAN - TRN TH VN ANH

1/303

G

NH XUT BR BI HC QUC GIA H MI

WWW.FACEBOOK.COM/DAYKEM.QU

WWW.FACEBOOK.COM/BOIDUONGHOAHOCQU

B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N

W.DAYKEMQUYNHON.UCOZ.COM

ng gp PDF bi GV. Nguyn Thanh T


2/303

pffljfflffiPHAPHAi

PHQHISphApbADAKTLB0NHMSM

HMs LOGARrr

56bAj L NiPHK e TfBffl B PSHUG TB

:*&PKffltBrirS!sa. .BHHMH



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




3/303



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




4/303



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




5/303

DNG 1. NG THANG V m t PHANG.D AN H SONG SONG

NG THRNG vn MAT PHRNG

Kin thc co bni cong v Bg thng v mt phng iv: U rv 7 , - .

5



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




6/303

Cc dng bi tp thung gp

M un^tm |gia^ P^p Sp^ig J !: 'J ^


7/303

) Giao tuyn ca (AEF) v (ASG).

Trong mt phng (SBC) ta c EF nSN = (H}

M SN c (ASG) v EFc mp(AEF) => Hem p(AEF)vmp(ASG).

Vy AH l giao tuyn ca hai mt phng (AEF) v (ASG).V d 2:Cho t in SABC, gi I l mt im trn SA. (D) l ng thng

bt k trong mt phng (ABC) ct AB, BC, CA ln uDt t M, N, p.a. Tm giao tuyn ca mt phng (I, D) ln lt vi cc mt phng

(SAB), (SAC), (SBC).b. Gi Q l giao im ca (I, D) v sc . Chng minh ba ng thng IM,

QN, SB ng quy.Gii:

) Ta c: I e (I,D); e SA Pemp(SAC)

V IeSAcm p(SAC) =5> Iemp(SAC) /

Vy IP giao tuyn ca mp(L,D) v mp(SAC). / \ \ ____JGi Q l giao im ca s c v IP trong mp(SAC). yP' \ o /

Ta c: Q GIP c mp(I, D) => Q e mp(, D) A*I

Q s c c mp(SBC) => Q e mp(SBC) .y

VNe(D)cmp(I, D) => Nemp(I,D)

N eB C c mp(SBC) = )N e mp(SBC) H

Vy QN l giao tuyn ca mp(,D) v mp(SBC).

) Trong mp (I, D) ta c IM v QN ct nhau ti H.Mt khc IM c mp(SAB) v QN c: mp(SBC) v mp(SAB) nmp(SBC) =SB.

Do giao im H ca IM v QN phi trn giao tuyn SB.Vy M, QN v SB ng quy ti H.

7



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




8/303

V d 1:Cho tm din Oxyz. Trn Ox ly hai im A v A '; trn Oy ly

im B v B' v trn Oz y hai im c v c \ "a. Chng minh AB va A'B' ing ct nhau ti E; J3C v B'C thct nhau ti F;: CA v C'A' thng ct nhau ti G.

b. Chng minh E, F, G thng hng.Gii:

) AB v AB', BC v B'C', Cv C'A'ct nhau.Ta c: AeOxcmp(xOy);

Be Oy c mp(xOy)

=> AB c mp(xOy)

A 'eO x c mp(xOy);

B 'eOy c mp(xOy)=>- A'B' c mp(xOy)

Vy AB v A'B' cha ong (xOy) nn thng ct nhau ti E.Tong t, BC v B'C' cha ong (yOz) nn thng ct nhau ti F.

CA v A' cha trong (zx) nn thfng ct nhau ti G.

b) Ba im A, B, c khng, thng hng xc nh mt phng (ABC).Ba imA',B', C' khng ing hng xc nh mt phng (A'B'C').

Ta c: E e AB cz (ABC) => Ee(ABC)

E e A'B' c (A'B'C') => E e (A'B'C')

=> E l im chung ca (ABC) v (T')

Tng t, F v G l hai im chung ca (ABC) v (A'B'C')

Do E, FsG phi trn giao tuyn ca (ABC) v (A'B'C').

Vy E, F, G thng hng.

V d 1:Cho hnh chp S.ABCD c y l mt hnh bnh hnh. Trong mphng (ABCD) v ng thng d i qua D v khng song sng vi cnh ca hnh binh hnh. Gi CM mt im nm n cnh SB. Tthit din ca hnh chp b ct bi mt phng (d, C).

8



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




9/303

Gii:+ Gi s ng thng d ct cnh BC ti G v ct cnh. AB ko di ti H.

Ni HC ct cnh SA tai E. Thit din cn tm l t gic DGCE.+ Tong t , nu d ct cnh AB ti H v BC ko di ti G. Ni GC* cts c ti F. Thit din l t gic DHCF.

+.Nu ng thrig i qua im B thi ta c thit din l tam gic SBD.s

G. Gi E l giao im ca SA v HC, F giao im ca s c v GC\Thit in l t gic DECF.

V d 2: Cho hnh hp ABCD.ABCD5. I, K l hai im bt k thucAB\CC\ Xc nh thit din ca hinh hp ct bi mt phng (IKP) khi:a. p im di ng n AD.

b. p l im i ng ong t gic ABCD.Gii:

Dng tit in ca hinfa. hp ABCD-ABCD ct bi mt phng (TKP)- Trong mt phng ABBA, tr I kIE//BB ct AB ti E. Ta c IE//KC(cng song song vi BB) nn IE,KC xc nh mt phng IECK.Trong mt phng ECK do IK ctIC7/EC nn IK ct EC. Gi G lgiao im ca IK v EC.

9



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




10/303

- Trong mt phng ABCD: Do Gnm ngoi EC : ko di GP ct DC tiH, ctABtiM.

- Trong mt phig ABBA: Ko diMI ct AA ti"N, ct BB ti Q.

- Trng mt phng BCCB\- Ni QK

ct BC ti R. Ta c thit in l lcgic KHPNIR.Trng hp c bit:+ Khi p trng D th H trng D, ta c

thit din l ng gic KDNIR:+ Khi p trng A th M trng A, H

trng G (G l giao im ca GAvi DC) ta c thit din l ng gicKGAIR.

V d 3:Cho hnh lng tr tam gic ABC.DEF c AD//BE//CF. Gi H l trungim ca cnh DE. Tun giao tuyn d ca haimt phng (AEF) v ODBC).Xc nh thit din ca mt phng (H, ) vi lng tr ABC.DEF cho.

Gii:Gi 0 l tm ca hnh binh hnh ABED.

Gi O l tm cua hi bnh, hnh ADFC.Hai mt phng (AEF) v (DBC) c hai imchung l o v O. Vy giao tuyn cn tm

ca hai mt phng (AEF) v (DBC) l ngthng 0 0 . Mt phng, (H, d) chnh l mt

phng (HOO). Mt phng ny cha HO//BEv OOV/BC nn song song vi mt bn(BCFH). ng thng qua 0 song song viAD ln lt ct AC v DF ti M, N.

ng thng HO ct AB ti K. Vy thit din ct hnh lng tr bi mtphng (H, d) l hnh bnh hnh MNHK c cc cnh song song vi cc

canh ca mt bn (BCFE).Vi d 4: Qua inh Bi ca hnh lp phng ABCDAiBiCiDi, dng mtphng ct canh'BG, AB v to vi y ABCD mt gc bng a. Thitdin c to thnh mt tam gic cn. Tnh din teh thit din, nucnh ca hnh lp phng bng a.

10



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




11/303

B,Gii:

c. BiS #|\ V /J JI \ \

! V \ ,Ai

I >

/

c - -

D iV

b,

'C,

thnh l tam gic cn c cnh y l MN hoc BiM hoc BiN (trong haikh nng cui ch cn xt mt kh nng l ).Trng hp 1:Ta GMBj = NB1=> MB = BN, BD i- MN. Theo nh l

3 ng vng gc th BjE JLMN => BXEB = a .

_ SMBN _ B.ME_ BE2 _ a2arc tan2 a _ a2COSaMRN " * "

^ cosa cosa cosa cosa sin a

Trng hp 2: Ta c MBj = MN.

V AMBjB = AMBN, nn BBj = BN => N = c .

K BE_LMG,B1EB = a. t X= CM =>MB = Vx2- a2, BE = arctana.

T y c: Vx2- a .a = x.actan a; X= astna_..V-cos 2a

SMEtC=-M C.B1E = - ^ = . ^ = 2 2V~cos2a sin a 2V- cos2a

Vy din tch thit din ci tm l: a Csa hocsin ain2a 2v-cos2a

K dl:Cho im s nm ngoi mt phng (ABC). Gi 0 l mt im thayi nm trn (ABC). Tim qu tch cc trung im M ca on thngS0 \ s :

Gii:

+Phn thun :o l im c nh bt k trnmt phng (ABC) v gi M l trung imB ca on so. Khi im 0 thay i nm trn

mt phrg; (ABC) thi trung im M caon SO lun lun nm trn mt phng (a) qua im M v sng song vi mt phng(ABC) vi ta lun c MMV/OO .

11



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




12/303

+ Phn o:Gi O l mt im bt k thuc mt phng (a7) i qua truim M ca on s o v song song vi mt phng (ABC). ng thSM ct (ABC) ti O. V (a)//(ABC) nn ta c M M 7/0 0\ Ta suyM l trung im ca SO.

+ Kt lun: Qu tch, cc im M l mt phng (a) i qua im(trung im ca on s o vi o mt im Q nh bt ki tmp(ABC)) v song song vi mp(AB C) cho trc.

Vi d 2:Cho hnh chp S.ABCD c ABCD l t gic vi B v CD khsong song.M mt im di ng n SB. Mt phng (ADy) ct s c ti a. Xc nh giao im N.b. Tun tp hp giao im ca AM v DN.

Gii:a) Xc nh giao im N ca mt phng

(ADM) vSCGi o l giao im ca AC v BD.Ta chn mt phng cha s c l (SAC).Xt hai mt phng (SAC) v (ADM) c:A l im chung th nht. Trong mt ping (SBD),SO ct DM ti I. thy rng I l im chung th hai. Do AI l giao tuyn ca hai m

phng (SAC) v-(DM) . Trong mt phng (SAC), A ko i ct s cN. Ta c: N thuc s c ?N iuc AI m AI nm trong mt phng (ADVy N l giao im- cn tm.

b) Xc inh tp hp giao im ca AM v S Thun:Ta c K l giao im chu

ca hai mt phng (SAB) v (SD c nh, do K di ng trn g

\ tuyn ca chng:\ Gii hn: Khi M trng vi s ti

trng vi s. Khi M trng vi B: Ko l giao im ca AB v CD. V

1 khi M chy t s n B thi K chy

SnKo.o:Gi K l im bt k trn on SKo- Khi AK ct SB ti M v s c ti N. Lc K l giao im ca AM v DN.

s

12



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




13/303

Vi d SCho t din ABC. I, J ln ltl hai im c nh n AB, AC v ukhng song song vi BC. Mt phng(P) quay quanh IJ ct cc cnh CD, BDti M, N.

A

a. Chng minh MN lun i qua mtim c nh.

b. Tim tp hp giao im ca IN v JM.-' DGii:

a) Chng minh MN un i qua mt im c nh.Gi K giao im ca IJ v BC. Do u v BC c nh, suy ra K c nh.K thuc u nn K thuc mt phng (IJM), K thuc BC nn K thuc mt

Gii hn: Gi Mo thuc CD sao cho JM0//AD. Gi No l giao im camp(IJMo) v BD. Ta cng c IN0//JM0 vi nu khng ,giao im cachng s thuc giao tuyn A ca hai mt phng (ABD), (ADC). Suy raAD v JMo c im chung. iu ny mu thun vi cch chn Mo- KhiMo trng c th N trng B,do L trng A. Kh M chy trn Mo c th Nchy trn BNo, giao im L chy trn tia A. Khi M trng D th N trngD o L trng D. Khi M chy trn M(jD thi N chy trn No D, giao

im L chy trn tia Dx. Vy khi M chy n CD th L chy trn ngthng AD b i khong AD.

phng (BCD), do MN l giao tuyn ca

hai mt phng (IJM) v (BCD) nn Kthuc MN. Vy MN lun qua im cnh K.

B'b) Tun tp hp giao im ca IN v JM:Thun: Ta c giao im L ca IN vJM l im chung ca mt phng(ABD) v mp(ADC) nn L di ng ngiao tuyn AD ca hai mt phng ttn.

L

13



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




14/303

QUAN H SONG SONG

Kin thc c bn

phng- " THai ng thng gi Iacyo^nha^^ ,Hai ng thng, gi la son& song^iu- chng ng phang .va khng c

im'chung. v i ;./'2. Hai ng thng sng song -

Tnh cht 1:Trong khbng ^gi^ qua mt im nm ngoi mt ng

thng, c mt v chi mt ng ting song song vi q^g M ng Bo lTinh cht 2:Hai ixng thng p&ii: bit cng song song vi mt ng

thng th ba th song song vi nhu^ .

nh l(gi tuyn p]a ba nt pjhiang): J I. 2' f u :Nu ba m ^ phhg ^;mt .ct nha-theo ba gao tun jpHan biet thi fe

H qu:Nu hai mt phng cat nhau ln it diquahaij ng thrigsong song th giao tuyn ca chng song sng vi hai ng ,ang 4 (hoctrng vi mt trong ha ng thng ). ' !

Ac#' uc II uM iHun ' ri rrH/f

Bnh l 1:Nu rn^thng a khngsong vi mt ng thng iic nm trn mt phrig (P) tii .a; song srgvi mt phang (P). , '^ '- V.. / .

;



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




15/303

L g a U g V U ' l i V ^ V JL* S ; U i O l g ; l ^ U U ; : IX V JL J 3 ; JJU W i i i a g . . ; , ; " .

H q '2: p h l n | j ^ p S ^ ^ D n g ^ ^hi giao tuyn c chung cng son^sng vi cmg iiig ;

nh ngha:Hai .mt phng gi l song song nu chng khng c imchung. " v 2. iu kin hai mt^harig srg sng

nh M l: Neil mtîng {PXola hi ng thng a, b ctnhau^

. Tnh c h t \ ^Ti^^cM/^Câttîem iram ngimt int phang, c; mt v ch .xQt imthrig song song vi rimtphang .V .



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




16/303

16



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




17/303


18/303

Cch 3:S dng tnh cht y giao tuyn c ba mt phng-.2. Chng minh rng thng sng song vm tphng.

Phng php chung: chng minh ng thng song song vi mtphng, ta thng s dng cc phng php sau:

Cch 1:S dng nh nghat e m p ( P )

VCch 2:S ng ph l : :


19/303

b. Mp(MNQP)// mp(CDFE)

c. Xc nh gc ca AB v DF. Khi AB vung gc vi DF th MNQP lhnh g?

Gii:Hai hnh bnh hnh ABCD v ABEF cho:

AD//BC v AF//BE.M AD v AF c (ADF)

BC v BE c (BCE)

Vy (ADF)//(BCE).

Tac M P//AB//CD,nn = S -AD AC

Do : ^ = M PQ//DFAD AF

M MP v PQ c (MNQP). CD v DF c (CDFE),Vy mp(MNQP) // mp(CDFE)

Ta c MP// AB v PQ//DF. Vy gc ca AB v DF l MPQ.

Ta c MP//NQ//AB, nn MNQP l hnh thang.

Khi AB1 DF th M P lP Q o MPQ = 90

Suy ra PQN - 90 v M P//NQ.

Vy MNQP l hnh thang vung ti p v Q. d 3:Cho hnh lp phng ABCD.A'B'CD'.

a. Chng minh hai mt phng ( AB'D') v (C'BD) song song.

b. G E, F, G ln lt l trung im ca AA', BB', c . Chng minh ba

mt phng (ABCD), (EFG), (A'B'C'D') song song.

c. Gi I, J, K ln lt l trung im ca AB, AD, T)'. Chng minh, haimt phng (IJK) v (BDD'B') song song-

d. Gi I, H ln lt l trung im ca B'C' v DD'. Chng minh hai mt

phng (ILH) v (C'BD) song song.Gii:

Ta c AB// = C'D' => ABCT)' l hnh binh hnhlDo AV/ BC(1)Ta c BB7/=DD' => BDD'B' lahnh bnh hnh. Do BX)' // BD (2)

19



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




20/303

T(l) v (2) ta c (AB'D')//(BD),

b) Ta c EF v FG ln lt l cc ng trung bnh, ca hai hnh vuABB'A' v BCA'. Do EF//AB//A^f v FG//BC//B'C' ;Vy mp(ABCD) // mp(EFG)// mp(A'B'C'D')

c) Ta c IJ v JK ln lt l cc ng trung bnh ca tam gic ABDhnh vung ADD'A'. Do IJ//BD v JK//DD'. -Vy mp(IJK) // mp(BDD'B')

d) Gi M l trng im C'D'. Ta c ML v IJ ln lt l hai ng trbinh ca tam gicB'C'D'v ABD.Do ML//B'D' v IJ//BD (3). M BD//BT)', nn ML//IJ v xnh mt phng IJML. Ta c MH ng ung bnh ca tam gDC'D',nnMH//C'D (4).

Ta c JL l ng trung bnh ca hnh bnh hnh ADB'(v AD// = B ' ) .Nn D/JL => MH//JL => He(IJML)

T (3) v (4) ta c (IJML)//(C'BD) hay (ILH)//(C'BD).

V d 4:Cho hnh lrig tr ABCjVB'C' Gi I v r ln lt trung im BC v B'C'; M, N, p ln lt l tm ca cc mtbn ABB'A', BCB' CAA'C'.a. Tm giao im ca IA 'v mt phng (B'C') *

b. Tm giao tuyn ca cc mt phng(AFC) v (BA') ; mt ph.ng (AB

v (ABC). . c. Chng minh ba mt phng (ABC); (MNP) v (ATB') song song

a) Hai ng cho ca hnh binh hnh AII'A' ct nhau ti o.Ta c 0 e AI' c mp(AB'C') v O eA l.

Vy 0 l giao im ca IA' v mt phng (AB')

b) Ta CC' (AB') v e (BA').

M l giao im ca AB' v BA'. M AB' c (AB'C') .=> Memp(AB'C') v Me(BA'C') c'

Vy M l giao tuyn ca mp(AB'> A

nhau.Gii;

v mp(BA')-p l giao im ca AC' v A'C.M AC' c mp(AB'C') v

A'C


21/303

=> PempCAB'C') vmpCABC).

Theotrn M mp(AB'C') v M A'Bcmp(A'BC) => Memp(A'BC).

Vy MP l giao tuyn ca mp (AB') v mp (A'BC).

c) Ta c MP v MN ln lt l ng trung bnh ca hai tam gic (A'BC)

v B'AC. Do MP//BC v MN // AC. Vy mp (ABC) // mp(MNP).

V d 5:Cho hnh hp ABOD-AB'C'D'.a. Chng minh hai mt phng (BDA') v (BT)'C) song song.

b. Chng minh ng cho AC' i qua trng tm Gx v G2 ca hai tamgicBDA' v BTD'C.

c. Chng minh hai im G1 v G2chia ng cho AC' thnh ba on

bng nhau.

Gii:a) Ta c BB' // = D => BDD'B' l hnh bnh hnh => BD // BT)'V BC// = B'C7/ = AT)' ^ BCD'A' l hnh binh hnh.=> BA'//CD'. Vy mp(BDA')//mp(BT)'C).

b) Tac AA7/ = CC'=> ACA' l hnh bnh hnh.=> AC f = A'C

C' Gi o v O' n lt l tm ca hai yABCD v A'B'C'D'.

Ta c: AO = = = 0'C'.c : ,2 2

Trong- mt phang ACC'A', A'o v c o ' ctA B AC' ti Gj v G2.

AO// A'C' o Q - = = => ! trng tm tam gic BDA'.GtA GtC AC 2

O'C lAC ^ = -=> G2 l trng tm tam gc BT)'C.G2C G2A AC 2

Vy AC' i qua ng tm v G2ca hai tam gic BDA' v B'D'C.c) Ta c 0C// = 0'A' OCA' l hmhbnh hnh => OA'//C0'.

Trong tam gic ACG2, ta c OGxCG2 v OGx qua trung im o caAC, nn Gi l trung im ca AG2 => AGL= G .^2 .Tng t: C'G2= . Do : AGj = GtG2 =G2C' .Vy G:> G2 chia on AC' thnh ba on bng nhau.

21



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




22/303

Bi tpvn dng

1. Cho t in u ABCD c cnh bng m. Gi I l trung m BC, a v bln lt l hai ng thng ln lt nm trong hai mt phng (ABC) v

(BCD) v cng vung gc vi BC.

a. Xc nh gc ca ng tfiang a v ng thng BD. Tnh gc .

b. Xc nh gc ca ng thng a v ng thng b. Tnh gc . .2. Cho hnh chp S.ABCD c y ABCD l hnh thang, vung ti A v Dvi AD = CD v AB = 2 CD.a. Tim giao tuyn ca hai mt phng (SAB) v (SCD).

b. Gi E l trung im AB. Tm giao tuyn ca cc cp mt phng(SAD) v (SCE); (SDE) v (SBC).

3. Cho t din ABCD c AB = AC = 2a, CD = a . Gc ca AB y CD bng30. Gi M l mt im trn cnh AC sao cho AM = x; (P) l mt

phng qua M v songsong vi AB vCD, ct BC, B v AD ln lt tiN,E, F.a. Thit din MNEF l hnh g?

b. Tih din tch s ca thit in theo a v X.c. nh v tr ca M trn AC s ln nht.d. nh v tr ca M trn AC s bng in tch ca mt hnh vung,

cnh m. .... .4. Cho hnh vung ABCD cnh a. Gi s l im ngoi mt phng ABC

sao cho SA = SB = s c = SD = a-v/3 . Gi E v F ln lt l trung mca SA v SB; M l trung im trn cnh BC. t BM - X. Mt phng(MEF) ct AD ti N.

a. Chng minh MNEF l hnh thang cn.

b. Tnh COS CBS v FM.c. Tnh din tch ca MNEF.

5. Cho t din ABCD vi BCD l tam gic u canh a, trng tm ; AOvung gc vi CD v AO = 2a. Gi M l mt im n ng cao BH

ca tam gic BCD.a. Kho st thit din ca (p) v t din ABCD.b. t BM = X. Tnh din tch ca thit in.c. Tnh X thit din c din tch ln nht. Tnh gi tr ln nht .

22



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




23/303

y ca mt hnh ng tr ng ABCDA1B1C1D1 hnh thoi ABCD cgc A = 60. Tt c cc cnh ca hnh lng tr u bng 1. F trung imca DC, M nm n ng thng A]F. Hy xc uL v tr M sao cho tngdin tch ca cc tam gic MBB], MCCi cc tiu v tnh tngBy.

. Hnh chp SABCD c cnh bn SA = 5 ng thi l ng cao ca hnh,chp; y ABCD l hnh ch nht c cc cnh AB = 2,AD = 4. Qu

cnh bn di nht v mt im nm trn cnh y, ng mt phng cthnh chp sao cho thit din to thnh c chu vi b nht. Hy tnh intch ca thit din .

. Trong hnh chp tam gic SABC, tt c cc cnh u c di bng 1.

Trn cnh SA y M sao cho SM = , n SB ly m N; ly im p

trn mt phng (ABC). Hy tm gi tr b nht ca tng cc di onthng MN v NP.

Hung n v p S

. a) Tm gic ABC u nn trung tuyn AI cng l ng cao.

AI-LBC v AI = I .2

Ta c a _LBC v nm ong mp(ABC), nn AI/ /a .

Gi J l trung im CD. Ta c IJ l ngtrung binh ca tam gic u BCD, nn:

TT.. B D _ TT_ mIJ//= => IJ = 2 2

Vy gc ca a v BD l gc AIJ - a c

AJ l trung tuyn ca tam gic u ACD, nn AJ = = AI

Do tam gic AIJ cn ti A, nn ng cao AH cng l trung tuyn.

Ta c: IH = HJ = = . Tam gic vung AHI cho ta:

IH . V : :co sa = = a = arccos-.AI 6 6 -

) Ta c ID l ng cao ca tam gic u BCD, nn D BC v ID =

Ta c b XBC v nm ong mp(BCD), nn ID// b .

Vy gc ca a v b lgcAID = 3

23



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




24/303

Tam gic AID cho AD2 = AI2 +DI2 - 2AI. D I . cos{3


25/303

c) Ta c s ln nht khi x(2a - x) ln nht.M tng hai tha s l X + (2a - x) = 2a khng i.

' _ __ ACVy s lnnht khi X = 2a - X X= a AM = ~

Suy ra s ln nht khi M l trung im AC.d) Ta c s = m2 X2 -2ax + 4m2=0 (1)V M trn AC, nn 0 < X < 2a .

Ta c: A'> 0 -4m 2+ a2> 0 - 0 o VmeR; -0 = a >0

Af (2a) = 4m2>J3 Vme R; - 2a = -a < 0z

V m l cnh ca hnh vung nn m > 0.

Vy 0 < m < : Phng trnh (1) c 2 nghim 0 < < x2< 2a2

Do c 2 im Mj, M2n AC.

m = : Phtrong trnh (1) c nghim kp 0 < X = x2= a < 2 a .2

Do c mt im M l trung im ca AC.AB a

4. a) Ta c EF l ng trung binh ca tam gic SAB, nn EF // = - = ^

Mt phng (MEF) 3 -EF , nn (MEF)// AB = (SAB) n (ABCD).Do (MEF) ct hai mt phng(SAB) v ^

(ABCD) theo giao tuyn:MN // F // AB -Nn MNEF hnh thang. E / , Yp. V V

Ta c: SA = SB = s c = SD = aV v c

AD = BC = a nn hai tam gic SAB v SBC / r " " 1 /y *1 \ / Mbng nhau. A B

Do A = B = a . V M N//= AB nn AN = BM = X.0 4 a / _ _ _ '

Ta li c AE =BF = - . Vy haitam gic ANE v BMF bng2 2

nhau. Do NE = M F. Vy MNEF l hnh thang cn.b) V dng cao SI ca tam gic SBC cn ti s.

Ta c SI l trung tuyn. Do BI = = .

2 2

25



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




26/303

Tam gic vung BSI cho: COSCBS = COS CBS .SI 6

Tam gic BMF cho: FM2= BM2+ BF2- 2BM.BF. COSCBS

2 3a. _ l-JZ y/s _ 1 r 2 Q 2-X + 2x- o FM = V4x - 2ax + 3a4 2 6 2

c) V EH v FK vung gc vi MN

Ta c: NH = KM = (MN - HK) = .

Tam gic vung MKF cho: FK2= FM2- KM2= -(4 x 2 2ax + 3a2) -16

o FK = V l6x2- 8ax +1 la2

Vy din tch ca MNEF i: s =(MN + EF)FK = Vl6x2 -8ax + l l a 22 16

5. a) * Vi: 0 < X EF//CD V MG = (P) n (ABH) B

=> MG//OA. V OA CD, nn MG-LEF.

T a c ~ ~ = ^ = l M l trung im ca EF.

Tam gic EFG c MG l trung tuyn cng l g cao.Vy EFG l tam gic cn ti G.

* X7,. a>/3 _ a>/3* Vi _


27/303

Ta c EF//CD o = EF = ^CD BH 3

Ta c MG//OA = MG - 2x^3OA BO

Vy din tch tam gic EFG l: s =EF.MG = 2x2

Hnh thang can EFGK / - 3a)(3a -2xV&) = (4x73 3a)(6a - 4xj3)3 3

Vi < X< :Ta c (4xjs - 3) + (6a - 4xV3) = 3a3 - 2 .....

Vy s ln nht (4x\/3 - 3a)(6a - 4x^3) ln nht.

Do 4xs/3 - 3a = 6a - 4x-s/s o X = 3aV8

. 02 -

Gi tri s ln nht: s = .4

27



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




28/303

Gi s: MFC, ML ln lt l cc ng cao ca tam gic MBBi, MCMi hiih chiu ca M xung mt phng ABG. M'l'B = MK, MiC = M

+ 8^ - b b1 .mib + | c c 1.m1c = ( m1b +mic)

Tng MjB + MjC nhn gi tr cc tiu nu Mi l giao im ca ng thng AF v BE trong E l im 'i xng ca c qua thng A F. Ta k N _LAF; khi

TW 2SAD7 _ AD.DF.sinl20 sAP VaD2+ DF2- 2AJD.DF.COS 120 ~ 2V

CE = 2CH = 2N = ~ .... ;. ... .

MjB + MC = BE = v'BC2-f CE2- 2BC.CE. cos(60 +


29/303

Do ASjAK' - ASjBC, nn ta C:

s *Ta li c: SSKC= ; trong gc gia mt phang SKC v mt

costp

phng y. K AL _L CK. Khi ALS = cp, tan

MiPi, du1 4 1 1

\ V

, . . , l f 2 Vy gi tr b nht ca tng cc di on thng MN vNP l: 1+ 7yj

29



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




30/303

DNG2. QUAN H VUNG C C

Kin thc c bn

- nh ngti 1:

ng thrig A- v; &.xim ^q^^tM ra :v n |i ^ sa&|ra^,ioc^

2. H ng hhg: uhg g c

gia chngbng >0

L '.ritingha ng thng iwong gc miphig V : inh ngha 1: Mt ng "thng gi .l vunggc vi mt mt phng

Jiu ri vung gc vi mi ng thng nmongmtphng.. ^ :y .':

nh t 1: Neu ng thng d vung gt : .vi hai ng thng ct nhau b cng nam : . w /ong mt phng (P) th ng thng d vung Xgc vi mt phng (P), ~ I2. Cc tinh cht

a *Tnh cht 1: c dvcy vktmt mt phng (P) i qua

m dieni ^cha^ V mt cmg; / ; ;;S ' ' 'ng thng A i

30



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




31/303


32/303

. px/rf/^2.-:TK'-Sr. ;dii" tci' caf^ron^]vpDg;,E^ijv'S in tch hinh chu H ,ca H trn mt phng (P -) th S';= S.CS9. trng:


33/303

III

j'ic',."r'-i-.-Vi>'*'

V&

'

nh ngha 3 ;.

Hnh lng tr ngLr Mnh ng . c cnhbn vng gc vi mt. y.

' r t ' r-v

\ ^ ' K

ffi/3 \ L d^ S . '

':

&.MHnh lng tr u

. :L-ffink:g\4yng: c::M&QM

;"'A3:1\

_...; ...;. : f*l

|0w!'tg.

Vh;: ZN.,.1.';(P As''!.-'.;'':~.i .:*Y>:

i i S

1 P P X " >

i

Ki;*\i

| #

Hnh hp ngL hnh lng tr ng c

: yM hnh b n h a

l l l i l

* *\/ 1 \

v / A \ < ''\ s ' y r ' ' * r\ ,*' s*V ';

;: ri^

&^ m iM s S S

1B1I1I1P'tS-n.SUs


34/303

. Hnh lp ptawng:. : / i /L m 0^chM ; c t:c cc cm hbm gnhm ^.x' J .

* 111 -'. 7

nh ngina^ Mot hinhri Vchp c gi l hnti ' chp .u nu y c ^ r l a --gic u v cc cnh bn:bng nhau.

ng thng vung gc^vi mt y k t. nh gi ng cao ca hih chp. ....-

inh ngha 5:Khi ct hnh chop u bi mt/? !\\ mt phng song song. y y c mt hnhchop ct th hnh chop ct c gi l hnh

A'TT5 t r ' chp ct u.on ni tm ca hai y c gi l ng

cao ca hiih chp ct u.

A" ' B ^IV. Khong cch. Khong cch t mt im n mt mt phang, n mt ng thng

nh ngha lKhong cch t im M n mt phng (P) (hoc nng thng A ) l khong cch gia hai im M v Hs trong H l hinhchiu ca M trn mt phng (p) (hoc trn ng thng A).

M . M

2.

Khong cch t im M n mt phng (P) c, k hiu l d(M;(P.

Khong cch tr m M n ng thng c k hiu l d(M;(A)>Khong ck gia r thngV mt phngsong song, gia hai mt phng song songnh ngha 2:' Khong- cch gia hai ng

thng a v mt phng (P) song; sng vi l khongcch t mt im no c^ en mp(P).

34



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




35/303


36/303

V d 1:Cho t din BCD c AB = 6cm, CD = 8cm - Gi I, J, K ln ll trung im cc cnh BC, AC v BD. Cho bit JK = 5cm - Chng mrng AB _LCD v IJ X CD.

Gii:

tam gic ABC v BCD.

Ta c: IJ //AB;IJ = = 3 cm;2

IK//CD;IK = = 4cm; JK = 5 cm; 2

Do : IJ 2+ IK2= 9 + 16 = 25 = JK2=> AUK vung gc ti I.Suy ra JIK = so ';I JIIBA, IK//CD. Nn (ScD) - JIK = 90\

Vy AB C.Ta c: IJ/ /A B . Vy u 1 CD.

36



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




37/303

p

b) BD_LSC .

c) V AH JLSD. Chng minh AHc = 90 .s Gii:

y /l^ s . a) Ta c: SA J_ mp(AE

/ AD l hnh chiu c

ADJ^CD => C D li s o 'O v V ^ ' (nhl ba ng w

c Tng t, ta cng cT A C A 1 ,* . / A T P T V l

Vi d 2 Cho hnh vung ABCD. Trn ng thng vung gc vi mtphng ca hnh vung ti A ly im s. Chng minh rng:a) C D lSD v B C1SB.b) BD1SC-

c) V AH JLSD. Chng minh AHC = 90s Gii:

a) Ta c: SA _Lmp(ABCD)

AD l hnh chiu ca SD trn mt hng (ABCD)

SDvung gc).

B ^ Tng t, ta cng c: BC-LSBb) Ta c SA 1 mp(ABCD)

AC hnh chiu ca s c trn mt phng (ABCD), AC _LBD (ung cho

vung gc).=> BD.LSC (nh l ba ng vung gc).Ta c CDX AD v CD _LSD i CD 1 (SAD)

DH = ch CH/(SAD) v D HlA Hc(SA D)

=> CH-LAH (nh l ba ng vunggc). Vy AHC = 90 .Vi d 3:Cho t din SABC sao cho SA = BC = a, SB = CA = b, s c = AB = c .

Trong mt phng ca AABC dng AMNK sao cho MN, NK, MK nhn

A, B, c lm trung im.a) Chng minh rng: SM, SN, SK i mt vung gc nhau.

b) Tnh VSABC theo a, b, c.

(i hc Thy sn 1998)Gii:

a) Ta c: SA = BC (gt); MN = 2BC (Tnh cht ng trung bnh)

c)

Do SA - MN2

ASMN vung ti s => SM JLSN

Chng minh trig t SM -1 SK, SN _LSK .b) Theo cu a) t gic SMNK t din vung ti s.V SABC v SMNK l 2 hnh chp c chung

ng cao k t s v s ^ c = .4

N n V ^ = V SMNK= S M . SN. SK

37



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




38/303

Ta c:

SM2+ SN2= MN2= 4a2SM2+ SK2= MK2= 4c2SK2+ SN2= NK2= 4b2

|SM2= 2(a2+c2- b 2)SN2= 2(a2+ b2- c2)SK2= 2(b2+ c2- a2)

=> SM . SN. SK =yj8(a2 +c2- b2)(a2+ b2- c2)(b2+ c2- a2)

= > V SABC = + 2 " + 1)2 c 2 ^ b 2 + 2 a 2 )

F/ / 4:Cho hnh chp SABCD c cnh sA = X v cc cnh cn li ubng 1 .a) Chng minh rng SA s c .

b) Tnh th tch hnh chp. Xc nh X bi ton c ngha.(Phn vin Bo ch v Tuyn truyn 1998)

Giia) Ta c ABCD l hnh thoi cnh 1 => AC = BD

SO-LBD.Xt 2 ASOD v AOBC c:

= 90V.SD = BC = 1, OB=ONn ASOD = ACOB

=> SO = OC = 2

ASAC vung ti s-----------j-------------c => SA 1 s c .

b) V SB = SC = SD nn s nm trn trc ca ABCD=> hnh chiu vung gc ca s xung mp(ABCD) nm n c o .

SA . SC X

ASBD cn tai sS

V SABCD = 3 S H S A BC D- ASAC vung =

=AC . BD vi AC = V ?"+ 5

BD = 2BO = 2n/bC2 - o c 2 = V - X2

SH =AC' x*+l

JABCD

Do V,SABCD3 V: , 7 * 9 - ^x"+l 2

2 + 1 . V 3 - X 2 =

Bi ton c ngha o 0 < X < .Vi d 5:Ch hnh lp phng ABCD.ABC D c cnh bng 1. Cc im

M, -N thay i ln lt trn cc on AB5v BD sao choAM = BN = a (0 < a < y2). Chn h trc Oxyz sao cho o trng vi ,nh B (1 ; 0; 0); D(0; 1 ; 0) vA(0; 0; 1).

38



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




39/303

a) Vit phng trnh ng thng MN. Tm a MN vung gc vi ABvBD.

b) Xc nh a n MN c di nh nht.Gii:

Vit phng trnh ng thng MNChn h trc Oxyz nh sau: Trc Ox i qua AB; Trc Oy i-qua AD

Trc Oz i qua AA; Khi : A(0; 0; 0); B(l; 0; 0); D(0; 1; 0); (0; 0; 1),

AM = a => M

BN = a=> N

a V2 a / i N;

2 ; 2 J\ _ * . _ ;0

2 2

2 2=?MN =

Suy ra phng trnh ng thng MN c dng:3^2 aV2

x 2_ _ y = z 21 - a.\2 a 2

~ Tm a MN vung gc vi AB v BD

Ta c: B = (l;0;l);D = (-l;l;0)rM N = l - aVr

theo gi thit MN vung gc vi AB v B khi v ch khi:

AB.MN = 0

BD.MN = 0--- = 1o a = 0.

, [; Siy2 A 2 3- l + aV2 n - = 02

Xc nh a on MN ngn nht

Ta c: MN= l - W2)2+Y + Y =s/? -2aV2 +1 M N 2=3a2-2a>/2 +1

- L* 1L' 2V2 >/2 1 'ngn nht khi a = = - => min(MN) = .

Vy minMN = khia = 3 3

39



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




40/303

V d 6:Cho hnh hp ch nht ABCD.ABCD c 3 kch thc AB = AD = b, AA = c.vi 0 < a < b < c. Gi I v J ln lt l trung im c

cnh AB v CD v cc im M, N tho mn iu kin: AM = kAD

BN = kBB1 vi 0 < k < 1 (1).

a) Tnh khong cch t A n mt phng (ABD).b) Chng minh rng vi mi gi tri ca k tho mn iu kin (1), bn i

I, M, J, N thuc mt mt phng. Tm cc gi tr ca k MNi- u.

a) Ti khong cch t A n (ABD)Chn h trc Oxyz; Trc Ox i qua AB; Trc Oy i qua ATrc Oz i qua AA\ Khi : A(0; 0; 0); B(a; 0; 0); C(a; ; 0); D(0; a; 0C(a; b; c ) ; D(0; b; c).

Phong trnh, mt phng (ABD): + 7 + - = l o bcx + acy + abz - abc =* ri' r

kIJ = IM + IN => IJ =IM +IN => IJ,IM,IN ng phngk kVy: bn im I, J, M, N cng nm trn mt mt phngTm k MN-LIJ:Ta c: MN = (0; - kb; kc);J = (0; b; e) => MN _LIJ

MN.IJ = 0 o - k b2+ kc2= 0 k(c2- b2) = 0 o k = 0(vb c)

Vy k = 0 th MN X u.

Gii.

b 2 . VbV + a V + a2b2

abc

b) Ghng minh rang: I, M, J, N cng thuc m0' mt phng:

40



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




41/303

Vi d 7:Cho hnh lp phng ABCD.ABCD7c cnh bng a.a) Xc nh ng vung gc chung ca cnh A

BB v ng cho AC.b) Mt mt phng (P) i qua irng cho AC5 B

cat BB5v DD ln lt ti M v N. Thit

in AMCN l hnh g? Hy xc nh mtphng (P) thit din AMCN c dintch nh nht. Tnh din tch .

Gii: B'a) Xc nh ng vung gc chung ca BB v AC

Ta c: BB7/CC => BBY/(ACCA) ; AC v BD ct nhau ti O

Ta c: BD _L(ACCAO. Gi I l trung im ca BB\ J l trung im

AC OJ//A A ; OJ = ;B I//AA ; BI = 2 2

=>U//B'0-; M B'O'-L(ACC'A') =>IJ ^(ACCAO^U JLAC'Vy IJ l on vung gpUig ca BB v AC v

B'D' _ aV2 2

IJ = BO =

b) T gic AMCN l hnh g?Ta c: (AASBB)//(CCDD) v (BBCC)//(AADD)Mt phng (P) ct 2 mt phng (AABB) v (CCDD) theo 2 giao

tuyn song song => AM C N .Mt phng (P) ct 2 mt phng (BBCC) v (CCDD) theo 2 giao tuynsong song => CM // ANVy t gic AMCN l hnh bnh hnh

=> SAMCN = 2Sa mc -= 2 . - .AC.MH = aT-MH(MH J_ AC) . Suy ra

Samcn nht khi MH nh nhto MH l on vung gc chung caBB v AC o MN = J M= I v H = J mp (P) - mp(AIC) vi I

l trng im BB. Khi : min = aV,IJ = aV3.^-^- - a ^ .

V d 1:T din OABC c cc cnh OA, OB, oc i mt vung gc vinhau. Gi H l im thuc mt phng (ABC) sao cho OH vung gc vimt phng (ABC). Chng minh rng:a) BC vung gc vi (OAH).

41



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




42/303

b) H l trc tm ca AABCX X _ 1 1 1c) ----T ---- + __ + n

OH OA2 OB o c 2Gii:

a) OH 1 mp(ABC), BC c mp(ABC) => BC JLOH (1).OA _LOB v OA _Lo c o

nn OA mp(OBC) ()M BCcmp(OBC) => BCIOA (2)

(1), (2) => BClmp(OAH).b) BC (OAH) m AH c mp(OAH)

=> BC1A H (3)Tng t ta cng chng minh c:AC _Lmp(OBH) m BH c mp(OBH) B

=> AC _!_BH (4). T (3), (4) => Hl trc tm ca AABC.

c) Gi A' vung OAA' vi OH l ng cao ta c:1 = 1 1 .. .

OH2 OA2 + OA'2 'Ta c BClmp(OAH) => BC X OA'c mp(OAH)Trong tam gic vung OBC vi OA' l ng cao ta c:

_ J _ = J L _ J _ (toOA'2 OB2 + OC2

Thay gi tr ca (6) vo (5) ta c: ~

OA'2 w w _ OH2 OA2 OB oe2V d 2\Cho hnh lp phng ABCDA'B'C'D'. Chng minh:) AC' 1 mp(A'BD) b) BD' 1 mp(B'CA)c)CA'mp(B) d) DB' 1 mp(D'AC)e) Gi I, J, K, L ln lt l trung im ca AA\ BB\ c c \ DD'. Chng

minh IK v JL ln lt l trc ca cc ng trn (BDD'B') v(ACCA).

Gii:

a) Gi a l canh ca hnh lp phng, ta c:AA' = AB = AD = aC'A' = C'B = C'D = a.y/2 (ng cho hnh vung) :.

Do AC' l trc ca ng trn (BDA').

Vy AC' -L mp(A'BD)

42



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




43/303

C'

,/^S-

/9 '>

/

n

V

b) Tng t ta cng c.BD' i rc ca ng trn (CAB').

Vy BD'-Lmp(CAB')

c) CA'l trc ca ng trn (BDC')-Vy CA' _Lmp(BBC'). .

d) BD' l trc ca ng trn (ACD')- Vy DB' mp(ACD').e) Bn tam gic vung IAB, IAD, LVB', AT)' c:

XA= IA' = = - v AB = AD = AB' =vAT>' = a .2 2

Nn chng bng nhau.Do : IB = D = IB '= ID '.Tcmgt: KB = KD = KB' = KD'Vy IK l trc ca ng trn (BDD'B').

Chng minh tng t, ta cng c JL c ca ng trn (ACC'A').

Vi d 3:Cho tam gic u ABC cnh a. Trn ng thng d vung gc vimt phrxg (ABC) ti A ly im M. Gi H l trc tm ca tam gicABC, K l trc tm tam gic BCM.a) Chng minh rng: MC_L mp(BHK) v HK_!_ mp(BCM)

b) Khi M thay i n d, tm gi tr ln nht ca din tch t gic KABC.Gii:

a) Chng minh rng: MC1 mp(BHK) v HK X mp(BCM)Ta c: BH 1 AC; BH -LMA v MA -Lmp(ABC)

BH -Lmp(MAC) => BH JlMC; M BK _LMC (ng cao)=>MC-L mp(BHK) => MC1 HK. Mt khc: C JLAI v BC J.MA

=> BC -Lmp(m a i) BC _LHK => HK -i. mp(BCM) (pcm). -

b) Tm gi tri ln nht ca din tch t gic K.ABCV KO1 I, BC1 mp(MAI) => BC KO => KO -L mp(ABC)

VK = -S abc.KO => Vln nht khi KO ln nht

c: HK -Lmp(BCM) HK1 KAHK1 vung ti K, gi J l trung im ca H

-_ . . . HI 11 AT_1 aV3=> KJ = -r- = -r-AI =2 2 3 6 2 12

M KO < KJ = (ng vung gc v ngxin).12



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




44/303

Vi d 4:Cho hii lp phng ABCD.ABCD c cnh bng 1. Trn ccnh BB\ CD, AD ln lt ly cc im M, N, p sao choBM = CN = D P = a (0 < a < 1). .a) Chng minh rng: MN = -a AB + AD + (a - 1)AA .

b)*Chng minh rng: AC _Lmp(MNP).Gii:

a) Chng minh: MN = -aAB + AD + (a - 1)AA'

Chn h trc Oxyz nh sau: Gc 0 = ATrc Ox i qua AB; Trc Oy qua AD; Trc Oz i qua AAKhi : A(0; 0; 0); B(l; 0; 0) ; D(0; 1 ; 0) , A(0; 0; 1), (l;1:1)=> M(l; 0; 1- a), N(1- a; 1 ; 0), P( 0; 1- a; 1) .=> MN = (-a; 1; a - 1);B = (1; 0; );D = (0; 1; 0);V= (0; 0; 1)

=> ~aAB + AD + (a - 1)AA' = (-341; a -1)

=> MN = -aAB + AD + (a - 1)AA' (pcm).b) Chng minh: AC1 mp(MNP)

Ta c: MN = (-a; 1; a -1) = 0;C = (1; 1;1); MP = ( - 1; 1 -a ; a)

AC\MN = -a + l + a - l = 0 [AC'MN nnLm\.Vy: AC -J_ (MNP)

aC .MP = 1 +1 - a + a = 0 lAClM P

Vi d 5:Cho hnh chp S.ABC c ABC l tam gic u cnh a v

SA-Lmp(ABC). t SA = h.a) Tnh khong cch t A n mt phng (SBC) theo v h.

b) Gi I l tm ng trn ngoi tip tam gic ABC v H l trc tm tgic SBC. Chng minh rng: IH -Lmp(SBC).

Gii:-a) Tnh khong cch t A n mt phng (SBC)

Chn h trc Oxz nh sau:Gc trng vi trc x vung gc ACa ^ . a . ntrc Oy i qua AC, trc Oz i qua SA. Khi : A(0; 0; 0);B

C(0; a; 0) ; S(0; 0; h); SB = ;-h ;SC = (0;a;-h)V 1 ) ,

=> Pp vect ca mt phng (SBC):

= [ s ^ ] = y hay(li;W3;"a73)

=5>Phomg trinh mp(SBC): hx +h-jsy +a%/3(z - la)= 0

44

2 2



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




45/303

v/b2 + 3h2 + 3a2 V412+ 3a

b) Chng minh rng: IH-lmp(SBC)Ta c: AABC u Tm I cng l trc tm ca tam gic ABC=>ICAB; MIC 1 SA =>IC1 mp(SAB) => IC1SB.M CH _LSB => SB X mp(ICH) => SB JLIH (1). Ta c: BC _LAM=> BC1 mp(SAM) => BC _LIH (2). T (1) v (2) IH J_ mp(SBC) (pcm).

o x + li>/3y + aVz - ah>/3 = 0

V d I: Cho hnh ch nht ABCD c AB = a, BC = 2a . Trn ng thng

vung gc vi mt phng (ABCD) ti tm o ca hnh ch nht ly im s3a ' sao cho SA = -r- Gi M v N ln lt l trung im ca AB v CD.2

a) Chng minh mt phng (SMN) vung gc vi hai mt phng (SAB)v mt phng (SCD).

b) Chng minh rng mp(SAB) v mp(SCD) vung gc vi nhau.Gii:

a) Ta c OA = OB = (ACD l hnh ch nht)

=> SA = SB (on vung gc v on xin) => Tam gic SAB cn ti s,nn trung tuyn SM cng l ng cao: SM -L AB.Ta c MN l ng trung bnh ca hinh ch nht ABCD, nn:MN//BC = 2a . M BC1 AB, nn M N lA B .Do : mp(SMN) J_ AB; AB c mp(SAB)

Vy mp(SMN) _Lmp(SAB). / ; y V \Tng t mp(SMN) _LCD; CD c mp(SCD) //, c

Vy mp MSN l gc phng ca nh dinto bi hai mt phng (SAB) v (SGD). Tam gic vung SAM c:

SM2 = SA2 - AM2 = - = 2a2. Tng t: SN2 = SM2 = 2a2.4 4

Ta li c MN2= 4a2. Do MN2= SM2+ SN2

45



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




46/303

=>Tam gic SMN vung ti s MSN = 90 .Vy mp(SAB) 1 mp(SCD).

Cch khc:Tam gic SMN vung ti s SM SNTheo trn SM -L AB//CD SM-LCDDo rap(SCD) _LSM;SM c mp(SAB). Vy mp(SCD) X mp(SAB).

Vi d 2:Cho tam gic u SAD v hnh vung ABCD cnh a nm trong 2mt phang vung gc nhau. Gi I l trung im ca AD, M l trung imca AB, F l trung im ca SB.a) Chng minh rng mt phng (CMF) Xmp(SIB).b) Tnh khong cch gia 2 ng thng AB v SD, gia CM v SA.

=> Php vect n2= [S, IB] = (-2; 1; 0)

=> = 2v3 - 2V3+ 0 = 0 = > a iln 2.Vy: mp (GMF)-Lmp(SIB) (pcm).

b) Tih khong cch gia AB v SD l khong cch t A n mt phng (SGD)Ta c: => AB // CD => AB//mp(SCD) cha SD:=> khong cch gia ABv SD khong cch t A n mt phng (SCD).

Gii:

Mt phng (SIB) c cp vec ch phng



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




47/303

=> Php vect n = ESC,SD] = (2>/; 0; - 2)

=>Phng trnh ca mp (SCD):( A '

2>/3x-2 Z - = 0 o 2 v / 3 x - 2 z + aV 3= 0

V 2 J -|aV3 + aV3| a /..=>d(AB,SD) = d(A,SCD)= ^ =^ 2 j

Ta c: SA/ / MF => SA//mp (CMF) cha GM=> Khong cch gia SA v CM l khong cch t A n mt phng (CMF).

Mt phng (CMF) c php vc t l: rtj (*J Phng trnh ca mt phng (CMF):

>/3^x + j + 2>/3y + z = 0 73x + 2V3y + z + - ^ - = 0 J

V3 SLys--- ---

d (SA, CM) = d (A, CMF)= =

Ta c: SC = i - | ; a ; - ^ j h a y ( - l ; 2 ; - > / 3 ) ; S D = ( l ; 0 ; - ^ )

s .

, ^l 4rftt 5: Cho t din ABCD c CD = 2a cc cnh cn i u bng V2.

a) Chng minh rng CAD, CBD u bng ,90,b) Tnh s tp ca t din ABCD.

c) Chng minh rng: mp(ACD) i. mp(BCD).(i hc Vn ha H Ni 1998)

Gii:p dng nh l Pythagore ta chng; minh c AACD, ABCD vung

cn ti A, B.stp= + SACDH-Sgcu -r'S^jj trong AABC, AABD u cnh: bng a.AACD, ABCD vung cn c cnh gc vung l a

0 a2>/3 2 a2(V3 + 2)=>s,, +. =- ~ -K trung tuyo AM, BM ca AACD; ABCD

^ => AMB l gc phng ca nh din (ACD,BCD)IBM CD 5 y .

47



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




48/303

Vi d 1:Hnh hp ch nht ABCDA1B1C1D1c cnh AB = 5V2 , AD = 3a/2AAX=8 . Dng mt phng i qua ng cho Bi v ct 'cnh AAi scho thit din to bi mt phang v hnh hp c chu vi b nht., Tm g

gia mt phng ct v mt y ABCD.Gii:

! DiMBN l t gic to bi mtphng hnh lp phng

V DiN // MB, DiM // NB suy N DMBN l hnh bnh hnh. R rng chu

ca n cc tiu nu DjM + BM cc tiu.

quay mt AAlDjD nh ong hnh v

thy rng: tng DjM+ BM cc tiu khi l giao im ca DiB v AiA.

Do AABM~ADBD,=> = 5 l =>AM = 5.B DB

Kh MB = 5-v/3,MD1= 3& .

VI DjB = Vl32 SDiMBN30>/2;SABCD= 30,SDiMBNcosq> =

Jcos = => = 45. Trong l gc gia mt ct v y.

zVy gc gia mt phng ct v mt y ABCD l


49/303

Gii:Gi s SABCD \hrv chp d io anh ). M \ N \ nmg dm catca cnh bn SA v SB vi thit din. K, Q v p l cc im gia ca

AB, MN v JEF tng ng,. QKP - a, QPK = (3. D thy t gic ENMF l2

hnh thang cn vi cc y l EF v MN. EF = a, KP = a.

Trong tam gic KQP: =-55-sin(i - (a - P)) sin a sin p

Hnha.T ' / \ T I 2asma TTT y QP = - ;KQ

Hnh b._ 2asin3

3 sin(a + p) " 3 sin(a + (3)Do cc tam gic ABS v MNS ng dng vi nhau, nn:

M N _ S Q _ S K - K Q _ 1 K Q ^ ^ _ g r_ _

AB SK SKMN = A B l -^ | = a

Khi : s

SK

EF + NM

1 -SK4 sin p cos a

2cos a

,tac:

ENMF

3 sin (a + p)

_ 2a2sin a J 2sinacosp I9s in(a + p) sin(a + p) J

Li gii ny tng ng vi gi thit khi Q thuc on thng SK (k ctrng hp c bit, khi Q 3K v Q =S). Trong trng hp ny, B thayi t o (Q - K) cho n arctan(3tana) (Q - S)

Ta xt trng hp: 3e arctan(3 tan a) ;~

(hnhb). .

y QPO = p,QTO = a,PT =. Khi , t tam gic PTQ,o

QP= as ma - v TQ = asin |3 sin(P - a) 3sin(P-a)

49



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




50/303

Do cc tam gic DCS v NMS ng dng vi nhau nn

7EFMN

P Q _ CA J 2sin a cos p

9sin(pa) s in(pa)Ch cn li phi xt trng hp, khihnh chiu thng gc ca thit dinthuc vo phn nh trong hai phn cay chp, tc thuc hnh ch nhtEFCD(hinhc).

Trong trng hp ny PQT = p,pe 0;^z

Li gii ny tng t nh trong trng hp trc.p s:Nu hnh chiu thng gc ca thit in thuc vo phn to ca y chp,

9 . / _ . \

_ 2a sina 2sin a~o_ Yk _ + /Oth: s = TTT 1+ 7 r vi Be 0;arctan(3tana) ,Gsinfa + P)^ sin(a + p)J Lv;J

0 2a2sin a f, 2sinacospS) j r . _, _/. \ 7ts ---- T--1 7--- vi Be arctan(3tana):9sin(a + p)[ sin(a + p) J L 2 .

Nu hnh chiu thng gc ca thit din thuc phn nh ca y chp, th:~ 9 . r -

s = 2a2 sin a 2+ 2 sin a cos p9sin(a + 3)^ sin(pa)

Vi d 3: di trung on ca mt bn trong mt hnh chp tam gic bng k. Ct hnh chp bi 1mt phng cch u tt c cc nh ca hnhchp. Tnh din tch thit din to thnh nu cnh bn ca hnh chp tovi mt y ca n mt gc p.

Gii:

Trang hp I. Thit in i qua ccim gia ca nhng cnh bn hnhchp (hnh v).SABC l hnh chp cho

SD = k, SCO = 0.



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




51/303


52/303

D n g5 C cb i ^oantirihg0c/;khoangcacii|^ , ,

Vi d : Ch hnh chp t gic SABCD c y ABCD l hnh thang vuti A v D;AB =AD = a; CD = 2a. Cnh bn SD vung gc vi mt

v SD = a,a) Chng minh rng tam gic SBC vung. Tnh din tch tam gic SBC

b) Tnh khong cch t A n mt phng (SBC) .. Gii:

a) Chng minh rng: ASBC vung.Tnh in tch tam gic SBC.Cho h trc Oxyz nh sau: Gc 0 =Trc Ox i qua DA. Trc Oy i qua DCTrc Oz i qua DS. Khi : D(0; 0; 0);A(a; 0; 0), B(a; a; 0), C(0; 2a; 0), S(0; 0; a)=> SB = (a; a; a) =>SB = a-\/3

=> BC = (-a; a; 0) => BC =L-J

=> BC.SC = - a 2+ a2= 0 ASBC vung ti B1 2fc

SBC= -BC.SB = .SBC 2 2

2. Tnh khong cch t A n mt phng (SBC)

SB = (a; a; - a) hay (1;1; - 1);BC = (-a; a; 0) hay(-l; 1; 0)=> Php vect: = [SB,BCj = (1; 1; 2) => phng trnh mt phng (SB

qua s l: X + y + 2 (z-a ) = x + y+ 2z-2a==0

Khong cch t n (SBC) cho bi: d(A,SBC)= =~= =

Vi d 2:Cho hiih lp phng ABCD-ADCD c cnh bng a. Gi s v N ln lt l trung im ca cc cnh BC v DD.a) Chng minh rng: MN // mt phng (ABD).

b) Tnh khong cch gia 2 ng thng MN v BD .Gii:

a) Chng minh rng: MN li(AB)Chn h trc Oxyz nh sau: Gc 0 = ATrc Ox i qua AB; Trc Oy i qua AD; Trc Oz i qua AA\Khi : A(0, 0; 0); B(a; 0; 0), D(0; a; 0), A(0; 0; a)Phong trnh mt pbng (ABD) c dng:

X V 2+ + = lx + y + z - a = 0a a a

52



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




53/303

=>Php vect n = (1 ;1; 1)

M l trung im BC => MI a ;~ ;0 1;N 1 2

N l trang im D=s>NI 0; a;

=>MN = | - a ; | ; |

=> n.MN = ~ a + - + - = 0 = > n l MN2 2

Vy: MN // mp(ABD).b) Tnh khong cch gia 2 ng thng MN v BD

Ta c: MN // mp(ABD) cha BD

Khong cch gia MN v BD l khong cch t M n mt phnga + s

(ABD)=>d(MN,BD) = d(M,A'BD) = -------.>/3 2v3 6

Vi d 3:y ca hnh chp SABCD l hnh thang cn ABCD c y bAB = asgc nhn bng


54/303

___ AO 9 _ 2 ______OBE = (p,snB = = - Khi (p -180 - a - arcsin. K OE XBC.

AB 3Ia/s"

T tam gic OBE, ta c: OE = OBsin = rsin

T y 12x = a v X = .K hiu ABO = 3,

Gc SEO l gc cn tm: tan SEO =

3

-21__a + arcsin

3 L 3 jSO _______3HOE = , ( . 2 YVoasin I a + arcsin I

Trng hp 2. im o nm ngoi hnh thang (hnh b). Gi nguyn cc

k hiu c, ta c: AK = KB = 7x,BO - x ) 2 - X 2 =4x^3.

AB = >/(8x)2+ (4xV)2 = 4xV, sin p = .

Cch gii tip tc ging nh trong trng hp 1 ta c kt qutan SEO = arctan

V3aarcsin . 2ya + arcsin 1V

V d 4: y ca hnh lng tr ABCAiBCi lm tam gic u ABC c cnhbng 2a. Hnh chiu vung gc ca hnh lng tr xung mt ABC l hnhthang c cnh bn AB v c din tch gp i din tch y. Tnh agcao hnh lng tr bit ABi = b.

Gii:

Hnh a. Hnh b.Trng hp .Mt AAiCiC vung gc vi mt phng y (hnh a.). V Tng din tchca hnh thang ABB2C2gp hai ln din tch tam gic ABC, nn dintch hnh bnh hnh BCC2B2v tam gic ABC l bng nhau. T y c

54



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




55/303

th kt lun rng CC2 = FCX=AC = a . Nhng lc AXF = A2C = a ,

c ngha l im A2 l im gia ca on AC. A2C2= A2C + CC2= 2a.Ngoi ra, A2B2= AB = 2av B2C2= BC = 2a, tc tam gic A2B2C2 l

am gic u.

Ta c: CB2= aV3, AB2= yjAC2+ c i f = V(2a)2+ (a>/3)2= aVr

T y = ^ A B -A B 22=Vb2 -7 a 2 .

Trng hp 2.Mt ( ABiCiC) vung gc vd mt phng y. Tng ttrng hp 1, c th chng minh rng im 2 l im gia cnh BC catam gic ABC. Khi AB2= a>/3 . T tam gic AB2B1c

B,B2=yjABf - A B =Vb2- 3a2 .

p s: Vb27a?hoc Vb2- 3a2

K t 5; Cho hinh chp S.ABCD c y ABCD l" hnh ..vung cnh a vSA-Lmp(ABCD) vi SA = h.a) Tm h thc gia a v h gc gia 2 ng thng AB v s c bng 60.

Cho h a V2, hy tnh khong cch gia 2 ig thng AB v sc.b) Chng minh rng gc phng nh din cnh-SC l gc t.

Gii:a) Tim h thc gia a v h gc (AB, SC) = 60 tz

Chn h trc Oxyz: Gc o trng A; s i.Trc Ox i qua AB. Trc Oy i qua AD; .Trc Oz i qua AS. Khi : A(0; 0; 0); / ;B(a; 0^0); C(a; a; 0)D(0; a; 0); S(0; 0;h) / h- . \ X .

Ta c: AB = (a;0;0); s c = (a;a;-h)/ fi-0- - V r

AB.SC / \ A=> COS! AB.SC) = . - B' \ /

AB.SC / y_

a2 _ . a

a\/2a2+ h2 y2a.2+ h2

M cos(AB,SC) = COS60 = => . a ... = = 0 2a = V2a2+ h22 v T ? 2 ;

o 4 a = 2a2+ h2 h = a>/2 . .* Tnh khong cch gia AB v SG .Ta c: AB // CD => AB // mp(SC) cha SC; h = a\/2=> S(a;0;aV2)

55



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




56/303

=> Khong cch gia AB v SG l khong cch t AeAB n mp(SCDMt phng (SCD) c cp vect ch phng l:s c = (a; a; - a-v/2) hay (1;1; - V2) ; SD = (0; a; - a-v/2) hay (0; 1; - 72)

o Php vect n = [SC,SD3 = (0; V2; 1).

Suy ra phng trnh mt phng (SCD) l : *j2(y - 0) + l(z -aV 2) = 0o ^ y + z-aV 2= 0^ j 2y +2 -aV 2= 0

=>d(AB,SC) = d(A,(SCD)) = t ^ = .n/3 3

b) Chng minh rng gc pkng nh in cnh sc l gc t:Nh din cnh sc to bi 2 mt phng mp(SCB) v mp(SCD)

Ta c: s c = (1;1; - >/2); SB = (a; 0; - a>/2) hay (1; 0; - V2)

=> Php vect nx= [SC,SB] ={-42;0; - 1) ca mt phng (SCB)Php vect ca mt phng (SCD) l: n2= [SC,S] = (;-J2;1).Gc phng nh din canh sc cho bi:

cosa - ^ ' ^ . - = r=w= = l a i gc t. X ,,. Jni|.|n2| V3.V3 3 * '

Vi d 6:Trong hnh hp ch nht ABCDA1B1C1D1c AB = a, BC = a-J,gc gia cc ng thng CBi v BDi bng 45. D.

Tnh cnh AA.

Trn hnh a th hin hrtrong bi:AB = a, BC = a-v/.Gi s AAj = D= X.

B Hnh aTrn ng thng AD- ko i, ly on DK = AD = aV.

C CBj //KDj, ngha l gc gia cc ng thng CBi v BDi bng ggia ng thng KDI, v BDI.BD2= AB2+ D2= 8a2,BDj = BD2+ D)f = Sa2 +X2.DjK2- Btc 2= BC2+ BB = 7a2-hx2

BK2= AB2+ AK2= 29a2- By gi, s dng nh l hm csin i vtam gic BDiK, ta c: BDj + DjK2- 2BDj-DjKCOSBD K = BK2.

56



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




57/303

y, vic phi xt 2trng hp khc nhau l do:Trong iu kin ca bi ton, khng ni r, gc no trong 2 gc BDKhay gc k b vi n bng 45. T cc trng hp nh vy, s c cc

phng trnh tng ng sau:

1) 8a2 + X2 + 7a2 + X2- 2. V(8a2 + x2)(7a2 +x2) = 29a22

2) 8a2 + X2 + 7a2 + X2 + 2. >/(8a^T3?)(7a^+x^) = 29a22

Gii cc phng trnh ny ta c AAi = a hoc AAl = a V2 .Vi d 7:Cho hnh chp t gic S.ABCD c y ABCD i hnh thang vung

ti A v D. AB AD = a, CD = 2a. Cnh bn SD vung gc vi yABCD v SD = a.

a) Chng mnh rng tam gic SBC vung. Tnh din tch tam gic ny.b) Tnh khong cch t im A n mt phng (SBC).Gii:

a) Chng minh tam gic ABC vung. Gi I l trung m DC C)

=> ID - IC = a => ABID l hnh vung => IB = a =

=> ADBCvung=>DB1 DC 'M SD1 BC=> BC X mp(SDB) => BC1 SB => ASBC vung (pcm)

Ta c: BD = aV2,BC = aV

SB2= SD2+ DB2=a2+ 2a2 =3a2 =.SD = aV3=>SSBC =SB.BC = ~ a V3 ,aV2 = A .

b) Tnh khong cch t A n mt phng (SBC). AI ct BDti o =>OA//BC =>AO//(SBC) =>d(A,SBC) = d(0,SBC)

V OH 1 (SBC) => d(0,SBC) = OH .T gic ODSH ni tip

=>BH.BS = BO.BD. Tnh BH OH2= OB2- BH2=>OH = r -6

=>d(A,SBC) = .

V d 8:Cho hnh chp t gic u SABCD c i cnh bn v cnh ycng l a.a) Chng minh rng SA _Ls c .b) Tnh th tch hnh chp SABCD theo a.c) Tnh khong cch gia AD v mp(SBC).

(i hc Cn Th khi D-1998)

57



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




58/303

Gii:

a) ABCD l hnh vung cnh a => AC = a>/2 .Gi thit cho SA = sc = a . Do ASAC vung cn ti s => SA 1 s c .

b ) V SABCD = g S 0 S ABCD = 3 * 2 A C SaB CD = 6 ' =

c) V AD//mp(SBC) nn(AD,SBC) = d(I,SBC). K IH1 SJ (1).

TaC BCXU } ^ B c j-mp (s u )

BC_LIH (2)(1) (2) (SBC) J.IH => d(I, SBC) = IH .

Ta c:

SSIJ = - I J . SO = IH . SJ => IH = IJg SO = - 2 =aM = r -SIJ 2 2 SJ aV3 3 3

2

Dng 6: Cc bi ton dng khc

Vi d 1: Cho t din u S.ABC cnh a. Gi H l hnh chiu ca s xungmp(ABC).a) Chng t H l tm ng trn ngai tip tam gic ABC. Tnh SH theo a.

b) Gi I l trung im ca SH. Chng minh tam din IABC l tam inba gc vung. sGii:

a) T din S.ABC u nn SA = SB = s c .HA, HB, HC i hnh chiu ca SA, SB, sc / ; Jnn:HA = HB = HC /=> H l tm ng trn ngoi tip tam Av ; - - -/gic ABC.

Trong tam gic vung SAH: SH2= SA2- AH2 (1)Trong tam gic u ABC, H cng l trng tm, trc tm nn:__2 a . J z _ a-s/3

Thay cc gi tr bit vo ():SH2= a2- = => SH = i -

b) V HA = HB = HC nn IA = IB = IC . Trong tam gic vung AIH:

58



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




59/303

A '

Nn AAIBvung I => AIB = BIC = CIA = 90

=> Tam din IABC l tam din 3 gc vung.i 2:Cho hnh lng tr ng ABC.ABC c y ABC l tam gic cn

ti A, gc ABC = a . BC hp vi y (ABC) mt gc p . Gi I l trung

im ca AA. Bit BC = 90a) Chng minh rng: tam gc BlC vung cn.

b) Chng minh rng: tan 2CC+ tan2p = Gii:

Chng minh rng ABIC vung cn

Ta c IA l on vung gc IB, C l onxin AB = AC => IB = IC=> ABIC vung cn

Chng minh rng: tan2a + tan2-P = 1 '

Ta c: BG l hnh chiu ca BC ln (ABC)

C*BC = p.Tac: tanp =CCBC

B' /

\ X Si\ X

, or*

Goi H l trung im BC=> AHJ_

BC v IH = .2Ta c: tn a = AH

BH

M AH2= IH2- IA2=

AHBC2

BC2

, 2 . _2 o 4AHtan a + tan p =

2AHBC

C '4

c-c2

o 4 A H 2= BC2-C C2

BC2- C C2+CC*2= l(pcm ).

BC2 BC2 BC2

i d 3:Cho t din S.ABC c ABC l tam gic vung ti A, scJL(ABC)v SC = AB = AC = aV2. Cc im M thuc SA v N thuc BC sao cho.AM = CN 1(0 < t < 2a).a) Tnh di on MN, tm t on MN ngn nht.b) Khi MN ngn nht. Chng minh rng Msri ng vung gc chung:

ca BC v SA. .



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




60/303

. - Gii: . .. .. 'a) Tn Vi MN theo t. Tm t MN ngn nht: Chn h trc. Oxyz nh sau:

Gc o - A; Trc Ox i qua AC.Trc Oy i qua AB; Trc Oz // CSKhi : A(0; 0; 0) ;B(0; a^2 :0); C(aV2 ; 0; 0) ;S(aV2; 0; a-s/2 )

=>MN2= (a-s/2- tVI)2+ + -

T\/r\T _ Q+2 O o2

tV22

= 3t2 -4 a t + 2a2

Vy: MN = V3t2 -4 a t + 2a2 .b) Chng minh rng: MN ng vung gc chung ca BC v SA

r r 2Ta c MN ngn nht khi t =

M n;aVT |,N 2a^2 a /2 0 liII aV2 a>/2 a-J '

3\ 3 i 1 3 ; 3 ; 3 ; 3 ; 3 J

m = - ; 0; ) . r s j - ? - . .

MN.AM =2a

;CN =

2a2

3;0

= 05 3 MNXAM2a2 2a2 ^ 1mNJ.CN

MN.CN = - + - 7- = 0 L9 9

=> MN l ng vung gc chung ca SA v BC.V d 4:Cho hnh lp phng ABCD.ABCD cnh bng a (a >0). Tr

BD v AD ln lt ly cc im M v N sao ch DM = AN = k(0


61/303

b) Tm k MN ngn nhtt f(k) = 3k2 - 2a%/2k + a2 (vi 0 < k < aV2)

f'1(k) = 6k - 2aV2;f''(k) =0 k =* 3

Bng bin thin:k0 aV aV2

3f(k) 1 0 +f(k) 2 3a2

3 /

Vy: min(MN)2 = min(MN) = -^=khi k = .3 -v/3 3

K 5; Cho hnh lp phng ABCD.ABCD cnh bng 1. Ba im M,N, I di ng trn AA\ BC, C5D tng ng sao cho AM = BN = Cl = a(0 < a < 1).

a) Gi (Pa) l mt phng qua 3 im M, N, I. Chng minh rng mtphng (Pa) lun lun song song vi nhau khi a e [0;l].

b) Tnh khong cch t a n (Pa) v tnh din tch tam gic MNI theo a.Xc nh M in tch tam gic MNI nh nht.

Gii:a) Chng minh rng: (Pa) lun lun song song vi nhau

Chn h trc to nh sau:Gc o = A; Trc Ox i qua ABTrc Oy i qua AD; Trc Oz i qua AA. B ' / 3?1..............c>Khi : A(0; 0; 0); B(1;0; 0); D(0; 1; 0);A(0; 0; 1); M(0; 0; 1- a ) ; N(1; a; 0);1(1 - a ;l ;a ) . _

=> MN = (l; a; a -1); MI = (1- a; 1; a)=>Php vect n = ^MN,MlJ

= (a2 - a + l;-a2+ a - l ; a 2 - a +1) hay (1; -1 ; 1)

=> ptmp(MNI) :x - l - ( y - a ) + z = 0 (Pa) : X- y + z + a -1 = 0

V php vect n = (l; -1 ; 1) khng ph thuc a nn (Pa) song song vi

nhau

61



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




62/303

b) Tnh khong cch t A n mp(Pa) v tnh smi.Xc nh M snh nht:

T a c :d (A ,P j = t p = 2 .= > = ( l; a ;a - l ); B = ( l- a ; l; a )

=> [mN,M] = (a2 - a + l;-a 2 + a - l;a2 - a +1)

[M N . m ] = ( a - a + l f = i[M N ,M l] = ( a 2 - a + l)

MNI

I f 3a - +

2 J 4

=> minSjvjNj = ^ k l i ia - = 0 o a = M l trung im AA\

Vi d 6:Cho hnh ip phcmg ABCD.ABCD cnha. Hai imM v Nchuyn ng trn 2 cnh BD v BA tong ng sao cho BM = BN = t

Gi a v p ln lt l cc gc to bi ng thng MN vi cc ngthng BD v BA.a) Tnh on MN theo a v t Tm t on MN ngn nht Khi tnh

a v p .

b) Trong trng hp tng qut, chng minh rang: cOS2 a + COS2 3= 1

Gii:a) Tnh MN theo a v t: Chn h c Oxyz

nh sau: Gc o =A;Trc Ox i qua B;Trc Oy i qua AD; Trc Oz i qua AA\Khi : A(0; 0; 0) ; B(a; 0; 0); C(a; a; 0);D(0; a; 0); A(0; 0; a); B'(a; ; ); C'(a; a; a)

MN2=t2-aV2.t + 2 =

a2 t , , XT a v 2 n i i av2MN >--=> min MN = khi t - 0 C5* t =

2 2 2 2M l trung im BD v N l trung im AB\ Khi :

W2 tyBM = t => M a - ;B'N = t =x>N^a

0 ; ; a - i => MN=Vt2 - a>/2.t + a22 2 I

0; t>/2

62



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




63/303

f

MB.MN ~ 1MB|.|MN| ~ aV2 a>/2 ~ 2

2

2

i (!)

Hi)

= -r a = 60

t -a --- ;0 ;a -

2 2AN =

1 2 2

"COS p =

______a* ' AN.MN T 1 L =>cos6 = r | 7 1- I = = = - o B = 60 .

ANUMN /2? 2V 4 V 4

_LTng qut: Ta c: COS a - . %-- = ------- d=

J t 2 f W2 Y /^J-t2 tV 2 f

. - S f . t S l2 J 1 2

cosp ---------------

-----I = d = = -----j =====

4 - 1# ; v T - f :

t* f t v ;2 f .-+ a o 2 2 2 X

=> cos a + cos 3 = p --------- =f = (pcm).

*H)] d 7:Cho hai na mt phng (P) v (Q) vung gc vi nhau theo giaotuyn (). Trn () ly on AB - a-(dio trc). Trn na ng thng"At_L(A) v trong (p) ly im M vi AM = b (b > 0). Trn na ngthng Bz vung gc vi CA) v trong (Q) ly im N sao cho

BN =

63



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




64/303

a) Tnh khong cch t A n mt phng (BMN) theo a v b.b) Tnh MN theo a v b. Vi gi tri no ca b th on MN ngn nht.

Tnh di ngn nht .Gii:

a) Tnh khong cch t A n (BMN)Chn h to Oxyz nh sau:Gc o trng vi B; Trc Ox i qua BA.Trc Oy song song At; Trc Oz i qua BN

BJU____ , Khi : B(0; 0; 0), A(a; 0; ),

/ -----------------------y / M(a; b; 0), N 0;0;^-_____b M / ^ b

Mt phng (BMN) c cp vect ch phng l:

BM = (a;b;0);BN = o,*0;-^

P : = B M , B J = 2; ;1; ;0J

Phng trnhmp (BMN): x - ^ y = 0 bx -a y = 0.D

=>

Khong cch tr A n mp(BMN) cho bi:

b) Tnh di MN theo a v b:

o |;b | 0;;0 ;s| ;D 0 ; ;0j 1 3 ) V ) V )

- a ; - b ;^ => MN = Ja 2 + b2 = I ^ a 4+b4+ a2b2b V b2 b

Vy: min MN > \2>khi b2= ~ o b= a .

bBi tp vn dng

1. Cho hnh hp ch nht BCD.A,B,CD c AB AD a,AA = b (a > 0, b > 0). Gi M l trung im cnh cc\a) Tnh th tch t in BDAM ieo a v M.

b) Xc Dh t s - 2 mt phng (ABD) v (MBD) vung gc vi nhau.h

Ta c: MN =

64



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




65/303

2. Cho t din ABCD c cnh A vung gc vi mt phng (ABC) vAC = AD - 4,B = 3, BC = 5. Tnh khong cch t im A n mt

phng (BCD).3. Cho hnh lp phng ABCD.ABC'D c cnh bng a.

a) Tnh theo a khong cch gia 2 ng thng AB v BD.b) Gi M, N, p ln lt l cc trung im ca cc canh BB, CD, AD\Tnh gc gia 2 ng thng MP v CN.

4. Cho hnh lng tr ng ABCD.ABCD c y ABCD l hnh thoi

cnh a, gc BAD= 60. Gi M l trung im cnh AA v N l trungim cnh CC.a) Chng minh rng 4 im B, M, D, N cng thuc mt mt phng.

b) Tnh cnh AA theo a gic BMDN l hnh vung. ' .

5. Gi o l tm ca hnh thoi ABCD cnh a vi OB = . Trn ngthng vung gc vi mt phng (ABCD) ti o ly im s vi SB = a.a) Chng minh rng tam gic s AC vung v s c _LBD.

b) Tnh gc phng nh din cnh SA v tnh khong cch gia SA v BD.6. Cho hnh p phng ABCD.ABCD c cnh .bng a. Trn cc cnh

AA\ BC, CD ln lt y cc on AM = CN = DP = X.a) Tnh din tch tam gic MNP theo a v X. nh X din tch ny nh

nht.

b) Chng minh rngkhi Xthay i, mt phng (MNP) lun lun songsong vi mt mt phng c nh.

7. Cho hnh chp S-ABCD c y ABCD l mt hnh vung tm 0, cnh a,

SO_Lmp(ABCD) v so = Trn cnh c ly mt im M vi

SM = X(0 < X< a). Mt phBg (ABM) ct cnh SD N.a) Chng minh rng ABMN l mt hnh thang cn. Tnh din tch hnh

thang ny theo a v X.

b) nh X mt phng (ABMN) vung gc vi mt phng (SCD).8. Trn 3 cnh ca mt tam din vung nh o ta ly OA = OB = o c = a.a) Chng minh rng: tam gic ABC u v tnh din tch ca n.b) Tnh di ng cao h t o xung mt phng (ABC). Gi D l

im i xng ca H qua o. Chng minh rng ABCD t din u.9. Cho hnh lp phng ABCD.ABCD cnh bng a. Tnh s o gc

phng nh in [B,A'C,D].

65



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




66/303

10. Cho hnh chp SABC y l tam gic u ABC cnh a. M e SA,

AB, BC ti I, K.a) Chng minh rng AIBK vung.b) K SH 1 BC. Nu SH _LIK v SB = sc = a thi hy tnh th tch hnh

chp NBIK. theo a.

11. Cho hnh chp SABCD c y ABCD l hnh vung cnh a.SA = h, SA 1 ABCD. M thay i trn CD. t CM = X.a) K SH -L BM. Tnh SH theo a, h, X.b) Xc nh v tr M th tch t din SABH t gi tr ln nht. Tnh

gi tr ln nht .

b) Tnh mp(ABD) _Lmp(MBD): Php vect ca mp(ArBD):b

Iij = [A *B, A ' DJ = (ab; ab; a2 ) hay (b; b; a).

NeSB, p [ A 1B, A D j = (ab; ab; a2)

=> Php vect ca mp(MBD) cho bi:

66



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




67/303

n2 =[MB>M D ] = i^ ;* ;- a 2ihay(b;b;-2a).

(ABD)-L (MBD) . = 0

o b +b -2 a = 0 o a =b o a = b (a >0,b>0). Vy= 1.

Tnh khong cch t A n mt phng (BCD).Ta c: AB2 + AC2 = 9 + 16 = 25 = BC2=>AABC vung ti A.Chn h trc Oxyz nh sau:Gc O sA ; Trc Ox i qua ABTrc Oy i qua AC; Trc Oz i qua AD;Khi : A{0; 0; 0), B(3; 0; 0), C(0; 4; 0),D(0; 0; 4). Phng trnh mt phng (BCD) l:

+ + = 1 4x + 3y + 3z -12 = 0;3 4 4

d(A(BCD))= d n S12

1+ 9 + 9 >/3 4 'a) Tnh khong cch gia AB v BD

Chn h trc Oxyz nh sau:D' GcO^ A. Trc Ox i qua AB

Trc Oy qua AD; Trc Oz i qua AA.Kh : A(0; 0; 0), B(a; 0; 0), D(0; a; 0),

A (0; 0;a),B (a; 0; a) _

Gi (P) l mt phng cha BD v // A*B=> Cp vect ch phng l:

A B = (a; 0; -) hy (1; 0; -1);

B D = (-a;a;-a) y (-l ;l;- l)

=>Php vector: n=> Phng trnh mt phng (P) qua B:

x - a + 2 y + z -a = 0x+2y-f-z-2a-0.Khong cch gia AB v BD l khong cch t A n ;.ot phng (P)

d(A 'B 3'D j = d(A',P) = i ^ L i

Tnh gc gia MP v CN. Ta C:Ml trung im BB=> M^a;0;~j.

p l ung im Ai> => p tt ; ;a j ; C(a; a; b), C(a; a; a).

67



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




68/303


69/303

=> DM = DN = |a2 + => B' M = B N = DM = DN =>BMDN l hnh thoi.

BMDN l hnh vung o DM.DN = 0

- + = 0 o i 2 = 2a2o h = aV2 . Vy: AA' = h = a-s/2 .4 4 4

5. a) Chng minh tam gic SAC vung v s c BD

aOA=Va B2 -O B 2 =Jaz = ------

V 3 3

SO = VSB2-OB 2 = J a 2- =V 3 3

SO = OA = OC=>ASAC vung ti s.Ta c: BD J_ AC;BDJ_SO

BD_L(SAC)=>.BDSC.

/* \ N/ / I \/ \/ 9 1 \

/ / \ H....... \ - - A -/ * N . i ' \ a/ / 's /

/ ' \' o \

b) Tnh gc phng nh din cnh SA D A' AGc o l tm hnh thoi; Ox i qua OA ; Oy i qua OB; 0z i qua OS.

Khi : A =*/.;oj ;b o ; ;0 ;Sf

0; ;D(0 ^ . o

) 1 3 V 3 V 3 s a J ^

a-v/6 a ->/6

n2 =[SA,SD] = - ^ ; ^ - :- 2 ^ j h ay(-V2;2;-V2).

GC phng nh din cnh SA cho bi: COS a = = 0n, . n.

Vy: a = 90* Tnh khong cch gia SA v BD: Gi H l trung im ca SA, ta c:OH -L SA (v ASOA vung cn).. V OH J_ SBD (vi BD J_ (SOA))=> OH l on vung gc chung ca SA v BD

Ta c: SA = On/2 =^6 rr 2aV =>- OH = =

2

Vy: d(SA,BD) = OH =lyS

69



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




70/303

6. a) Tnh din tch tam gic MNPChn h trc Oxyz: Gc o trng vi DTrc Ox i qua DA; Trc Oy i qua DCTrc Oz i qua DD\ Khi : D(0; 0; 0),A(a; 0; 0), B(a; a; 0), C(0; a; 0), D(0; 0; a)=>M(a; 0; x), N(x; a; 0), P(0; x; a).

Ta c: MN = (x - a; a; -x)

=>MN =yj(x- a)2+ a2 + ? = V2X2-2ax + 2a2

MP = (-a; x; a - x) => MP = V2x2 -2ax + 2a2

NP = (-x; X- a; a) => NP = 'Jzx2-2ax + 2a2

=> MN = MP = NP => AMNP uMN2.V

MNP = (2x2 -2ax + 2a2)s/3 __(x2 - 2ax + a^)V3

Ta c: X -a x + a = IX--T- 3a 3a+ - > -4 4

=>minS = k h ix - = 0 x = . Khi M l trung im AA\8 2 2

N l trung im ca BC v p l trung im ca CD.b) Chng minh khi Xthay i, mt phng (MNP) lun song song vi 1 mt

phang c nh. Php vect ca mt phng (MNP) cho bin = [MN,MP] = (x2 - ax + a2; X2 - ax + a2; X2 - ax + a2) hay (l; 1; 1)

Mt phng (a) qua o c php vect: n = (1; 1; 1) l X + y + z = 0Vy, mt phng (MNP) lun lun song song vi mt phng c nh (ct)c phng trnh: x + y + z = 0 ' +2

7. a) Chng minh: ABMN l hnh thang cn. sTa c: AB // CD=>AB // mp(SCD)

mp(ABM) ct mp(SCD) theo giao tuynMN // AB=> MN // CD => CM = DN=> AAND = ABCM => AN = BM

Vy: ABMN l hnh thang cn N - - y=^ . MN SM ^ aV2 1/Ta c: = 77; OA = oc = / ' K _ \ Z l

CD s c 2 \________________________ -------- D T

M SC=VS2+OC2=,fel + l = a =>MN=T V 4 4

70



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




71/303

Gi I l trung im AB: J l trung im MN; K l trung im CD. ASCD u

a>/3=> SK = ; ASBC u => SCB = 60

BM2 = CB2 + CM2 - 2CB.CM. cos_60.

= a2 + (a - x)2- 2a(a - x).= X2- ax + a2.

V MHJ-AB=> BH =

2

a X

Ta c: MH2 = MB2 - BH2 = X2 - ax + a2 - = .: l ; 2 j 4

ySx2-2ax + 3a2=>MH = -------- -----------.

2 -

SABMN= -(AB + MN).MH = -(a rx )v 3 x 2 - 2ax-r3a2.

nh X mp(ABMN) _Lmp(SCD).Chn h trc Oxyz nh sau: Gc 0 l tm hnh vung ABCDTrc Ox i qua OA; Trc y i qua OB; Trc Oz i qua OS.ySA v

2K h i : C - ~ ;0;0

r

0; 0;

\ J \ /

] ^ s c = i - ; 0 ; - ^2 J [ 2 2

0 0

XM T

yM=(0) = 0 M

SM = s ca

ZMV2 _ x aV2 _ _ W 2

V

Mt phng (ABMN) c cp vect ch phngf X \

AB = -a-s/2 aV2

2 ;-~2 ;0 kay(-l;l;0).

BM = - ^ ; ^ ; ( a - x ) ^ j h a y ( - x ; - a ; a ~ x ) .

=>Php vect rtj = a , m ] = (a -_x; a -x : a + x ).

71


WWW.FACEBOOK.COM/BOIDUON

Documents

PHƯƠNG PHÁP GIẢI TOÁN TỰ LUẬN HÌNH HỌC KHÔNG GIAN - TRẦN THỊ VÂN ANH