Tuyển tập 500 bài toán hình không gian chọn lọc phần 2 nguyễn đức đồng

Taong tif do tinh doi xuTng : iSOB la mat phan giac nhi dien S B SOC la mat phan giac nhi dien SC SOD la mat phan giac nhi dien SD Do do S O la giao tuyen tifong ufng trong ycbt dpcm.

Di rn

g BH

1 IE =

>

NH _L I

E (d in

h ly b

a dt fd ng v uo ng g oc )

=> g oc c ii

a ma

t ph Sn

g th ie

t di en v

a ma

t da

y la :

(p =

N KB

( ho ac S Ft )

NB = p nc ta => B d g on vu H NB A Do

HB (1 )

TiT e ac

h di Tn

g ta s uy r

a

N la t ru ng d ie

m cu

a BB ' NB = —

Va : ab

HB

^

(1

=> ta nt

p

=

' BE

^ 2V

a2

cV a^

~

b ab

1

^

^ c ( a + 2 )

2 CO

S (p

CO S(

V

Di nh l

y ch ie

u hi nh c ho t

a : S'

= S.

co si

p (v

i

S la d ie

n ti ch t hi et d ie n)

( yc bt ).

co si

p

4

Uo ai S : MA

T Pf lA

« GI AC C UA N flJ D it

N

I PHUONG PHA

P

Co s

o cu

a ph if an

g ph ap d i/

ng m at p ha

n gi ac t he

o di nh

ng hi

a ca

n ph ai t hi fc h ie

n ba

h xid

c ca

b an :

• Bi : Di fn

g go

c nh

i di en a

^ cu

a nh

i di en ( a;

d

; )

• B2 : Di rn

g ti

a ph an g ia

c Mc c ii

a aM fe

• B

3 : Du ng m at y = ( d;

M c) v di d = a r>

(5 Th

i

y la m at

ph an g ia

c ci ia a;

P ).

O Gh

i ch ii

:

Mo

t di em tr en m at p ha

n gi c

6 kh od ng e ac

h

de

n ha

i ma

t nh

i di en b dn

g nh au

VL

CA

C BA

I T O A

N CO B AN

p S.

AB CD c

6 la t am g ia

c SA

B de

u va A BC

D la h in

h vu on

g

Gi

a si

f ( SA

n gi ac c ii

a ca

c nh

i di en ( B;

A D;

S ).

C de u.

Ke A G n S

B

=

M (A MN ) o S

C

=

N

Th eo t in

h ch at g ia

c tu ye

n so ng s on

g ta c

6 MN //

C D.

De y A

D

± (S AB )

! A D

1

S

A

[A D

1

S

B

Do ng th cf

i AM l

a ph an g ia

c

SA

B

Do d

o mp (A MN D) l

a mo

t ph an g ia

c go

c nh

i di

^n ( S;

A D;

B ) (y cb t)

Biti 179

Cho hinh chop tuf giac deu S.ABCD ChiJng minh rkng SO la giao tuyen cua cac m at phan

giac tucfng ufng vdi cac cap nhi dien (SA), (SB) va (SB), (SC) va (SC), (SD) va (SD), (SA)

Giai

Do tinh chat doi xtifng ta chi can xet mot cSp nhi dien (SA), (SB) la

=> O M la tia phan giac cua I5MB =

=> (SOA) la mat phan giac nhi dien ( S A )

Taong tif do tinh doi xuTng :

i(SOB) la mat phan giac nhi dien ( S B )

(SOC) la mat phan giac nhi dien (SC)

(SOD) la mat phan giac nhi dien (SD)

Do do S O la giao tuyen tifong ufng trong ycbt (dpcm)

tail G: XAC Dpjfl MAT CAU tiQl Tli? HlNfl CHOP DA GlAC Dt u

LPHirONGPHAP

Co so cua phuang phap thuomg sijf dung cho hinh chop

tam giac deu qua ba budc :

0 B i : Xac dinh true difdng tron d cija da gidc day (thong

thifdng d qua dinh S)

tCr W den cac mat ben hay day cua hinh chop la ban kinh

r cua mat cau

tiep hinh chop bdng the tich V va di&n tich todn phdn Stp

Goi G la trong tam AABC deu => GB = GD = GC ( 1 )

Tuf dien deu ABCD canh a => AB = AD = AC ( 2 )

Tir ( 1 ) va ( 2 ) cho ta AG la true dudng tron ngoai tiep ABCD

Diftig mat phan giac (y) cua goc nhi dien canh CD la

ly

difdng phan

giac

\k

dien tich

O Cac

h kha

c : Ta c6 : V = -S,„

r =

> r =

3V

Trong

d6 :

tp

1 aV

3 2

2 ;

(1)

=

V3 a2

V =

iBh=

l 3 3

—.a ,2 2

j

(aV3

Y

f 1 aV3

12 ^ aV6

^ = -a

(ycbt)

Stp ^

2 2 :.2a =

tich hinh

chop tuf gia

c deu

1/ Ngifcfi ta xem 6 mat trung triJc ciia 6 canh cua mot tuf dien C h i i n g to rkng 6 mSt ay giao

nhau tai mot diem

2/ Goi r va R Ian lifot la ban k i n h ciia dUdng tron npi tiep va cua du&ng tron ngoai t i e p ciia

mot mat tam giac tuy y ciia tijf dien tren

ChiJng m i n h r k n g : ~ ^ —

R 2

G i a i

1/ Cho til dien A B C D G o i I la giao d i e m cua ba di/dng t r u n g

true ciia t a m giac B C D

=> I la t a m difdng t r o n ngoai t i e p (BCD)

Hai mat t r u n g trUc cua h a i canh BC va C D c6 d i e m

chung I , nen p h a i cftt nhau theo giao tuyen d

^ d 1 (BCD)

Hai mat t r u n g trUc ciia canh B D va C D cung

^ (d') ^ (d) ^

Vay ba m a t phSng t r u n g trUc cua ba canh cua t a m giac (BCD) c&t n h a u theo dUcrng t h i n g

d vuong goc v6i m a t p h d n g (BCD) t a i I

Canh A B x i e n goc so vdi m a t p h a n g (BCD), nen m a t p h a n g t r u n g triTc ciia canh A B cSt

difcfng t h i n g d t a i m o t d i e m O

=> 0 la diem chung ciia bon m a t t r u n g trUc cua bon canh BC, C D , D B va A B

OB = OC = O D = OA Cac d a n g thiic OC = OA va O D = O A c h i i n g to O cung dong thcfi

thuoc cac mat t r u n g trifc A C va A D

Vay O la giao d i e m cua sau mat t r u n g triic cua sau canh cua t i i dien da cho (dpcm)

• C a c h k h a c ta c 6 the l|Lp l u ^n dcfn g i a n nhii s a u :

Ro rang mot d i e m tuy y thuoc dong t h d i ca sau m a t p h a n g t r u n g trifc ciia sau canh ciia t i i

dien k h i va chi k h i d i e m do each deu b o n d i n h ciia t i i d i e n

2/ Gia sii goi AABC la t a m giac can xet t r o n g gia t h i e t va goi : Diicfng t r o n noi t i e p AABC c6

tam Oi va ban k i n h r ; con dudng t r o n ngoai tiep AABC c6 t a m O2 va ban k i n h R

• Diing t a m giac A i B i C i c6 cac canh d i qua cac d i n h ciia AABC va song song v d i cac canh

cua \ABC

• Diing cac tiep tuyen cua dadng t r o n t a m O2 (ban k i n h R) t a i cac t i e p d i e m T j , T2, T 3 va

song song v d i cac canh ciia t a m giac A i B j C i sao cho tiep tuyen A2 B2 song song vdfi A i B j va

tiep diem T2 nhm t r e n cung A B (chiia d i e m C), ta difcfc t a m giac A2B2C2

« =

- 2R

r ^ 0

<K

>

<= > — ^ —

Theo

each

dUng tre

n day, mie

a

trong cua mien trong tam giac

A 2B 2C

2 v

a

AA IB ,C

I 00 A A2 B2 C2

Goi the

m R, l

a ba

n ki'n

h dUcfn

g tron noi tie

R

Mat khac, t

y so dong dang cua ban kin

h cu

a diTdn

g tr

c l

a :

Ri ^ =

2 «

Ri = 2r<=>

2r<R » —

^

•

Ca ch k ha

c

:

Dat

() ,(

)2

= /,

2Rr

[R >

0 (hie

n nhien)

I A

- MIEN

B AC

- 1972)

Cho mot kho

i tu

f die

n de

u SABC Go

i S

H l

a du dn

g ta

m T ciia tam giac ABC va tam hin

b/ Tinh ban kin

h cii

a hinh cau noi tie

n ngoa

i tie

p

AABC va

= > 1< E

IH = I

b/ Go

i O la tam hin

i O

nam tre

n IH, don

g thcf

i O nkm tre

i die

n can

h AB, tuf

c l

a O la giao diem cua

IH

va dirdng phan giac goc IMH, v

a r = O

H l

a ba

n kinh hin

h

cau noi tie

p kho

i tu

f die

n lABC

Ta

CO

: SH = A / SA

^~

MF =

^ I H

=> Cac tam giac AIB, BIC va CID deu vuong tai I , trung tuyen c6 dp dai bang nCfei canh do

=> A I , BI va CI doi mot vuong goc nhau (dpcm)

iBai 186 (DAI HOC KHOI B - M I E N B A C - 1972)

I Cho hinh thang ABCD, vuong o A va D, AB = AD = a, DC = 2a Tren du&ng t h i n g vuong goc pp6i mat phing ABCD tai D lay mot diem S sao cho SD = a

a/ Cac mat ben cua hinh chop SABD la nhufng tam giac nhu the nao ?

b/ Xac dinh tam va ban kinh hinh cau di qua cac diem S, B, C, D

[d Goi M l a diem giiira SA Mat phang DMC cat hinh chop SABCD theo thiet dien gi ? Hay

tinh dien tich cua thiet dien do

G i a i

1/ Mot each dan gian chung ta se chuTng minh duac cdc mat ben cua hinh chop la bon tam

jp^c vuong

Doc gia tu giai Chang ban ta t i n h duac :

SC = VSD^TCD^ = Va^ + (2a)2 = aVs

• SB = V S A ^ T A B ^ = V(aV2)2 + a^ = aVs ^ SC^ = SB^ + BC^ = Sa^

BC = aV2

=i> ASBC vuong 0 B (dpcm)

G

oi

O la trung die

m cu

a

SC

De

y thay h

ai ta

m

gia

c SD

C vuon

g va SB

C vuon

g th

n ta c6 :

OS = OD = OB = OC =

n m

at cau tam O, ban

kinh:

„ SC

i ma

t phin

g (DMC)

va (SAB) c6 die

m

M chun

g nen

chun

g cfit nha

u the

o m

ot gia

o tuyen M

N (N nkm tren SB)

va

vi

AB // C

D ne

n M

N / / CD (v

a MN //

AB)

=> CDMN la

hin

h

than

g,

do la thie

t die

n cua

hin

h

chop v

di mS

t phAng DMC Ma

t die

n CDM

N la

hin

h

than

g vuong t

ai

D

va ta

i M (ycbt)

Do M la trung die

m SA nen t

a c

6 : D

M

=

i SA =

; M

N

=

- AB = -

n t

ic

h

thie

t die

n CDM

N bSn

g :

S

= CD +

MN D

M

=

2a+ - /r 2 aV2

I HO

C KH OI A

- MI EN B AC

- 19 73 )

(ycbt)

Cho

hin

h

chop tu

T gia

c deu S

ABCD din

h

S, da

y l

a mo

t h

in

h

vuon

g ABC

D can

n la

m vd

i ma

t phan

g da

y vd

i mo

t goc (p T

a dU

ng ma

t phSng pha

n gia

h BC trong

hin

h

chop (tilc la go

c nh

i die

n cu

a

hin

h

chop x

ac din

h ba

m

at phan

g phan gia

c na

y cM

eo

thu

f t

iT la trung die

m cu

a BC, AD Va d

o

hin

h

chop S

ABCD la

ASIL la tam gia

c can m

a goc

a

day S

it

= SL)I =

c p

va SH la dudng

o BD

Ma

t phan

g phan gia

c (a) cu

a nh

i die

n can

h

BC chin

h l

a m

at phan

g d

i qua B

C va difdng pha

n

gia

c I

K ciia Si lt v

a

K la trung die

m cua M

N

Tif

do

de dan

g thay ran

g thie

t die

n BCMN la m

ot

hin

h

than

thang BCM

N du

SBCM

N

= -(

BC + MN ).

IK

Theo din

h ly ham sin trong ALK

I ta c6 :

IL

sin

Theo dinh l y h a m sin t r o n g A S K I , ta c6 :

Do (SKI) 1 ( B C M N ) nen difcfng vuong goc SJ ha t i f S xuong I K cung c h i n h \k dUdng cao

cua hinh chop S B C M N T r o n g t a m giac vuong S J I ta c6 :

a/ Chiifng m i n h r a n g t a m giac ASC la vuong

b/ Tinh the t i c h va dien t i c h toan p h a n ciia h i n h chop S.ABCD

Chutng m i n h rang goc n h i dien tao nen bdi cac mat SAB va SAD la vuong

G i a i

a/ Goi O = AC ^ B D : t r o n g t a m giac vuong O A B ta c6:

OA'' = AB^ OB^ = a'

-IV3J

2 a ^

g ta

= a' -

A = OS =

g tuye

n S

O

bkng niia canh day

C, AS

C 1

^ tam giac vuon

g ta

i S (dpcm)

b/ T

a CO : VK AB CU = ~ SO.dt(ABCD)

3

Tron

g d

o : dt(

ABCD) = OA.B

D =

> dt(

ABCD) =

V 3

n xe

t rkn

g cd

e ma

t be

n (SAB), (SBC), (SDC), (SAD) l

i S

A =

SC, S

B = SD va tCr gi

a thie

t : AB = B

C = CD = A

D t

hi 4 t

g iln

g bSn

g nhau

Ha BH

1 SA, ABS

n B

H cun

g l

a trung tuyen

Tron

g ta

m gia

c vuong OSA ta

eo :

SA'-* = OS^

+ OA'

^ =

4a'

SA =

2a

AH = SA

a

BH

AB'

- HA

^ =

Ja 2

a 3

ABCD) + 4dt(SAB) =

2a

^4

2 4a

^y f2

DH la dadng cao ciia tam giac ca

n DSA

Do vay

iJH

fi =

(

^ =

[(SABiTisAD) ]

Ta C O : DH^

= BH^

* =

2a^

DII^ + B H ' = 4a' = BD'^

3 3

=> ABHD vuong tai dinh H

=> Goc nhi dien canh (SA) la nhi dien vuong (ycbt)

Bai 189 (DAI HOC K I I O I B, N - 1975)

Cho mot tarn giac vuong can ABC, AB = AC = a BB' va CC cung vuong goc vcJi (ABC), d cung mot phi'a doi vdi mat phang do va BB' = CC = a

a/ ChiTng minh rSng tarn giac AB'C la tarn giac deu

d Chufng minh rang nam diem A, B, C, C , B' ciing nSm tren mot mat cau Tim the tich ciia

hinh cau tifang ijfng

Do OH la dtfdng trung binh cua tam giac BCC

^ OH = C C =

-2 -2

Do do O la tam hinh cau di qua nam diem A', B, B', C, C va the tich Vc ciia hinh cau do bkng:

V , = - 7 : R 3 = -n

3 3

i 190

(DA

I HOC KHO

A HO

C KHOI

A 1977)

-Cho AABC deu canh

a, no

i tiep trong mot difdn

g tron tam

O chiif

a trong mpt mft

t phSn

g

(P) Go

i D la di6

m xuye

n ta

m do'

i cu

a A tre

n dudn

g tron nay, co

n S

D la m

a a va vuong goc v(J

i (P)

1/ Chufn

g min

h SA

C v

a SA

B l

a nhOTn

g tam giac vuong

2/ Tinh die

n tic

h toan phSn cua hin

h cho

p SABC

3/

Djnh tam ciia hin

S,

A,

B, C, D

(theo djn

h ly

3 difdn

g vu6n

g goc)

a A

JA D 1 B C

(tai

H )

^ [S

D

± (P )

va

A B 1 B

D

AB _L SB =i

> ASA

B vuon

g ta

i B (dpcm)

Ly luan ti/ang

t

ii

ta eung c6 ASAC vuong t

ai C (dpcm)

2/ Go

i

S |p la die

n tic

h toa

n phan cua hin

h cho

p SABC, ta c6

S,p = dt(

AABC) + dt(ASAB) +

dt(SAC) + dt(ASBC)

dt(AABC) =

Tron

2 SD

2

= + a

SB^

= 3a^

9

2 12a

^

+ a =

SB =

2aS

Ne

n : dt(ASAB) =

~ AB.S

B =

- a.SB

Xot:

S H^

= H D^

+ S D^

faV3

-.SH

^

=

^

^-

36 SH

=

1 V 39

36 3

6

=>dt(ASBC)

=lBC.SH=

—

— + — +

= (1

1 +

Vl3 ) (ycbt)

3/ D

e y den ba tam giac SAB, SAC, SA

g ta

D, S nkm tre

n ma

t cl

u, bd

n kinh

R =

cua hin

h ca

u la trung die

m I ciia doan SA (ycbt)

SA V(aV3 )2 + a

Bai 191 ( D A I H O C K H O I A - 1976)

Cho mot tarn dien ba mSt vuong Oxyz Ngifdi t a l a y I a n lugt t r e n Ox, O y , Oz c a c d i e m P, 'Q, R cung khac d i e m O Goi A , B , C theo t h i i tiT l a d i e m giuTa ciia c a c doan P Q , QR, RP

a/ Chifng m i n h r i n g c a c m a t cua k h o i tuT dien O A B C l a nhOfng t a r n giac b a n g nhau

b/ Cho OP = a , OQ = b , OR = c T i n h t h e t i c h k h o i tuT dien OABC

c/ Tim tarn ciia m a t cau ngoai tie'p k h o i tiir dien OABC

il Chijfng m i n h r a n g t o n t a i m o t m a t cau tie'p xiic v 6i c a bon m a t cua k h o i tuf d i ^ n O A B C

Tim tarn m a t cau do

fi Chijfng m i n h r a n g ne u c a c goc n h i die n O A cua k h o i tuf die n O A B C l a g6c n h i die n vuon g

ihi hai goc B v i C cua t a m giac A B C thoa m a n h e thufc : t a n f i t a n C = 2 v a nguac l a i

=> Cac m a t cua t a m dien O A B C b a n g nhau (ycbt)

b' Goi H la cha n dirdng cao cua tiif die n O A B C h a t i f

0 va h c h i n h l a daotng cao cua tiJ die n OPQ R h a t i f O

=> giao d i e m I c h i n h l a t a m m a t c a u n g o a i tie'p tuf d i e n O A B C , ( v i l A = I B = I C = l O = R) d/ Tif I h a I O 3, l O , I a n lUcrt v u o n g goc vdri m a t p h l n g ( A O B ) v a (OAC)

=> 0:t; O 4 thuf tiT l a t a m v o n g t r o n n g o a i ti6'p t a m g i d c A O B v a AOAC

V i bon t a m g i a c O A B , O B C , O A C , A B C b a n g n h a u n e n c a c diTcrng t r o n n g o a i tie'p c h u n g cung b a n k i n h R j

I O2 O

ta c6 :

lOi =

I O2

lA = 10 = R

OiA = O 2O = R

= lOi

; IO

4

= IO

2

= > lO

i =

I O2 = IO :J

= IO

4 = r

(tinh

chat h&

c

cau)

= > I chfnh

la ta

m m kt

cau tiep xuc vd

Ha tiJ C : CH

1

OA ^ CH

_L

(AOB)

Noi H

B ta

C O

Ha ti

r B : BH'

1

OA = > BH' =

CH, v

a O

H = BH'

Do AAB

C = ACOA = ^ C = CAD

OH

tan

6 ta

n C =

HC H

C HC

^

AH O

H

AH O

H

Xet de

n :

O A^

= ( OH + A H) '

= O H'

* + A H'^

+

2AH.0H

Trong tam giac vuon

g H BC

ta c6 :

O A' = C' = H C ' + H B '

= H C ' + H B ' + H H ''

= 2HC'

+ ( AH

- O H ) '

= 2HC

' + A H ' +

O H'

20H.

-AH

Tir (2 ) v

a (3 ) t

a c

6 : H C ' = 2AH.

OH

Thay vao (1) t

a dua

c :

tanfi.tanC =

2

Ngugc lai, gi

ha AH 1 O A.

Ta CO : tanCJO

A = ; tanCX

O =

OH A

± OA

Va

AC O

A

= AB AO : C

H

= BH ';

O

H

= AH '.

Ta c6:

H B '

= BH '' + H H '' = B H' ' + (A

H

- O H ) '

^ HB '

= C H' + A H' + O H' -

2AH.0H

= A H' + O H'

=> C H ' + H B '

= C H ' + A ' + O H '

Mat kha

c t

a c6

: BC '

= O A'

= ( OH + H A) '

EC '

= A H' + O H' +

2AH.0H

=

H' + O H' + CH '

Tom lai t

a c6:

B C' = C H ' + H B '

AH B C

vuong tai

H <=

>

C H

1

B H

Ta d

a c6

C H

_L

OA B) =

>

(C OA )

1

( AO B)

< => nh

i die

n ( OA )

la nhi die

n vuon

g (dpcm)

(1) (2) (3)

Trong mot mat phang ( P ) ngifcri ta cho mot tam giac vuong can ABC v d r i AB = AC = a Bu

Cv ngUcfi ta Ian lUcrt lay nhCfng diem M va N di dong sao cho tam giac A M N vuong goc tai M

1/ Hay tinh y khi x = a; tinh dien tich tam giac A M N , suy ra cosin ciia goc nhon hop b d i mat

phang (AMN) v d r i ( P )

2/ 1 la trung diem cua BC Chiifng minh rSng goc A M I la goc p h l n g cua nhi di$n c6 canh la

MN va CO cac mat p h i n g BMNC va AMN Tinh gia t r i ciia goc nay k h i x = a

3/ Chufng minh :

al Bon diem C, I , M , N cung nkm tren vong tron

hi Nam diem A, C, I , M , N cung nkm tren mot mSt cau Hay xac dinh tam cua hinh cau nay

Cho hinh chop V.ABC c6 cac mat ben hop vdi mat p h i n g day ABC thanh nhifng goc nhi

dien bang nhau c6 goc p h i n g la (p

1/ Chutng minh rang chan cua dii&ng cao hinh chop xua't phdt tif dinh V la tam cua dUdng tron npi tiep trong tam giac ABC

2/ Tim diem O each deu cdc mat ben va mat day hinh chop

3/ Goi r la ban kinh difdng tron noi tiep trong tam giac ABC va R la bdn kinh hinh cau noi

tiep trong hinh chop V.ABC, tinh R theo r va ip

1/ Goi

Giai

I la h i n h chieu cua V xuong (ABC)

J , K, M la h i n h chieu cua V len cac canh A B , BC, CA

V J l A B

Ba tam giac vuong VIJ, VIK, VIM bkng nhau (vi c6

chung canh goc vuong V I va c6 goc nhon (p bkng nhau)

IJ = I K = I M

Vay I la tam ciia dacfng tron npi tiep cua tam giac

ABC (dpcm)

2/ Gia sijf da t i m difac diem O each deu 4 mat

cua hinh chop V.ABC Goi Oj, O 2, O3 Ian lugt la

hinh chieu ciia O tren cac mat (VAB), (VBC),

(VCA) theo thOf tif do

Ta CO : 01 = OOj = O O 2 = O O 3

01 = 0 0 ] => O d tren phan giac ciia

01 = OO

Vay chi ca

n O la chan dudng phan gidc cii

a X'J

l l

a d

u (ycbt)

3/ Tin

h R theo

r v

a 9

Ta

CO

: R =

O2

;

IJ = I

K = IM = r

9 - = OKsin O2 O => K 2 OO A Trong

Trong AOIK =^

OK =

IK cos—

2

Thay (4) va

o (3), t

tan—

(ycbt) 2

(3) (4)

n—

—2 =

rtan

(DAI HO

C Y

- NH

A DtfCJC

- QUA

N Y TP.HCM

- 1980)

Trong khong gian cho mot doa

n thin

g IJ Tre

n mo

t dUdn

g thin

g d vuong goc vd

m A, B vdi A

I = IB = a

m C, D vdi C

J = JD = b Ca

• C va

ABII

J

JC = J

D

CDIIJ^

y r

a A

C = BD (dpcm)

Lap luan tifcfng ti

f t

a c

6 : AD = B

C (dpcm)

2/ Go

i O la tam ciia hinh cau ngoai tie

p tij

f die

n ABCD V

O pha

i d

tren dudng thing

IJ

3/ Ta

CO :

= OA

^ = 10^

+ lA

^ = + O

F

R2

= OC' -

Oj 2

+ JC^

= +

IJ -

Olf

R2

= + (c -

-(1)

(2)

TC r (1) v

a (2), su

Bai 195 (DAI HOC SLf PHAM - NONG NGHIEP - TONG HOP - K H O I B - 1981)

Cho goc tam di?n Oxyz vdi cdc goc p h i n g d dinh x 6 y = 60°; y 9 z = 90°; z 6 x = 120°

Tren Ox, Oy, Oz lay cac diem A, B, C vdi OA = OB = OC = a

hi Chijfng minh rSng trung diem H cua doan AC la hinh chieu vuong goc cua O len mat

d Xac dinh tam I va ban kinh hinh cau ngoai tiep til di?n OABC

=> H la hinh chieu cua O xuong mp(ABC)

Luc do the tich V cua hinh ch6p OABC la :

c) OH la true (d) cua AABC

Tam I cua hinh cau ngoai ti6'p til dien OABC la giao dilm cua OH vdi mSt trung trUc (a)

cua OC

Ta c6: 10 = l A = IC => AlOC c&n c6 Xdl = 60° => AIOA diu

=> 10 = l A = IC = a = R (bdn kinh hinh cau) (ycbt)

Tr on

g ma

t ph an

g (P ) ch

o nOf

a du dn

g tr on d ud ng

k in

h

AB = 2R Gi

a suT M

a hi nh c hi eu v uo ng goc cii

a M l en

m nuf

a du dn

g th an

g Mu v uo ng g6c vd

i ) ta

i M, g ia t hi et M

u lu on l uo

n d v

oi vd

i ) kh

i M t ha

y do

i tr en M

u la

y di em

S vd

i SM = MH

a/ Ti nh do d

ai ca

c ca nh c ua tuT

ie

n SA BM

b/ Ch uf ng t

o ra ng goc nh

i di en c an

h AB t ao b

di cd

c ma

t ph

^n

g (P ) va ( SA B) c

i M t ha

y do

i

d X

ac d in

h

vi tr

i t

ar

n

hi nh c au n go

ai ti ep tiif di en S AB

M Ti nh b an

k in

h

hi nh c au

a tr

i n ao c ua

x th

i b an

k in

h

ay c6

gi

a

tr

i I dn n ha

t ?

Gi

ai

(X em D

l DA

I HO

C SP

- KT

- KT

- NN

- KH OI

A 1 98 2)

P K

T K

T N

N K HO

-I A

- 19 82 )

Tr on

g ma

t ph ln

g (P ) ch

o nuf

a dUcfn

g tr on d Ud ng

k in

h

AB = 2R Gi

a thC

t M l

n, kha

c vd

i A v

a

B, gp

i H l

a hi nh c hi eu v uo ng goc ci ia

M tr en A

g MN v uo ng goc vd

i (P ) ta

i M.

G ia t hi et M

N lu on l uo

n d v

oi vd

i ) kh

i M t ha

y do

i, tr en M

N la

y di em

S vd

i SM = MH

a/ Ti nh do d

ai ca

c ca nh c iia tuf di en S AM

B

b/ Ch uf ng t

o ra ng goc nh

i di en c an

h AB t ao n en b

di ca

c ma

t ph in

g (P ) va ( SA

g do

i kh

i M t ha

y do

i

d X

ac di nh v

i tr

i t am h in

h ca

ii ng oa

i ti ep tuf di en S AB

M

Ti nh

ban

k in

h

hi nh c au

a tr

i n ao c ua

x th

i b an

k in

h

ay c6

gi

a

tr

i I dn n ha

t ?

Gi

ai

a/ He t hd

c li io ng t ro ng t am giac c ho :

H X

A

B = 2R(2R

- x)

BM

= V2R(2R

- x)

MH

^ = A

H

X

B H = x(

2R

- x)

MH

= Vx

(2R

- x)

MS

= M

+ MA

^

= x(

2R

- x) + 2 Rx = x(

4R

- x)

=>

S

A = Vx(4R

- x)

SB

^ = M B ' + M S' = 2R(2

R x ) + x (2

R x ) = 4

-R'

- X'

=^ S

B =

V

4R

2 - X

^ ; A

B = 2

R (y cb t)

b/ Go

c nh

i di en t ao b

di (P ) va ( SA B) k ho ng d oi

S

M 1 ( P)

M

H 1 A

SHM

( AS MH v uo ng c an )

SH

1 AB ( di nh l

y 3 d Ud ng v uo ng gdc)

=> (p = — = const (ycbt)

4

c/ True cua A A M B l a dudng t h S n g (d) J_ (P) t a i O ( t r u n g d i e m A B )

Luc do t a m I ciia h i n h cau ngoai tiep tijf dien S A M B la giao d i e m I

cua (d) va mat phang t r u n g triTc (a) cua S M

Ban k i n h R, cua mSt cau l a : l A = I B = I M = I S = R i

2

Bai 198 ( D A I H O C K I E N T R U C - T O N G H O P - N O N G N G H I E P - 1984)

Cho ba nijfa difcfng t h a n g Ox, Oy, Oz k h o n g cung n k m t r o n g m o t m a t p h a n g va m o t d i e m A nam tren Oz H i n h chieu vuong goc cua A xuong m a t p h a n g (Oxy), Ox, Oy I a n lUcft l a A ' , B , C

1/ Chufng m i n h rftng neu x O z = y O z = a, t h i OA' l a ducfng p h a n giac cua x6y \k B C l a

triTc giao v d i OA

2/ Cho biet x O z = y5z = a (45° < a < 90") va x O y = 90° Goi (5 l a goc hgp b d i difcfng t h i n g

Oz va m a t p h a n g (Oxy); y l a goc n h g n hgp bdi m a t phSng ( A B C ) va (Oxy) T i m he thiifc giCfa

tanP va tany; giOa cos^ va cosa Suy r a bieu thiJc cua tany theo t a n a T i n h P neu a = 60° 3/ Cung gia t h i e t nhiT d p h a n 2, h a y xac d i n h t a m cua h i n h cau ngoai t i e p tijf d i e n O A B C va xac dinh t i e p dien cua h i n h cau nay t a i A Cho b i e t OA = a va a = 60° H a y t i n h dien t i c h thiet dien cua m a t phSng (ABC) v d i h i n h cau t r e n

1/ Theo d i n h l y ba dudng vuong goc

=> D i e m A' n S m each deu 2 canh cua x O y

=> A' nftm t r e n dudng p h a n giac A t cua goc do

=> OA' l a dudng p h a n giac ciia x6y (dpcm)

De y AOBC can, c6 OA' l a dUcfng p h a n giac => OA' 1 BC

: AA' ± (OBC) =

> AA'

21

Ta

CO :

Nen

AOX'

= p = [Oz

; (xOy) ]

y (2)

2

Tir (1

) v

a (2) =

> OA'tany

O

B = OA.cos

V2

(dudng ch6o cu

a hin

h vuon

g OBA'C)

=

V2

OA.cosa <= > cos

P =

V2

cosa

a, P la hai g6

tan^y = 4tan^p

= 4

^cos^p -1 2cos^a

<= > tan

^ = 2(tan^a

- 1) v

i y 1^

1 + tan^a) -

Neu

a = 60' ,0 cosa

= — =>cosP

=

V2

cosa =

2

2

P = 45° (ycbt)

3/ X6

t :

ABt )

= Act ) =

90°, ne

n B, C n^m

Ta biet tie

g (T) vuon

g go

c \(s\g kinh O

t O

y ta

i C

Mat phan

g (T) =

(AB'C)

1 O

A ta

i A nen

n6 \k

tiep dien phai x^

c din

h (ycbt)

a

Ng'u OA = a va

a = 60°

OB = O

C =

BC =

^,A

B

= A

C = ^

Goi M 1^

AAJM

CO AAB

H

AJ A

M

AB ~ A

(3)

kinh OA' c6 day BC 1 OA' nen OA' cSt BC tai trung diem H cua BC Ro r^ng OA' la dUcfng

trung true cua BC nen OB = OC Hai tam gidc vuong OAB va OAC c6 OA chung OB = OC nen ehiing bang nhau

Vay menh de : "BC trUc giao vcJi OA t h i xOz = la dung" (ycbt)

Bai 199 (DE BACH KHOA - K I E N TRUC - TONG HOP TP.HCM - 1 9 8 5 )

Cho hinh chop tuf giac deu S.ABCD c6 dinh la S; canh ben la b; mat ben hcfp vdri day g6c a 1/ Tinh canh day, dien tich toan phan va the tich cua hinh chop

2/ Trong trudng hop tam cua hinh cau noi tiep trung vdi tfim cua hinh cau ngoai tiep hinh

chop hay chufng minh : cos a = V2 - 1

3/ Cho biet cos a = 1 ; va goi (p la goc hap vdi canh ben va mat ddy

a/ Tinh tancp

b/ Trinh bay day du each difng thiet dien tao bdi mot mat phang (a) di qua dinh A, song

song vdi difcfng cheo BD va hgp vdi canh AB mot goc 3 0 "

G i a i 1/ Goi I la trung diem canh ddy BC => S l ^ = a

Trong tam giac vuong SHI => H I = Sl.cosa

Mat khac : SI -L BC va trong tam giac vuong SIC ta c6 :

^ a

Vl + cos^a

2bcosa

Tif d

o : BC = (ycbt) 1 + co

Ta goi die

dt(ABCD) =

4 SI.B

+ cos^

a 2bcosa

a =bsina

1

+ co

s a

Vay : V^,,,™

2

2/ Gi

a sij

f O la tam hinh cau noi tie

Ki e

SI

fOH = OK , O

S =

OC

Theo gia thie

O + OH = O

C + OH = H

tana = = tan

HI

2 +

tan2^

Mat khac => cosa = 2 , 1 - ^ + 1 V^ _ i ( d p c m )

3a'Ta c6 ; SCfi = ip Mat khac theo (1), thi :

Vdi gia thiet cosa = ^f2 - 1, da c6 duoc : tan^— = V2 - 1

=> BK' nkm tren mat phang vuong goc vdri BD Goi do la mat ph^ng (BB'y)

Gia siJ MN o BB' = I; By n Ax = E ^ (a) n (BB'E) = IE va K' e IE

Tarn giac vuong AK'B c6 ICAB = 30° ^ KB = - AB

2

n gid

c vuong AEB

c6

B

E =

AB yf2

K 'B

2

=>

ABKE vuo

ng

can c6

BElt'

= 4 5°

AB V2

AE

BI vuon

g ca

n c

6 : B

E = B

I = —

2

B^n

g phan tich tren

day ta

su

y r

a cd

ch difn

g nhu sa

u :

• Trong

m

at

phin

g S

BD

difn

g B

B' 1 B

D,

tre

n B

B' lay

diem

I sao cho:

AB V2

BI

= B

E = —

Iz c^

t S

B ,

S D

tUang ufn

AF ca

AMQ

N l

a t

hie

t dien

ca

n d

Ung

•

De y

thay v

di

gia

thie

t

da ch

o,

thie

t die

n luon luon difn

g duac ThU

g v

di

FH <

S

H

Nhifn

V V2 +1

AB V2

C BACH KHO

A TP.HCM

- 1988)

Cho m

ot hin

h chop ta

m giac de

u D.ABC

c6

canh da

y a

, ca

c m

at ben nghien

h

cac

di§

n tic

h

s v

a S cu

a ca

c m

at cau n

oi

tie

p, ngoa

i tie

p vd

i hin

h chdp

an

S ,p cu

a h

in

h chop theo

a v

a a

b/ Bie

u die

n S

t p

theo

s va S

g tam AAB

C

=> DH

la

difdng

ca

o h

in

h chop

Go

i O

, I, r, R Ia

n lifg

t l

a tam v

a ban

kin

ai tie

p hin

h chop D

o t

in

h d

oi xilng

cua hin

h chop deu

nen d

De

y thay:

itEt)

=

CJETD

=

-

; lA = I

D = R

2

r =

O

H = HE.tan —

= —

s

=

AT IT

=

2V

3

tana

AHIA

=>

lA

^ = AH^

+

IH

^ = AH^

+

(D

H DI)^

-hay: = tana - R R = a(tan^ a + 4)

4V3tana

Vay d i$n ti c h SR cua m at cAu ng o ai tiep

Mat khac : D E =

48tan^a

HE a

12tan^a cos a 2V3cosa

Stp = S „ + Sd = 3SnBC + SABC = 3 | a D E + ^ a - " ^

a^Vsd + cos a)

4 cos a 3S

2 cos a(l - cos a)

Cho h i n h cho p deu S.A BC c6 canh d ay la a, dUdng cao SH = h

1/ Tinh a theo h v a b an k i n h r, R cua m at cau np i tiep , ng o ai tiep cua h i n h ch6p

2/ Khi a co d i n h v a h thay d o i, xac d i n h h de t i so — d at g ia t r i \dn nha't

R

G i ai

1/ Goi H la trifc ta m (ho ac tro n g tarn) A A BC

=> SH = h la dUcfng cao cua h i n h cho p S.A BC

i tie

p deu nSm

tren

SH

SH ± (ABC)

SM ± B

C

BC

_L

SH va BC ± S

M B

C ± (SAM)

= > (SBC)

1

SM

thi (o

J ± (SBC) A

Do (o

H = (oJ = r nen

a

o

T

«J Sc

o SH

HM

SM

SM

(h -r )

aV s

h

~

r

= > r =

12 h2 +

a

Trong tam giac SAM, trun

SO

SI S

SA SH 2SH

SA

2 9h 2+

> R = (dpcm)

2/ Xe

t :

y = — = 18h

6h 6h

6a h^

R (a 2+

3h 2)(

a + Va '+

12 h2 )

<= > y =

1 2

12—) 3 I 3

Dat : 12

X

(4

+ t an

^ x)(

l + V

l + t an

^x )

•» y =

,s in

^

X

CO S^

X

2sin^

x co

x)

Hinh tron tam O b i n kinh a noi tiep trong hinh thang ABCD Ian lUdt tai M , N , P, Q

Day nho DC = 2

x =

a

Ddy Idn AB = 8

x = 4a

Canh xien BC = D

A = 5x = (ycbt)

5a

dt(ABCD) = -(AB + CD).M

P = -(4a + a).2

a =

2/ Tac

6 S0_L(ABCD)1

Q i DA

; S

M 1 AB; S

N ±

BC

Do

OM = O

N = OP = O Q

= a = >

SM = S

N = SP = S

Q

S„

=

SM.A

(AB + B

C + CD + DA

S„,=

i V4 a2 + a2 5a

53^1 + a + — 4a

+

2)

= -aV5.10a

= 5V5a

Sd =

5 ^/

=

5 a2 2a =

BC

1 O

N BC

1 (OSN)

=t > (SBC) J (OSN) the

o gia

o tuye

n SN, ne

n h

a ti

f O, O

H 1

SN

= > O

H ± (SBC) =

> O

H = d[0; (SBC)]

Lap luan tifcfng tU

, h

a O

K 1 SP; O

I 1 SQ va OJ

1 S

M

= > OK, 01, O

t (SDC),

(SDA) v

a (SAB)

De y den bon tam gi^c SON, SOP, SOQ, SO

J cun

g

bang nhau

> E

e SN, v

a w

E // OH

hinh

ch6p S.ABCD (ycbt

K

AOSN OH^

ON^

4a^

<= > 0

H = OS^

a

^

2aV5 4a

^

4 a=

^

ASwE 0 0 ASOH : caE Sco S O - c o O S O - c o E

Cho h i n h ch6p tuT gidc deu S A B C D , c6 canh My hkng a XSfe = a

1/ Xdc d i n h t&m -vk hAn k i n h m a t cau ngoai tiep h i n h ch6p

3/ Chthig m i n h r a n g h a i t a m m a t cau d6 t r u n g nhau k h i vk c h i k h i a = 45°

G i a i 1/ Trong m a t p h ^ n g (SAO) m a t t r u n g trifc (P) cua SA c^t true difdng t r 6 n ngoai ti§p ( A B C D )

Trong (SHO), m a t p h a n g i i c cua n h i d i ^ n (AB)

chia doi g6c S f i t ) se cat SO t a i I

- 2 2

SO = V SH

OH

^-^

SO Vcos

a

2sin

—

2

ta np =

OH

+

2si

n

— t

- Vcosa =

ct

2 a

a =

-s in

— + c os

- V2 co

a J :

2^4)

Vcosa cosa

+ cos- = f tan-!^ o : Luc d

a n— si cos

r =

-J ^

^ (*

) (ycbt)

C O

K =

> AIO

B = AIKB

^

= Act ) (cun

g cha

n cun

g AB)

< =>

a = —

4

Kiem tra sir dun

g dS

n cu

a (4) d

e dan

g nh

d (*

) (dpcm)

Cho hinh ch6p S A B C c6 day A B C la mot tam giac can c6 A B = A C = a, ( S B C ) ± ( A B C ) ;

SA = SB = a, S C = X

1/ Chijfng to rang B C la ducrng kinh ciia difdng tron ngoai tiep tam giac S B C

2/ Hay xac dinh tam va ban k i n h cua hinh cau ngoai tiep hinh chop tren

Vay B C la diTcrng k i n h cua dudng tron ngoai tiep A S B C (dpcm)

21 Goi O la giao diem gifla difdng trung trifc cua A B vdi A I

Cho tuf dien A B C D

1/ ChuTng to r i n g cac difdng t h i n g noi moi dinh vdi trong tam ciia mSt doi dien dong quy tai diem G

2/ Neu di6m G triing vdi tam hinh cau noi tiep, chuTng to r i n g cac mat cua til dien la bang nhau

Giai

1/ Goi M la trung diem D C ; A' va B' Ian lifcft la trong

ABCD va A A C D

tam

g AABM, t

a go

i G

la giao di^

m cu

a AA'

v dr i

BB'

Theo

tinh

chat cii

a trong tam, t

a c

6 :

AM B

M

MB'

AG MA'

BG

= 3 ^ A

B' //

AB

AB A

M

Vay : B

B' c&t

AA' ta

i G

co dinh

Tuan

g tu, ca

c dirdn

g thSng noi C

v6i t

rong tam

C

cua AAB

D no

i D v

di tron

g ta

m D' cu

p luan

do (dpcm)

2/ Gi

a sij

f G cung la tam hin

h ca

u no

i tiep tuf dien

Dimg : C

H 1 AB; D

K 1 AB.Xet: SAB

C =

SABD

=>

Goi M;

N; O la trung die

m cu

a KH

; CD;

HD t

hi :

AMHC = AMD

K MC = M

DK =>

A

B IM

O

NO //

CH

ABI

NO

CH =

N 1 A

D d

i qu

a

trun

g die

m N cua C

g die

m M

cua A

B

jMA = M

B

^ [UK = M

H

AC HB = AD KA B

C = A

D

Ly luan tifan

g t

ii :

C = BD;

A

B = C

D

Vay cac ma

g nha

u (dpcm)

Bai 20

6

(D AI

HOC

T ON

G HOP

Khoi A

- B 1994)

AK = B

H

ir mh

ABC la tam giac ca

n AB = A

C = a; (ABC) 1

(SBC) va

ot ta

m gia

c vuong

2/ Xa

c dinh tam

v

a ba

n ki'n

h hinh cau ngoai tie

p hinh chop, bie

t S

C =

x

Gi

ai

(Xem De DAI HOC TAI CHANH K

E TOAN

- 1993)

Ba

i 207

(DA

I HO

C Y DLfOC

TP

HC

M 1994)

-Tron

g ma

t ph

^n

g (?

) ch

o mo

t di/crn

g thang (d) v

a mo

t die

m A ngoai (d) Mo

t (d) ta

i B va

C

Tren diXctng than

g qu

a A va vuong g6c vd

K l

a ca

c hinh chieu vuong goc cu

a A tre

n S

B va

SC

1/ Chilng min

h

A, B, C,

H, K thupc ciin

g m

ot ma

t cau

2/ Tinh ban kin

h ma

t ca

u tren biet AB = 2, A

C =

3, BAC

= 60"

3/ Gia

S Lt

tam giac ABC vuong t

ai A ChuTn

g minh rkn

g ma

t ca

u ngoa

i tiep

K luo

n

di qu

a mo

t dudn

g tron

co dinh k

hi S tha

y doi

Gi ai 1/ Goi AA' la dtfcrng k in h dudng tr on ngoai tiep

Vay A; B; C; H ; K ciing thuoc mSt c l u

tam W difdng k in h AA' (dpcm)

cau ngoai tiep kh oi da dien A B C H K di dong tr en (d) co dinh

Vay kh i S thay doi, mat cau chOfa di/dng tron qua A nhan (d) la m true (dpcm)

Bai 208 ( D A I H O C Q U O C G I A H A N O I - 1 9 9 5 )

Cho h in h tiir dien AB C D c6 canh AB = x H a i mdt (ACD ) va (B CD ) 1^ nhflng ta m giac deu canh a Goi M la tr u n g diem cua canh AB

a/ Xac dinh x k h i D M la dircrng cao ciia h in h tiit dien AB C D

^ B C D

G i a i a/ Do D M la diidng cao tijf dien AB C D (hay hinh chop D.ABC)

Ma ta CO :

= S ACAB =

SADCA = SADB

=> St

p = SDAB + SCA

B + S^CA + SDB

C = (

yc bt )

Goi r la ban kin

h m&t

cau n

2 V3)

Ba

i 20

9 (D

AI HO

C NGOA

I THl/

dN

G D

-E 3

- 1994)

Cho hinh chop tut gia

c de

u S.ABC

D din

h S, can

n kinh

R v

a r cua ca

c hinh cau ngoai tie

h hinh chop, Vj, l

a th

e tic

h hinh cau ngoai tie

p

Vj la thi t

c dinh quan h§

giijfa a va

a tr

i Idn nhS't

b/

dat gi

a tr

i \dn

nha

h vuon

g d

day va H

K l

a doa

n trung trUc cu

a S

C tron

g ASOC, K

KD

Kh

i do, ba

n kinh hin

h ca

u ngoa

i tie

p l

a :

R = S

K

SK S

O

SC

^ 2.S0

,2 (1)

+

Suy r

a ti

r (1) :

R = ^ (ycbt)

4h

Goi M la trung die

dSf e

trong AOMS

n va

m

at dd

y cu

a hinh chop deu

bang

nha

u, ha

y N la t

^

^ S0 _

^ 2h MO

a SMt> n ta D = tanNM t = Dat:

=> h(t) = h.t^ + a.t - h = 0 ; 1 ^ t > 0 (c6 : Ah = + > 0)

U =

t2 =

2h Va^ + 4h^ - a

• The tich hinh clu ngoai tiep: V j =

• The tich hinh cAu npi tiep : =

( 2 h 2 + a 2 ) a + V4 h 2+a=

4h2

n3

+ 1

ti r:

HC

M 1995)

-Cho goc tain dien Sxyz din

h S

v6i

5cSy = 120°

, zSy

=

60°

, 5c§

z = 90°

Tr

m

A, B, C sao cho SA = S

B = SC = a

1/ Chufn

g t

o rang ABC la m

pt tar

n gid

c vuong Xa

c dinh hin

h ca

u no

i tie

p ti

l die

n SAB

C the

o a

2/

Tin

h go

c phin

g cu

a nh

i die

n can

h AC

Gi

ai

1/ Ta C

O : AC =

V SA

Ja'^

2a^

cos = a

^ = A C^

+ BC

^ =

3a^

=>

AABC

vuong t

ai

C (ycbt)

Goi H la trung die

m

AB

=> S H I A

B (1 )

Dim g

: M

H / / B

1

S

M

Nen A C

1

( SM H)

AC

1

S H

(2)

Tir(l)va(2) =:>

SH

1

(ABC)

Vay

H la hin

h chie

u cii

a S xufi'ng

m

at phin

g

(ABC) (ycbt)

Ta c6:

Suy ra : r =

^(V3

+V 2

~ 2(V

3 +

V2 + 1)

ta

n cp

= SH

2 1

n

= — = 1 = > a

^n

g nh

i die

n can

h AC bSng cp = — (ycbt)

4

Bai 211 ( D A I H O C B A C H K H O A H A N O I - 1995)

1/ ChuTng minh r^ng tro ng mot hinh ti i dien, 4 doan thi ng no i d inh vdi trpng tam ciia mSt doi dien dong quy tai 1 diem, diem nay chia moi doan thftng ay theo ti so 3 : 1 tinh tiJ d inh

va hinh cau noi tiep, ta deu c6 R > 3r

Dau d ing thiifc a (2) xay ra khi va chi khi hinh cau ay trung vdi hinh cau npi tiep

Phep dong dang phoi canh (vi ti i tro ng khong gian) tfim (G; - — ) bi§'n hinh til di^n A BCD

o

thanh hinh ti l dien A 'B'C'D'

Hinh cau ay c6 c^c diem chung A ' ; B'; C; D ' vdi ca b6'n mSt hinh ti l di^n A BCD

R

=> — > r « R > 3 r (dpcm)

o

Bai 212 ( D A I H O C K I N H T E Q U O C D A N H A N O I - 1997)

Cho hinh ch6p tam giac deu S A BC c6 dtf&ng cao SO = 1 va day A B C c6 canh 2-76 D i l m

M, N la trung diem cua canh A C , A B tUang ilng Tinh the tich hinh chop S A M N va ban kinh

mat cau noi tiep hinh chop d6

Trong ACOM vuong tai M

c6 :

g ti

J die

n deu

S AB

C

« AS N

A

= AS MA

= > SsN

A = SsMA

can tai

S ,

goi

I

la trun

g digm

M

N

=>

M N

= S M '

- MI '

= SM ' -

2

= SM ' -

= > = — r

Ba

i 21

3

(D AI H O

C Q U O

C GI

A H A

NOI

- KH O

I D

- 1998 )

Cho dircfn

g tr6

n tar

n O ban kinh

va A

co dinh), S

A =

h ch

o trUdc, da

p rngt dUdng tr6

h ch6p)

Trong

A SA O

, goi Ky

X h.

tia trung trifc cu

a SA

Goi : O' = Ky

n

Ox =

>

O S

= O

A = O'B = O

C = O'D

vuong tai

O

=> A O'

= R' =

+ 4 R

- ^ SA.SABC

D

SABC

D -h.AC.BD

Nen : 3

max(VsABC

D ) < ^

o AC va B D dong thcfi la difdng k i n h dudng t r 6 n t a m O

Ma : AC 1 B D Do do k h i A B C D la h i n h vuong n o i t i e p dudng t r d n t S m O t h i t h i t i c h

hinh chop dat gia t r i I d n nhS't (ycbt)

Bai 214 ( D A I H O C DLfOC H A N O I - 1999)

Hinh chop S.ABC c6 do dki cac canh ben bkng / cac mat ben lap vdi mat day goc a (0° < a < 90")

1/ Chilng m i n h h i n h chop la h i n h ch6p deu

2/ Tinh theo I va a cac ban k i n h R, r cua cac mSt cau ngoai t i e p , n p i t i e p h i n h ch6p

3/ ChuTng m i n h ^ < — vk dS'u dang thiifc x i y ra k h i va chi k h i h i n h ch6p Ik h i n h tiJf dien d4u

Vay S.ABCD la h i n h ch6p deu (dpcm)

2/ Trong ASAA', gpi O la giao diem ciia S H va d i t o g

trung tnfc K O cua SA

tich hinh chop

)

2(1 + 3 co s^

a)V

l

+ 3c os

^a

Di?n tich toAn phdn cua hin

^ cosa(

l

+ cosa )

ABC

1 + 3cos^a

/sin 2a

3/ T

a la

i c

6 : 2V

/sin 2a 2sina

(ycbt)

^ 2V

l + 3

co s2 a (l + cosa)

l^|l + 3

cos^a

2c os a (l - cos a ) 2cosa(l -

cosa) -2co

s a + 2cos

a

R (

1 + cosaXl +

3cos^

a) 1 + 3cos

- 2t

^ + 2t

3t 2 + 1

^ ' (3 t2

a

(1)

t X -1

3 R

3

Ba

i 21 5

(DAI HOC MO BAN CON

G TP.HC

M KHO

-I A & B

2/

Chilng minh nam diem S; A

; M

; N

; P cung nkm tren

Giai

BD IAC (vi AB CD 1^ hin

h vuong )

BD ISA (viSAi.(ABCD

)3BD )

BD ± (SAC)

; ma: A

N c (SAC)

• Ti/cfng tU, sijf dung cac k h a i n i e m

vuong goc ta suy r a :

2

Tir (1) va (2) => ba d i e m P; M ; N nSm t r e n mSt

cau (S) dudng k i n h SA

Vay S; A; M ; N ; P cung n k m tren mSt cau (S) (dpcm) ^

Bai 216 ( D A I H O C D l W C H A N O I - 2000)

Cho hai dUcfng t h ^ n g cheo nhau (d) (d') nhan doan AA' = a Ikra doan vuong goc chung (A

e (d), A' e ( d ) ) Goi (P) mSt phAng qua A' va vuong goc vcJi ( d ) ; (Q) 1^ m a t p h i n g di dong nhUng luon luon song song v(Ji (P) va cat (d); ( d ) Ian lifat d M , M ' Con N la h i n h chieu vuong goc cua M tren (P); x la k h o a n g each gifla (P) va (Q); a la goc giOa (d) va (P)

1/ Tinh the' t i c h hinh chop A A ' M ' M N theo a, x, a

2/ Xac d i n h t a m O cua h i n h cau ngoai tiep h i n h ch6p t r e n ChiJng m i n h r S n g k h i (Q) d i dong thi 0 luon thuoc mot du&ng t h a n g co d i n h va h i n h cau ngoai tiep h i n h ch6p A A ' M ' M N cung luon chiifa mot difcrng t r o n co d i n h