Các dạng toán và phương pháp giải Hình học giải tích 12 - Trần Đình Thì

8/10/2019 Cc dng ton v phng php gii Hnh hc gii tch 12 - Trn nh Th

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Trn inh Thi

12 Bin son tho ni dung v'chng, trnh SG K phn ban mi* Bi dng v rn k nng gi c dng ton in hnh thhg gp n tp v chun b cho C k thi quc gia (ttnghip, tuyn snh...}doB GD& Tt chc - :

K1(EG

NBA XH BN I HC QUC GIA H NS

WWW.FACEBOOK.COM/DAYKEM.QU

WWW.FACEBOOK.COM/BOIDUONGHOAHOCQU

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W.DAYKEMQUYNHON.UCOZ.COM

ng gp PDF bi GV. Nguyn Thanh T


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Cung mt tac gi I- Phn dng & phngphp gii bi tp hnh

hc 11 *

- Phn dng & phng

php gii i s - gii

tch 11

- Thit k bi ging Hnhhc 11

-Thit k bi ging i

s - Gii tch 11

- Rn k nng gii ton

trc nghim 12

- Cc dng ton &

phng php gii Giitch 12

- Cc dng ton &

phng php gii Hnhhc 12

- Phng php gii bi

tp phng trnh lnggic

- Dng HH Gii tch gii PT, BPT, HPT...



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NH XUT BN I HC u c GIA H NI16 Hng Chui - Hai B Trng - H Ni

T (04) 9715013; (04) 7685236. Fax: (04) 9714899

Chu trcli nhim xut bn:

Gim c PHNG QUC BO Tng bin tp NGUYN b t h n h

B i n tp ni dungMINH HI

Sa bi

HONG NGUYN

Ch bn CNG TI ANPHA

:Trnh by ba

CC DNG TON & PHNG TRNH GII HNH HC GII TCH 12M s: 2L-131H2008In 2.000 cun, kh 16 X 24 cm ti Cng ty TNHH in Bao b Hng PhS xut bn:293-2008/CXB/4i-54HQG HN, ngy 08/04/2008Quyt nh xut bn s: 131 LIC/XBIn xong v np lu chiu qu IV nm 2008.

SN Ki tc lin kt xut bn

CNG TI ANPHA



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LI NI U

Nhm nng cao cht lng dy v hc ca b mri tn ti cc trng

THPT. Cuh sch ' Cc ng ton v phng php gii hnh hc gii tch

12" c bin son theo chng trnh phn ban nm 2008 - 2009 (Ban khoa

hc t nhin v Ban c bn).

Cuh sch c phn theo cc ch v cc dng ton trong cc ch

. Mi ch tc gi^nu bi tp mu c phn theo cc dng v c bi

tp nng cao gip hc sinh lm quen vi phng php t hc, t nghin

cu, chim lnlv kin thc. Cu'n sch l ti liu tham kho gip cc em rn

luyn k nng gii ton v iu kin hon thnh cc bi thi trong cc k

thi Quc gia (Tt nghip, tuyn sinh vo i hc) do b Gio dc v o

to t chc. Tc gi cm n s ng gp nhn xt ca cc ging vin, ng

nghip trong qu trnh bin son b sch hon thnh.

Qu trnh bin son c 'th c nhng khim khuyt mang s ng gp

ca bn c, ca ng nghip.

Mi kin ng gp xin lin h:

- Trung tm sch gio dc Attpha

225C Nguyn Tri Phng, P.9, Q.5, Tp. HGM.

- Cng ti sch - thi b gio c ANP HA .

50 Nguy n Vn Sng, Qun Tn Ph, TP.HCM

T: 08.62676463, 38547464.

Email: [email protected]

Xin trn trng cm n!

Tc gi



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CHSTGr 1: PIII IH fl IIOTI V P in ts*

S 6 n S O lR O in E H SG u m

PHN 1: PHP DI HNH TRONG KHNG GIAN

A. K in thc c b n :

I. Ph p bin hnh:

1 . (+) nh ngha 1: Php bin hnh trong khng gian l quy tc ng vimt im M xc nh c mt im duy nh t M. im M gi l nh ca M qua php bin h nh .

(+) K hiu: php bin hnh l f th:

f: R3 -> R3

M > M = f(M)

(+) Nu H l hnh no th tp hp tt c cc im M = f(M) viM e H to thnh hnh H. Th H gi l nh ca hnh H qua phpbin hnh f v vit: H = f(H).

2 . Tnh ch't ca hai php bin hnh:

a) nh n gha 2: Cho ha i php bin hnh f v g trong khng gian theos

Php bin hnh f bin im M thn h M v qua php bin hnh g thM bin thnh M. M gi l tch ca hai php bin hnh f v g theo

th t k h iu g.f:M = gf (M) vi M e H

3. Php bin hnh ng nht:

Php bin hnh f gi l php bin hnh ng nht nu qua php bin hnh mi im M G H bin th nh chnh'n. K hiu I: R3> R3

M > M = I(M)

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C h o f : R 3 - R3 Ifim M thuc H v qua php bin fM = M = f (M) th im M gi l im bt ng. (Hay cn gi l imbt bin). Vy php bin hnh ng nht mi im thuc hnh H u

bt ng.Hnh bt ng: 'Hnh (H) gi l bt ng qua php bin h nh f kh i f(H) = H

Php bin hnh ngc:Cho f l php bin hnh song nh (f gi l song nh: nu vi V M e H th c duy nh t nh M = f(M) G H v ngc li vi M e H c tonh duy nht M e H)Th tn ti php bin hnh ngc (f)_1cng song nh m M f(M)

= >A:= f " W )

Php di hnh:'nh ngha 3: Php di hnh l mt php bin hnh khng lm thay i khong cch gia hai im bt k ngha l f l php di hnh:

f: R3 -> R3 V-

Vi VA : A -> A = f(A)

VB : B -> B = f(B ).

Tho mn AB = AB th f l php di hnh.

Tnh cht:a) Php di hnh bin 3 im thng hng thnh 3 im thng hng::

Ngh a l V A, B, c thn g hng , B, C thn g hng.

b) Php di hnh bin 3 im , B, c khn g thng h ng thnh 3 imA!, B, C khng thng hng.

c) Php di hn h b in mt tam gic A ABC thn h AABC: vAABC = A ABC. -

d) Php di hnh bin 1 t din ABCD thnh t din ABCD v t

din ABCD = ABCDe) Php di hnh bin mt mt cu thnh mt mt cu c bn knh

bng mt cu cho trc.g) Php i hnh bin hai mt phng song song thnh hai mt phang

song song, hai m t phng ct nhu thnh hai mt phng ct nhau.h) Php di hnh bin hai ng thng song song thnh hai ng

thng song song, hai ng thng ct nhau thnh hai ng thng ct nhau.



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B. Bi tp:II. Bi tp mu:

IDng. * Cc bi ton lin quan ti php di hnh

* Bin ha im AB thnh AB v khong cch gia AB = AB

Bi 1: Cho t din ABCD chng minh rng f l php di hnh bin mi im ABCD th nh chnh n th f l php' ng nht.

Gii:

Gi s php bin hnh f l php di hnh

f: A = f(A) = A

. B > B = f(B) = B

c -C = f(C) = C

D > D = f(D) = D

Ta chng m inh rng vi mi M e R3th f(M) M

Tht vy gi s 3 M0 G R3 sao cho f(M0) M0

Khi

AM0 = AM0BM0 = BM0'

CMn :CM0

:DM

ABCD

[CM0 ---- xvx0

ABCD thuc mt phng trung trc .....M0M0 iu ny v l v ABCD lt din. Vy khng c im no m M0 = f(M0) ^ Mo

Vy f l php bin h nh ng nht.

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Bi 2: Cho t din ABCD .v ABCD c c cnh tng ng bng nhau AB = AB ; AC = AC; AD = AD ; BC 4% C; BD = B C; DC = D C.Chrig minh rng khng vt qua mt php di hnh bin cc im A, B, c, D ln lt thnh cc im tng ng A, B, C, D.

Gii:Dng; phng php phn chng chng minh. Gi s c tn ti haiphp di hnh f1v f2vi fj 5* fz sao cho: .

^(A) = A f(A) = Af|(B) = B f2(B) = B

f(C) = C f2(C) = Cf](D) = D f2(D) = D

V f, 7: f2nn tn ti mt im M sao cho fi(M) f2(M). Gi s fi(M) = Mj

U M ) - = M Z Theo bi ton th A, B, C, D nm trn mt mt phang trung trc M:M2

iu ny tri vi gi thit ABCD l mt t din, nn = f2 (pcm)[M = A 'M 1 (Theofj)

(A ,B trung trc MjM2v) , _[AM = A M2 (Theof2)

Bi 3:Trong khng gian R3 cho hai im A, B v php ddi hnh f bin A thnh A, B thnh B\ Chng minh rng php di hnh f bin miim nm trn AB thnh chnh n.

Gii:Gi s php di hn h f: R3 > R3

A - A = f(A)B B = f(B)

M = f(M)

Ta xt M e AB ta c

T (1) v (2) =>M = M.

A = A = > AM = A M - ( 1)g = g |BM = BM' (2 )

m t. Tmp lp ffiism M ho mn mt s u kin (rim qu lch)

Bi 4: Cho t din ABCD php di hnh bin ABCD thnh chnh n(Tc l mi nh ca t din thnh mt nh no ca t din). Tm tp hp cc im M trong khng gian sao cho f(M) = M.Trong cc trng hp sau:a) f(A) - B ; f(C) = A ; f(B) = c ; f(D) = Db) f(A) - B ; f(B) = A ; f(C) = D ; f(D) = c



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a)

;=> AM = BM = CM

b)

Gii:

Gi m trong khng gian qua php di hnh f(M) = M

fAM = BM

|CM = AM :

Nn M nm trn ng thng vung gc vi mt phng ABC i qua

tm ng trn ngoi tip tam gic ABC v i qua D (V ABCD l tdin u).

Gi M trong khng gian qua php di hnh f(M) = M

ff(M) = MTa c

|f(A) = B

f(M) = M

f(C) = D

>BM = AM (1)

Gi s M e (ABC) ta c =>

AM = AM

BM = BM'CM = CM' M' = M

CM = DM (2)

Vy t (1) v (2) =>M l giao tuyr hai m t phn g trun g trc AB v

mt phng trung trc CD, (ng i qua trung im AB v trung im DC).

B i 5:Cho tam gic ABC mt php di hnh f bin tam gic ABC thnh chnh n: f(A) = A; f(B) = B; f(C) = c. Chng minh rng php di hnh f bin mi im M thuc mt phng (ABC) thnh chnh n.

Gii:

M' = f(M)

A = f(A)

B = f(B)C = f(C)

(V ABC khng th ng hng) (pcm).

II. Bi tp luyn tp - n tp:Bi 6 : Trong cc mnh sau mnh no ng:

1) Php chiu song song ln mt mt phng l php di hnh.

2) Php ng nht l php i hnh.

3) Nu php di hnh bin im A thnh im B th bin im B thnh im A.

4) Php di hnh bin im A thnh im B v bin im B thnh im A th php di hnh bin trung im AB thnh trung im AB.

Bi 7: Cho php di hnh f ng thng a v mt phang p ln lt cnh qua f l a v p. Chng m inh rng:

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1) Nu a vung gc vi p th a vung gc vi p\2) Nu a song song vi p th a song song vi p.3) Nu a ct p th a ct p \ 14) Gc (a,p) = (a,p): ,

Bi 8 ' Tm mi php di hnh bin mt tam gic u thnh chnh n.Xt tch cc php di hnh .

Bi 9: Tm mi php di hnh bin mt hnh ch nht cho (khng phi hnh vung) thn h chnh n. Xt tch cc php di hnh .

Bi 10: Chng minh tch hai php di hnh l php di hnh.Bi 11: Cho php di hnh f bin hai mt phng song song p v Q ln

lt thnh hai mt phng P v Q. Chng minh rng khong cch gia p v Q bng khong cch gia P v Q.

1) Cho a -L p

III. Hhg dn gii bi tp t luyn phn 1

Bi 6 : * Mni (1) sai v nu im A v B song song vi phng chiut h i A ^ B .* Mnh (2) ng v V A: A = f(A) v V B: B = f(B)

AB = A B* Mnh (3) sai v ch ng khi php l php i xng tm* Mnh (4) ng v php php i xng tm, trc l trung trcAB, mt trung trc AB.

B i 7:

a l b e p

a i c e p

b x c e p

a' = f(a)

b' = f(b )e p'

c' = f(c) e p r

a' -Lb'

a ' c ' => a ' -L p ' (pcm).

b'x c'2) Nu a // p => a '// p '

Tht vy a / /p = > a/ /b ep qua

f : a -> a ' = f (a)

b >b' = f(b) e p'

a/ /b => a' //b' => a'/ /p' (pcm).

Vi

a b

a c =>

b X c

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3) Gi s a X p

Gi b l hnh chiu ca a ln mt phng p. Khi (a,p) = (a,b) qua f : a > a 1 p 1

b - b' e p'

V do (1)N u b l h nh f chiu ca a ln m t phan g p

(a,b) = (a,p) m (a,b) = (a,b)=> (a,p) = (a,p).Bi 8 : Xt cc php di hnh bin tam gic u ABC thnh chnh n trong

cc trng hp sau:1) Trng hpl :

f: A ?B > cc - A

Php quay quanh trc vung gc mt

phng BC i qua tm ng trnngoi tip tam gic ABC

Q A , 1 2 0 = f .

2) Trng hp 2:Xt g: (a): A A (Vi ( a ) mt phng trung trc BC)

B * cc B

3) Xt f.f: A -> c

B > Ac - B4) Xt g.f: A > c

B -*B

c AA A

(f2= Qa, -120)

g.f = p . Vi p mt phng trung trc AC

5) g2= I B .

c

Bi 9: Cho hnh ch nht ABCD, xt cc php di hnh bin ABCD thnh chnh n.Hng dn gii:

Xt cc trng hp sau:1) Trng hp 1:

f: A BB -> Ac -* DD c

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Khi f = a . Vi 0Cj l mt trung tri c AB .v CD.

2) Trng hp 2: I

g: A DB c

c > B => g = .a ( a l mt trung trc AD v BC)2 2

D A r3) Xt f.g = g.* A >c

C-> AB DD >B

Khi f.g = g.f = ....... i xng trc A , A J_ (ABCD) i qua tm hnhch nht.

4) ^ = ^ = 1.B 10: Chng minh tch hai php di hnh l php di hnh.

Gii:

Xt ha:i php di hnh f: R3 > R.3

g: R3 - R3

Vi VA, B : A -> A =.f(A)

B -> B = f(B) => AB = A'B' (1)

g: A > A = g(A)B> B = g(B) =>A"B" = A"B" (2)

iu ny chng t g.f: AB AB tho mn AB = AB

. Nn g.ila php di hnh (pcm).

B i 11: Cho p // Q, A e p k AH _L Q

A e p H e Q

f: A -> A e P

H -> H s QV A H - L P v A H X Q

Theo bi ton 1 th AH_LP v AH Q

f l php di hnh => AH = AH (pcm).

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PHN 2: CC PHP DI HNH THNG GP

TRONG KHNG GIRN

BI I. PHP TNH TIN

A. Kin thc c bn:

1) nh ngha:

Trong R3 cho vect V. Mt php bin hnh trong khng gian R3 bin

mi im M thnh M sao cho MM = V. Php bin h nh gi

php tnh tin theo vect V.M'

K hiu: T : R3 -> R3

M - > M t h o m n V = M M 1 .

2) Tnh cht php tnh tin:

a) Php tnh tin l php di hnh.

Tht vy xt T : K3 - R3

A > A => AA ='V .

B B => BB' -= V

Vy AABB l h nh bnh hnh ==> AB = AB -

(iu ny chng t T V bo ton khong cch h ai im nn n l)hp di hnh) - ,

b) Php tnh tin c tt c cc tnh cht ca php i hnh.

c) Php tnh tin bin mt mt cu thnh mt mt eu bn g mt cu

cho trc.

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8. Bi tp:1. B i tp m au:

Dng 1: Cc bi ton linlim vec tilth n.

- Cho hnh H v php tnh in Tv lm H = Tv CHP h n g p h p c h u n g :

Da vo inh ngha v tnh cht ca php tnh tin.

B i. 12: Cho hai ng thn g a // a tm t t c cc php tnh tin bin ng th ng a th nh a.

Gii:Trn ng thng a ly im ATrn ng th ng a ly im A

Xt php T V vi V = AA'Khi Ty: a -> a.Vy vi cch lm nh trn c v s' php tnh tin nh trn theo

vect T vi A e a A e a bin ng thn g a thnh a.Ch : Mun tm nh ca mt ng thng a qua php bin hnh f (php di hnh) ta thng ly hai im A, B e a c nh l A = f(A)B = f(B) ng thng nh a = f(a) i qua AB.

B i 13: Cho hai php tnh tin T v T V. Chng m inh rng tch ca hai

php tnh tin T-Ty l php tnh tin theo vect + V. Tch c

t n h c h t g i a o h o n ( T -T V = T V -T ) .

Gii:Trong khn g gian ly A b't k

T V : A - A k h i A A = V

T : A - A kh i AA =

T-T: A -> A th " = V +

A"

14



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Vy T u T v = T u + v : A >A (pcm)

V do php cng hai vect c tnh cht giao hon nn:

T u - T v = T V + T u = T u + V -

B i 14: Php tnh tin theo vect V? 0 bin ng thn g a thnh ng

thng a trong trng hp no th:a) a trng a.

b) a song song a.

c) a ct a.

a) a tr ng a kh i a // V

Tht vy A 6 a th A = T V(a)

AA = V => AA // V => A e a => a = a

b) => A a => AA // V

B a = > B B / / V

=> AB // AB => a // a.

c) a khng th ct a. (Theo chng minh cu b)

Dng 2: ng pltp lnl tin clHg minh mi s tinh cftt titan iip__________P h n g p h p c h u n g :

Bc 1: Xc n h php tnh tin bin H th nh H Bc 2: S dng tnh cht bt bin ca p hp tnh tin xc inh nuH c tnh cht A th H c tnh cht A v ngc li.

Bi 15: Cho hnh thang ABCD. Cc cnh AB song song vi BC. im M l giao im ca phn gic gc A v B. N l giao im ca phn gic

gc c v D. Chng minh 2MN = |AB + CD BD A D | .

Gii:

Gii:

Theo gi thit m l tm ng trn (S) tip xc vi AB, AD, BC.

N l tm ca ng trn (L) tip xc vi CD, BC, DA.

Thc hin php tnh tin N M . A D' D

D D

c CDC tip xc v i ng trn (S)

15



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17/246

Khi ( t din ABCD ngoi tip ng trn (S)Vy AB + DC = AD + BC Hay AB + DC = AD + BC (1) r

BC - MN = BC' (2)M ta c ^

[ AD - MN = AD (3)

T (2) v (3) ta c:

BC + AD - 2MN = AD + BC (4)T (1) v (4) ta c:

BC + AD - 2MN = AB + DC

=> |BC + AD AB DC = 2M N (pcm).

Bi 16: Cho t din ABCD ni tip trong mt cu (S) c bn knh R = AB.Mt im M thay i trn mt cu. Gi c \ D, M l cc im sao cho:

CC' = DP = MM ' =B

Chng minh rng BCDM l mt t din th tm mt cu ngoi tipt din nm trn mt cu (S).

Gii:

Xt php tnh tin theo vect AB

T a b : A - Bc - CD - D

M - M

Vy t din ACDM > t din BCDMTrn mt cu (S) > (S)

Gi tm mt cu s l (O) th T a b -O >

0 0 = AB = R => O (S) (pcm).

Dng 3: Diig pip t ti gi co bi ton sn p IP eist. Tia qu eh)

Phng php chung: '

Bc : Tm mi lin quan gi a i m d i ng v cc i m cn tm thc hin ph p bin hnh f: M >M'.

Bc 2:M e H th M'= T (M) => M e H'= T V(H).

B i 17: Cho tam din vung Oxyz. A s Qz c' nh OA = 1. At l tia songsong cng chiu Oy. M, N l hai im ln lt chuyn ng trn Ox v At.1) Tm qu tch trung im I ca MN khi: OM. + AN = 2 .2) Tm qu tch I l trung im ca MN khi: OM.AN = 4.



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18/246

Gii:

Chiu vung gc xung mt ph ng Oxy I > J

J = - N N = = V2 2

Vy xt T_: J > I2OATm qu tch I th ta tm qu tch im J

Trong hai trng hp trn:

(1) Tm qu tch trung im

I ca MN khi OM + AN = 2

Tht vy x t im J(x,y)

t OM = 2x > 0

ON = 2y > 0

OH = -O M = X > 02

OK = ON' = y > 02

OH + OK = (OM + OISP) = (OM + AN)2 2

X + y = (OM + AN) = 1 x + y = l x + y 1 = 0 l ng thng2trong mt phng J c to (x, y)

Tho mn phng trnh X + y 1 = 0

1 1Vi X > 0, y > 0 nn 0 < x < , 0 < y <

2 2

(2) Xt trng hp OM.AN = 4 kh i J(x, y)

x.y = (OM.ON)

x.y = OM.AN = 1 vi X > 0, y > 0 4

Vy x.y = 1 th J thuc nhnh di ca hyperbol c phng trnh y ==.

N

17



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19/246

Bi 18: Cho mt mt phng a, hai ng hng A v A' cho nhau ct

( a ) ti o v 0 . Gi mt phng p ch ^(A ) v song song vi A \ Mt

ng thng di ng song song vi mt ph ng ( a ) hoc cha trong

( a ) ct A ti A ct A ti A v im M nm trn ng thn g y sao

cho - = k (k l S chia trc, k 1). ng thn g song song vi OMA

v t M ct mp p t i M. Tm trng hp M kh i A di ng trn .

Gii:

Dng mt phng (3 chaA song

song A qua o 4s: A ! // A

=> A t 6 p.

T A k AK // 0 0

=> AK ct p ti K.

AK // ( a ). - (1)

MM // ( a ) => MM // AK

=> AMAMK thuc mt mt phng

_ M ' , M'K=> = k =

MA MA

V ta thy (A ) v ( A j) c nh, MK // (d) vi d = (a ) o (P) c" nh

=> MK c phng khng i. M cha AK theo t s k ^ 1Vy M chy ng thng c' nh i qua o chia AK // d theo t s'. M Kk =

M A

18



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20/246

BI l. PH P I X NG MT

A. Kin thc c bn:

1) nh ngha: php xng qua mt phngCho m t phng p. Php i xng qua mt phn g p l php bin hnhbin mi im M /thnh. M sao cho m t phng (P) l m t ph ng trung

trc ca MM. . . . f M M ' l p ;K hiu: p: M > M tho mn __ ti H.rn i r MH = HM 2) Tnh cht:

a) Php i xn g qua mt phng p l php di hnh Tht vy V A,B e R3

Dpi A >AB - B

AA, BB Jl p v p l trun g trc AA v BB

V vy: AABB l hn h tha ng cn.

Vy: AB = AB.

Ib) Nhn g im M e p th: M = M = f (M).

c) Php i xng qua bin ng thng a thn h ng th ng trong cctrng hp sau:

+ Nu a -L p th a = a+ N u a kh ng vu ng g vi P:

- a ct p a cat a ti M e p.

- a // p => a // a. -

6 p=> a-= a. -

() Php x xng qua m t phang tho mn t t c tnh ch t ca php di. hnh. '

B. Bi tp:I. Bi t p mu:

Bi 19: Tm tt c cc mt phng i xng ca hnh hp lp phng.

19



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21/246

Gii: J

Bi ton nu: Tm mt phn g p m ^(H ) (H)

Ta xt cc trng hp sau:

1) Php i xng

a) p: A > A

B -> BD > B => p l mt phn g trung trc ca BD v BD.

b)

c cp: A B

D >- c

c > DB - A

p l mt ph ng trung trc AB v CD.

c)

d)

].: A > cD D

B > B

c > Ap: A -> D

D -$AB > c =c B

C'

p l m t phng tru ng trc AC.

p l mt phang trung trc BC v AD.

2) Php i xng p bin hnh vung ABCD th nh mt m t phng cmt cnh chung.

a) ABCD bin thnh mt c .cnh chung AB l m t ABB A.

AB AB

c -> B

D -> AA -> o

B - cC -> C

D -> D

mt (P): (ABCD)

20



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22/246

r

b) Tng t ABCD bin th nh m t c chung AD ngh a l m t ADDAc mt ! phng xng (ABCB).

c) ABCD c cnh chung BC l m t BCBC th c mt phng i xng(BCAT)).

d) (ABCD) bin thnh mt cnh chung vi mt DC DCTh m t ph ng i xng (DCBA).

3) M t phng (ABCD) bin thnh mt phng khn g c cnh chung:(ABCD) l mt phng trung trc AA, BB, CC, DD.Vy c tt c 9 mt i xng bin (ABCD)(ABCD) thnh chnh n.

Bi 20: Cho hai ng thng di v d2v m t phng p. Chng minh rngnu ct d2th c nh i = p(dj) v d2 = p(d2) ct nhau .

Giai:Gi s di ct d2= A => p : A AVA 6 d] => A e d / (1)

A 6 d => A' 6 d2 (2)T (1) va (2) => A - (d) n (d2) (pcm).

Gii:Tng t bi 19 th t din ABCD c 6 mt trung trc cc cnh cat din. Mi mt cha 1 cnh v trung trc cnh ! din.V d: Mt cha AB v ung gc vi DC (Trung trc DC)

Dng 2: Tm tp hp imP h n g p h p c h u n g:Tm mi quan h im M thay i vi cc im cho, v M' = p

(M) nu M e H - M e H \Bi 22: Cho mt phng p v 2 im

AB cng pha i vi mt phang p,mt im M sao cho tng cc khon g cch AM + MB nh nht.

Gii:Ly A i xng vi A qua p.N i A vi B ct p ti M im M l im cn tmTh t vy ly M thc p ta c:M + MB = MA + MB= > M A + MB = AB.MA = M AMB = MBMA + MB = MA + MB > AB

= MA + MB.iu ny chng t MA + MB b nht.

21



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23/246


24/246

Php bin h nh trn gi l php quay quanh trc d gc quay a

+ Nu gc (X = (180) (hoc 180) php quay quanh trc (d) l php

a) Php quay quanh mt trc v gc quay a ] php di hnh.

b) Tt c cc im thuc ng thng u bt bin qua php quayquanh trc d, gc quay a .

B. Bi tp:I. Bi tp mu:

Dng 1: Cho php quay quanh m! trc, gc quay CC. Tm nl! ca cc 'im

cho, tim nh ca mt hnh cho.Bi 23: Cho hnh lp phng ABCD.ABCD. Tm nh ca AC v AB

qua php quay quanh trc BD mt gc 120 hng dng BD.

Gii:

Cho hnh lp phng ABCD.ABCD

Trc quay BD.

a c

Tht vy: AABD = ACBDCH _LB'D

^ [ A H - L B ' D

=> Mt phn g (AHC) vung gc vi BD ^

Xt A AHC c: AH = CH = 1V2

(a = AB) (1)

D

r" HB /

D'

ir

c

C'

23



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25/246

AC = aV2 (HA, HC) = 120 (2)

T (1) v (2)

^ a c

B > C

Vy nh ca AC l CD

AB c nh l CC.

Dng 2: Tm trc quay, gc quay bicn hnh H thnh hnh H*

Bi 24: Cho t din u ABCD. Tm php quay quanh, mt trc v gc

quay bin t din ABCD thn h chnh n.

Gii:

+ Xt ng thng i qua A v vung gc vi mt BCD. Ti o (O l

tm ng trn ngoi tip tam gic BCD).Xt php quay quanh trc AO

Chiu dng OA, gc quay 120 yY

^ :B c

c DD B

A AVy ABCD -> ACDB

Tng t c 4 trc quay v gc quay 120

+ Nu x t gc 120, OA chiu dng th

a : B D

D -> c c B

A A => ABCD -> ADBC.Bi 25: Cho hnh lp phng ABCD.ACBD. Tm gc quay quanh trc

AC bin ABCD.ABCD thnh chnh n.

Gii:

Da vo bi 23 ta x t php quay quanh trc AC ch i dng CA gcquay 12 0 th:



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26/246

D^: A A A

B > D

c > D

D -> A B

A -> B

B cC - C

D - B B'

=> ABCD.ABCD bin thnh ADDA.BCCD.

Tng t hc sinh xt php j ! f th hnh hp lp phng ABCD.ABD

bin thnh chnh n.

III. Bi tp t luyn n tp:

Bi 26: Cho hnh hp ABCD.ACBD. Gi E, F, G theo th t l trung im ca AA, AB, AD. o l giao im AC v A C. Tm nh cua tdin AEFG qua:

1) Php tnh tin theo vect D C .

2) Php i xng qua tm o.

Bi 27: Cho hnh lp phng ABCD.ABCD. Gi E, F, G theo th t l

trung im ca AA, AB, AD. I l giao ca AC v BD. I l giao im ca C v BD.

1) Tm nh ca t din AEFG qua php i xng trc II.

2) Tm nh ca t din AEFG qua php quay quanh trc BD gc quay120 hng dng BD.

Bi 28: Cho hnh hp ch nht ABCD.ABCD c y l hai hnh vung ABCD.ABCD tm o, O. Tm nh ca hnh chp AABD qua php

a) I vi I trun g im 0 0 (I i xng tm I).

b) TAB (Tnh tin theo vect AB).

c) Q*00,1900) h ng dng t o n 0 .

Bi 29: Cho hnh chp u SABC, ng cao SH. Gi M, N, p l trung

im cc enh BC, CA, AB, thc h in php Q*s 120o (hng dng HS)

Tm nh ca:

25



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27/246

1) SABC qua php quay QhS120o .

2) Hnh chp S.APM N qua php QJjg J 0o

Bi 30: Cho hnh lp phng ABCD.ABCD. Tm mt php xng mt phng bin hnh chp AABD thnh hnh chp CCBD.

Bi 31: Chng minh rng php tnh tin theo vect V c th xem nhkt qu thc hin lin tip h ai php xng qua mt phng.

B i 32: Gho trong mt phang ( a ) mt ng trn (C) tm o , bn k nh R.

ngoi mt phng ( a ) cho mt I c' nh. Gi O l im i xngca o qua I. Mt im M tu ca (C).

Xt j: M Mj

o.: M 1 -> M

Khi M v trn (C) tm tp hp M.B i 33: Cho hnh lp phng ABCM .DEFN vi D = B = CF = M N .

Cho bit:

(ABED): M M i

(BCFE): > M2

(EFND): M2> M3

Tm php di hnh bin M thnh M3.

Bi 34: Cho hnh lp phng ABCD.EFGH (AE = BF = CG = DH). Gi

M; N; p; Q ln lt trung im ca HG, HE, FG, BC. Chng minh

1) Ba hnh chp GABCD; GABFE; GAEHD bng nhau.

2) Bn t din DHFM; CGAQ; CGEP; CGEM bang nhau .

Bi 35: Cho hai mt phang p v Q song song vi nhau. Gi A, B l hai im nm v ha i pha ca hai mt phn g p v Q. Tm trn p v Q haiim M, N tng ng v MN vung gc vi PQ v tng ca khong cch AM + MN + N B nh nht.

26



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28/246

HNG DN GII

B i 26: l)T m nh ca AEFG qua T )Q ly F i xng vi F qua B

Xt php tnh tin theo vect V = DCT dC : A -> B

F - F

E >p (P Trung im BB)

G > I (I Trung im BC)Vy AEFG -> BFPI.

2) Tm nh ca AEFG qua php i xng tm o .

Xt 0:F > L trung im DC

G > Gj l tru ng im BC

E > Ej l trung im CC

A -> C

Vy (AEFG) -> (CEjFxG,).

Bi 27:

ABCDABCD

E, F, G l trung im ca AA, AB, AD

Tm nh ca AEFG qua r:

Gi E i xng E qua II J

=> EE _L II

nn xt trong hnh ch nht ^AACC => EE l ng

trung bnh A'=> E l trung im ca CC

/ / G* / 1' / 1 / / i K

- - 31

I/ **/ "

/

\

D

Tng t gi G l im i xng ca G qua II th G tru ng im caBC, A v c i xng qua II

AEFG bin thnh t din CEFG.

27



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29/246

B i 28;

a) ,: A -> C

A - c

B D

D - B

Nn AABD bin thn h CCDBb) Xt TB ?A B

A B

B Bi (BB^ = B )

D -> C

Vy t gic AABD -> BBBjC.

c) Xt trong php quay quanh trc 0 0 , gc quay 90

-(*00\90) A -> B

A -> B

B - C

D A

Vy AABD bin thnh t din BBCA.

Bi 29: Hng dn da vo bi tp 24.

B i 30:

Tm php xng mtBin AABD CCBD

Mt i xng (DB BD).

B i 31: T V

Cho M mt phng P i V v P l V

Khong cch t p n P .

HH - V2

Vy ta xt lin tip hai php ixng m t p v P l php tnh tin

t h e o v e c t V

Tht vy T V: M > M

Mftp = v

A

A

M

H

M,

H'

wM '

28



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30/246

Xt lin tip hai php i xng mt p v P

p: M -> M, : MH + HM, = 2HMX

+

MM' = 2CHM, + MjH') = HH' = V

Vy MM' = V - T- : M>M' (pcm).

Bi 32:

j: o O

M -> Mj

0.: M M

Ta thy 10 = M M '.2

=> MM' = 0 0

Vy M l nh ca M qua T 0 0 ' (Php tnh tin theo vect 0 0 ' )

=> c -> C = T o o 1nn C e mt phng (P) i xng vi ( a ) qua I.

B i 34: Hng dn gii:

Ch :+ Chng minh H bng H c phng php chung tm ti php di hnh bin hnh H thnh H \

+ Thc hin cc bc chng minh: Tm php di hnh bin hnh H thnh hnh H \

1) Chng minh 3 hnh chp GABCD; GABFE; GAEHD bng nhau.

Bi 33:

Mxv M xng qua (ABDE)

A l trung im ca M v M

Tng t Mi v M2nh n I l

trung im v MjM2 _L (BCFE)MaM2M3 nhn G lm trung im v M2M3 -L DNFE.

Do vy M, M3 xng qua E,

E l trung im ca

N F

e: M ->M 3.

29



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31/246

Chng minh t din GABCDbng t din GABEF

Xt php i xng m t(ABGH) th G > G

A -> AB > B

F -> C '

E -> D => T din GABCD - GABFE

Nn t din GABCD = t din GABFE

Tng t xt php i xng mt ADGF

th G G

A > A

B E

c ^ HD D => GABCD = GAEHD.

2) Chng minh bn t din DHFM; AGEN; GECP; GCEM bn g nhau.

Gii:

Gi I, J 1 tm ca EFCfH v ABCD

Xt php quay quanh trc IJ gc quay 90

Xt t gic DHFM bin thnh t gic ( u i 9 0 o ) : D > a

H E

F G

M - N

Xt t gic GECP

(IJ,90): G

E > F =

p M =

Xt t gic GECP

(IJ,90): G H

c > DE > F

M > N

DHFM -> AEGN

HFDM = AEGN

GECP -> HFDM

GECP =3 HFDM

GCEM HDFN

(2)

30



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32/246

Xt (HFDB):

H -> H

F -> F=> HF DN = HFDM (3)

D > D

N - M

T (1), (2), (3) th bn t gic bng nhau .

Bi 35:

T bi ton trn ta c th s dng trong bi ton hnh hc phng:Cho hai ng thng dx// d2 (song song vi nhau). Hai im AB nm v hai pha ca hai ng thng. Tm trn dj, d2 hai im M, N (lnlt thuc d 1; d2) v MN vung gc dj, d2 sao cho AM + MN + NB b

Ta thy khong cch MN a =const

MN _Ld dng php t nh tin A > A theo vect Ni A vi B ct d2t i N, NM = a th M, N cn tm.

Th t vy g i s c MN ln lt thuc dl7 d2MN vung gc d!, d2

Th AM + MN + NB = AN + MN + N B

M MN = MN

AN + N B > AB iu ny chng t

AM + MN + NB > AM +MN + NB (pcm)

Tng t hc sinh gii bi ton 3 i vi hnh hc khng gian.

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33/246

PHN 3 : PHiPv TR vn pip NG DN

TRONG KHNG GIAN

A. Kin thc c bn:

. nh ngha v nh cht ca php v tr1. nh ngha:

Trong khng gian cho mt im o v mt s thc k 0. Php bin

hnh f tng ng mi im M vi mt im M sao cho: OM' = kOM:.Php bin hnh gi l php v tr tm o, t.sk

Kh i u: V kM M tho mn M1= -kM .. \

2. Tnh cht:

a) Nu M, N c nh tng ng M\ N qua php v tr tm o t sk th:

M N' = kMN ' :

Nu k > 0 th M'N' cng hng vi MN

Nu k < 0 th M' N' ngc hng i MN .b) Php \ tr tm o t so k

(1) Bin ba im thng hng thnh ba im thng hng v m bo theo th t ca chng.

(2) Bin mt ng thng d thnh d xy ra cc trng hp sau:

- Nu d i qua o d = d- Nu d khng i qua o =>d//d

(3) B in mt m t png p thnh P tho mn:

- Nu p cha o ->P = p

- Nu p khng cha o P//P.

(4) Bin mt gc thnh mt gc bng chnh n.

(5) Bin mt t din thnh mt t din.

(6) Bin, mt mt cu bn'knh R thnh mt mt cu c bn knh |k| R .

(7) Phtip v tr VDk(O) - 0 .

IL nh nciha v tnh ch php ng:dng: :1. nh ngha:

Php bin hnh f trong khng gian c gi l php ng dng nu hai im bt k M, N c nh MN = kMN. Vi k sdng cho trc.

+ S k gi l t s" ng dng ca php ng dng f.

32



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

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1

0

0

0

B

T

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34/246

2 . Tnh cht:

a) Php ng dng bin'ba im thng hng thnh ba im thng hng v bo ton theo th t nh.

b) Bin mt ng thng thnh mt ng thng.

c) Bin m t m t phng thnh m t mt phng.

d) Bin mt tia thn h m t tia.e) Bin na mt phng thnh na m t phng.

f) Bin mt gc thnh mt gc bng gc cho trc.

g) Bin mt tam gic ABC thnh tam gic ABC m tam gic ABCng dng vi tam gic ABC.

h) Bin t din thnh mt t din.

k) B in m t mt cu bn knh R thnh mt m t cu bn knh kK.

m) Php di hnh, l trng hp c bit ca php rg dng k = 1 .

n) Nu thc hin lin, tip 1 s hu hn php ng dng th n l php ng dng. . '

II. C c hnh ng dng.1. nh ngha:

Hai hnh c gi l ng dng nu c mt php bin l ng dngbin hnh ny thnh hnh kia.

B. Bi tp:

I. Bi tp mu:Dng I: Xc nli nh ca m! hnh qua php v tr, hoc ng dng

+ Cho php v tr Vok v hnh H hy xc nh H qua php v tr.

+ Cho php ng dng t sk v hnh H hy xc nh nh ca H quaphp ng dng.

Ph ng php g i i chung:

Tm nh m t im xc nh hnh H suy ra H' xc nh qua cc im nh.

V d: T din ABCD c 4 im xc nh t din ta tm nh ABCD th (ABCD) = f(ABCD).

B i 36: Cho hn h lp phng ABCD.ABCD. o l giao im ca ccng cho. Gi I, I l tm ca hai hnh vung ABCD.ABCD. Tmnh ca hnh chp u IABCD qua php, V

(I-.)

33



B

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O

N

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Gii: 'j

Qua o v m t phn g (P) song song v | (ABCD)T 7 v / -N - | N , _ __ ___ J h '


36/246

1P h n g p h p g i i:

chng m in h hn h H ng dng H ta cn ch ra th c hin lin tipmt s hu hn php ng dng bin H thnh H ' thng thng thchin ph p di, ph p v tr bin H thnh H, Hi bin thnh H \

B i 38: Cho t din u ABCD, gi A1; B 1; Ci, Dj tng ng l trun g tmca cc tam gic BCD, ACD, ABD, ABC. Chng minh Ax, B 1; Cj, Dt lt din u.

Gii:

Theo bi ra ta thy:

AAj, BB1; CCi, DD] ng quy

(HS t CM) theo hnh v bn

GBj _ GAt B 1A 1 1

GB GA ~~ AB ~ 31 _ _ '

Xt php v ^

Vy v j : ABCD > AjBjCjD! nn A 1B1C1D 1 l t din u

AJE5, = AB .1 1 3

Dng 3: Tm tp hp im

P h n g p h p g i i:

Tm tp hp M th ta tm mi quan h gia M v M' e H ' qua php bin hnh f : M > M' khi M e H m f(H) = H'.

Xt V : C -> C

F D

E - c

G B

Vy thc hin lin tip v


37/246

ti H. Tm tp hp im N i xng ca H qua A khi M v trn m t cu.

Gii:

Thc hin php Vj B > o

M HKhi rSt cu (S) tm o, bn knh 4R thnh mt cu ngknh AO = 2R.

H e mt cu ng knh AO.

+ Xt a =>(0, 2R)

- (O'j, 2R) ng knh AC.

Nn N thuc mt cu ng knh AC.

B i 40: Cho hai im AB, ng thng d cho nhau vi ng thn g AB.Ly trn d mt im c. Dng hnh bnh hnh ABCD. Tm qu tch trung im M ca AD khi c chy trn ng thn g d.

Gii:

Xt ABCD l hnh bnh hnh

=> C) = B

Xt T B php tnh tin theo vect BA

T : c - DVy c e d

D e T B (d) = d// dA

Xt V | : D M vy M e V | (d) = d // d i qua trung im AB.

Qu tch M l ng d.

II. B P P DNG V N P(1) bi:

Bi 41: Cho hai on thng AB v CD hy tm php ng bin A v Btheo th t thnh c v D.

B i 42: Cho php V0k(php v t tm o t s' k) v php VQk' (php v t

tm 0 , t s' k). Chng m inh rng k .k = 1 th php bin h nh thchin lin tip hai php v t trn l php bin hnh, php bin hnh l php tnh tin.

36



B

I

D

N

GT

O

N

-

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0

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Bi 43: Chng minh rng hai hnh lp phng bt k lun lun ng dng vi nhau.

Bi 44: Cho ABC ln lt l nh ca ABC qua php ng dng t s" k

khi v ch khi: A ' B '.A' c ' = k2AB.AC.

B i 45: Cho ABC theo th t ln lt l nh ca A, B, c qua php

ng dng. Chng minh rng nu AC = pAB (Vi p l mt s thc)th A'C' = p A 'B '.

Bi 46: Cho t din ABCD. Tm nh ca t din qua php V(GA). Trong3

G l trng tm ca t din.

Bi 47: Cho hnh chp ct ABCDE.ABCDE. Tm php v tr bin a gic ABCDE thn h t gic ABCDE.

Bi 48: Cho hai hnh t din ABCD v ABCD c cc cnh tng ng

song song vi nhau. Chng minh rng c mt php tnh tin hocphp v tr bin t din ny thnh t din kia.

Bi 49: Cho hnh lp phng ABCD.ABCD. Gi I v I l tm caABCD v ABCD. Trong mt phng DBBD gi M l giao im ca

DI v BB. Tm php bin hnh f: B'M A 'A .

(2) H ng d n g ii:

Bi 41: Trn tia CD ly im A sao cho CA = AB

Xt hai on thn g CA v AB thc hin php di hnh f: A > c

B -> A

Thc hin tip php v tr tm c t s" k = ( CD = kCA')(J A

Vy khi ta thc hin lin tip hai php di hnh v php v trbin A thnh c , B thnh D.

B i 42: V0k : M > Mj M^ = kM

Vi': M] M 7M = k7M^

Qua hai php V0k . v : M - M

m ta c MM' = MM; + MM' = OMi - OM

= M^-- G M T + k'TMT-CTM^k

MM 1= (1- )M^ + ( k 1 ) M



B

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N

GT

O

N

-

L

-

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Vi k.k = 1 tha y vo ta c: I

r n 1= ( l- i) M T + ( l - ) 7M; = - ) 1 v O v O (*)k k - .k

Qua hai php v tr o o ' c' nh a (*)

Chng t M = T(.k - _ OO ') M (pcm).k

B i 43: Hng dn gii cho ha i hnh lp phng ABCD.ABCD v AB 1C1D 1'1B'1.C,1D'1. T rn cc ti A j A j , A ^ D j , l y l n lt cc

im A 2,B 2,D 2 dng mt hnh lp phng A 2B2C2D2.AiE>2C2D2 sao

cho hnh AjjBjjCjDjj.AjBjCjDjj bng hnh lp phng AB CD .ABCD.

Khi gi f php di hnh bin:

ABCD.ABCD bin thnh AjB^Da-AiB'jCgDj

D ng php v tr tm Aj t s' k

k = th bin - t h n h hnh vungiu

Vy ta thc hin lin tip hai php di hnh, v php v tr bin ABCD.ABCD thnh A ^ C ^ .A ^ C ' ^ ; nn ABCD.ABCD ng

dng AxBjCiDj. AjBiC'iDi (ng dng).

B i 44: Gi f l php ng dng t s k kh i B ' c ' = kBC** (IFCO2 = k2BC2

(^C' - l ')2 = k 2( c - B)2r

^ 2- 2t C.A 'B '+ -7!^2= k2( c2- 2CB + B2

M ta c A'C '2=k2C

7B'2 =k2B

Nn VCVTb^ = k2C.B (pcm).

B i 45: Theo nh n gha php ng dng f t s" k th:

A'C' = kA C ,A rB*' = kB

A 'C = kC ^ TC1= k(pB) = p(kB)


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B i 46: Gi s G l trng-tm ca t din ABCD th:

GA + GB 4- GC + GD = 0 (1)

Gi Gj l trng tm tam gic BCD, mt i din vi nh A

Ta c: G^B + G^c + ~D = (2)

T (1) ta c: GGt + GjA + GGj + GjB + GGi + GLC + GGj + GtD ==

4GGt + GXA = 0 4GGj + GjG + GA = 0

3GGj .= GA GGj = GA (3)

iu (3) ny chng t V(G-!>: A > Gi3

Tng t G2, G3, G4 l trng tm ca cc tam gic ACD, ABD, ABC

th V(GjL)! B > G23

c -> G3

D > G4

Vy nh ca t d in ABCD qua V(G ) l t din GiG2G3G4 l trng3

tm cc mt t din.

B i 47: Gi s h nh chp ct ABCDE.ABCDE cc cnh AA, BB, c , DD, EE c t nhau o (ha i mt (ABCD) // (ABCD))

_ AB BC _ C D OA

^ A'B' B'C' C'D' OA'

Vy xt vk : A -

* B > B

C c

D > DE > E

Vy Vck: ABCDE ABCDE

Bi 48: Cho hai t din ABCD; ABCD c cc cnh tng ng song song vi nhau. Chng minh tn ti php tnh tin hoc php v tr bin tdin ny thn h t din kia.

39



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Gii:

t B[ trn AB on ABj = AB

t Cj trn AC on ACi = AC

t Di trn AD on AD! = AD

Khi ct qua php di hnh th

ABCD bin thn h ABjCjP j(V AB // AB\ AC // AC, AD // AD)

Suy ra cc gc BAC = B ' A ' c '

BD = RTTd"' => t din ABCD = ABjCjDj

CD = Cr7D'

Xt Vak : ABjCiDi -> ABCD trong t s'kA B ,

A'Bj:k


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PHN 4: N Tp CHNG 1

Bi 1: Cho hn h lp phng ABCD.ABCD, o l giao im ca AC v BD. Tm im X sao cho f(X) = c trong cc trng hp:

1) f l php tnh tin theo vect AB .

2) f l php xng qua mt phng trung trc AA.3) f l i xng tm o .

Bi 2: Cho hai on thng bng nhau AB v CD. Hy ch ra php binhnh khi thc hin lin tip hai php 'i xng mt bin A thnh B,

' bin c thnh D.

Bi 3: Cho hnh lp phng ABCD.ABCD. Tm php di hnh bit:

a) AB > AD.

b) AB BC.

c) AB - DD\

Bi 4: Chng minh rng php tnh tin theo V bin trung im M caon thn g AB vi A, B l nh ca A, B qua php tnh, tin .

Bi 5: Cho hnh lp phng ABCD.ABCD. Tm php di hnh bin tam gic ABD thnh tam gic CDB.

B i 6 : Cho hnh hp ABCD .ABCD c tm o . M, N, p ln lt l trungim ca AB, AD, AA\ Chng minh rng hai t din AMNP v CDBC ng dng vdi nhau.

B i 7: Cho php v tr tm o t s" k v php v tr tm O t s" k. Chngminh rng nu k.k ^ li th php bin hnh f = V. k-V(ok) l php v tr

c tm xc nh I, t s' v tr k.k.

Bi 8 : Cho ba tia Ox, Oy, Oz ng quy t i . Ba im A, B, c ln ltthuc Ox, Oy, Oz. Gi G l trng tm .tam gic ABC. Gi s A, B cTnh, im c di ng trn Oz

a) Tm qu tch cc trung im ca BC, AC.

b) Tm tp hp trng tm G khi c thay i trn Oz.

Bi 9: Trong khng gian sp xp 3 ng gic u ABCDE, ALNMB, v AEKPL chng minh cc ng AC, AN, AK vung gc vi nhau tng mt.

B i 10: Chng m inh rng php ng dng bo ton t s ca ha i onthng khc phng.

41



B

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GT

O

N

-

L

-

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A

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2

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0

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43/246

Hng dn gi: .

Bi tp trc nghim: I

1D 2D 3C 4D 5D 6B 7D 8A 9D 10D

11B 12D 13D 14D 15B' 16A 17D 18C 19D 20C

D cBi tp t lun:Bi 50:1) 'B = CXS> f(X) = C

2) a : C

> Vi a l trung trc AA

3) 0: A -> c.

Bi 51: Cho hai on thng.bng nhau AB v CD. Hy ch ra php bin

hnh khi thc hin lin tip hai php i xng mt bin A thnh c,bin' B thnh D.Gii:

Xt m t phng trung trc AC l ( a ) BLy i xng qua a th

A -> cB > B => AB = CB

Xt mt phang phn gic gc(CB, CD) l P ly oi xng

qua (P) th:

c - cB D

Vy thc hin lin tip hai php xng mt qua a v p th:

AB -> CD.B i 52:

a) Php di hnh bin AB thnh AD

Thc hin Q(AA,90")

A A

D - D

Vy php di hnh T A A '

Q : AB - AD(AA',90)

b) Tng t thc hin

T b B ' . Q (BB._90O) : AB > BC

42



B

I

D

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GT

O

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-

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c) T a d ;Q(DD.j90'>) : AB DD

Bi 53: Ta c AB - AB

th A A

B B

v A = BB = V = MM1

M = MM1+ M7^ +T '

MB = MM~' + M'HT' + W B

MA + MB = 2v + M ' - M 'B + (-2v)

= M 'A + M'B

iu ny chng t M l trung im ca AB.

B i 54: (Hc sinh t v hnh). Xt 0 : A ABD > A CDB

Trong o l tm ca hnh lp phng.

Bi 55: Xt php tm o

q : A cM M e DCN - N e CB

p -> P e CC

Xt php v (c,2) :

C -> M - D

N B

F -> c

Vy qua V(C-2) 0 : AM NP bin t din CDBC nn AM NP ng dng CDBC t s' ng dng l k = 2 .

B i 56: Phng hng gii: f = V0'k'-V0kvi k.k 56 l th f = VI>kk. (*)

Bc 1: Chng m inh qua tch ca hai php v tr c im b t bin

N gha f(I) = I.

ha y Vokl -> li (1)

V ov l i -> I (2)

T (1) => = k

(2) =>T = k ' T ; (3)

-XO'

43



B

I

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N

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O

N

-

L

-

H

A

C

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2

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0

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45/246

Xt (*) => 01 - 0 0 ' = k *(OIj - 00 ') = k (kOI - 0 0 ')

=> OO = kk 'OI k'O O 1

(] - kk ')0I = (1 - k ')0 .0 '

(4) iu ny chng t I xc nh duy nht.

Bc 2: f = V(jjkk.)

Gi M qua VokM Mj

VovMx -> M

Ta c M-= KM

OM = kOMj

Ta c :

vT = - l= k' (M^-i) - I

= k '(kM - *') - c v

= k ' kM - k 1*' - 7!

= k ' k( + M) - k ' CT' - l

= k ' kM + k k ! - k ' (K' - 7!

= k 'kM + k 'k - k OCV - (CTO + )

= k ' kM + (k ' k - 1) + (1 - k ')CT'

m t (4) th (k k -1 )0 1 + (1 - k ') 0 0 ' =

Vy IM-' = k ' kM (5)

Vy t (5) => M = V(i,kk) (M).

B i 57:

b) Xt V : M - G m M e O, z

Vy G thuc V (OV) = 0 z M e V - (Oz)( A )

Vy M e 0 z

o

y

X



B

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O

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47/246

C in tS G 2: KH I OA M S

BI 1: KHI A DIN

A. Kin thc c bnI. Khi nim hrth a din

Hnh a din l mt hnh tq bi mt s' hu hn cc min a gic th mn h ai tnh cht:

1. Hai'min a gic phn bit c ba kh nn g

Hoc khng giao nhau v

Hoc c mt nh ch ungHoc c mt cnh chung.

2 . Mi cnh ca mt min a gic no bt k cng -r,. chung cang hai min a gic. Mi a gic gi l mt ca hi.h a din. Cc nh, cnh ca cc min a gic theo th t gi l cc cc cnhca hnh a din.

II. Khi nim v khi a din ,

1 . Khi a d i n kphn khng gian c gii hn bi mt hnh a din;

k c hnh a din .2 . Nhng im khng thuc khi a din c gi l im n goi ca

khi a din, nhng im thuc khi a din nhng khng thuc hnh a din gi l im trong.

* Tp hp cc im ngoi kh a.din gi l m in ngoi ca kh a din;

* Tp hp cc im trong ca khi a din gi l min trong ca khia din. .

3. Mi khi a din hon ton c xc nh bi hn h a din ng vi n

v ngc li.4. Khi a din l i: Khi a din (H) c gi l li nu on thng n'i

hai im bt k ca (H) lun lun thuc (H), khi hnh a din xc nh (H) gi l a din li.

5. nh l le:Tl'ong mt hnh a din li th: d - c + m = 2

Trong d, c, m ln lt l s" nh, s cnh v s mt ca khi a din li .,

46



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O

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1. n h ngha: Khi a din li c gi l khi a din u loi (p, q)

'; 3 t l a) Nu mi mt l a-din u p cnh

b) Mi nh l nh, chung q mt

2 . Tnh cht:Trong khi a din u th cc mt l nhng min a dinu bng nhau.

3. n h l: Ch c 5 loi khi a din u loi (p, q) sau y:

a) Loi (3, 3) l t din u

b) Loi (4, 3) l hnh lp phng

c) Loi (3, 4) gi l khi 8mt u

d) Loi (5, 3) gi l khi 12mt u

e) Loi (3, 5) gi l khi 20 mt u.B. Bi tp

!. Bi tp mu

Dng 1: Chng minis mt s tnh cht lin quan n s' nh, s' cnh, s mt camt khi a din.

B i 60: Chng m inh mt khi a din c t nh t 4 m t (s' mt: ) > 4)

G i i

Gi Mj l m t mt khi a din (H), v Mj l rrtin a gicnn c tnht 3 cnh c1; c2, c3. Vy c mt mt M2chung cnh C!vi M, (M2 5* M,).Gi M3c chun g cnh c2vi Mi (M3 5* Mj) v Cl e M2, Cl e M]

c, M3. Gi M4c chung cnh c3vi M]M4khng cha cnh cxv

c2nn n khc Mj, M2, M3. Vy (H) c t nht 4 mt.

B i 61: C hng minh rng trong bt k khi a din th:

1) S" mt lun lun nh hn s" cnh (m < c).

2) S nh nh hn s cnh (d < c).G i i

1) Gi s" nh l d, s" mt l m, s cnh l c, mi mt t nht 3 cnh,mi cnh c hai mt nn ta c:

m 2 .3m = 2c =


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g 2:PSi Giia hoc lp gSip khi a din,' psHi php gii C9 a din HPtncha tsinitH,,Haa,...HKhianm (ffi)u(ff~)u(ff3)u ...u (ffB) = H

' Nh ph p di hnh chng m inh H j=H2=H 3=...= H n* S dng mt s tnh cht xt (Hj) suy ra cc tnh cht ca khia din H

Bi 62: Cho hnh lp phng ABCDEFGH, AE = BF = CG = DH. Hy phn chia hnh lp phng thnh 6hnh chp bng nhau.* Gii

Xt o l tm ca hnh lp phng. Xt 6 hnh chp OABCD; OBFGC,OGCDH, ODHEA, OEABF, OFGHE. Xi cc php xng tm o th 0 : (OABCD) -* (QGHEF) ;

(ODCGH) (OFEAB);

(OADHE) -> (OGFBC)

=> (OABCD) = (OGHEF) (1)

(OADHE) = (OGFBC) (2)(ODCGH) = (OFEAB) (3)

Xt php quay quanh trc IJgc quay 90 chiu dng IJ

V o r ODCGH _> 0 HEOADHE - QBAEF

Vy ODCGH = OBAEF (4), cch lm tng t 6 hnh chp cn li,ta c: (OABCD) = (OBFGC) = (OGGDH)

= (ODHEA) - (OEABF) = (OFG HE).

B i 63: Cho hnh lp phng ABCDEFG H, (F = BF = CG = DH ) . Hy

chia hnh hp ra 3 hnh chp bng nhau.Gii

Ly G lm nh, xt 3 hnh chp(G.ABCD), (G.ADHE), (GABFE).Xt mt phng (ABGH). Xt php i xng qua mt (ABGH) th

A -4- AB BF-C => GABFE GABCDE - D

^ ^ B ~ ~ "Tng t GADHE 'i xng GABCD qua mt (ADGF)

Vy GABFE = GABCD = GADHE.

48



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B i 69: Chng minh rng khn g tn t i mt hn h a d in li c s' nh,s' mt, s' cnh u l. I*

Bi 70: Chng minh tm ca cc mt cu mt hnh t din u l ccrnVi ca mt t din u.

Bi 71: Chng minh rng tm ca cc mt ca hnh tm mt u l cc nh ca hnh lp phng.

Bi 72: Chng minh rng tm cc mt ca hnh lp phng l cc nhca hnh tm m t u.

Bi 73: Cho t din u ABCD, gi M, N, p, E, F, H ln lt l trung im 6cnh DA, DB, DC, AB, BC, CA. Chng minh kh'i a din u NMEFPH l khi 8m t u.

Bi 74: Cho hnh bt gic u EABCDF, gi M, N, p, Q, R, s, H, L ln lt l trung im cc cnh EA, EB, EC, ED, FA, FB, FC, FD.

a) Khi a din MNPQRSHL l hnh g?

b) Tnh cnh khi MNPQRSHL bit mi cnh ca khi bt gic u l 2a.

II. Hng dn gii bi tp t rn luynBi 66 : Gi d, c, m ln lt l s" nh, cnh,, m rn a hnh a din u

(p, q) (mi mt p cnh, mi nh c q mt)

Ta c pm = qd = 2c

Theo rih l le t a c d c + m = 2

T (*) suy ra 2p + 2 q - p q > 0 ( p - 2) (q -2 ) < 4

Theo bi ton 60 th p > 3; q > 3 nn (p, q N* ) ch xy ra cc trng hp

d m c d + m - c _ 2 _______4pq4pq (*) - + - - 2 p + 2 q - p q 2 p + 2 q - p qq p 2 q p 2 2pqq p 2 q + p 2

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fp 2= 2(4) < => p = 4 , q = 3 Khi lp phng

[q - 2= 1

ip 2 = 3(5) -I = > p = 5 , q = 3 Kh 12 m-t u

|q 2= 1

Ta c bng tng ng: ____________________Loi Tn Tnh cht ccmt

S'nh(d)

S' cnh(c)

S mt(m)

{3,3} Khi t din u Tam gic u 4 6 4

{3,4} Khi 8mt u Tam gic u 6 12 8

{3,5} Khi 2mt u Tam gic u 12 30 20

{4,3} Khi lp phng H nh vung 8 12 ^ 6

{5,3}Khi 12 m t u N g gic u 20 30 12

Bi 67: 1) Hnh t din (6 cnh)2) Hnh chp t gic (8 cnh)3) Hnh 9 cnh

A

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Hnh (a din ghp bi hai hnh t din ABCD v EBCD.

Bi 6 8 : Xt hnh chp c mt y l a ic li k cnh, khi khi adin c 2k cnh.

Bi 69: Hng dn s dng nh l le

Gi s cc mt l, cc nh l, cc cnh l

=> d c + m l s l, iu n y v l v d c + m = 2.

Bi 70: Gi cc trng tm cc mtAABC, ACD , AADB, ABCDl Gj, G2, G3j G4 th ta c:

GjG4 = | a d , g 2g 4 = a b ,

GoG, = CD , G2G3 = ^ B C3 1 3 2 3 3

Vy t din G^aG sG l t din u.Bi 71: Nhn xt hnh bt gic unhn cc mt (ABCD) chia hai hnh chp bng nhau i xng qua mt(ABCD), cc mt AFCE, mt (SDEB)i xng.

Tht vy: AE = BE = CE = DE = AF = BF - CF = DF nn (ABCD) mttrung trc EF

Tng t (DEBF) trung trc AC(AECF) trung trc DB

Vy AC -L DB m ABCD l hnh thoi => ABCD l hnh vung. Do gi cc: trng tm tam gic AEBA, AEBC , AECD , AEDA l Gi, G2, G3,G4 l nh 4 hnh vung. Tng t trng tm cc tam gic AFAB,AFBC , AFCD, AFDA la Gs, GS; G7 G8i nh hnh vung

Bi 72: Gi I, J l tm mt phng(ABCD) v A'B'C'D', IJ l trc ixng hnh lp phngGi G), G2, G3, Gln lt l tm ccmt (ABB'A ) , (BCC'B'), (CDD'C'),

(ADAT)'), do tnh cht i xng

nn G,G3 _L G2G4 v GJG2G3G4 l hnh \-ung. A'

I

\ y y/ ' ::B

G / / -" ' yy

J/

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Tng t IG!JG3l hnh vung

Vy IG! = IG2 - IG3= IG4= JG! = JG2= JGS= JG4. Hnh IJGxGaGgG*l hnh bt gic u.

Bi 73: Theo tnh cht ng trung bnh tm tam gic HEF, HEM,HMP, HPP, NEF, NEM, NMB,

NPF l tm tam gic u, mi

cnh bng AB v ta thy EP2

l ng n! trung im ca ABv CD (hai cnh i); FM lng ni trung im ca haicnh AD v BC ca mt t dinu nn EPXFM. Suy ra EFPMl hnh vung.

Vy NEFPMH l khi 8mt u.

B i 74:

Ta nhn thy hnh MNPQ l hnh vung, mi cnh

MN = AB = a2

Tng t RSHL l hnh vung v mi cnhRS = SH = HL = LR = a

EF_L(ABCD)

=> EF J_ (MNPQ), m MR,NS, PH, QL song song EF

EF

2PH, QL vung gc vi haiy (MNPQ) v (RSHL).

Vy hnh MNPQRSHL l hnh hp ch nht, hai y l hai hnh vung c cnh b ng a.

MR = NS = PH = QL = , EF = 2a>/2

1 '

/ /: N

\ : L / \ IH \ /

N\ / / B

-

v QL = MR, NS,

o MR = NS = PH = QL = aV

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BI 2: KHI NIM v THE TC KHl A DIN u

A. Kin thc c bn

I. Cng thc tnh th tch1. Th tch khi chp V = B h (B din tch y, h l ng cao hnh

2. Th tch khcfi chp ct V = (B + B' + 7B') .

(Trong B, B l d in tch ha i y, h l ng cao chp ct ). .

3. Th tch khi, ln g tr -V = Bh (trong B l din tch y, h l

ng cao).4. Th tch khi, hp ch nh t V = Bh = abc (a, b, c l di ba cnh

hnh hp ch nht).

5. Th tch khi hp lp phng V = a3(a l di cnh kh i hp lp

phng).

II. Ch :1. Mt khi a din (H) c phn chia thnh ha i khi a din (Hj) v

th-Vro-Vpy+Vp^Trong : V HJ l th tch, khi a d in (H)

' V(HJ l th tch khi a din (Hj)

V{H0 l t h tch khi a d in '(H2).

2 . T s' th tch ha i khi a din ng dng bng lp ph ng t s" ng

3. Trong mt s' b i ton thng phn chia khi th nh mt t din(hnh chp ABCD). Ta s dn g bi ton cbn:

Cho t din SABC, trn SA, SB, s c ly ln lt ba im A', B', C'

khc s th kh i : V (SA3-C-) S' SB' SCf(SA.BC) SA-SB-SC

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B. Bi tp

I. Bi t p m u

Dng 1: Cc bi ton lin quan n hnh chp

a) Tnh di cc cnh

b) Tnh din tch xung quanh, din tch ton phn, tnh din tch thit din

c) Tnh th tch h nh chp.

Phng h ng chung :

S dng cc kin thc: H thc lng trong tam gi c, trong t g ic,s dng tnh song song, tnh vung gc ca ng thng vi mt

phang, s dng nh l P itago

S dng cng thc tnh khi phn chia thnh tng khi hn h chp.

Bi 75: Cho hnh chp SABC, y ABC l tam gic u cnh bng a. Mt bn (SBC) vung gc vi mt y (ABC). Cc mt Bn (SAB),

(SAC) hp vi m t y (ABC) cng mt gc a = . Gi D l trung3

im ca BC, k D E vung gc vi AC.

1) Chng minh AD vung gc vi SD

2) Chng minh SE = 2DE

3) Tnh din tch ton phn ca hnh chp SABC \ \

4) Tnh th t ch hnh chp SABC.

Gii

Ta nh n xt: (SBC) -L (ABC) c giao tuyn BC n n mt ng th ng c

vi y mt gc a = 3

d e (ABC) -L BC => d ( A B (

=> AD JLBC => AD (SBC)

1) T (1) => AD SD

2) Do mt (SBA) v (SAC) hp

^ SD J_ (ABC) (2) B

T (2) theo nh l bang vung gcDE AC => SE _LAC

=> SED = a = - .3

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57/246

XtASDE ta c: SD = SEsin= SE-^5- / \-Z.3 2 bV> .

DE = SE sin = SE (*). ^ * . 6 2 *

:=> S = 2 D E = - ^ (pcm). c$> t - ? " CL/ ~ 2 2

3) Tnh th tch ton phn hnh chp SABC:

TP = ^ASBC + AABC + ASBA

Sasbc = - S D - B C = - -a = 2 2 4 8

a _ a2V^ A A B C _ 4

S a s b a = - S E - A C = ^ ^ - a = - ? - ^ASBA 2 2 2 4

3a2V3 . 3a2 6a2V s + 3 a 2 3a:

m D l t run g im BC suy ra : DE =

TP 4 8 8 ' 8- ( 1 + 2V3 ) (VDT).

4) Tnh th tch:

V = Bh3

B = sa'Vs

ABC - 4

a ( V s ) _ 3 ah = SI) =

4 4

V a2~^ _ a3>/3 (VTT)-.~ 3 4 ' 4 16

Bi 76: Cho hnh chp SABCD, y ABCD l mt hnh ch nht, mtbn (SCD) vung gc vi mt y. ASCD vung gc s v SCD = a .

Mt bn (SAB) to vi y mt gc p = 90 a . Gi SH, SE theo th

t l ng cao ca ASCD v ASAB. Bit SH + SE = m. Tnh:

1) Th tch hnh chp SABCD theo m v a,p

2) Tnh th tch hai mt bn SAD v SBC.

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58/246

Gii1) V (SDC) (ABCD)

=> SH DC => SH (ABCD)

HE AB => SE AB

(Theo nh l 3 ng vung gc)

Vy SH = P = 9 0 - a

SH = SE - sinp

= SE sin ^90 - a j =SE cosa

- , SH + SE = m 'M ta c: < o

ISH = SEcosa

SE = -m

cosa +1(1)

SH = _mcosa_(2)cosa +1

Ta thy: HE = AD = cosp = SE sina

HE = m sina (3)cosa +1

Ta nhn thy SDH = 90 a = p => DH = HE =

HC = SH cot ga = SH-^-Sa

msma cosa +1

sm a

HC:mcosa cosa _ mcos2a

c o s a + 1 sina s in a ( c Q s a + 1 )

DC = DH + HC =

DH =

msma mcos a" ~ + ...... .......

cosa +1 sina(co sa + 1)

m sin2a + mcos2a _ m(4)

s i n a ( c o s a + 1 ) s in a ( c o s a + 1 )

U.T. 1- o A _ Din tch (ABCD) SHVy the tch hnh chp SABCD = -----------------------

Din tch ABCD = AD - DC = m sina __m----------(cosa +1) sin a(cosa +1)

rv,2Din tch ABCD =

m

cos a + 2cosa +1,2

y _ 1 m2 mcosa _ 1 m3cosa (YTT)3(co sa + l (cosa + l) 3 ( cosa + 1)3

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59/246

2) Theo nh l ba ng vung gc th SDA vung gc D, ASCBI vunggc c

Din tch ASDA = SD DAz

M S D = - ^ - = ~ (theo 2)cosa cosa + 1

DA =HE = msi-a cosa +1

Vy din tch ASDA = smoc2 (cosa + 1)

Din tch ASCB = SC-B C =--SC-DA2 2

M s c = -7 - = mCQSa------; BC = DA = m s n -sin a sin a (cosa + 1) (cosa + 1)

Vy d in tch ASCB = s c BC = - -.-osa - .2 (cosa + 1)

Bi 77: Cho hnh chp SABCD l na lc gic u, AD = 2a, AB = BC = CD = a.Cnh SA 5=h v vung gc vi mt y. Mt phng p vung gc vis i) i qua A ct SB, s c , SD ti B, , D

1) Chng m inh rng AB CD l m t t gic ni tip2) Tnh th tch hnh chp SABCD.

Gii s

1) SA X (ABCD) => SA X BD

M ABCD l na lc gic u

nn ABD = IV

=> BD (SB A ) => BD AB

V (AB'C'D') SD SD -L AB'

Vy AB' 1 (SBD) => AB' B'D' (1)

Tng t AC ' C'D' (2).

T (1) v (2) suy ra AB'C'D' nit i p t r o n g n g t r n n g -

knh A D.

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2) Tnh th tch hnh chp V(SA-B,C,) = S A - SB ^ S _ SB'-SC'V (SABC) S A - S B - S C S B - S C

Xt trong tam gic vung SAB, ta c:SB' SB = SA 2 = h2

SC' s c = SA2 = h2V i2 V i2

SB' = -, SC' =SB sc

V i v f e h V(SABC) SB2 -SC2

o, VSAC.D, SC'-Sb' h4Tng t -" - = = - -

VSArn S C-S D SC2 -SD2VSACDTa tnh

SB2 = SA 2 + AB2 : h2 + a2s c 2 = SA2 + AC2 = h2 + 3a2SD2 = SA2 + AD2 = h2 + 4a2

V,SABC :h dtAABC = h . a . A = iL3 3 2 2 12;W3

VsACD - 2Vsabc2hV3

*SAB'C'V,

V SAC.D.

(h2 + a2)(h2 + 3az)V, 2h5V3

SABC 12(h2 + a2)(h2 + 3a2)h4 a2hV *hB>/3

(h2 + 3a 2)( h2 + 4 a 2)' 6 e ( h 2 + 3a2)(h 2 + 4a2)

a2h5V a2h5VsVy SAB-C-D- 1 2 ( h 2 + a 2 ) J h 2 + 3 a 2 + 6 Jh 2 + 3 a 2 J J l l 2 + 4 a 2

Bi 78: Cho hnh ehp ct u c cc y l hnh vung cnh l chiu cao chp ct l h. Tnh th tch hnh chp ct.

GiiVn dng cng thc

a v 2a,

V = (B + B' +VB7)h3

V = (a2 + 4a2 + Va4 )h 3

v = - 7 a 2h .3



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Bi 79: Cho hnh chp tam gic SABC, r tam gic ABC vung gc B. Cnh SA vung gc vi y AB = a, Bp = b, CA = c. Tnh k hong cch t A n m t phng SBC.

Gii

VSABC= SA dtAABC

SABC - 2 a c : -abc

Ta X : m hnh chp SABC c nh A v y ASBC . Gi khong ccht A n mt phng (SBC) l h

V = ^- h - (dtASBC) = abc .

Theo nh l ba ng vung gc suy ra:

dtASBC = SB BC m SB = Va2 + 2

dtASBC : Va2 + c2 b

622 b h = abc

1

6

=> h =ac

Va2+C2

Sns 2: Cc b osi c lin quan an kli ini r linli hp, hnh bp C1&nht,hnii bp lp phiang, tnh th tch, tnh din tch cc thit din.

Phng php chung:

S dng cc kin thc

H thc lng trong t g ic, ta m gic, hn h vung

nh l ba ng vung gc

Quan h song song gi a ng v m t phn g* Tm giao tuyn 2 m t ph ng

* Tm giao im ca m t v ng.

Bi 80: Cho hnh lng tr ng ABCAjBiCx, y AABC vung A, AC = ,

C = 60 . ng cho BCj ca mt bn (BCCjBj) hp vi mt bn (ACCiAO mt gc 30

a) Tnh AC!b) Tnh th tch h nh lng tr.



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62/246

Gii

a) Ta c

=> BA JL(AA1C1C)

Vy BC^ = 30

Xt tam gic AA BC: AB = AC tan60

AB = aV

Vy BC = 2a

Xt ABACj vung A

=> AC! = AB cot 30 = aV3 y3 = 3a

BA AC

BA _LA,A

A

Vy ACj = 3a.

b) Tnh th tch hnh lng tr

V = dtAABC CCj = a V3 a c c ,2

Trong tam gic vung ACCj => CCj2 = ACj2 - AC2

CCi =V9 a2 - a 2 =2aV2

V = a2 >/3 2aV2 = a3 V .

B i 81: Cho hnh lng tr ng ABCABC y ABC vung B, cc cnhbn AA', BB',CC'

a) Chng minh AAB'C' l tam gic vung.

b) Chng minh hai tam giacAAB'C'va AAB'C chia hnh lng trthn h 3 hnh chp c th tch bng nhau.

c) B it din tch y (ABC) l s , cnh BC = a, gc gia AC vi mty bng a . Tnh din tch t o n phn ca hnh chp AA'B'C'.

T (1) =5* C'B' -L AB' nn AAB'C' l tam gic vung B

b) Cc tam g ic AAB'C' v AAB'C chia hnh lng tr ra ba hnh chp:

B ABC , AA B C', C'B'AC

Giia) Ta nhn thy:

=> CB'B _L(AA'B'B) (1)

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63/246

Va b c a b'c = dtABC CC'

Gi din tch ABC l s, CC = h

Y a b c a BC' = k

V a a 'B 'C ' - -S h (2)

V c B'AC V , r.' ~ Vtj'iftc VAA.R.(yABCAB'C' ~ v B'ABG y AAB'C

= S- h - S - h - S - h 3 3

VC.B'AC=^ (3)

T ( 1 ) , ( 2 ) , ( 3 ) t a C: VB.ABC VC.B,AC ( p c m ) .

c) Hnh chp AA'B'C' c 4 mt: AAA'B', AAB'C', AAA'C', AA'B'C'. Tt c 4 mt trn l nhng tam gic vun g v ta c:

A'B' = AB = = BC a

A'C' = AC = VbC2+ AB2 = - Va4+ 4S2

AA' = BB' = CC' = AC tan a = Va4 + 4 S 2

aAB' = VBB'2 + AB2 = - Va4tan2a + 4S2(1 + ta n2o )

a

Vy :

(1) = s

= - AA' A'B' = - tan a /a4 ~l g 2 2S = Stan aV a 4+ 4S 2 - y a a2

(a4 +4S2)

(2) SAAA.g,

(3) S ^ ^ A A ' - A C *

2_1

2

2 a

1 t ana

2 a2

(4) S ^ c - = - AB' B'C' = Va4 tan2 a + 4S2(l + tan2a )2 a

= -y/aj ta n2a + 4S2(1+ tan 2a )2 .

Vy STp = + Saaa.b. + SAAA,C, + SAAB,C. .

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65/246

1) Tnh khong cch gia OjB v BjlC .1

Dng mt phng (BOjD) // BjC (v Oi|3 // B,C)

Tnh khong cch t c n mt phng (BO,D)Khong cch h = cKOjDjBiC)

Vn (ing bi ton 79. Xt hnh chp OjBCD

V0lBCD = d i (BCD) . = 1 . 1 . a2a = 1 . a3

V o , BCD = - ^ d t C O j B D ) h

M AOjBD l tam gic u c cnh BD = aV2

dtAQ1BD = - a V 2 - a^ ^ =^1 2 2 2

V0iBCD= a 2V3h = a 36

1 l'\/3 , ,=> h = J= - (pcm).

a/3 3

2 ) Nhn xt: N l tm ca hnh hp OBCD.O.BjC Dj

Xt hnh chp (OjNBBi) c:

Vnobi! = Vonbb, = 7r(dtBBiD) h

h l khong cch t N n OOjBjB => h = -

J B B , 0 V,1 a2 a a3

NO BB ,

3) Ta x t vi M bt k thuc OOj

V,MBCCjB!

h l khong cch t M nmt (BCCjB!) => h = a

dt(BCC1B1) = a 2

V,

A

D

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66/246

V a 3 0VMBCC1BI _ 3 _ 2 ___v = -f = -r (cmst) (pcm).OBCOjBiCi _a3

2

B i 84: Cho lng tr u ABCA'B'C' c chiu cao AA' = BB' = CC' = h v hai ng thng AB, BC vung gc vi nhau. T nh th tch h nh lng tr .

Gii

Trong mt phng ABB'A' k BAj //B'A

VyBC' -LBAj, BC' = BAj

=> ABC'Aj vung, cn t i B.

Gi cnh y l a th ta c:

AB '2 = h 2+ a2

Xt ABC'Aj ta c:

C'AX2 = 2BAj2= 2(h2+ a2)

Xt AC'AjA' ta c C'Aj2= a2+ 4h2

Vy ta c 2^h2+ a2) = a2+ 4h2

a2 = 2h 2 o a = hV2

Vy th tch VABCA.B.C. = h = (pcm).

B i 8: Cho hnh hp ch nh t ABCD.A'B'C'D'. Gi cc l gc phn g nh

din (B'.CD B ) .

1) Xc nh a v chng t rng SBCD = SB.CD cosa

2) Cho AA' = a v ng thn g B'D to vi m t phn g (ABCD) mt

g c p

ng thi B'D' to vi mt (BCC'B') mt gc y . Tnh th tch hnh

hp ch nht ABCD.A'B'C'D' theo a, a , p, Y .

V0BC.0)B1C1 = (dtOBC ) BB1 = I . a 2 . a = I a3

65

C'



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N



8/10/2019 Cc dng ton v phng php gii Hnh hc gii tch 12 - T

Các dạng toán và phương pháp giải Hình học giải tích 12 - Trần Đình Thì

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