Tuyển tập 500 bài toán hình không gian chọn lọc phần 4 nguyễn đức đồng
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V
A, B & cung
ben so v& i
(D)
D img A' d o i x i J ng vdi A q ua (D).
Luc do : A' v a B d kha c b e n so v6i
M A + M B > AB
(D),
n e n t r d v e t rUc f ng hop
tren :
<=>MA' + M A ^ AB .
min(MA + M B ) = min(MA' + M B ) = A B
ttfong Ofng : M a M Q = ( A ' B ) o
O Ket
•
Xa c
(D)
l u a n : v a y t r o n g m o i t r i f d n g h g p t a x d c d i n h di f ac M t h o a y c b t .
d i n h d i e m m t r e n d Uo T n g t h S n g ( d )
de
| MA
- MB |
T i f a n g tiT c a n p h a n b i e t h a i t rUc f ng h a p :
V A, B & cung
ben so v& i
|MA - MB | <
(D)
AB
M
ma x| M A ~ M B | = A B
Mo
(d)
tu a ng ling M = M Q = (AB ) r^ (D)
V A,B
d khdc
ben so v& i
(D)
| M A - M B | < |MA' - M B | ^
AB
yiy^
Vdi A' la h i n h doi xijfng ciia diem A qua (d), t h i A' va B
I
a cu ng p h ia (D).
ma x I M A - M B | = ma x J M A ' - M B | = A B
\
(D )
/Mo
A'
ta ong tjfng M s M Q = (A'B ) n (D)
O Ket lu a n : vay tr on g moi tri/dng hap ta da xac d in h diem M thoa ycb t.
n. Gl A l T OAN T HI
Bai
393 ( D A I H O C K H O I A M I E N B A C -
1972)
Cho mot kh oi t i l dien AB C D .
a/ Mot ma t phang song song vdri canh B C cat cac canh AB , AC, D C, D B d c^c diem M , N , P, Q.
Chufng m i n h rS ng tuT giac M N P Q la mot h i n h th a n g pha i thoa ma n dieu ki§n nao de tuT gidc
do la mot h i n h b in h h a n h ? la mot h in h c h a n h a t ?
b/ Cho b iet cac goc B AC , C AD , D A B la vu ong, con B C D la mot ta m g iic deu canh a. Tin h th^
tich cua kh oi tijf dien theo a.
c/ Cho b iet B C D la mot ta m giac deu canh a va c6 ta m la diem O. Ti n h doan OA theo a sao
cho m a t cau ngoai tiep ti l dien AB C D n h a n dudng tr on (B CD ) l a m mot ducfng tr on Idn. Tinh
dien tich m a t cau tr on g trU dng hap ay. Xac d in h vi t r i ciia d in h A tr en m a t cau ay de the tich
h in h t i J dien A B C D I6n n h a t.
G ia i
aJ Ta c6 : mp(P) // B C = (AB C) n (B CD )
=> M N // PQ
Vay th iet d i6n M N P Q la mot h in h th a n g (ycbt).
300
Muon cho M N P Q la h i n h b i n h h a n h ; tifang t u
tren ta phai c6 t h e m dieu k i e n N P // M Q , (P) // A D .
Vay dieu k i e n de M N P Q la h i n h b i n h h a n h
la
mat phang (P) p h a i song song d o n g thcfi v d i : BC
AD (ycbt).
Hcfn nOa de M N P Q la h i n h chfl n h a t t h i ta p h a i
c6 MN
1
NP.
Vi BC // M N va A D // N P =i> BC
1
AD
Vay dieu k i e n de M N P Q la h i n h chff n h a t la BC 1 A D (ycbt).
b/ Tuf dien A B C D la tuT d i e n v u o n g d A.
fBC = CD = DB
= a
AB = A C . A D = ^
=
2
^
2
Vay the t i c h k h o i tuT dien A B C D la :
V= - A B . - A C A D = i ( A B ) ^=i
2
3
6
24
6
(ycbt).
c) De y dUcfng t r o n ( B C D ) la m o t dudng t r o n I d n cua m a t cau ngoai t i e p tiir d i e n A B C D va c6
0 la tam ciia t a m giac B C D canh a, nen t a m O cua t a m giac deu B C D cung c h i n h la t a m cua
mat eau ngoai t i e p tuf d i e n A B C D .
^0 A =
O B = ^
Tir do dien t i c h Sc ciia m a t cau ngoai t i e p tuf d i ? n A B C D l a :
Se = 471 OA^ = 471
aV3
4
2
= — 7ca
3
Goi A H la dUcfng cao ciia tuf d i e n A B C D ha t i f d i n h A
suong mat day (BCD).
AHOC ( f i = 9 0 " )
=i> A H < O A
Va t i n h duoc the t i c h cua k h o i tuf d i e n A B C D b a n g :
V = ^ S , „c D . A H =
V3.a2
12
.AH <
I
1
aV3
2
2
Vs.a 2
12
.a
.AH
.OA
(1)
Dau dang thufc t r o n g (1) xay ra <=> l i = A ( h i n h ch6p A . B C D deu)
3maxV=
12
.OA
(ycbt).
301
Bai
394
(D A I H O C K H O I A - M I E N B A C -
1974)
Trong mat p hlng (P), cho hinh vuong ABCD c6 canh b ing a. Tren dUdng thi ng Ax di qua
A va vuong goc vdi mftt p hing P, ngucri ta lay mpt diem S tiiy y, ro i difng mat p hing Q qua A
c i t SB, SC, SD Ian lirat tai B', C, D'. Biet (Q) i SC.
a/
ChuTng minh rSng SB vuong g6c A B' va SD vuong goc vdi A D'.
b/ Tim c^c quy tich ciia B', C, D' khi S chay tren Ax.
c/ Xac dinh v i tri cua S tren Ax sao cho hinh ch6p C'ABCD c6 the tich \dn nha't va tinh thi
tich ay.
Gi&i
a/ Ta c6 :
CB 1 A B
C B± S A
CB ± (SAB) 3 A B' => CB 1 A B'
Mat khac ta c6 : SC _L A B' (vi A B' n^m trong mp(Q) ma SC 1 Q).
Do d6 A B' 1 (SBC) 3 SB => A B' ± SB
(dpcm).
ChiJng minh hoan to^n tUcfng tif ta c6 A D' 1 SD (dpcm).
b/ De y :
B' e (SAB)
A B ^ = 90° ; A, Bco dinh
=> Vay quy tich ciia nhiJng diem B' la dudng
tro n (trong mat p hlng (SAB) dtfdng kinh A B. (Doc
gia tif lam phan dao)
• Taang ti i ta c6 quy tich ciia nhufng diem D'
trong mat p hing (SAD) difdng kinh A D (ycbt).
• Tucfng tu ta cung c6 quy tich cua nhCifng diem
C la dudng tro n tro ng mat p hing (SAC) dadng
kinh AC (ycbt).
0/
Ha CO 1 AC. va thay OC // SA; SA 1 P n6n :
V = VC'ABCI) = — SA BCD - O C '
Vi hinh chU nhat A BCD la co d inh, nen the tich V se I6n nhat khi OC la Idrn nhat. C luon
luon nam tren diidng tro n dtfdng kinh AC. Vi vay OC se Idrn nhat khi no la ban kinh. (NhK
vay CO hai v i tri Co' va Co" doi xufng vdi nhau qua AC cung thoa man tinh chat d6 :
C'oC'o ± AC tai O; C'oC'o c (SAC).
Khi do OC = OA = OC =
hay nay tam giac vuong SAC co OC la difdng trung binh,
taang ling do do : AS = 20C = a-s^ .
Vay khi S (nam tren A x) each A mOt doan a ^/2 (co hai v i tri SQ va SQ' doi xufng vdi nhau
qua A) thi hinh chop C'ABCD c6 the tich Idn nhat, va the tich d6 la :
1V 2
V = i a ^ ^
3
3
a"-^V2
2
(ycbt).
302
Bai 395 ( D A I H O C Y - N H A - D L f d C -
1976)
Cho h i n h vuong A B C D canh a. Goi SA la doan thSng goc vdri mSt p h ^ n g ( A B C D )
vk M la m o t d i e m d i dong t r e n doan SD. DSt S M = x.
SA = a
a/ Mat phang ( A B M ) cat doan SC t a i N . Chitog m i n h tiif gidc M A B N l a mot h i n h t h a n g vuong.
b/ Dat y = A M ^ . T i n h y theo a va x.
d Khao s a t sif b i e n t h i e n va ve dudng bieu d i e n cua y = A M ^ k h i M ve t r e n doan SD.
Gi&i
a/ Ta CO : A B // C D => A B // (SCD); A B c ( A B M N )
^
Lai
( A B M N ) n (SDC) = M N // A B // C D
CO :
A B 1 (SAD)
AB 1 AD
AB
I S A
=> A B 1 A M => M N 1 (SAD)
=>
M N I A M
Vay A M N B l a m o t h i n h t h a n g v u o n g h a i ddy l a A B va
MN (ycbt).
hi Goi H l a h i n h chieu cua M xuong canh CD.
ADMH oo ADAS
MH
MD
SA
DS
aV2
-
X
M H
=
aV2
AAHM ^ AM^ = A H ' + HM^
aV2-X
HMD vuong can => H D = H M =
AH = AD - HD = a -
a -
n ^- A » , 2
(aV2-x)2
Do do: A M ' = — +
2
2
^
x^ +
+
- 2aV2.x
=
2
A M ^ = x ' - a V 2 x + a^
Vay y = x ' - a V 2 x + a ^ Vx e [0; a V 2 ] (ycbt).
d Mien xac d i n h cua y : Df = [0; a ^/2 ]
^ y' = 2 x - a V 2
( aV2
= 0
Bang bien t h i e n :
Do t h i :
tV2
X
0
2
a^f2
y
y'
a'
a.
2
aV2
aV2
x
Dudng bieu d i e n la cung Parabola A S B .
303
Ba i 396 ( D A I H O C B A C H K H O A - T O N G H O P -
1980)
Trong khong gian cho ba tia Ox, Oy, Oz tifng doi mot tao wdi nhau mot goc a (0 < a < 90°)
tren Ox, Oy, Oz lay Ian lucft cac diem A, B, C sao cho : OA = a, OB = b, OC = c.
1/ a, b, c phai thoa man h? thilc gi de tam giac ABC c6 goc A vuong ? Hay tim dieu ki§n can
va du rang buoc b, c, a de tim dUcfc a thoa man he thufc ay.
21 Gia sijf a co dinh (0 < a < 90°) v^ b = c co dinh. Xac dinh a de tam giac ABC c6 g6c A Wn
nhat. Gia tr i Idn nhat ay cua goc A bang bao nhieu.
3/ Vdi cac gia thiet cua 2. Hay tinh the tich cua tiif dien OABC ufng vdi gia tr i \6n nhat cua
goc A.
Gi a i
1) AABC vuong ta i A
»
BC^ = AB^ + AC^
(1)
Dinh ly ham cosin trong cac tam giac :
AOAB, AOBC, AOAC cho :
(2)
BC^ = b^ + c^ - 2bccosa
(1)
AB'^ = a^ + b^ - 2abcosa
AC^ =
+ c^ - 2accosa
(3)
Thay (2), (3) va (4) vao (1).
(1)
<=> b^ + c^ - 2bccosa = 2a^ + b^ + c^ - 2a(b + c)cosa
<=> a^ - a(b + c)cosa + bccosa = 0
<=> g(a) = a^ - [(b + c)cosa]a + bccosa = 0
(5)
De tim dtfcfc a thoa man (5).
<=> A = (b + c)^cos^a - 4bccosa > 0
<=> A = cosaKb + c)^cosu - 4bc] ^ 0 (0 < a < 90° => cosa > 0)
<=> (b + c)^cosa - 4bc ^ 0
(6)
(ycbt)
(6) dieu kien can va du rang buoc b, c va a de tim difac a thoa man (5).
21 Xet gia thiet : b = c
Goi HB 1 OA
«
<=> AOAB = AOAC
CH 1 OA
^
B H = CH
Xet hai tam giac can ABC va HBC; chung co canh chung BC.
JAB ^ H B
^
gAC < B Ht; => maxBAC = B Ht; tircJngilngA^H.
J A O H C
AOBH
jO H = bcosa
' H B = HC = bsina
BC'^ = OB'' + OC' - 20B.OCcosa
AOBC
<^ B C' = 2b ' - 2b'cosa
B C' = H B ' + H C ' - 2HB.HCcosH
(7)
AHBC
(8)
B C' = 2b 'sin'a - 2b'sin'acosH
So sanh (7) va (8) theo ve :
=> 2b^ - 2b^ cosa = 2b^ sin^ a - 2b^ sin^ a cos H
=> 1 - cosa = sin'a - sin'acosH
304
sin^acosH = sin^u + cosa - 1 = cosa - cos^o
cosH =
cosa.2sin^ —
c o s a d - cosa)
sin^a
H = arccos
cosa
4sin —.cos —
2
cosa
2cos22)
cosa
Vay gia tri \6n nhat cua A la : minA = H = arccos
(ycbt).
2cos2^
2)
3/ The tich V cua tiJ dien OABC la :
V = i dt(AHBC).OH = - . - .HB.HCsinH.OH
3
3 2
=>
V = — .b^sin^a.sinH.bcosa = — b^sina.sin2a.sinH
12
6
(ycbt).
Bai 397 (DAI HOC BACH KHOA - TONG HCfP - Y - N H A - D L f d C - 1982)
Tren canh AD ciia hinh vuong ABCD canh a, ngudi ta lay diem M v
vuong, ngUdi ta lay diem S \6i SA = y (y > 0).
a/ Chijfng minh rkng nhi dien canh SB tao bdi cac mat phang (SBA) va (SBC) la mot n h i dien
vuong.
b/ Goi I la trung diem cua SC, H la hinh chieu vuong goc cua I len CM. T i m quy tich cua H
khi M chay tren canh AD va S chay tren Ax.
c/ Tinh the tich hinh chop S.ABCM.
d/Vdi gia thiet x^ + y^ = ai^, t i m gia t r i Idn nhat cua the tich hinh chop S.ABCM.
Giai
a/ Ta CO
:
AD
1 (SAB)
BC ± (SAB)
BC / / A D
(SBC)
1 (SAB)
Vay nhi dien (SB) la mot nhi dien vuong (dpcm).
b/ Ta CO :
01 // SA
Mat khac I H 1 CM =>
^
O i l (ABCD).
OH 1 CM (dinh ly 3 difcrng vuong goc).
Vi M e (AD) va S G Ax nen H a trong iCCt). Vay H 0 tren cung tron O H Q cua dUcfng tron
dUdng kinh OC
HQ la trung diem cua CD, khi M e AD va S e Ax.
• Dao lai, lay mot diem H bat ky tren cung OHQ, ta c6 : O H 1 HC; CH n AD = M , tren nOfa
dudng t h i n g Ox' // Ax lay mot diem I sac cho CI c i t Ax tai S. Ro rang :
* I la trung diem cua SC.
* I H _L CM (dinh ly 3 dUcfng vuong goc).
Ket luan : Quy tich H la cung tron OHg cua dudng tron dudng kinh OC trong mat p h i n g
(ABCD) (xem hinh) (ycbt).
305
c/
The tfch hinh chop.
V(SABCM)
= - dt(ABCM).SA
3
= - . - .(A M + BOA B.SA
3 2
=
(x + a).a.y (ycbt)
6
d/ Xet : x^ + y^ = a^ «
y = Va^ -
V = V(SA BCM) = - ( x + aWa^ - x^
Ta CO ; maxV xay ra <=> max(3V)^ xay ra
maSV* = — (x + a)(x + a)(x + a)(3a - 3x)
36
(1)
Ap dung BDT Cauchy cho 4 so' khong am, ta c6 :
(1)
<=>3V^ <
(x + a) + (x + a) + (x + a) + (3a - 3x)
36
,2
81a'
V2
36.3 v2
.
36.3.16
Dau d ing thiifc tro ng (2) xay ra
3a^
„
V3
2
64
<=>
a + x = 3a - 3x <=>
(2)
a
X =
2
Do d6 khi M la trung diem A D thi the tich VSABCM ciTc dai va
73
maxV =
2
a
(ycbt).
8
Bai
398
( D A I H O C B A C H K H O A - T O N G HCJP - Y - N H A - D l / d C -
1983)
Trong khong gian, cho hinh chop S.ABCD, day ABCD la hinh cha nhat vdi A B = a, AD = b;
canh SA cua hinh chop vuong goc vdi day, AS = 2a.
a/ M la mot difim tren canh AS, vdi A M = x (0 ^ x < 2a). Mat ph^ng MBC cMt hinh chdp theo
thiet dien gi ? Tinh dien tich thiet dien ay theo a, b, x.
Xac d inh x sao cho mat phSng (MBC) chia hinh ch6p ra hai phan c6 the tich bkng nhau.
c/
Xac d inh x sao cho thiet dien tren c6 dien tich Idn nhat.
hi
G iai
Al Goi N la giao diem cua mat p hing (MBC) vdi SD. Liic do :
Mat phang (MBC) chiifa BC // A D.
Ma A D = (SAD) n (ABCD)
^
(MBC) n (SAD) = M N // A D // BC
Han nfla v i BC 1 A B va BC 1 SA
=> BC 1 (SAB)
BC 1 MB
Thanh thijf thiet dien MBCN \k mgt hinh thang vuong
(6
= 1^ = 9 0 ° )
B
306
ASMN
MN _ SM
ASAD
M N = AD.
AD ~ SA
SM
= b.
SA
2a-X
=
2a
b
"[^
2a,
Vi vay h i n h t h a n g vuong M B C N c6 dien t i c h :
b +
b
X
1
2a
= b
.^[^77
a^
(ycbt).
4a j
b/ Vi S > 0, S dat gia t r i lorn n h a t k h i
Taco:
=
dat gia t r i Idm n h a t .
r(4a-x)2(x^+a^) =
- f i x ) ; Vx e [0; 2a]
16a'
=> fix)
^
= 2(x - 4a)(x'^ + a^) + 2x(4a - x)^
f (x) = 2(x - 4a)(2x^ - 4ax
pv,
,
.
n
Cho f (x) = 0 <=>
+a')
X
= 4a
a ( 2 ±V 2 )
V
x =
2
Lap difgtcbang bien t h i e n cua h a m so fix) t r e n doan [0; 2a].
a(2-V2)
a(2 + V2)
0
9
9
f(x)
16a'
fix)
a ' ( 7 1 + 8V2)
4
a'(71-8V2)
DiTa vac b a n g b i e n t h i e n n^y t a t h a y :
Vay amaxS^ o
khi:
X
= AM =
3maxS = —
8
^^^iJ^
^ 7 1 + 8^2
k h i va chi
(yebt).
d Hien n h i e n h i n h chop S.ABCD c6 the t i c h :
V= iAS.AB.AD=
3
—
3
De t i n h the t i c h V cua h i n h iSng t r u t a m giac cut M A B . N C D , t a difng m a t phSng ((3) qua N
1 vuong goc vdi BC; t h i m a t p h i n g (P) cat A D va BC I a n lifat t a i K va L , (P) chia l a n g t r u
I cut thanh h a i p h a n : l a n g t r u t a m gidc dtifng M A B . N K L c6 the t i c h :
V , = M N . d t ( M A B ) = - abx 1
2
2a j
va h i n h chop d i n h N , day K L C D , c6 the t i c h :
Va = - N K . d t ( K L C D ) =
3
-bx^
6
307
t h a n h t h i l ISng t r u cut M A B . N C D c6 t h e t i c h :
V = V i + Va = - a b x 3 6
2a J
Yeu cau b a i t o a n can xac d i n h x ( h i e n n h i e n 0 < x < 2a) sao cho V = 2V'.
2a^b
1 .
= — abx 3 3
3
2a )
X
Phiicfng t r i n h nay c6 n g h i e m :
- 6ax + 4a^ = 0
= a(3 + V S )
x = a(3-V5)
V i 0 ^ X < 2a n e n c h i c6 t h e c h o n : x = a ( 3 + V S ) ( y c b t ) .
B a i 399
( D A I H O C K T - T H - SP - N N -
1983)
Cho tuf d i e n S A B C , day A B C la tarn giac vuong t a i A , A B = 2a, A C = 3a, canh SB vuong
g6c v d i day SB =
a/ C h i ro tarn v a b a n k i n h m a t cau ngoai t i e p tijf d i e n SABC.
b/ M l a m o t d i e m d i d o n g t r e n canh SC, dat M C = x. Goi H va K I a n liftft l a cdc h i n h chieu
vuong goc ciia M l e n cac m a t phSng (ABC) va (SAB). Mat p h l n g K M H , cat A B t a i L . ChiJng
m i n h r k n g : K M H L l a m o t h i n h chi? n h a t . V d i gia t r i nao cua x t h i K M H L la mot h i n h vuong.
c/
T i n h theo a va x do d a i dudng cheo M L ciia h i n h c h a n h a t K M H L . V d i gia t r i nao cua x
t h i M L C O do d a i nho n h a t ? l 7 n g v d i gia t r i da t i m difdc cua x, hay neu l e n dac t i n h h i n h hoc
cua dean M L .
d/
H a y t i n h theo a va x t h e t i c h V cua h i n h chop d i n h A , day PCMHL. K h a o sat sif bien thien
va ve do t h i ciia h a m V k h i M d i dong t r e n canh SC.
e/
Xac d i n h x sao cho : V =
4V3
27
Giai
a/ Goi O va I I a n lifat la t r u n g d i e m ciia SC va BC. Cac t a m gidc SAC, SBC theo thii ti(
vuong t a i A , B nen t a c6 :
0 1 l a t r u e dUcJng t r o n ( A B C )
|ma
Ola
trung diem S C
OA = OB = OS = OC
Vay O l a t a m cua m a t cau ngoai t i e p tuf d i e n SABC va
ban k i n h m a t cau l a :
R = OA = —
2
= 2a
That vay :
SC^ = S B ^ + B C ^ = S B ^ + A B ^ + A C ^
^
S C ' = 33^ + 4 a ^ + Qa^ = I G a ^
=> SC = 4a
=> R = 2a (ycbt).
308
b/ KMHL la hinh chuf nhat.
MK ±
(SA B)
A C 1 (SA B)
'
=> M K // AC ^
MK // HL // AC
M H // SB
M H // KL // SB
M H 1 (A BC)'
SB 1 (A BC) J
=c.
Vay tuT giac KM H L la mot hinh binh hanh.
De y den SB 1 AC => H L 1 LK : nhvt vay tit giac KM H L la mot hinh chff nhat (dpcm).
Dinh X de KM H L la mot hinh vuong. Ta c6 :
M H _ CM
SB
~
MH =
SC
MK
SM
AC
SC
MK =
SB.MC _ aVs.x _ V 3 x
SC
~
4a
"
4
A C SM _ 3 a ( 4 a - x) _ 3 ( 4 a - x)
SC
~
4a
~
4
Vay : KM H L la mot hinh vuong
«
d Ta
V3 x ^ 3 ( 4 a ^
4
4
CO
< ^ V 3 x = 3( 4a- x ) ^
: ML^ = M H ^ + HL^ =
3x2
9(4g _
16
16
x = 2 a V 3 ( V 3 - 1) (ycbt).
...
V3( x 2 - 6 a x + 12 a 2 )
V3(x - 3af
ML =
:
= -'^
2
M L>
9a'
3a
2
2
; Vx
G
3(x^ - 6 a x + 12 a 2 )
+ 9a^
^
;Vx
6
[0; 4a]
^^
[0; 4a]
( 1)
Dau dang thilc tro ng (1) xay ra <=> x = 3a
3a
=> minML = — , xay ra khi va chi khi x = 3a.
Ta CO : AB 1 (MKLH)
minML =
3a
A B± M L
= d[(AB); SC]
Vay khi M L nho nhat thi doan M L la doan vuong goc chung cua hai difdng th i n g A B va
SC (ycbt).
d/ The tich V hinh chop A .MHKL.
V = i d t(MHKL).A L = - M K.M H.A L
3
3
AB 1
Ta
CO
(2)
BS
: AB 1 ML
o
BS, M L, AC ciing vuong goc \6i A B.
AB 1 AC
<=> BS, ML, AC cung nkm trong 3 mat p hing song song.
Ap dung dinh ly Thales
AL
CM
AB
CS
—
4a
AT
AL = —
2
309
'2'
=^V=
1 3
xV3 X
- . _ ( 4 a - x ) - — =
3 4
4
2
Vs
o
—x^(4a-x)
32
(ycbt).
Sau day ta khao sat sif b ien th ien va ve do t h i cua V tr on g he true (OxV).
M ien xac d in h : D y = [0; 4].
"x = 0
=i> V = 0
Dao h a m: V = ^
(- 3x^ + S a x) = 0 <=>
32
27
3
V" — ( - 6 x + 8a ) = 0 «
=> V =
X =
—
4a^V3
Vay (C) la do t h i ciia V tr on g ycbt.
8V3a^
4^3
,
e/ D in h X de ; V = ~ ~ a^
27
Xet : V =
VS
<=>
27
'
(C )
4V3a^ ,
4^3
27
27
2,.
,
4V3
(4a - x) =^
— X
32
Rx) = x^ - 4ax^ +
De y thay : f
3
0
a
27
128
27 i
= 0
a^ = 0
4a
3
32
(4)
o
3
3
8a
4a
4a
x
(3)
- 4a
nen ta c6 :
4a
X =
(3)
O
9
8
x'^ - - a x
3
/
128ay
727
/
/ 0
8a
/
0
_
a"^ = 0
9
4a
X =
—
x = i(l±V^^
310
Vi X e [0; 4a] nen cdc n g h i ^m cua (3) la : x =
4 J3
Vay : V = ^ a ^ ^
t
«
4a
4a
4
x = ^
V X
=
- ( 1 + V3 )a
r-
v x = ^ ( 1 + Vs a) (ycbt).
Bai 400 ( D A I H O C K I N H T E T P . H C M -
1991)
Trong k h o n g g i a n cho doan OO' = h va h a i nijfa dtfdng t h ^n g O d , O'd' cung vuong goc v d i
00' va vuong goc v d i nhau. D i e m M chay t r e n Od, d i e m N chay t r e n O'd' sao cho ta luon c6:
OM^ + O'N^ = k^ (vdi k la m o t dp dai cho trifdc).
1/ Chijfng m i n h rSng do dai doan M N k h o n g doi.
2 1 Vdi v i t r i nao cua M t r e n Od va N t r e n O'd' t h i tuf d i e n O O ' M N c6 the t i c h I d n n h a t . T i n h
gia t r i do theo h va k.
3/ Muon M N tie'p xuc v d i mSt cau difdng k i n h OO' t h i h va k p h a i thoa m a n di4u k i e n gi ?
Neu each diTng M N t r o n g t r i f d n g hop do.
Goi P va Q t u a n g iJng la h a i d i e m nSm t r e n O d doan O'd'. G o i H l a h i n h chieu cua
diem giaa K cua doan OO' l e n PQ. H a y chufng m i n h r S n g k h i PQ t h a y d o i sao cho OP +
Q'Q = PQ t h i H n a m t r e n m o t d i f d n g c6' d i n h . H a y c h i r a dirdng c6' d i n h do.
Giai
1/ Chijfng m i n h doan M N k h o n g doi :
O'N l O O ' l
O'N l O M
ON
1 O O ' M => O N J. O M
Ta CO : tiT d i n h l y Pythagore
^
M N ' = O ' N ' + O'M'*
(1)
OM' = 0 0 ' + OM'
(2)
M N ' = O ' O ' + O ' N ' + O M ' =>
=> M N = Vh^
M N ' = h' + k'
= const (dpcm).
2/ Ta CO : O'N 1 (O'OM)
^ V = V, o o M N , = -dt(AOO'M) .OTSr = - O O ' . O M . O ' N = - O M . O ' N
^ V'= ii-.OM^.O'N" < —
36
36
36 v 2 .
(3)
144
Dau dang thufc t r o n g (3) xay ra <=> O M ' = O ' N '
o
OM' + O N ' = 20M' = k'
=> O M = O N =
kV2
Vay k h i chon M e d va N e d' sao cho ; O M = O ' N =
Ifa nhat : maxV = — O M ' =
6
kV2
t h i t i i d i $ n O O ' N M c6 the t i c h
(ycbt).
12
311
3/ Dieu k i e n de M N t i e p xuc v
O N
= IN
OM = IM
M N = O N + OM
« . M N ^ = O'N^ + OM^ + 2 0 ' N . 0 M
=> h ' + k^ = k^ + 2 0 ' N . 0 M
^
20M.0'N = h '
(4)
De y : ( O M - O'N)^ = O M ^ + O'N^ - 2 0 M . ON > 0
o
2 0 M . O N $ O M ^ + O'N ^ (Ap dung (4))
<=> h^ < k^ «
•
0 < h < k
(5) (ycbt).
Cach d u n g doan M N : Cho M e d m a O M = a. TiT (4) t a c6 : O'N =
T a t i m di/oc N e d ' difgc xac d i n h bdi O'N =
2a
2a
v6i dieu k i e n (5).
Chiitng m i n h H n k m t r e n m o t dUdng co d i n h :
Q
(d')
D a t OP = X, O'Q = y.
P Q = O P + O'Q = X + y
Ta
CO :
• [PQ = P H + H Q
PH + HQ = x + y
(6)
D i n h l y Pythagore :
KP2 = OK^ + 0P2 = H K ^ + HP2
KQ2 = O'K^ + 0 Q 2
= H K ^ + HQ2
Do O K = O'K ^ KP2 - KQ^ = OP^ - OQ^ = HP^ - HQ^ =
^
- y^
( H P - H Q K H P + H Q ) = (x - y)(x + y)
(8)
K e t hop (6) va (7) => H P = x va H Q = y
(7)
Tif (6) => H P - H Q = x - y
Qua O difng d i // d' => d' // (d'; d i ) va (d'; d i ) 1 (d; d i ) , nen goi Q' la h i n h chieu cua Q tren
(d; d i ) t h i Q' e d , va do (QQ'P) _L (d; d j ) nen goi H ' l a h i n h chieu cua H t r e n (d; di) t h i H' e
PQ'.
Tiir giac OO'QQ' la h i n h chif n h a t => OQ' = O'Q = y
Do H H ' // QQ' :
HQ
HQ'
H P
H P
Ma
X
"
'
(8)
OP
H P
OP
HQ'
OQ'
OQ'
O H ' la p h a n giac P O ^ ' =
(3^73^
Dieu nay chijfng to H n a m t r o n g m a t p h a n g co' d i n h (a) tao bdri OO' va p h a n giac Ot cua
goc hap bdi d va d i // d'.
Do : A O P K = A H P K ^ H K = O K =
- .
2
312
Vay H nftm t r e n diicrng t r o n co d i n h t a m K, ban k i n h — chufa t r o n g mSt p h ^ n g co d i n h
2
(a), xac d i n h n h u t r e n (ycbt).
Bai 401 ( D A I H O C SlJ P H A M T P . H C M -
1991)
Cho h i n h chop S.ABCD day A B C D , co A B D va CBD 1^ h a i t a m gi^c deu canh a. Canh SA = h
vuong goc v(Ji day. Goi O l a giao d i e m cua A C va B D , M 1^ d i e m d i dong t r e n A C , khdc A
va C; (Q) la m a t p h ^ n g qua M va vuong goc AC.
1/ Tuy theo M thuoc OC hay thugc OA hay chi ro each d i m g t h i e t dien m a (Q) c&t h i n h ch6p.
2/ Bat X = M C . T i n h dien t i c h t h i e t dien noi t r e n theo x, a, h . K h i nao d i e n t i c h ay lorn n h a t .
Giai
1/ Doc gia tU p h a n t i c h va chiifng m i n h va bien l u a n , d day t a x e t h a i k h a n a n g xay r a t u y
theo v i t r i cua M t r e n A C = A O ^ OC.
D C a c h dtfng t h i e t d i § n k h i M 6 A O
• Trong ( A B C D ) , qua M dUng di/cfng t h S n g song song
vdi cheo B D I a n lUOt gap A B , A D t a i N va G.
• Qua N , M , G d i i n g cac difdng t h S n g vuong goc v d i
(ABCD) Ian luot gap SB, SC, SD t a i L , E , F t a dUcfc
thiet dien muon t i m la ngu giac N L E F G gom h a i h i n h
thang vuong bSng nhau co chung day lorn M E (ycbt).
• Cach dUng thiet di$n k h i M e O C
• Trong ( A B C D ) qua M dUng diidng t h i n g
song song v d i B D I a n luat cSt C B , C D t a i N , G.
• Trong (SAC), qua M d u n g diTdng t h a n g song song
vdi SA cat SC t a i E. T a m giac E N G can t a i E l a t h i e t
dien muon diTng (ycbt).
21 De xac d i n h x = C M de dien t i c h S ciia t h i e t d i e n Idn n h a t t a x6t h a i k h a n a n g sau
• THi : M
NO // B D
ME // AS
N L // AS
6
OA : 0 < A M < A O «
0 < aVs - X
NM
AM
2(aV3 - x )
BO
AO
iV3
ME _ CM
ME
AS ~ CA
NL
BN
OM
< X
NM =
< a V 3 (1)
3a - xVs
xhV3
3a
X - •
A S ~ B A ~ OA
NL =
2x-a>^
NL =
aV3
Do do
<
S, =
2xV3h
3a
2xV3 - 3a
3a
.
x h V 3 Y 3a - x V 3
- h +
3a
A
313
•
Si = — ( x V 3 - a) ( 3a - x V 3 )
3a
S, = - ^ ( - S x ^ + 4 a V 3 x - 3 a ^ )
3a
S' l = — ( - 6 x + 4aV3) = 0 «
3a
1V3
2
X=
<
CM
CO
BO
3
< a
(2)
(3)
< aV3 - x <
2x
NM
N G // B D :
S2 =
X
2 a A/ 3
2
TH2 : M e C O : A O ^ A M < A C
NM
.
CM
ME
2xV3 a
3a
aVs
a V s <=> 0 < x <
xV3
2
X A / 3,
M E // A S :
ME =
AS " CA " aV3
xVs x V s,
Luc do :
( 0 < x < ^ )
2
3a
= 0
h
3a
x ^h
ih = : i ^
3a
2hx
S'2 =
<=>
X =
(4)
(5)
0
3a
De y den (3) va (5) va hai bieu thiifc S ] , S2 d (2) va (4), ta se lap difgc bang bien thien kep
nhu sau :
0
a
S'2
Si
S2
Si
S2
^
,
ah
Tu do, ta CO :
,
ma x fa = — , tifcrng iTng voi x =
0< x < a V3
2aV2
3
,
, ^
(ycbt).
3
B a i 402 ( T T D A O T A O va B O I D U O N G C A N B O Y T E T P . H C M
1993)
Cho hai diem A , B doi xilng nhau qua mat phang ( P ) , I la giao diem cua A B vdi ( P ) , 0 la
mot diem nam ngoai ( P ) , co hinh chieu vuong goc xuong ( P ) la H , con M la mot diem chay
tren dudng tron dUcmg kinh I H ve trong ( P ) .
1/ Chufng minh rang I M la diidng vuong goc chung cua A B va O M .
2/ Chufng minh rkng hai diem A, B luon each deu dudng O M .
3/ Cho A B = 2a, M H = x, M I = y. Tinh the tich tuf dien O M A B . Xac dinh vi tr i M de the tich
do Idtn nhat.
Gi a i
l/ Ta c6: AB l( P)
=^ AB I . I M (1)
M thuoc dudng tron difdng kinh I H
^
(2)
I M _ LM H
314
Mat khac : O H 1 (P)
=> I M 1 O H
(3)
TO (2) va (3)
I M 1 (OMH)
=> I M 1 O M
(4)
TO (1) va (4) => I M l a dudng vuong
goc chung cua A B va O M (dpcm).
21 Difng AR 1 O M ; B F 1 O M
JAE = d [ A ; ( O M ) |
^
[ B F = d[B; ( 0 M ) |
Trong m a t phang ( O M H ) t a diftig:
Ex 1 OF; F y 1 O F
f(AEx) ± O F
^
|(BFy) ± O F •
=> A E ; I M ; B F n a m t r e n ba m a t p h a n g song song
Theo d i n h l y Thales :
Nhung : A A M B can
3/ De' y tha'y :
Ma : \
ME
lA
M F
I B
=
=>
A M = M B
O H // ( A B M ) ^
M H
1
M I
' M H
1
A B
V = V o A i i M = V|| AMB =
1 => M E = M F
=> A A E M =
VQ.ABM =
1
M H
B
A E = B F (dpcm).
ABFM
^UMIM
( A B M )
— -MH.SAAMB
o
(ycbt)
(5)
+ y^ = I I I ^ = const
(6)
V = - M H . A B . I M = - axy
6
3
Xet : A I M H vuong
2
(1)
BDT Cauchy =>
2
= —.xly^ < —
9
9
Dau dang thufc t r o n g (6) xay r a
^
«
Luc do : maxV^ = ^
Vay
VoABM
36
X
+y
36
x2
= y'
(6)
X = y
A I M H vuong can t a i H .
<=> m a x V = — . I H
6
l
Biii 403 ( D A I H O C SLf P H A M T P . H C M - 1993)
Cho tuf dien SABC c6 goc p h i n g d d i n h S vuong.
V Chiing m i n h r a n g : Vs S A D Q ^ Sggc + SogA + S' S A C 2/ Biet rang SA = a; SB + SC = k k h o n g doi. D a t SB = x. Ti'nh the t i c h tuf dien SABC theo a, k ,
jva xac d i n h SB, SC de t h e t i c h tuf dien SABC \6n nhat.
3/ Cho A CO d i n h B v a C t h a y d o i sao cho SB + SC = k ( k h o n g ddi). T i m quy t i c h giao d i e m O
cua cac dUcmg cheo h i n h hop c6 ba canh l a SA, S B , SC.
315
1/
Giai
D i f n g : S I 1 BC va n o i A I , t a c6 :
sue
RBA ^
SAC
"
= -(BC^SI^ +SA^SB^
4
+SA2.SC2)
= - [ B C ^ S I ^ + S A ^ I S B ^ + SC^)]
4
= i(BC2.Sl2
4^
I S
+SA^BC2)
'
= - [ B C ^ . C S I ^ + S A 2 ) ] = -(BC2.Al2) = SiBc
4
4
(1)
Ap d u n g B D T S c h w a r t z :
( S S_l i C + S , S„B .A
+SsAc)
'SBC + ^I B A + S|AC )
2/ Goi t h e t i c h tiir d i e n S A B C l a V :
V = - AS.S„„,, = - S A . S B . S C = - a x ( k - x) ^
3
-
X
+ k -
X
(2)
24
Dau d a n g thijfc t r o n g (2) x a y r a
•»
x = k - x
« •
k
X =
—
.
2
Vay m a x V =
ak^
y
; t u a n g ufng : •
24
C
SB = SC = — (ycbt).
2
3/
c , / /
Goi O' l a h i n h chieu cua O xuong
(SBC).
T r o n g m a t p h a n g toa do (Sxy) = (SBC), t a c6 :
X
SB
x =
s
O' :
B
\
SC
S B + SC
^
X
+
y
=
V i SB + SC = k ^
= —=>
2
y = -x + —
2
A
0 < SC < k
316
\
0 < X < k
Quy tich O' la khoang (BC) :
De y thay : OO' = - SA = SG ; VB; C
2
0 < X < k
Vay quy tich diem O la khoang (B|Ci) : •
k
la hinh tinh tien cua (BC) theo
y = -X + •
vecto : u = — S A
2
(ycbt).
Bai 404 ( D A I H O C T O N G H d P H A N O I - K H O I A -
1993)
Cho tuf dien ABCD c6 A B = x va CD = b, cac canh con lai bang nhau va bSng a. Goi E, F
n luat la trung diem ciia A B va CD.
1/ Chiifng minh rSng : A B 1 CD va EF la ducfng vuong goc chung cua A B va CD. Tinh EF
theo X, a, b.
2/ Tim X de hai mat phang (ACD) va (BCD) vuong goc vdi nhau. Chiifng minh rang khi do tiif
dien ABCD c6 the tich \6n nha't.
G iai
1/ Ti/ gia thiet :
ma
x/
BC = BD => BF 1 CD
E
/
\
[CD 1 AB
Vay CD 1 (ABF)
Do BC = AC
A
AC ± A D => A F _L CD
\
B.
[CD 1 E F
•D
CE I A B ma A B 1 CD
a\
=> AB ± (CDE) => A B X EF
Vay EF vuong goc chung cua A B va CD (dpcm).
A F2 = A C ^ - C F ^
Ta CO
A F^ =
-
=> EF^ = a^ -
b^
x^
(= A F' - A E')
A E^ = i i EF
4a' - b-^ -
2/ Theo tren CD 1 (ABF)
EF =
(0 < X < V4a^ - b ^ ) (ycbt).
A FS =
. Goc nhi dien do vuong khi va chi khi
4a'-h'
AB
x
=
—
<=> X =
Khi do : VABCD = VcAiif• + VDAHF = - CF.SA RF + 1
*ABCD = - SA I)F(CF + D F ) = - SA BK.CD = -
D F. S'ABF
SABF
3
317
D o FA
VABCI) I d n n h a t k h i SABF I d n n h a t .
= FB= i Vi l ^
n e n VAHCD Id n n h a t
<=> SAUK = i F A . F B s i n X p ^ = - ( 4 a ^ - b 2 ) s i n A F B
(1)
3ma xi( 4a ^ - b ^ ) s i n A F B =
8
»
s i n A F f e = 1 <=> A f f e
(1)
i ^ ! — ^
8
= 90" ( yc b t) .
1994)
Ba i 405 ( D A I H O C B A C H K H O A T P . H C M -
Tr o n g m a t p h S n g ( P ) c h o ta r n g i a c A B C vu o n g go c t a i A , A B = c; A C = b. T r e n diJdng
t h a n g vu o n g go c v d i ( P ) t a i A , l a y m o t d i e m S s a o ch o S A = h ( h > 0). M l a m gt d i e m d i dpng
tr o n c a n h S B . Gp i I , J I a n lu o t l a t r u n g d i e m c i i a B C v a A B .
1/ T i n h do d a i d o a n vu o n g go c c h u n g c i i a h a i d u d n g S I v a A B .
2/ T i n h ty s6 ' giOTa t h e t i c h c^ c h i n h c h o p B M I J v a B S C A k h i do d a i d o a n vu o n g go c chung
c iia h a i dUcfng A C v a M J d a t g i a t r i Id n n h a t .
Giai
1/ D e y t h a y A B 1 ( S A C ) v a S I c6 h i n h c h i e u x u o n g
( S A C) l a S K .
Lu c do :
J l K 1 A C => I K 1 ( S A C)
' AN ± S K; N e S K
J AN 1 I K
K h i do
' AN ± S K
AN 1 ( S I K )
N E / / AB , E e S I
Dung :
E F / / AN . F e
EF 1 ( S I K )
AB
=> EF 1 S I
A B 1 ( S A C ) => A B 1 A N => A B 1 EF
Do :
=> EF l a d o a n vu o n g go c c h u n g c u a A B v a S I .
N h i fn g : EF = A N , n e n t a t i n h do d a i A N .
I
A S A K vu o n g
l
AN ^
l
AS^
1
1
1
4
+ •
AN ^
AK^
b^_ +4h^
«
bh
AN =
AN ^
Vb ^ + 4 h 2
D o d ^ i d o a n vu o n g go c c h u n g c u a A B v ^ S I l a :
S
EF =
2/
,
Vb ^ + 4 h 2
( yc b t) .
D e y A C ± ( S A B ) 3 M J , t r o n g m a t p h l n g ( S A B ) , d ito g A H 1 M J
=> A H l a d o a n v u o n g go c c h u n g c i i a A C v a M J .
T a CO : A H < A J =
M.
B
318
maxAH = - k h i va chi k h i H = J
2
<=> M la trung diem cua SB.
Luc do ti/ang ufng : —m i.
=
(ycbt).
8
^BSCA
Bai 406 (DAI HOC K I E N TRUC TP.HCM - 1995)
Trong mat phfing (P) cho tarn giac OAB vdi OA = OB, AB = 2a v^ difcfng cao OH = h. Tren
dudng thang (d) vuong goc vdi (?) tai O, lay diem M vdi OM = x. Goi E va F Ian luat la hinh
chieu vuong goc cua A len MB va OB; N la giao diem cua EF va (d).
1/ Chijfng minh MB 1 NA v^ MA 1 NB.
21 Tinh BE, BF, EF, AF va the tich tuf dien ABEF theo a,h v^ x.
3/ Tim vi t r i cua M tren (d) sao cho t i l dien MNAB c6 the tich nho nhat vk t i n h gia t r i nho
nhat nay.
Giai
M O l AF
1/ De y :
Ma :
Ta
CO
1
AF
[ B O l AF
(MOB) ^ A F 1 MB
AE 1 MB => MB 1 (AEF) --
M B 1 A N (dpcm).
: AF 1 (MOB) => AF _L NB
Mat khac, F la trUc tarn cua AMNB
M F J_ NB
Vay : N B 1 ( A F M ) => N B 1 A M (dpcm).
2/ Ta CO : A H K B co A H A O
HK
HB
HA
HO
HK.h = a^
HK =
OK = h
n
u2
h
„2
- a
h
Trong tiJ giac noi tiep B F K H ta c6 :
I ^
= FtHF
A O F H CO A O K B
OF
OH
OK
OB
OF.OB = O H . O K
OF.OB = h' -
OF =
a'
(vdi : B 0 = V h 2
)
Khi do : B F = O B - O F = V h ^ + a ^ V h ^ + a^
319
3/
Vay :
B F =
2a^
,
(ycbt).
T u a n g t i f , t r o n g tur giac n o i t i e p M O F E t a c6 : E S J F ' = E l O ^
ABEO
ABFM
=>
=
B F
to
,
EF
AOMB
BE.BM = BO.BF
B M
BE =
T a CO : A E F B
2a^
=
(ycbt).
B F
O M
B M
EF =
2a2x
,
V(a2 + h 2 ) ( a 2
(ycbt).
+x2)
2ah
T u a n g t a , t a c6 : A F . O B = O H . A B
AF =
,
(ycbt).
The t i c h tut d i e n A B E F :
VABEF = ^ A F . S B E F = - A F . E F . B E
=
„,
„
,
+ n )(a + n
^
+ "
(ycbt).
'
S^ ^ j , =- |^ A F . M N . O B
The" t i c h tiJ d i e n M N A B l a : V M N A B = ^^ - A
A FF. S
Nen : 3 min(V^J^AB '
3niin(MN)
N h U n g : A N O F co A B O M = > —
=
OB
A p dung B D T Cauchy :
^
= > O M . O N = O F . O B = h^ - a^
O M
M N = M O + ON > 2 V h ^ - a^
3 m i n ( M N ) = 2 V h ^ - a^
OM = ON
2ah-\/h^ - a ^
I
Vay : 3 m i n V M N A B <=> O M = V h ^ - a^ va minVMNAis =
(ycbt).
3
B a i 407 ( D A I H O C T O N G H O P T P . H C M - 1995)
Cho tarn giac deu O A B c6 c a n h bSng a > 0. T r e n difdng t h i n g (d) d i qua O vuong goc vdi
m a t p h i n g (OAB) l a y d i e m M v d i O M = x. Goi E,F I a n l u g t l a cac h i n h chieu vuong goc cua A
len M B , O B . D i f d n g t h i n g E F cSt d t a i N .
1/
Chtifng m i n h r S n g A N 1 B M .
2/ Xac d i n h x de t h e t i c h t i l d i e n A B M N l a nho n h a t .
Giai
1/
T a CO :
Ma :
A F I B O
[AF
I d
AE 1 M B
=> A F 1 ( M O B )
^
=> A F 1 M B
M B ± (ANE)
2/ De y : A F X ( M B O ) s ( M N B )
^
^ M B ± N A (ycbt).
A F l a chieu cao h i n h chop A . B M N
320
=> V
=~AF.S
ABMN
2
=-AF.BO.MN
MNB
Q
iV3
AF =
Trong do : BO=a
MN =MO+ON= +ON
X
Do do : 3m
in(V
.„„„) o 3min(x +ON)
AnMN
Mat khac, ta c6 : ANOF w ABOM NO OF
BO OM
« OM.ON = BO.OF « x.ON = — = const
2
Apdung BDT Cauchy, ta c6 : x + ON ^ 2 J — = aV2
Dau dang thutc trong (*) xay ra <=> x = ON =
3rain(x +0N) = aV2
(*)
1V2
x = ON = ^
Vay the tich tut dien ABMN nho nhat khi va chi khi x =
'
(ycbt).
2
1 Bai 408 (DAI HOC XAY Dl/NG HA NOI - 1995)
Trong mat phang (P) cho hinh vuong ABCD vdi AB = 2a. Tren mat phang chuTa BC va
JTuong goc vdi (P) lay diem E sao cho AEBC la tarn giac deu; diem I nkm tren doan BC, dat :
|B1 = X. Kla hinh chieu vuong g6c cua diem E tren dUdng thang AI; O la trung diem cua AE.
1/ Timquy tich ciia diem K khi I chay tren doan BC.
2/ Tinh do dai 0 1 theo a va X.
13/ TimX de do dai OI Idn nhat, be nhat.
Giaima [(P); (EBC)] = 90"
11/ AEBC_deujien trung tuyen cua no EF _L BC
:=> EF 1 (P)
EK : dudng xien
FK : hinh chieu
ma EK 1 AK => FK 1 AI (dinh ly 3 difdng vuong goc)
luon vuong, AF co dinh nen K di chuyen tren tron (C) dUdng kinh AF.
321
GicJi han khoang chay:
I
s
C => K ^ H; H e AC; FH // BD
Dao lai K e
BH c
(C)
^ EK ± A I
Vay quy tich K la cung BH c (C) (ycbt).
2/ Dinh ly dirdng trung tuye'n cho :
_ 2AI^ + 2 E I ^
- AE^
4
U F ^ =5a2
AE^ = AF^ + EF^ = 8 a ^ AI^ = 4a^ +
=> 01 = V x ^ - a x + 2a^
(ycbt).
3/ Ta Viet : OI^ = fix) = x^ - ax + 2a^; Va e [0; 2a]
=> f (x) = (2x - a) = 0 <=> X = 2
Difa v^o bang bien thien, ta c6 :
m a x f ( x ) = f(2a) = (2a)^
0
2a
+00
f'(x)
min f(x) = f
^a^
0
m a x 01 = S(2a) = 2a
0« x< 2a
min OI = S
^a^
0< x« 2a
B a i 409 (DAI HOC D A I CUONG - 1996)
Cho id dien ABCD c6 AB = CD = 2x va 4 canh con lai deu c6 do dai bkng 1.
1/ Tinh dien tich toan phan (tdng dien tich. cua 4 mSt) cua tur dien theo x.
2/ Xac dinh x d6 dien tich toan phan dat gia t r i Idn nhat.
Giai
1/ Nhan thay, cac mat cua tijf dien la cac tam gidc bkng nhau.
Suy ra, dien tich toan phan ciia til dien la :
= 4.8^^^ =
2.AI.CD
(1)
Vdri A I la dudng cao cua ACAD can tai A; ta c6 :
AI = V l - x^ ;
(0
322
^
= 2.2x.Vl -
2/ N h a n t h a y :
= 4 x V l - x ^ ; (0 < X < 1) (ycbt).
3max(Stp)
2x/
3 max x V l - x 2
^
<=>
J
2ri
..2
B'
3 max(16x^[l - x'^)]
sD
Ap dung B D T Cauchy :
2x
Stp = 1 6 x ^ ( 1 - x ^ X
16
= 4
Dau d^ng thuTc trong (2) xay ra <=>
Vay v d i x = —
= 1 - x^
(2)
x =
t h i dien tich toan phan cua tif dien dat gi£i t r i Idn n h a t la
maxStp = 2 (ycbt).
Bai 410 (DAI H O C Q U O C GIA T P . H C M -
1996)
Cho tuT dien SABC c6 goc p h i n g d dinh S vuong.
1/ Chufng minh rSng VS.SABC ^ SSAB + SSBC + ^SAC
2/ Biet rftng SA = a; SB + BC = k. Dat SB = x. Tinh the tich tuf dien SABC theo a; k; x va
xac dinh SB; SC de the tich tiif dien SABC Idn nhat.
Giai
(Xem
D A I H O C SU P H A M T P . H C M -
1993)
Bai 411 ( D A I H O C Q U O C G I A T P . H C M - K H O I D -
1997)
Cho tiJ dien deu ABCD canh a. Goi H la hinh chieu vuong goc cua A xuong mat phSng
(BCD) va O la trung diem cua A H .
1/ Tinh the tich V cua tuf dien theo a.
2/ ChOfng minh rang AB 1 CD. Tinh khoang each gifla hai difdng th^ng AB, CD theo a.
3/ ChOfng minh rang cac dudng th^ng OB, OC, OD ti^ng doi mot vuong goc nhau.
4/ Xac dinh diem M trong khong gian sao cho MA^ + MB^ + MC^ + MD^ dat gia t r i nho nhat.
Giai
1/ Do ABCD la tuf dien deu canh a va H la hinh chieu vuong goc ciia A xuong (BCD)
H la trong tarn ABCD
=>BH=2.aV^
3
2
A H
aV^
= V A B ^ - B H ^
Vay: VABCD = -
=
\ ^ ^ ~
.AH.S„CD
323
2/ T a
1 aVe
V,„c n = 3
-
y
a^Vs
-
^
=
a^V2 ,
-^( y c bt ) .
CO :
B H l a h i n h c hi e u c ua B A l e n ( B C D )
BH
i C D
=> A B 1 C D
(dpc m ).
Do A B C D l a tiJ di e n deu, ne n B H se c&t C D t ai
t rung di e m I v a B I = A I .
Go i J l a t rung di e m c ua A B , t h i t a c6 : I J 1 A B.
T uan g t i f : J D = J C => J I 1 C D .
=> I J l a do an v uo ng goc c hung cua A B v a C D.
a^
3a2
I J =V BI 2
3/ T a
CO :
A H
- B J 2
=
V
aV2
(y cbt ).
4
= V A B^ - B H ^
l^/6
= ja^ ^ ^
3
3
O e A H (t rue dudng t ro n) => O B = O C = OD
2
2
Ma: OB^ = OH ^ + H B^ = - A H ^ + H B^ = — +
4
6
OB = OC = OD =
—
3
i V2
N h a n t h ay : OB^ + OC^ = a^ = BC ^ => A BO C v uong t ai O.
T i f ang tU, A BO D v a A C O D v uo ng t ai O.
V ay O B; O C ; O D t ifng doi mot v uong goc nhau (dpcm).
4/ Go i G l a t ro ng t a m t i l di e n A B C D => G A + G B + G C + G D = O
f
Xet :
I
= MA^ + MB^ + MC^ + MD^ =
2
^
f ~> \ 2
+
MA
MB
1-
+
> N, 2
MC
^
—> ^ 2
f ^
=> S
=
2
+
MG+GA
2
+
MG+GB
(
+
MG+GC
+
J
MD
V
/
2
_>
MG+GD
V
>
= 4 . M G2 + G A ^ + G B ^ + G C ^ + G D ^
S
+ 2MG GA +GB + GC+GD
= 4 .MG2 + G A ^ + G B ^ + G C ^ + G D ^
I = 4 .MG2 + G A ^ + G B ^ + G C ^ + G D ^
324
