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

D i r n g B H 1 I E => N H _L I E ( d i n h l y b a d t f d n g v u o n g

go c)

=> g o c c i i a m a t p h S n g t h i e t d i e n v a m a t d a y l a :
(p = N K B ( h o a c S F t )
D o A N B H v u o n g d B => ta n c p =

NB
HB

( 1)

Ti T e a c h d i Tn g t a s u y r a N l a t r u n g d i e m c u a B B '

H B^

BI '

BE ^

c Va ^ ~ b

1

—

2Va2 . b^
^ ^ ^ c 2 ( a 2 + b 2 )
2

ab

NB =

ab

Va :

( 1)

=> t a n t p =

CO S

a^b^

(p

ab
CO S ( p

=

V a V + b V + c V
D i n h l y c h i e u h i n h c h o t a : S ' = S.co s ip
^

S =

S'

cosip

(vi S l a d ie n tich th ie t d ie n )

= - VI v 7 b V7 ^

(ycbt).

4

U o a i S : M A T P f l A« G I A C C U A NflJ D i t N
I . P H U O N G P H AP
Co

so cu a ph ifan g

phap

di/ n g m a t

phan

giac

th e o

dinh

n g h i a c a n p h a i t h i f c h i e n b a h xid c ca b a n :

B 3 : D u n g m a t y = (d; Mc ) vdi d = a

•

B2 : D irn g tia p h a n giac M c ciia

•

B i : D ifn g go c n h i d i e n

•

a ^

c u a n h i d i e n ( a ; d ; (5).
aMfe
r> (5. T h i y l a

m at

p h a n giac ciia ( a ; P ) .
O

Gh i
d en

VL

c h i i : Mot
ha i

C A C

m at

BAI

d iem

nhi

T O A N

t r en

d ien

m at

bdng

C O

p ha n

g ia c

c6 k h o d n g

ea ch

nha u.

BAN

B a i 178
Ch o h i n h ch o p S . A B C D c6 l a t a m gia c S A B d e u v a A B C D l a h i n h vu o n g. G i a s if ( S A B ) 1 ( A B CD ) .
X a c d in h m a t p h a n giac ciia cac n h i die n ( B ; A D ; S ) .

Giai
Go i G la tro n g t a m A A B C deu.
Ke A G n S B = M

( AMN ) o S C = N

Th e o t i n h c h a t giac tu ye n s o n g s o n g

t a c 6 M N // C D .

De y AD ± (SAB)

1

! AD

SA

[ AD 1 SB
D o n g th c fi A M l a p h a n g i a c

S AB .

D o do m p ( A M N D ) l a m o t p h a n gia c go c n h i d i ^ n ( S ; A D ; B ) ( ycbt) .

10 0


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)
De y den nhi dien (SA), dUng BM 1 SA

OM

la

i SA

=> 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).
• B 2 : Xac dinh mat phan giac (y) cua mot mat ben tiiy y va day.
• B 3 : (d) n (y) = W la tam mat cau can t i m va khoang each
tCr W den cac mat ben hay day cua hinh chop la ban kinh
r cua mat cau.

Sau nay ta c6 the tinh ban kinh inft cdu ngi
tiep hinh chop bdng the tich V va di&n tich todn phdn Stp

0 Ghi chii :
nhiC sau:

r =•

3V

n. C A C B A I T O A N C O B A N

Bai 180
Xac dinh tam va ban kinh mat cau noi tiep tiir dien deu canh a.
Giai
Goi G la trong tam AABC deu. => GB = GD = GC

(1)

Tuf dien deu ABCD canh a.

(2)

=> AB = AD = AC

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
^ ynAG = W
Taco:

W e AG ^ d[W; (ABC)] = d[W; (ACD)] = d[W; (ADB)]
We y

^ d[W; (ACD)] = d[W; (BCD)]

i W la tam mat cau npi tiep tuf di§n deu A B C D (ycbt)
[WH = W G - r la ban k i n h mat cau noi tiep tijf di?n

101


O

D i n h l y difdng p h a n giac \k d i e n t i c h cho: I W =

2 A I . I G c o s ( \^)
AI + IG

= W G = V W ^ ~ G F = J ^^' - ^^' . c o s ^( ^) ^G I ^
K ( A I ^^G I ) ^
Doc gia t i f g i a i => r =

(ycbt)

C a c h k h a c : T a c6 : V = - S , „
t p .- r => r =

2

.2

aV3

1

3V

(1)

= V3 a 2

;

T r o n g d6 :
V = i B h = l —. a .
3
3 ,2
(1)

Do do :

r =

12
VSa^

2

^ aV6
"

(aV3 Y
j K 2 J

f1
,3

aV3 Y
j

-Jla^
12

2

(ycbt).

12

Bai181
T r e n dudng v u o n g goc v d i t a m giac deu A B C canh a t a i triTc t a m l a y m o t d i e m S sao cho

S H = 2a. Xac d i n h t a m v a b a n k i n h m a t cau n o i t i e p tuf d i e n SABC.
Hi^dfng d i n
Tifang t i f b a i t r e n : W l a t a m va r = W H l a b a n k i n h m a t cau n o i t i e p
tuf d i e n S A B C . V(Ji S,p = 3 i . B C . S l V - . B C . A I

U

j

Va V = - B h = 3
3 2
Do do : r = | ^ = - a
Stp

Bai

2
:.2a =

2

(ycbt).

^

182
xac dinh t a m va ban k i n h mat cau noi tiep tijf dien A B C c6 AABC deu canh a cho SA = SB = SC

va goc n h i d i e n (BC) la a.
UiXdng

din

Doc gia t i f g i a i , t u o n g t u b a i t r e n .
Bai

183
T i m b a n k i n h m a t cau n p i t i e p h i n h ch6p tuf giac deu c6 canh ben b a n g c a n h day l a a.
Giai
T a CO : S,p = a^ + 4

aV3l

fl
2

= (V3+l)a^

2

2

SO =

I

,2.

2 >

2

4

Goi V l a t h e t i c h h i n h chop tuf giac deu.

102


V= l Bh = l . a ^ - ^ ^ = ^ a ^
3
3
2
6

Do do : r =

3V
S,p

(V3 - l ) a ^

2(V3

(ycbt).
4-1)

nLGlAlTOANTHI
Bai 184 ( D A I H O C C O N G A N T R U N G LfONG -

1970)

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
Giai
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

diem

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
CO diem chung I , nen p h a i cftt nhau theo giao tuyen (d') qua I va d' 1 (BCD).
^ (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 c6 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 .
Chi CO m o t d i e m duy nha't c6 t i n h c h a t n a y , do la t a m cua h i n h cau n g o a i t i e p t i i d i e n
=5 (dpcm).
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 d u d n g 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.

103


Theo each dUng t r e n day, m i e n t r o n g t a m giac A i B j C i
nkm

han

AAI B, CI

a

t r o n g cua

mien

t r o n g t a m giac A2 B 2 C2

va

0 0 A A 2 B 2 C2

Goi t h e m R, la ban ki'nh dUcfng t r o n n o i t i e p t a m giac
A,B,C,.
=j> R, < R.
M a t khac, t y so dong d a n g cua ban k i n h cua diTdng t r o n
noi tiep h a i t a m giac dong d a n g A i B j C i va A B C bSng t y so
cac canh ciia h a i t a m giac do, tilc l a :
Ri ^
r
•

= 2

«

R i = 2r<=>

2r
»

AB

— ^ - (ycbt).
R
2

Cach khac :
D a t ( ) , ( ) 2 = / , tii cong thiJc Euler : P =
«

=

- 2Rr ^ 0 <K>

- 2Rr.

[ R > 0 (hien nhien)
R - 2r > 0

<=> — ^ — (dpcm).
R
2

B a i 185 ( D A I H O C K H O I A - M I E N B A C - 1972)
Cho mot k h o i tuf d i e n deu SABC. Goi S H l a dudng cao ciia k h o i tuf d i e n d6 va I la t r u n g
d i e m cua S H .
a/ ChuTng m i n h r k n g d i e m I , t r p n g t a m T ciia t a m giac A B C va t a m h i n h cau ngoai tiep k h o i
tuf dien l A B C t h a n g h a n g .
b/ T i n h ban k i n h ciia h i n h cau n o i t i e p k h o i tuf d i e n l A B C theo canh a cua tuf d i e n deu SABC.
c/

Chufng m i n h r k n g ba dudng t h a n g A I , B I va C I tiTng doi mot vuong goc v d i nhau.

Giai

a/

De y do t i n h chat deu ciia tuf d i e n S A B C nen S H chufa I la true dadng t r o n ngoai tiep

AABC va T = H
Goi r l a t a m m a t cau ngoai t i e p h i n h chop.

=> 1=>

I H = IS

r, I , I I t h a n g h a n g (dpcm).

b/ Goi O l a t a m h i n h cau n o i t i e p tuf dien l A B C , t h i O
n a m t r e n I H , dong thcfi O n k m t r e n m a t p h a n g p h a n giac
ciia goc n h i dien canh A B , tufc l a O la giao d i e m cua I H
va dirdng p h a n giac goc I M H , va r = O H la ban k i n h h i n h
cau noi t i e p k h o i tuf d i e n l A B C .

Ta CO : S H

= A/ SA^~MF =

a^-

(2

aS_

3"

^

I

H

=

^

=

2

^

2

6

Xet t a m giac vuong I M H ta eo :
tan

^

I H

(1)

fi

aV3

M H

O
-- r

A/ 3

M

H

104


2

Mat khac :

IM= VMH^ + I H ^ =

Theo cong thilc nhan :

(i>

a

i

4"

6

,

6
V

2

/

2tanM

r ^ = V^
1 - tan^^Mj

o

V2 tan^ M j + 2 t a n M i -

tanMi =

J-LJE.

tanMj =

-1-V3
= — < 0 (loai)

> 0

=0 «

<=> t a n M j = t a n
2

"

V2

Xet tam gi ac vu o n g OMH, ta c6 :

r = OH = M H t a n O M H = ^

.^zA =
6

V2

^^^^^^

6V2

,
V
pO Ghi chu : Doc gld c6 the tinli r = ——, bcti loan se dan gian hem nhieu!
3.

3.

£L

Nhif ta da tinh difcfc trong cau b) : I M = — tuang t u ta c6 : I N = —, IP = —.
2
2
2
^ IM = ^ A B , I N = - B C , I P = - A C
2
2
2
=> 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.
Giai
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)
105


hi

G oi O la tr u n g d iem cua SC. De y th a y h a i ta m

giac SDC vu ong va SBC vu ong theo thuf t u ta i D va
B nen ta c6 :
OS = OD = OB = OC =

SC
2

2

Vay S, B, C, D na m tren mat cau ta m O, ban kinh:

„

SC

K =

aVs
=

2

.

2

c/ Hai mat p h i n g (D MC) va (SAB) c6 diem M chung nen
chung cfit nhau theo mot giao tuyen M N (N n km tren SB) va
vi AB // CD nen M N // CD (va M N // AB).
=> C D M N la h in h tha ng, do la th iet dien cua h in h chop vdi mSt phAng D MC . M a CD J. CM
=> Th iet dien C D M N la h in h th a n g vu ong ta i D va ta i M . (ycbt)
Do M la tr u n g d iem SA nen ta c6 : D M = i SA =

2

; M N = - AB =

2

2

-

2

Vay dien tich th iet dien C D M N bSng :

S =

CD + M N

.DM =

2
Bai

187

(DAI HO C

2a + /r
2 aV2
2

2

KH O I A - M I E N BAC

-

(ycbt).
8

1973)

Cho h in h chop tuT giac deu S.AB CD d in h S, day la mot h i n h vu ong AB C D canh b ang a, cac
ma t ben l a m vd i ma t pha ng day vd i mot goc (p. Ta dUng ma t phSng pha n giac cua goc nhi
dien canh BC tr on g h in h chop (tilc la goc n h i dien cua h in h chop xac d in h b a i ma t BCS va
ma t B C AD ); ma t pha ng pha n giac nay cMt SD d M va SA d N . Ti n h the tich cua h in h chop
S .B C MN theo a va ip.
Gi ai
G oi I , L theo thuf tiT la tr u n g diem cua B C, AD . Va do h in h chop S.AB CD la h in h chop tiif
giac deu => ASIL la ta m giac can ma goc a day S i t = SL)I = cp va S H la du dng cao cua hinh
chop. H = AC o B D
M a t pha ng p h a n giac (a) cua n h i dien canh
BC ch in h la ma t pha ng d i qua BC va difdng pha n

Silt

giac IK ciia

va K la tr u n g diem cua M N . Tif

do de dang th a y r a n g th iet dien B C M N la mot
h in h th a n g can.
D ien tich h i n h th a n g B C M N duoc ti n h theo
cong thufc :
SBCMN =

- ( B C

+

MN).IK

IL

Theo d in h ly h a m s in tr on g A L KI ta c6 :

sin

106


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

SI

SK

- 9
^^^^2

singlk

singKl

SI

g.^3(p

Nhifng :

MN
AD

=> SBCMN =

=

SK

=

SL

SK

^^^^
SK
M N = AD.
SI

SI

• 9
^ '"2

= a

. 3 (0
sin—

• 9
sin—
sincp
2
—a 1 +
.a
. 3(p
Sep
sin—
sin-

s i n — + sin—
2
2;

sin

9
(pcos—

2 - 2
a sm

sincp

• 2 3q)
sin
2

2

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 :
• 9
asm—
SJ = S l s i n ^
2

=

^
2 c o s (p
9 . 0
(p
. (I)
a s i n (pcos— a s m —

V s .BCMN= - S ( B C M N ) . S J =

o

2

3

^ .^ 2 3
2

2C0 SCP

a^sin^cp

(ycbt).

^2sin^^
coscp

Bai 188 ( D A I H O C K H O I B - 1974 - M I E N BAC)
2a
Trong mat p h a n g P cho h i n h t h o i A B C D v d i A B = a va B D = -—j= . T r e n difdng vuong goc
V3
vdi mat phang P va d i qua giao d i e m h a i dUdng cheo cua h i n h t h o i t r e n , ngUcri ta lay d i e m S
sac cho SB = a.
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 r a n g goc n h i dien tao nen bdi cac mat SAB va SAD la vuong.
Giai
a/ Goi O = AC ^ B D : t r o n g t a m giac vuong O A B ta c6:
2a^

OA'' = AB^ - OB^ = a' -

IV3J

107


T r o n g t a m giac vuong OBS t a c6 :
OS^ = S B ' - OB^ = a ' -

N h U vay : OC = OA = OS =

2a'

[43
a42

Noi each khac t a m giac ASC c6 dudng t r u n g tuyen SO
bkng niia canh day A C , ASC 1^ t a m giac vuong tai S (dpcm).
b/ T a CO : VKABCU = ~ S O . d t ( A B C D )

3

T r o n g do : d t ( A B C D ) = O A . B D => d t ( A B C D ) =

aV2

2a^ _ 2^f2a^

^/3 • V3 "
V =

1 a>/2 2y[2a^
3

V3 •

3

4a^ ^ ,
= — ^ (ycbt).

9V3

Trade h e t t a n h ^ n x e t r k n g cde m a t ben (SAB), (SBC), (SDC), (SAD) l a bon t a m giac bkng
nhau.
T h a t v a y , v d i SA = SC, SB = SD v a tCr gia t h i e t : A B = BC = C D = A D t h i 4 t a m gi^c d6 c6
c^c canh t a o n g ilng bSng nhau.
H a B H 1 SA, ABSA la t a m giac can ( B A = BS) n e n B H cung l a t r u n g t u y e n cua t a m gidc
AS
BSA, turc l a H A = H S =
T r o n g t a m giac v u o n g O S A t a eo :
SA'-* = OS^ + OA'^ =

AH =

SA

BH

4a'

SA =

2a

a

- HA^

AB'

= Ja

2

a

V3

3

a
aV2
dt(SAB) = - SA.BH = H A . B H = ^ . ^
2
V3 V3

a^yf2
3

Cuoi cung t a ducte :
Sip = d t ( A B C D ) + 4 d t ( S A B ) =

4a^ y f2

2a^ 42

3

3
=> S,p = 2a^V2 (ycbt).

e/ N o i D v(Ji H => D H l a dadng cao ciia t a m giac can D S A .
Do vay i J H f i = ( ^
Ta

CO

= [(SABiTisAD)]

: D H ^ = BH^* =

2a^

108


4a'

DII^ + B H ' =

= 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.
hi Tinh the tich cua hinh chop c6 dinh la A va day la tuf giac BCC'B'.
d Chufng minh rang nam diem A, B, C, C , B' ciing nSm tren mot mat cau. T i m the tich ciia
hinh cau tifang ijfng.
Giai
a/ Cac tam giac ziABC, ABAB', ACAC vuong can.
=> BC = AB' = A C =

a>/2

=> B'C = A C = AB' = a
=> AAB'C la tam giac deu (dpcm).
hi Ha AH 1 BC. Va thay BB' 1 (ABC) => BB' ± A H
Lai CO :

A H _L BC

A H 1 (BCC'B)

A H 1 BB'J

Dieu do chufng to A H la difcfng cao ciia hinh chop
A.BCC'B'.

aV2
Trong A vuong can ABC : => A H =
Vay

VAHCCIV = ^ A H .dt (BCCB') = \

o

3

.(a.aV2)
=
^
2

(ycbt).

c/ Goi O la tam hinh chuT nhat BCC'B'. Ta c6 :

OB = OB' = OC = OC =

aVs

Trong tam giac vuong HOA => A H =

(1)

1V2

Do OH la dtfdng trung binh cua tam giac BCC.

^ OH = - C C = -

2

^

2

OA = V A H 2 + O H ^

Tif (1) va (2)

1V3

=

OA = OB = OB' = OC = OC =

(2)
iV3

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

S1

(ycbt).

109


B a i 190 ( D A I H O C K H O A H O C K H O I A -

1977)

Cho AABC deu canh a, n o i t i e p t r o n g mot difdng t r o n t a m O chiifa t r o n g mpt mftt phSng
(P). Goi D la d i 6 m xuyen t a m do'i cua A t r e n dudng t r o n nay, con SD l a m g t doan t h i n g cd
chieu dai la a va vuong goc v(Ji (P).
1/ Chufng m i n h SAC va S A B la nhOTng t a m giac vuong.
2/ T i n h d i e n t i c h t o a n phSn cua h i n h chop SABC.
3/

D j n h t a m ciia h i n h cau d i qua 5 d i e m S, A , B , C, D.
Giai

1/ Ta CO : D l a xuyen t a m doi cija A
JA D
^

1

B C (tai H )

[ S D ± ( P) va A B

1

(theo d j n h l y 3 difdng vu6ng goc)
BD

A B _L SB =i> ASAB vuong t a i B (dpcm).
L y luan t i / a n g t i i t a eung c6 ASAC vuong t a i C (dpcm).
2/ Goi S|p la d i e n t i c h toan p h a n cua h i n h chop SABC, t a c6
S,p = d t ( A A B C ) + dt(ASAB) + dt(SAC) + dt(ASBC)
dt(AABC) =
T r o n g ASDB

SB^

=

BD2

3a^

SB^ =

9

SD 2

2

+ a

=

+a

12a^

SB =

=

2aS

N e n : dt(ASAB) = ~ A B . S B = - a.SB.
2
2

a^Vs

dt(ASAB) =

= dt(ASAC)

Xot: SH ^ = H D ^ + SD ^

-

faV3

a^ =

3a^

2

SH =

39a''

+a =
36

. S H ^ = ^ ^ 36

36

1V39
6

= > d t ( A S B C ) = l B C . S H = i . a . ^
2
2
6
Luc do : S,p = — —
3/

+

—

=

^
12

+

=

(11 + Vl3)

(ycbt).

De y den ba t a m giac S A B , SAC, SAD chung canh SA va Ian luat vuong t a i B, C, D nen

n a m d i e m A , B, C, D , S n k m t r e n m a t c l u , b d n k i n h R = S A

V(aV3) 2 + a^ = a va t a m

cua h i n h cau l a t r u n g d i e m I ciia doan SA (ycbt).
110


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 neu c a c goc n h i dien O A cua k h o i tuf dien O A B C l a g6c n h i dien vuong

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 .

Giai
a,' Tinh chat dudng t r u n g b i n h ciia t a m giac.
AB =

PR

2

'

BC =

P Q

2

'

AC =

Q R

2

Do cac l a m giac POQ; QOR; ROP l a v u o n g t a i O:

PQ
=> OA =
^

QR

; OB =

2

2

; OC =

P R

OA = B C ; O B = A C ; OC = A B .

=> Cac m a t cua t a m dien O A B C b a n g nhau (ycbt).
b' Goi H la chan dirdng cao cua tiif dien O A B C h a t i f
0 va h c h i n h l a daotng cao cua tiJ dien OPQR ha t i f O.
De y 4 t a m giac Q A B , B C R , C B A , A P C l a b a n g nhau.
dt(AABC)=

-.dt(APQR)
4

Vo.ABC= | h . d t ( A A B C ) = ^ . h . ^ . d t ( A P Q R ) = ^ VQ.PQR

Mat khac t a c6 t h e t i n h t h e t i c h t\jf dien vuong OPQR l a :
Vo.i'QK=

^
c/

- P O . dt(QOR) = - a . - b e = - abc
3
2
6
3

-V,
V,o A u r =
4
OPQR

1
= —.
— abc = 2 4 abc (ycbt).
4 g - 4 6

Goi O i l a t a m dUcfng t r o n n g o a i tie'p t a m g i a c A B C ; O2

la tam d u a n g t r o n n g o a i tie'p t a m g i a c O B C .
TCr O] t a k e dirdng t h a n g v u o n g g6c v d i m a t
(ABC) v a t il 0-2 t a k e dtfcfng t h i n g v u o n g

phlng

goc v d i m a t

phang (OBC) : do l a h a i t r u e d u d n g t r o n ( A B C ) v a ( O B C ) .
H a i di/crng t h i n g n a y c S t n h a u t a i I ( v i c h u n g c i i n g n a m t r o n g m a t p h l n g t r u n g trifc c u a

I

doan BC).
=> 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 ( O A C ) .
=> 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 .

111


Xet tam gi^c vuong lOjA va

IO2O

ta c6 :

l A = 10 = R
OiA = O2O = Rj

ChuTng minh hoan toan tuang t u ta c6 : IO 3 = l O i ;

lOi =

IO2

IO4 = I O 2

=> lOi = I O 2 = I O :J = IO4 = r (tinh chat h&c cau)
=> I chfnh la tam m kt cau tiep xuc vdi bon mat ciia tijf dien OABC (tam mftt cau noi tiep) (dpcm).
e/ Vdi dieu kien ban dau goc nhi dien canh OA la vuong.
Ha tiJ C : CH 1 OA ^
Noi HB ta

CO

CH _L (AOB).

CH 1 I I B

=> AHBC vuong tai H .
Ha tir B : BH'

1 OA => BH' = CH, va OH = BH'.

Do AABC = ACOA =^ C = C A D ; 6 = C O A
=> tanfi.tanC = tanCOA.tanCAD
Lai xet cac tam giac vuong HCO va HCA ta c6 :
t an CAt )

HC

tanCOA =

AH
tan 6. tan C =

HC
AH

HC

HC
OH

HC^

(1)

AH.OH

OH

(2)

Xet den : O A ^ = ( O H + A H ) ' = O H ' * + A H '^ + 2AH.0H
Trong tam giac vuong H B C ta c6 :
O A'

=

B C '

= H C '

Tir (2) va (3) ta c6 :

+ H B '

= H C '

+ H B '

+ A H '+ O H ' -

= 2HC'

+ ( AH

= 2HC'

-

+

H H ''

O H ) '

(3)

2 0 H . AH

H C '= 2AH. O H

Thay vao (1) ta duac : tanfi.tanC = 2
Ngugc lai, gia sijf da c6 tanfi.tanC = 2
Vdi

chu

y

: AAB C

=

ACO A

= COA;

^

C

=

CAD

tanCOA.tanCAO = 2.

^

Tif C ha A H 1 O A . Ta CO : tanCJOA =

;

tanCXO =

OH
Tif

B

ha

Ta c6:

B H ' ±

H B '

=

Va

O A.

BH ''

+

ACO A

H H '' =

=

AB AO

BH ''

:

AH
C H

+ ( AH

-

= BH ';

O H

=

AH '.

O H ) '

H B ' = C H ' + A H ' + O H ' - 2AH.0H = A H ' + O H '

^
=>

C H ' + H B '

=

C H '

+ A H '

+

O H '

Mat khac ta c6: B C ' = O A ' = ( O H + H A ) '
E C ' = A H ' + O H ' + 2AH.0H = A H ' + O H ' + C H '
Tom lai ta c6:

B C '

=

vuong tai

A H B C

Ta da c6

C H

1

O A

C H ' +
H

H B '

1

<=> C H

B H .

_L ( O A B )

=> C H

1

=> ( C O A )

( AO B)

<=> nhi dien ( O A ) la nhi dien vuong (dpcm).
112

Bai 192

( D AI

H O C Si/ P H A M T P . H CM H E 4 N AM

- KH O I A -

1976)

Trong mot mat phang ( P ) ngifcri ta cho mot tam giac vuong can ABC v d r i AB = AC = a. Bu
va Cv la nhOfng niifa dUcrng th^ng goc v d r i ( P ) tai B va C va 0 cijng mot phia doi v d f i ( P ) . Tren 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.
Dat: RM = X va CN = y.
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)

vdri ( 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 A M N . 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.
HUdng dSn
Doc gia ta giai va dap so nhtf sau :
4Q

1/ y = 2a; dt(AAMN) =
Bai 193

( D AI

2

1

21 = 30°.

;' cosa = V e '

H O C KT - TC -

KT -

SP TP . H CM - KH O I A -

1979)

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.
Giai
1/ Goi

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.

Theo dinh ly 3 difcfng vuong goc, ta c6 :
Ti/ong tif, I K ± BC va I M 1 AC =i.

V I 1 (ABC)
V J l AB

I J l AB

= V l ^ = VS^ = n?

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 =

OO2 =

O O 3.

01 = 0 0 ] => O d tren phan giac ciia
113


.

OI = O O 3 ^ O d tren phan giac cua \

.

01 = OOz => O d tren phan giac cua

Vay chi can O la chan dudng phan gidc ciia X'Jl la du (ycbt).
3 / Tinh R theo r va 9 .
Ta

CO

Trong

: R = O O 2 ; I J = I K = I M = r.
9

A O O 2 K => O O 2

= OKsin-

Trong AOIK =^ OK =

IK

(3)
(4)

cos—
2

• 9

sin—
= rtan^
Thay (4) vao (3), ta c6 : O O 2 = I K — 2
(p
2
cos—
2
Vay : R = r t a n — (ycbt).
2
B a i 194 (DAI HOC Y - NHA - DtfCJC - QUAN Y TP.HCM - 1980)
Trong khong gian cho mot doan t h i n g I J . Tren mot dUdng t h i n g d vuong goc vdi I J va di
qua I ; lay ve hai phia cua I hai diem A, B vdi A I = IB = a; tren mot difcfng t h i n g (d') vuong
goc vdi I J va di qua J ; lay ve hai phia cua J hai diem C, D vdi CJ = JD = b. Cac dudng t h i n g
(d) va (d') khong n i m trong ciing mot mSt phlng.
1/ Chijfng minh r i n g : AC = BD va AD = BC.
2/ ChuTng minh r i n g tam cua hinh c l u ngoai tiep tiJ dien ABCD n i m tr&n diXdng t h i n g I J .
3/ Tinh ban kinh R cua hinh cau tren theo a, b va c = I J .
Giai
1/ Ta CO

AI = I B
:

AB I I J
JC = J D
CDIIJ^

A va B do'i xihig vcJi nhau qua dtfdng t h i n g I J .

• C va D doi xijtng vdi nhau qua di/dng t h i n g I J .

Do do doan t h i n g AC va BD doi xilng nhau qua I J , suy ra AC = BD (dpcm).
Lap luan tifcfng tif ta c6 : AD = BC (dpcm).
2/ Goi O la tam ciia hinh cau ngoai tiep tijf dien ABCD. V i I J Ilk mpt difdng trung trifc cua ca
AB va CD nen tam O phai d tren dudng t h i n g I J .
3/ Ta CO

:

= OA^ = 10^ + l A ^ =
R2 = O C ' - O j 2 + JC^ =

R2 =

+ (c - 10)2 =

+OF
+ (IJ-

(1)

Olf

+ c % 0 1 ^ - 2C.0I

(2)

+ O r = b^ + c^ + O r - 2c .OI

TCr (1) va (2), suy ra :

114


01 =
2c

Thay 01 =

^

vao (1)

2c

= a^ +

b^+c^-a^^'
2c

Vay : R = — V a ^ + b " + c" + 2a2c2 + 2hh^ - 2a%^ (dpcm).
2c
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.
a/ Chiing minh rSng tam gidc ABC vuong.
hi Chijfng minh rSng trung diem H cua doan AC la hinh chieu vuong goc cua O len mat
phang ABC. TO d6 suy ra the tich cua tiJ dien OABC.
d Xac dinh tam I va ban kinh hinh cau ngoai tiep til di?n OABC.
Giai
a/ Ta CO :

T A A O B can => A A O B deu => A B = a
S A B O C vuong can

BC = aV2

. A A O C => A C = a-s /3

AC^ = AB'^ + B C ^ = 3a^ => A A B C vuong tai B (dpcm).
OA

b) De y : OH =

a

BH =

2

2

OB^ = O H ' + H B '
Tif:

OH 1 H B
OH 1 A C

A C ^ aV3
~

2

OH 1 HB.

O H 1 (ABC)

=> H la hinh chieu cua O xuong mp(ABC).
Luc do the tich V cua hinh ch6p OABC la :
VQABC = ^ d t (AABC) X O H

o
= -ABxBCxOH
6
=> V =

aS/ 2
12

(ycbt).

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).
115


Bai

196 ( D E A 2 - 1 9 8 1 )
T r o n g m a t p h a n g ( P ) c h o nOfa d u d n g t r o n d u d n g k i n h A B = 2 R . G i a suT M l a m p t d i e m t u y

y t r e n nufa difcrng t r o n , k h a c vdi A v a B . G o i H l a h i n h c h i e u v u o n g goc c i i a M l e n A B , d a t x =
AH.

X e m nufa d u d n g t h a n g M u v u o n g g6c v d i ( P ) t a i M , g i a t h i e t M u l u o n l u o n d v e c u n g m p t

p h i a doi v d i (P) k h i M t h a y doi t r e n M u lay d i e m S v d i S M = M H .

C h u f n g t o r a n g goc n h i d i e n c a n h A B t a o b d i cdc m a t p h ^ n g ( P ) v a ( S A B ) c6 gia t r i k h o n g

b/

T i n h do d a i cac c a n h c u a tuT d i e n S A B M .

a/

doi k h i M t h a y doi.

d

X a c d i n h v i t r i t a r n h i n h c a u n g o a i t i e p tiif d i e n S A B M T i n h b a n k i n h h i n h c a u a y t h e o R va

X . V d r i g i a t r i n a o c u a x t h i b a n k i n h a y c6 gia t r i I d n n h a t ?
Giai
(Xem D l D A I H O C SP - K T - K T - N N - K H O I A Bai

197 ( D A I H O C S P - K T - K T - N N - K H O I A -

1982)

1982)

T r o n g m a t p h l n g ( P ) c h o nufa dUcfng t r o n d U d n g k i n h A B = 2 R . G i a thCt M l a m p t d i e m t u y
y t r e n nufa t r o n , k h a c v d i A v a B , g p i H l a h i n h c h i e u v u o n g goc c i i a M t r e n A B , d a t A H = x.
X e m nufa dUcfng t h i n g M N v u o n g goc

v d i (P) t a i M . G i a t h i e t M N l u o n l u o n d ve cijng mpt

p h i a doi v d i (P) k h i M t h a y doi, t r e n M N lay d i e m S v d i S M = M H .

C h u f n g t o r a n g goc

b/

T i n h do d a i cac c a n h c i i a tuf d i e n

a/

SAMB.

n h i d i e n c a n h A B t a o n e n b d i cac m a t p h i n g ( P ) v a ( S A B ) c6 gia t r i

k h o n g doi k h i M t h a y doi.
d

X a c d i n h v i t r i t a m h i n h c a i i n g o a i t i e p tuf d i e n S A B M . T i n h ban k i n h h i n h c a u a y t h e o R \k

X . V d i g i a t r i n a o c u a x t h i b a n k i n h a y c6 gia t r i I d n n h a t ?
Giai
a/

H e t h d c l i i o n g t r o n g t a m giac cho :

A M ^ = A H X A B = 2 R x = > A M = V2Rx
B M ^ = B H X A B = 2R(2R -

x)

B M = V2R(2R - x)
M H ^

= A H

= x(2R -

X B H

x)

M H = Vx(2R - x)
M S = M H = Vx(2R - x)
SA^

= MS^ +

M A ^

= x(2R - x) + 2Rx = x(4R -

=> S A = V x ( 4 R -

x)

x)

S B ^ = M B ' + M S ' = 2R(2R - x) + x(2R - x) = 4 R ' - X '
=^ S B = V 4 R 2 - X ^ ; A B = 2R
b/

(ycbt).

Goc n h i d i e n t a o b d i (P) v a ( S A B ) k h o n g d o i .
S M

1

(P)

S H
M H

1

=> ip =

1 AB

( d i n h l y 3 dUdng vuong

gdc)

A B

[CT5r(S5B71 =

= SHM

( A S M H

vuong

can)

116


=> (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
R,2

x(2R-x)^^.
4
5R'

/

+ R^

= I M ^ = lO^ + O M ^ =
1

Si

rx + 2 R - x ^ '

+ R2

(d)

^

/MLJ

I

\

4
(1)

o

B

Dau d i n g thuTc t r o n g (1) xay ra<=>x = 2 R - x < = > x = R

V5
Vay X = R t h i m a x R j = —
2

R (ycbt).

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 .
4/ Menh de "Neu BC trifc giao vdi OA t h i

5c9z = y9z"

c6 dung khong ? T a i sao ?

Giai
1/

Theo d i n h l y ba dudng vuong goc

A-B
A'C

1
1

Ox
Oy

AAOB = AOAC
AB = A C va OB = OC
Hai dudng x i e n A B = A C c6 h a i h i n h chieu A ' B =
AC ti/cfng ling.
=> 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.
117


: AA' ± (OBC) => AA' J_ BC.
BC 1 (OAA) v4 OBA'C la hinh vuong.
21 Ta CO

Nen

BClOA

:

A O X ' = p = [Oz; (xOy)]

(2)

AAHA' => A A ' = H A ' t a n y = - . O A ' . t a n y
2

(1)

AAOA' => A A ' = OA'.tanp

Tir (1) va (2) => - OA'tany = OA'tanp hay tany = 2tanp
2
(4)

AAOA' => O A ' = OA.cosp

(3)

AAOB => O B = OA.cosa

=> OA' = OB V2

(dudng ch6o cua hinh vuong OBA'C)

Ket hcfp vdi (3) vk (4) : OAcosP = V2 OA.cosa <=>

cosP = V2 cosa

a, P la hai g6c nhon nen => tany = 2tanp
tan^y = 4tan^p = 4

^cos^p

-1

- 4 = 2(1 + tan^a) - 4
2cos^a

<=> t a n ^ = 2(tan^a - 1) vi 45° < a < 90°
tana > 1, vi y 1^ goc nhon => tany > 0

tany = V2(tan^ a - 1) (ycbt).
Neu a = 60',0

cosa = — =>cosP = V2 cosa =
2

P = 45° (ycbt).
2

3/ X6t : ABt) = Act) = 90°, nen B, C n^m tren mSt cau difdng kinh OA. Vay tarn cua hinh
cau ngoai tiep tuT dien OABC la trung diem I cua OA.
Ta biet tiep dien ciia hinh cau t a i diem A la mat p h l n g (T) vuong goc \(s\g kinh OA
tai A. Do do thiet dien dUcfc xac dinh nhu sau :
Qua A ve mat ph^ng vuong g6c vdi Oz, cat Ox tai B', c^t Oy tai C.
Mat phang (T) = (AB'C) 1 OA tai A nen n6 \k tiep dien phai x^c dinh (ycbt).
Mat p h i n g (ABC) cat hinh cau theo hinh tr6n (P) ngoai tiep AABC.
OB = OC =

a

Ng'u OA = a va a = 60°
BC = ^ , A B = AC =
2 '

^ ^
2

Goi M 1^ trung diem AB. Ke difdng trung trifc cua canh AB trong tam giac can ABC n6 cat
difcfng cao HA tai J t h i JA la ban kinh hinh tron ngoai ti§'p tam gidc ABC.
AAJM CO AABH

AJ

AM

AB ~ A H

AB^

AJ =

(3)

2AH

118


AH^ = AB^ - B H ^

=

aV3

lOa^

AH =

16

iVio

3a^

Thay vao (3) => A J =

3a

2a->/l0

Dien tfch thiet dien : S = TI.AJ^ =

(ycbt).
40

4/ O B X ' = C J C ^ = 9 0 " nen ttJ gidc OBA'C noi tiep trong h i n h tr6n dudng k i n h OA'. Neu
BC i OA va theo tren BC ± AA' t h i BC 1 (OAA') tiirc la BC ± OA'. Trong hinh tr6n dUdng
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.
Suy ra : AOB = AOC hay x9z = y9z.
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 ".
Giai

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 :
IC2 = S C2 -SI^
119


Vi IC = I H , ket hop lai ta c6 :

HI^ = SC^ HI =

HI^
cos^a

cos^ a
' 1 + cos^ a

Tif do : BC =
Ta

CO

: SI =

2bcosa
.
1 + cos a

bcosa
V l + cos^a
(ycbt)

HI
cosa

V l + cos^a

Ta goi dien tich toan phan ciia hinh chop la Stp thi :
S,p = 4dt(ASBC) + dt(ABCD) = 4. - . SI.BC + BC^
2

S,p =

^

Stp = 4.

^

bcosa
V l + cos^a

V l + cos^a

cosa(4 + 4cosa)b
1 + cos

2bcosa
V l + cos^ a

(ycbt)

a

Trong tam giac vuong SHI

=>

SH = Sl.sina =

bsina
1 + cos

Vay : V^,,,™ = 1 SHdt(ABCD) = A .

^^--cos^

a

(^.^t).

( l + cos^a)2
2/ Gia sijf O la tam hinh cau noi tiep va ngoai tiep hinh chop, va K i la chan diTdng cao ha
O tdi (SBC) ^ K i e SI.
fOH = O K , O S = O C
Theo gia thiet :
2
Xet tam giac vuong O H I

=> O H = H I . t a n — .
2

Xet tam giac vuong O H C

O C = V O H ^T H C ^

O C = j H i n a n " - +2 H r

= H L 2 + t a n ^a - .

Tir do : S H = S O + O H = O C + O H = H I t a n — + ./:2 + t a n 2 2
\

(1)

Lai xet tam giac vuong S H I va ( 1 ) .
SH

^
= tan

tana =

2 + tan2^

HI
120


2tan2

2 + tan2^

= tan

l-tan ^ l + tan^ ^

2C
o tan —.

2 + tan2^
1- t a n ^ ^

2)

o tan*—+ 2tan ^

1= 0

2
2

t a n ^ " = - l + V2
2

tan2- = - l - V 2

Mat khac => cosa =

« tan2^ = > ^ - l
< 0 (loai)

^

l-tan ^ ^
2 . , 1 - ^ + 1 . V^ _ i ( d p c m ) .

3a'Ta c6 ; SCfi = ip. Mat khac theo (1), thi :
l2 + tan2-

SH= HI t a n -

(2)

maHC = HI. V2
Lay

(3)
SH

(2)
(3)

tancp =

HC

V2

t a n — + J ;2
2 \

+ tan2-^

(4)

Vdi gia thiet cosa = ^f2 - 1, da c6 duoc : tan^— = V2 - 1
2

(3)

tancp =

V V 2 - 1 + 7 7 2 + 11

(ycbt).

b/ Gia sLf thiet dien ciia mp(a) va hinh chop S.ABCD da difng dugc.
Vi (a) // BD =>(a) n (SBD) = MN // BD
Mat khac neu gpi K' la chan dUcrng vuong goc ha tif B xuong (a)
=> <^

ma BD // MN => BK 1 BD
1 MN
=> BK' nkm tren mat phang vuong goc vdri BD. Goi do la mat ph^ng (BB'y).
BK'

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

121


Tarn gidc vuong A EB c6 BE =

AByf2
AB
EB ^ - B K'2

=

=

K'B

2
=> A BKE vuong can c6 BEl t ' = 45°
ABV2

AEBI vuong can c6 : BE = BI =

— .

2
B^ng phan tich tren day ta suy ra cdch difng nhu sau :
• Trong mat p hing SBD difng BB' 1 BD, tren BB' lay diem I sao cho:
ABV2

BI = BE =

— .

2
Tii I ke Iz // BD (tro ng mp(SBD), Iz c^t SB, SD tUang ufng tai M , N v^ c^it SH d F. Noi A
vdi F (trong (SAC), A F cat SC tai Q. Tijf giAc A MQN la thiet dien can dUng.
• De y thay vdi gia thiet da cho, thiet dien luon luon difng duac. ThUc vay dieu do ti/ ang
S H
^ A Bf V V 2- 1 + VV2+1
duang vdi FH
<
SH.
Nhifng
ta
da
c6
:
2
FH < SH (ycbt).
F H

= B I

ABV2

Bai 200 (DA I HOC BA CH KHOA TP.HCM - 1988)
Cho mot hinh chop tam giac deu D.ABC c6 canh day a, cac mat ben nghieng vdi day m6t goc a.
a/ Tinh cac di§n tich s va S cua cac mat cau no i tiep, ngoai tiep vdi hinh chdp va di§n tich
toan phan S,p cua hinh chop theo a va a.
b/ Bieu dien Stp theo s va S.
Giki
aJ Goi E la trung diem cua BC. Goi H la tro ng tam AABC.
=> DH la difdng cao hinh chop va A E I ) = a.
Goi O, I , r, R Ian lifgt la tam va ban kinh cac mat
cau no i tiep va ngoai tiep hinh chop. Do tinh doi xilng
cua hinh chop deu nen de thay O va I e DH.
De y thay: i t Et ) = CJETD = - ; l A = ID = R
2

r = OH = HE.tan — = — ^ .tan —

2

Vay

:

s

=

ATIT^

=

2V3

2

9 a
•tan^ —

Ta c6: DH = HEtana =

(2)

tana
2 V3

A HIA => lA^ = A H^ + IH^ = A H^ + (DH - DI)^

122


hay:

tana - R

=

a(tan^ a + 4)

R =

4V3tana

(3)

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

S = 4itR^ = 471

Mat khac : D E =

a^(tan^a + 4)2

7ta^ (tan^ a + 4)^

48tan^a
HE
a

cos a

12tan^a

2V3cosa

Stp = S „ + Sd = 3SnBC + SABC = 3. | a . D E +

^ a - " ^

a^Vsd + cos a)
hi Tif(2) tac o :

=

(4)

4 cos a
3S

(5)

•. Thay vko (5) ta duac :

Titan' a
I

sVSsd + cosa)cos^ "
2
X
• 2 2
4 sin a cos a

(5)

„

„„o^^2

/^^

oV3s (1 + cosa)
47t
cosa(l-cosa)

(6)

TCf(l) d en (4) suy ra :

(S
fS _ R _ 2a(tan2a + 4)V3
^ «
\ s = rr = . /r^
4V3tana.a.tan—
2

l + 3cos^a
2cosa(l-cosa)

S _ 1 + 2 cos a + cos^ a - 2 cos a + 2 cos^ a
s
2cosa(l - cos a)

VS
s
o

1+

(1 + cos a)^ - 2 cos a(l - cos a)
2cosa(l - cos a)

(1 + cosa)
-1
2cosa(l - cosa)

(1 + cosa)^
2 cos a(l - cos a)

Tir (6) v a (7), ta c6 :

Stp =

3V 3s

2n

(7)

-i f

(ycbt).

Bai 201 ( D A I H O C N O N G L A M T P. H C M - 1991)
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.

123


Do hinh chop deu nen cac tam co va O Ian lifat cua cac
mat cau npi, ngoai tiep deu nSm tren SH.
Ta

CO :

SH ± (ABC)l
A H 1 BC

BC _L SH va BC ± SM

SM ± BC
BC ± (SAM)

=> (SBC) _L (SAM) nen tiT w ha « J 1 SM
thi

(oJ ± (SBC).

A

Do (oH = (oJ = r nen co la giao diem ciia SH va diTdng

phan giac cua
o

T

«J

ASwJ t/ 5 ASMH =>

Sco

=
HM

SH~(OH

=
SM

SM

aVs

( h -r)

=> r =

6

i^f3

h ~ r

6
12

ah
Va 2 + 12 h 2

(ycbt).
+a

Trong tam giac SAM, trung triTc canh SA gap SH tai O :
ASIO CO ASHA

SH

SA

SI

SO

SA
2SH

SA2
9h2+3a2
Sh^+a^
SO = 2SH =

=
6h
18h

^ 3hUa\^
<=> R =
6h

^
(dpcm).

6ah^

2/ Xet : y = — =
R

( a 2 + 3 h 2 )(a+ Va'+ 12 h 2 )

<=> y =
1

2

I

1

2

(1 + 3 — ) ( 1 + J l + 1 2 — )

3

Dat : 12

3.

I

*V =

t^^x = 2-73 —

^ ^ ^ ^

tanx > 0
2 V3

2tan^ X

Luc do : y =

(4 + t a n ^ x ) ( l + V l + t a n ^ x )
,sin^

X

2sin^ x c o s x

COS^ X

•» y =
4 +

• 2 ^
sm x

(1 + 3cos x ) ( l + c o s x )

1 +-

cos^ x

cosx

124