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JCgi noi ddu Cuon sach "Boi dtiditg hoc sinh gidi Hinh hoc 10" la mot tai lieu tham khao mon toan dung trong truong trung hoc pho thong, nh^m muc dich giijp cho hoc sinh ren luyen cac

^JSJBm. NGUYEN TRONG TUAN (CHU BIEN)

ThS DANG PHUC THANH - NGUYEN TAN SIENG

(Gido vien cliuy§ii toan va nang khieu)

h i , ; ' U i t

Bdi duong hoc sinh gidi

Biensoan thgochuongtrlnhnidi

K H

THIT VIEWTIfv 'HeiN 'H THUAW

NHA XU^T BAN T6 NG H0P THANH PH^ Hd CHi MINH

JCgi noi ddu

Cuon sach "Boi dtiditg hoc sinh gidi Hinh hoc 10" la mot tai lieu tham

khao mon toan dung trong truong trung hoc pho thong, nh^m muc dich

giijp cho hoc sinh ren luyen cac ky nang giai cac dang toan theo yeu cau

Chuan Kien thuc - Ky nang cua chuang trinh, tren ca so do tai lieu con

giiip cho cac em tiep can, giai cac de thi Dai hoc, Cao dang va cac bai toan

Noi dung cuon sach gom 3 chuang, tuan tu theo sach giao khoa Hinli

hoc 10 hien hanh Moi chuang gom cac chuyen de tuong ung vai cac bai

hoc (§) duoc trinh bay theo cau true sau

B Mot so dang toan

C Luyen tap

Cuoi nidi chuang la phan cac bai tap on cudi chuang bao gom cac bai tap

tong hap kien thuc trong chuang

Trong hai phan B Mot so dang toan va C Luyen tap, cac bai tap dugc

trinh bay theo tung dcing, tu dan gian den phiic tap, ngoai cac bai theo

dang Chuan Kien thuc - Ky nang, chung toi da trinh bay them cac bai tap

mai, dang kho, phiic tap phuc vu cho doi tugng hoc sinh kha - gioi nham

phat huy kha nang tu duy linh boat, sang tao, doc lap a moi em

De hoc tot tai lieu nay, hoc sinh can nSm vung l i thuyet, cac dang

toan va cac bai tap c6 lai giai mau Voi tinh than t u hoc mot each nghiem

tue, khoa hoc, chiing toi hi vong rang tai lieu nay c6 the giiip cho cac em cai

thien dugc nang luc hoc toan cua ban than va vuan len trinh do kha, gioi

Dii da CO gSng rat nhieu, nhung thieu sot la dieu kho tranh khoi, rat

mong cac thay, c6 giao va cac em hoc sinh gop y de cuon sach nay dugc

dieu chinh, bo sung, hoan thien han trong Ian tai ban

Cdctdcgid Wtd sdch Khang Viet xin trdn trgng giai thieu t&i Quy doc gid vd xin

idng nghe moi y kien dong gop, de cuon sdch ngdy cang hay horn, bo ich hem

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Chirofng 1

V E C T O

§ 1 KHAI NIEM VECTQ

A TOM TAT LI THUYET

m

• Vecta la mot doan thMng c6 huong, nghla la trong hai diem miit ciia dpan

thang, da chi ro diem nao la diem dau, diem nao la diem cuoi

• Ki hieu vecto c6 M la diem dau va N la diem cuoi la M N Nhieu khi nguoi

ta dung ki hieu a de chi mot vecto AB nao do

• Vecta CO diem dau va diem cuoi triing nhau dugc ggi la vecta - khong, ki

hi?u la 0

2 Jiai vecta cung phuang, ciing hitang

• Gia ciia vecto AB: Cho AB khac 0

Duong thJing AB dugc ggi la gia ciia A B

• Hai vecto cung phuang: Hai vecta dugc ggi la cung phuang neu chiing

CO gia song song hoac trung nhau

• Neu hai vecta cung phuang thi hoac chiing cung huong, hoac chiing ngugc huang

Chu y Vecta - khong A A c6 gia la mgi duang thJing qua A; 0 cimg

phuang va cung huang voi mgi vecta

Tren hinh ve ta c6 cac vecto AB, CD, EG ciing phuong vai nhau, trong do

AB, CD cimg huong, EG ngugc huang vai cac vecto AB, CD

3 Jiai vecta bdng nhau

• Dg dai ciia vecto AB : Dg dai ciia doan thang AB dugc ggi la do dai ciia vecta A B , k i hi^u la I AB I

Hai vecta a va b ggi la bang nhau neu chiing cung huang va cung dg dai

ta viet a = b

Boi liuchig IISG Hinh hoc 10

B MOTSODANGTOAN

i)ang 1 So sdnh cdc vecta Sit dung cdc dinh nghta vehai vecta cung phuong, cung huong, bang nhau

^di 1 Cho ba d i e m A , B, C phan biet Cac menh de sau day d i i n g hay sai ?

c) D u n g v i neu A B = "BC thi A B , B C cung h u o n g va AB = AC Suy ra A,B,C

th^ng hang theo t h u t u do va A B = A C Vay B la trung diem doan A C

d) Sai v i v o i A , B phan biet t h i cac vecto nay nguoc h u o n g

<Bdi 2 Cho tam giac ABC Goi M , N Ian luot la trung diem ciia A B va A C Hay

so sanh phuong, huong, dp dai cua cac cap vecto

a) AB va A C ;

c) A N va N C ;

e) BC va N M ;

b) B C va M N ; d) M A va M B ; g) C A va N C

G i a i a) H a i vecto A B va A C khong cung p h u o n g ;

b) H a i vecto B C va M N cung h u o n g va B C

d) Hai vecto M A va M B cung phuong, ngugc huong va ciing do dai;

e) H a i vecto B C va N M ciing phuong, ngugc huong va BC - 2 N M

g) Hai vecto CA va N C cung p h u o n g , ngugc h u o n g va C A = 2 N C

'£)ang2 Xdc dinh diem thod man he thuc vecta

'Bdi 3 Cho hai d i e m A va B phan bi^t Hay t i m diem M sao cho:

G i a i a) M la t r u n g d i e m doan AB >

b) M la diem doi x i i n g ciia A qua B ,v '

117 r,>

<Bdi 4.Cho tam giac ABC Hay xac d i n h cac diem D va E sao cho

Hai vecto AB va BA ciing phuong, ngugc huong va cung do dai

1.2 Cho hinh thang A B C D v o i A B // CD va C D = 2AB Goi M la trung diem

CD, N la trung diem BC Hay dien dau " x " vao cac cot u n g v o i cac t u "ci^mg

p h u o n g " , " c u n g h u o n g " , "ngugc h u o n g " , "bang nhau" de dien ta moi quan he giira cac cap vecto vao bang sau :

J-Iaang dan giai

a) Vecto A B ngugc h u o n g voi vecto DE H J-!r>.r!)0 '•.I'lfiv '

b) Vecto BF ngugc h u o n g v o i vecto E D j.^, | '\ , c) Vecto A E bang vecto FD -la^i >:V'V' > ' d) Vecto DE cijng h u o n g v o i vecto BA a f ' v

e) Vecto FE cung h u o n g v o i vecto BC v/

-h) Vecto EC k h o n g cung p h u o n g v g i vecto BF 4^^,, :,,

Boi tluimg HSG Hinh hoc 10

1.3 Cho ba diem A, B, C Co nhan xet gi ve ba diem do neu :

c) A B = |AC| va AB, AC khong cung phuong i •

a) Diem A nam tren duong trung tryc doan BC r'j' *

b) Diem B trung voi diem C ,

c) A la trung diem doan BC VJA

1.4 Cho ba diem A, B, C M$nh de nao sau day diing ?

a) AB = 0<=>A = B; b) AB = CD <=> A = C va B = D ; y j

c) Neu AB =|AC thi B = C; d) Neu BA = CA thi B = C

-f''"'•• " ' i '•'•'v^^!v'• Jilcang dan gidi

a) dung b) sai c) sai d) dung • v

1.5 Cho ba diem A, B, C phan biet Ket luan gi ve ba diem A, B, C neu :

a) Vecto A B cung phuong voi vecto BC;

b) Vecto AB cung huong voi vecto AC ;

c) Vecto AB bang vecto AC

Jiicang dan gidi

a) Ba diem A, B, C thang hang

b) Ba diem A, B, C thang hang va B, C nam ve cung mot phia doi voi diem A

c) Hai diem B, C trung nhau

1.6 Cho hinh binh hanh ABCD Goi E, F Ian luot la trung diem cac canh AB,

CD Duong cheo BD cat AF tai G va cat CE tai H Chung minh rang :

Mat khac, AADG = ACBH => AG = C H

Nhu the AGCH la hinh binh hanh Tu do A H = GC

Chu y:ABCD la hinh binh hanh <^A,B,C khong thang hang va AB = DC

CUj TNHHMIV DVVII Khnug Vict

1.7 Cho tijf giac loi ABCD Co ket luan gi ve tu giac ABCD neu

A B = DC va AB A D

Jilzang ddn gidi

1.7 Vi ABCD la tu giac loi va AB = DC nen ABCD la hinh binh hanh

Hon nira t u dieu kien c6n lai ta tha'y hinh binh hanh nay c6 hai canh lien tie'p bang nhau Vay ABCD la hinh thoi

a+b

§2 TONG CUA HAI VECTa

A TdMTATLITHUYET

1 ^inhnghla '

• Cho hai vecto a va b

Tu mot diem A ba't ki dung cac vecto A B - a , BC = b

Khi do vecto AC duoc goi la tong ciia hai vecto a va b

• Tinh chat trung diem :

M la trung diem doan AB => M A + MB = 0

• Tinh chat trong tam tarn giac :

G la trong tam cua tam giac ABC => GA + GB + GC = 6

B MOTSODANGTOAN

( b c )

t)ang 1 Chung minh dang thuc vecta

Su dung qui tac ba diem Va qui tac hinh binh hanh

5Sdt 5 Cho 4 diem A, B, C, D tuy y Chung minh rang

AB + CD = A D + CB

7

Boi dudng HSG Hinh hoc 10

Giai Theo qui tac ba diem, ta c6: AB + CD = AD + DB + CB + BD

Ap dung tinli chat giao hoan ciia phep cpng, ta c6 :

AB + CD = A D + CB + DB + BD = AD + CB

<8di 6 Cho t i i giac ABCD Goi O la trung diem ciia AB

Chung minh rang OD + OC = AD + BC

Chu y rang O la trung diem AB nen OA + OB = 0

V i t h e t a c o OD + OC = AD + BC

OV^JHl U

'bang2 Xdc dinh vd ttnh do dai cua vecto ,^ ~, ,5, ^, j ; •

• Dm vdo quy tac ba diem vd cdc ttnh chat cua tong hai vecto, ta rut gon

kei qua phep todn tong cdc vecto

• Khi ttnh do dai vecto ta thuang xem do dai do la canh cua mgt tarn gidc

vecto: A D + AB, OA + OC, OB + BC, AB + AC

Giai

Truochettaco

AC=BD=aV2, OA = OB = OC - OD = —

2 Theo qui tac hinh binh hanh, ta c6 ,,,|

AC = A C = a ^

A D + AB = A C : AD + AB

Vi O la trung diem AC nen

OA + OC = 0 => OA + OC = 0 = 0

Cty TNHH MTV DWH Khang Vil-t

• Theo qui tac hinh binh hanh: OB + BD = OD:

• Dung hinh binh hanh BACE Khi do

OB + BD OD =OD = .72

AB + AC = AE: AB + AC AE = AE

Ta CO tam giac ADE vuong tai D biet AD = a, DE = 2a

Vay AE^ = AD^+DE^ =5a^ => AE = a>y5

'bang 3 Chung minh tinh chat hinh hoc

chi khi voi moi diem M , ta c6 M A + MC = MB + M D

Ap dung qui tac ba diem, ta c6

M A = MB + BA

MC = M D + DC • M A + MC = MB + M D + BA + DC, vdi moi M (*)

Ta c6: ABCD la hinh binh hanh <=> DC = AB o BA + DC = BA + AB = 0

Do do tir (*) ta suy dugc:

ABCD la hinh binh hanh o M A + MC = MB + M D , voi moi diem M

bang 4 Tim tap hop diem thod man dang thuc ve tong cdc vecto

^di 10 Cho doan thang AB Tim tap hop cac diem M thoa man dSng thuc :

Boi dumig HSG Hinh hoc 10

a) Sai, ch3ng han trong truang hop b,c khong cung phuong

b) Diing Neu I la trung diem doan M N thi M I = I N , do do:

1.10 Cho tam giac A B C deu canh a

a) Xac djnh va tinh do dai cac vecto u = AB + AC , v = CA + BA

b) Goi M , N Ian lugt la trung diem ciia BC va AC Xac dinh va tinh do dai

vecto A M + BN

Jiuang ddn gidi

a) Dung hinh binh hanh ABDC Do tam giac ABC deu nen A D 1 BC

a>/3 Vay A D = 2AH = 2 — = aVS

Cty TNHH MVV }) VVU Khnitg Viet

1.12 Cho 4 diem A, B, C, D Chung minh rang AB = CD khi va chi khi trung diem ciia hai doan thang AD va BC trung nhau -

Goi I va J Ian luot la trung diem cua AD va BC Khi do , j / j ^

1.15 Cho tam giac ABC Goi D, E va F Ian lugt la trung diem ciia cac canh BC,

Jiuang ddn gidi Cdch i;Ta c6 :

AD + BE + CF =AB + BD + BC + CE + CA + AF

= (AB + BC + CA) + (BD + CE + AF)

Dira vao tinh chat duong trung binh trong

AABC, ta CO BD = FE; CE = DF; AF = ED

Boi dumtg HSG Hhih hoc 10 Cty TNHH MIV I > VVH Khang Viet

1.16 Cho tarn giac ABC Tir cac diem A, B, C ta dung cac vecto bang nhau tiiy

y A A = BB' = C C Chung minh rang "^^"'^ '""^'^ ' '

1.17 Cho ngu giac deu A B C D E CO tarn O m<\i 6b f/3 rinfiri rinM

a) Chiing minh cac vecto u = O A + OB va v = OC + OE ciing phuong voi

vecto O D

b) Chung minh rang OA + OB + OC + OD + O E - 0 v * ^ , ,* ' , u / ^ i ;

Jiuang ddn gidi

a) Truoc he't chii y rang do tinh chat doi xiing cua ngu giac deu nen OD 1 AB

Dung hinh binh hanh AOBF Khi do li - OA + OB = OF

Do OA = OB nen AOBF la hinh thoi Nhu the OF 1 AB Suy ra u = OF, OD

cung phuong Chiing minh tuong tu ta cung c6 hai vecto v = O C + OE,

OD cung phuong

b) Nhan xet rang OA + OB + OC + OD + OE la mot vecto cung phuong vol

OD Chung minh tuong tu ta cung c6 OA + OB + OC + OD + OE cimg

phuong voi OB (vi vai tro hai vecto OB, OD nhu nhau)

Vay chi c6 the tong cac vecto tren la vecto - khong

1.18 Cho hai vecto a va b khac vecto 0

a) Chung minh rang l a + b l < l a l + l b l ; dau " = " xay ra khi nao ?

i;

ii

b) Ap dung : Tim tap hop cac diem M thoa man dieu kien

I A M + MB 1=I A M I +1 MB I, voi A, B la hai diem cho truoc

' Jiu&ng ddn gidi Hit vsr i

a) Tu mot diem A bat ki ta dung AB = a, dung BC = b

+ Neu ABC la mot tam giac ta c6

AC < AB + BC, do do I a + b I < I a I + I b I + Trong truong hop A, B, C thSng hang thi

AC < AB + BC khi AB va AC ngugc huong ; ' a + b _ ,

AC = AB + BC khi AB va AC cijng huong

Vay la + b l < l a l + l b l ; dau " = "xay ra khi a va b cijng huong ; ' b) Dua vao ket qua cau a) ta thay rang

I A M + MBI=IAMI + IMBI ^ I AB 1 = 1 A M I + I MB I , hay AB = A M + MB, dieu nay xay ra khi va chi khi A M va MB cung huong, hay M nam tren doan thang AB

Vay tap hop cac diem M la doan thang AB ' " '*

• Neu a + b = 0 thi tanoi a la vecto doi ciia vecto b vanguoclai '

• Vecto doi cua vecto a (ki hieu - a ) la mot vecto ngugc huong voi vecto a

va CO cimg do dai voi vecto a , <!• 4' , ^

, 'u • i d V • OA •-• i ; {j-y-jv •Mv:uinm :>(./

Jiieu cua hai vecto:

a) Djnh nghia: Hieu cua hai vecto a va b , ki h i | u a - b la tong ciia vecto a

a - b = a + (-b)

^) Cach ve vecto a - b : Cho cac vecto a va b

Tu diem O bat ki, ta ve OA = a, OB = b 4 ,

Ta CO BA = a - b

^) Quy tac ve hieu vecto: Voi ba diem M, N , O tiiy y thi ta c6: KS^J = - OM

1 3

Boi duang HSG Hhih hoc 10

T u (1) va (2) ta CO dang thiic can chiing minh

<Bdi 12 Cho 6 diem A , B , C, D , E, F tiay y Chung m i n h dang thuc sau

^q.ng2 Xdc dink vd tinh do ddi cua vecto

• Rut gon kei qua cdc phep todn tong, hieu cdc vecta

• Khi tinh dp ddi vecta ta thuang xem dg ddi do Id canh ciia mot tam giac

ndo do

<^dil3 Cho tam giac ABC

a) Hay xac dinh cac vecta u = AB - AC, v = BA - BC - CA

b) Xac djnh diem M sao cho M A + M C - M B = 0

c) Xac dinh diem N sao cho A N = AB + AC - CB

a) Taco u = A B - A C = CB

v = B A - B C - C A = C A - C A - 0

b) Taco

M A + M C - M B = O o M A + BC = 0 o M 4 = BC D

Do do M la d i n h thii t u cua hinh binh hanh ABCM

c) Goi D la diem do'i xung ciia A qua trung diem cua BC Khi do tu giac

ACDB la hinh binh hanh N h u the AB + AC = A D

14

Cty TNHH MTV DWH Khang Viet

Vay A N = AB + A C - C B = A D + BC = A D + A M Vay N la dinh thu tu cua hinh binh hanh A D N M Cac diem M , N dugc xac dinh n h u tren hinh ve « »

^ d i 14 Cho h i n h thoi A B C D c6 tam O, AB = a va ABC = 60"

Xac d i n h va tinh dp dai cac vecta: AB + A D va AB - A D

Giai

Theo quy tac hinh binh hanh, ta c6 : " '

AB + A D = A C Do do :

1 AB + A D 1 = A C = a (vi AABC deu)

Theo quy tac ve hieu cua hai vecta, ta c6

V i ABCD la hinh thoi CO AB = a va ABC = 60° nen BD = 2BO = a x/3 (vi AABC la tam giac deu)

Vay I A B - A D I = D B = a>y3

©ang- 3 Chitng minh tinh chat hinh hoc

^di 15 Cho t u giac ABCD thoa man dong thoi cac dieu kien sau day :

Do do tir dieu kien ii), ta dug-c A C DB => AC = B D

V i hinh binh hanh ABCD c6 hai duang cheo bang nhau nen do la hinh chu nhat

^at 16 Cho hai vecta a va b khong cimg phuong Chung m i n h rang : Ne'u I a - b I = I a + b I thi vecta a va b c6 gia vuong goc vai nhau

Giai, Vai hai vecta a va b khong cung phuang, t u diem O bat k i , ta ve OA = a,

OB = b D y n g hinh binh hanh OACB, k h i do

BA = a - b v a O C = a + b Dodo l a - b I = l a + b l o A B = OC OACB la hinh chij nhat

B

Boi duang HSG Hinh hoc 10

Suy ra OA J_ OB

Vay neu I a - b I = I a + b I thi a va b c6 gia vuong goc voi nhau • ?

'£>ang 4 Tim tap hap diem thoa man dang thiic vehieu cac vecto

Bai 17, Cho tam giac ABC Tim tap hap cac diem M thoa man dang thiic:

IBA + M B I = i C M - C B I , , , n

• • ' Giai ,< j ^ , '/^ -tip

Ta c6: IBA + MB 1=1 CM - CB 1^1 BA - BM 1=1 BMI

• "' <=> I M A I = I B M I <::> M A = M B Vay tap hop cac diem M la duang trung tryc doan thang AB

C.LUYENTAP

1,19, Chon khang dinh diing trong cac khang dinh sau

a) Hai vecto doi nhau c6 dO dai bang nhau

b) Neu hai vecto AB va AC doi nhau thi A = C |

c) Hai vecto doi nhau thi cung phuong

d) Vecto doi ciia vecto a - b la vecto a + b , ' ) ^ ( <

Jiuong dan gidi

a) Dung vi hai vecto doi nhau c6 do dai bang nhau j , ,

b) Sai vi khi do A la trung diem BC

c) Dung vi gia ciia chiing trimg nhau

d) Sai vi vecto doi ciia a - b l a - a + b

1,20, Cho tam giac ABC va diem M tiiy y Cac khang dinh sau day dung hay sai?

c) Khang dinh diing v i CA = M A - MC = BA - BC M A - BA = MC - BC

1,21 Cho 4 diem A, B, C, D Chiing minh rang

a) AB + CD = A D - B C ; ; b) A B - C D = A C -M

Cty TNI in Ml V DWH Khmig Vii Jiuang ddn gidi

a) AB + CD = AD + DB + CD = AD + C B = AD - BC : " ' =' ' b) AB - CD = AC + CB - CD = AC + DB = AC - BID

BCGH, CDIJ, DAKL

a) Chiing minh rang KF + EH + Gj + I t = 0

b) Chiing minh rang EL - FH = FK - GJ - i , >

Jiicang ddn gidi

a) Sii dung ti'nh chat ciia hieu hai vecto ta c6:

KF = AF - AK, EH = BH - BE, GJ = CJ - CG, I L = DL - DI Tir do

KF +EH + Gj + I L = A F - A k > M - B E + CJ-CG + D L - D i

= (AF - BE) + (BH - CG) + ( C | - Dl) + (D L - A K ) = 0 ' b) Ta CO

EL = EF + FK + KL = BA + FK + A D , H I = HB + BA + A D + D I

=> EL - H I = FK - HB - D i = FK + CG - CJ = FK + JG = FK - Gj

sau day a) M A - M B = CA + BC; b) M A - M B + MC = 0 ; ,

Vay M la dinh thii t u ciia hinh binh hanh ABCM

^) Gia sii M la diem thoa man

M A

<=> M nam tren duang trung true doan A B

Gia sii M thoa man B A - B M = I B A - B C M A CA <=>MA = CA

<=> M n a m tren dirniiv lirin hi in A, ]}j\f\

17

Boi tiuaiig use, lliitli hoc 10

1.24 Cho hinh chu nhat ABCD c6 AB = 2a, AD = a Hay xac dinh va tinh do

1.25 Cho hinh thang vuong ABCD c6 hai day AB = a, CD = 2a, duong cao

AD = a Hay xac djnh cac vecto sau va tinh do dai ciia chiing:

CD - BA, AC - BD, DA - AB - CD, AB - EA, AC - DA

1.26 Duong tron noi tiep ciia tarn giac ABC tiep xiic voi cac canh BC, CA, AB

Ian lugt tai M , N va P Cho A M + BN + C P = 0 Chung minh rang tarn giac

ABC la tarn giac deu

CUj TNHII Ml/V DVVII KItiuig Vi

Dung mpt tarn giac DEF c6 DE = AP, EF = BM, F D = C N , ta c6 AABC dong dang ADEF = ^ — = — = — (1)

2 'Tinh chat

• Voi moi vecto a, b va mpi so thuc k, 1 ta c6 :

2) (k + \)a = ka + la ; 3) k(a + b) = ka + k b ; k ( a - b ) = k a - k b ; 4) ka = d <=> k = 0 hoac a = 0

• I la trung diem doan AB o M A + MB = 2 M I , voi mpi diem M

• Ne'u G la trpng tarn tarn giac ABC thi vol mpi diem M ta luon c6 :

M A + M B + M C = 3 M G

3 ^ieu kien de hai vecta ciing phuang

• b Cling phuong a (a ; t 0) o 3 k e M : b = k a

• Ba diem phan biet A , B, C thang hang o 3 ke K : AB = k A C

4 S jeu thi mdt vecta qua hai vecta khong cung phuang • '

Cho hai vecto khong cijng phuong a va b Khi do mpi vecto c deu c6 the bieu thj dupe mot each duy nhat qua hai vecto a va b , nghla la eo duy nhat cap so'm va n sao cho c = ma + n b

1

Boi duong MSG Hinh hoc 10

B.MpTSODANGTOAN

i)ang l.Dung vecta

• SM' dung djnh nghta

• Cac diem can xdc dinh nen la diem ciioi vecta

<Bdi 18 Cho tarn giac ABC Hay xac dinh cac diem D, E, F sao cho :

c) Dung hinh binh hanh ADFE Diem F la diem duoc xac djnh (nhu hinh ve)

^ang2.Xdc dinh diem thod man he thuc vecta

• Neil CO the nen thu gon he thuc

• Su dung mot soky thuat sau de nit gon :

- Neu CO M A + MB thi got I Id trung diem AB de co M A + MB = 2MI

- Neu CO M A + MB + MC thi goi G Id trong tarn tam gidc ABC de c6

- Su dung hieu cua hai vecta ciing diem ddu M A - MB = BA

^dj 2 9 Cho tam giac ABC Xac djnh cac diem M, N thoa man ding thuc sau :

<=> M nam tren doan BC va MB = 2MC

b) Goi I la trung diem doan AB

Khi do N A + NB = 2NI

Taco: N A + NB + 2NC = 0.<i>2NI + 2 N C - 0

<=> N I + NC = 0 <=> N la trung diem IC

Cty TNHH MTV DWH Khang Vic

i)ang 3 Bieu dim vecta theo hai vecta khong ciing phuong

• Su: dung qui tac ba diem, qui tac hinh hinh hanh debieh dot vecta

• Nen chon hai vecta ca so (khong cung phuong)

(Bdi 20 Cho tam giac ABC Cac diem M, N tren canh AB va P, Q tren canh AC

Boi liuihig HSG Hinh hoc 10

Do do tu giac A H C G la hinh binh hanh, do do : ,

B H = H G = 2 H O (voi O la giao diem cua A C va BD) ' ' '

Suy ra H la trong tam cua AABC, ma A M la duong trung tuyen

nen A M = - A H ' " '* ,'•>,• \

2 Chung minh tuong tu ta ciing c6 N la trong tam cua A A D C , ma A M la

3 duong trung tuyen nen A N = — A G

D o d o : A M + A N = | ( A G + A H ) = - A C

^ang 4 Chung minh ha diem thang hang

• De'chiing minh ba diem phdn biet A,B,C thang hang ta chung minh hai

vecto A B , A C cilng phuong, nghid la AB = k A C

Trong nhieu trudmg hofp ta bieu dim cdc vecto A B , A C theo hai vecto

khong ciing phttong

<Bdi 22 Cho tam giac ABC

a) Xac djnh diem I sao cho l A + IB + 2iC = d ' r

b) M va N la hai diem thay doi tren mat phang sao cho

M N = M A + MB + 2 M C Chung minh rang M, N, I thang hang

G i a i

a) G Q I J la trung diem doan AB Khi do lA + IB = 21]

Do do l A + IB + 2rc = 6 o 2 (ij + IC) = d <=> I la trung diem doan JC

b) Ap dung qui tac ba diem, ta c6

NW = M A + MB + 2 M C = 5 ^ + IA + Ml + iB + 2(N«

' = 4Mi + l A + IB + 2 I C , vi l A + FB + 2rc = 0 nen

22

M N - 4 M I

T u day suy ra M, N, I thang hjiig

Chtl y Kci ludn a cdu b) cd thcdien dqt each khdc nhxt Id : Dwang thang MN ludn

di qua mot dic'nl codinh

Cty TNHH MTVDWH Khang Viet

^ai 23 Cho tam giac A B C N la diem tren canh A C sao cho N C = SNA M la

diem tren canh B C voi B M = kBC Goi I la trung diem A M a) Tinh cac vecto B I , B N theo cac vecto B A , B C

b) Xac dinh k sao cho ba diem B, I , N thang hang

Giai

1 a) Ta CO A N = - A C => A N = - A C va B N = kBC

'£>(ing 5 Chiing minh tinh chat hinh hoc, hai diem trimg nhau

24 Cho tam giac A B C va diem G Chung minh rang

a) Neu G A + GB + G C = 0 thi G la trong tam cua tam giac A B C ; b) Neu CO diem O sao cho O A + O B + O C = 3 0 G thi G la trong tam ciia tam

B o i iditnig use Uinli hoc 10

jVhdn xet Tir lai ^iai cua hai loan nay ta c6 cdc kct qua sau :

G la tning tam cua tain i^idc ABC <=> GA + GB + GC = 0

> < » O A + OB + OC

^di 25 Cho tam giac ABC Goi A' la diem doi xung cua A qua B, B' la diem

doi xi'mg cua B qua C, C la diem doi xiVng cua C qua A Chirng minh rang

hai tam giac ABC va A'B'C c6 cimg trong tam

Giai

Goi G la trong tam cua tam giac ABC Khi do GA + GB + GC = 0

Ta can phai chung minh G A ' + GB' + G C = 6 va tu do suy ra rang G cung

la trong tam ciia tam giac A ' B ' C <

Suy ra G la trong tam cua tam giac A ' B ' C

Vay hai tam giac c6 cung trong tam """I J

^ang 6 Tap hap diem

• A M = kv v&i A codinh va v khong doi thi tap hap cdc diem M Id duong

thdng qua A vd cung phuattg vai gid vecto v

• I M A 1=1 MB I vai A, B codinh thi tap hap cdc diem M Id duaitg trung true

doanAB • >

• I A M 1=1 v I vai A codjnh, v ^ 0 vd c6 do ddi khong doi thi tap hap cdc

diem M Id duong tron tam A, hdn kinh bang I v I

Suy ra hai vecto M A , BC cung phuong Vi vecto BC c6' djnh nen tap hop

M la duong thang qua A va song song voi BC

b) Gia su M la diem thoa man M A + (1 - k)MB + kMC = 0 Ta c6

l 2 M A - M B - M C I = a khongdoi

Goi K la diem duoc xac djnh boi KC + 2KD = 0 Khi do K co djnh

Gia su M la diem thoa man dang thuc da cho, ta c6 "

1-27 Hinh ben c6 4 vecto a, b, c, d

Ian luot nam tren 4 duong thang song song Hay xac djnh cac so k, 1,

Boi tlumtg USG Ilhih hoc 10 Cty TNHH MTV DWH KItaiig ViSt

1.28 Cho a khac 0

Cac khang dinh sau day dung hay sai?

a) Hai vecto a va-3a ciinghuang;

b) Vecta-3a c6 do dai gap ba Ian do dai vecto a; ^ ,

3 c) Cac vecto -—a va5a cung phtrong ;

-d) Cac vecto 2 a va-2 a C O do dai khac nhau ' ' ' ' j ' " ! ^ '' ' ' ' ^ ^ ^ j *

Jiuangddngidi >'^> [ '""-^ ^f'-^J^! ' a) va d) sai; b) va c) dung

1.29 Diem M tren doan thang AB sao cho 3MA = 2MB Hay xac dinh so k de

dugc dang thiic dung

Khido - O A + - O B = OM + ON = OP

- O A + - O B

2 Vay:

• Tuong tu nhu tren ta c6

I'H) DHA " i f i i ; ^ r.';>:.f t'>.] j

• Chmig minh he thiic vecto

1.31 Cho tu giac ABCD c6 I, J la trung diem cua hai duong cheo AC, BD Chung minh rang

5 b)MA = - ^ A B c)MB = - A M

b) BM = - B A

5 e) AB = - A M ' 2 g) BA = — A M 6 ; 2

A B + A D + C B + C D

= 2 (AJ + Q ) = -2 (JA + JC) = - 4 j i = 4IJ

1.30 Cho tarn giac OAB vuong tai O voi AB = a va A = 60" Hay dung cac

vecto sau day va tinh do dai cua chiing:

Tu do suy ra dang thuc can chung minh b) Do M , N la trung diem cua AB, CD

nen lA + IB = 2 i M va IC + ID = 2iN Vay IA + iB + iC + i D = 2 ( l M + IN) = 0

1-33 Cho doan thang AB Goi M la diem dxxqc xac djnh boi M\=kIvB (k^^-l) Chung minh rang voi mpi diem O bat ki trong mat phang ta luon c6

O l v l = 0 ^ - k O B

1 + k

27

Boi duimg HSG Hiiih hoc 10

Jiu&ng dan gidi

Voi diem O bat ki, ta c6

M A = kMB <=> OA - O M = k (OB - O M ) (1 - k ) O M = O A - kOB

V i k^l nen O M = O A - k O B

1 - k

1.34 Cho tam giac ABC Goi M , N Ian lirgt la hai diem tren hai canh AB, AC

sao cho A M = 2MB va 2AN = 3NC Goi I la trung diem doan M N '

1 ^ 3 a) Chune minh A I = —AB + — A C

I la diem dugc xac djnh nhu hinh ve

b) Goi D la trung diem AB

K h i d o JA + JB = 2JD

T u d o j A + jB + 3jC = 0<=>2jD + 3jC = 0

gf-Nhu the J la diem nam trong doan CD sao cho 2JD = 3JC (hinh ve)

c) Ta CO 2KA + KB - KC = CA o 2KA + CB = CA o 2KA = CA - CB

1 2KA = BA o A K = - AB <=> K la trung diem doan AB

Cty TNHH MTV DWH Khang Viet

J) Goi E la trung diem BC Khi do LB + LC = 2LE

Ta CO 2LA + LB + LC = 0 « 2LA + 2LE = 0 » LA + LE = 0 <=> L la trung diem

a) De tha'y I la diem doi xung ciia A qua B

va J la diem nam tren doan AC sao cho 3JA = 2JC

1.37, Cho t u giac ABCD Goi G j , G2 Ian luot la trong tam cac tam giac ABC

va BCD Chung minh rang:

Boi ditaiig MSG Hinh hqc 10

1.38 Cho tarn giac A B C G(?i D la diem tren canh B C sao cho DC=3DB, E Ici

,v diem tren tia doi ciia tia BA sao cho AB = 2BE Dat A B = a, A C = b

a) Tinh cac vecto A D , D E theo cac vecto a, b

b) Goi I la trung diem A C Chung minh rang D, E, I thing hang va D la trung

diem E I

A P - A B - A C

c) Xac dinh diem N tren mat phang sao cho S N A + 2NB + 5 N C = 0

d) Tim tap hop nhirng diem P tren mat phang sao cho

So sanh ta thay D I = - D E => D, E, I thang hang va D la trung diem EI

c) Goi J la trung dm B C Ta c65NA + 5 N C = 5 ( N A + N C ) = l O N I

D o d o 5NA + 2NB + 5 N C = 0 c 5 2NB + 10NI = 0 c > N B + 5NI = 0

Vay N la diem tren doan BI sao cho NB = 5NI

d) Goi E la dinh cua hinh binh hanh A B E C Khi do A B + A C = A E

= k o

A M - A B - A C -k o A P - A E EP = k o EP = k

Vay tap hop diem P la duong tron tam E ban kinh k

1.39 Cho tu giac A B C D va diem M tuy y Chiing minh rang cac vecto sau

khong phu thuoc vao M

a) a = M A - 2 M B + M C ; b) b = - M A + 4MB - 3 S 4 C ;

c) c = M A + 3MB - 2MC - 2MD ; d) d = M A + MB + M C + 3DM

Ta chung minh cac vecto a, b, c, d bang mot vecto co'djnh

Cty TNHIl A;/ V UVVii Killing

1,40 Cho hinh binh hanh A B C D c6 tam O Goi M, N Ian luat la trung diem ciia cac canh B C va C D Dat AB = a, A D = b

a) Tinh cac vecto B N , A M theo a, b b) Goi I la giao diem ciia B N va D M Chung minh rang: IB + I C + ID = 0

3 r

c) Ggi K la diem xac djnh boi A K = - h • Chung minh N la trung diem doan MK

Jiuang dan gidi

- I r

a) A M = a + - b , B N = - - a + b b) I la trong tam cua A B C D c) Ta CO A N = A D + D N = - a + b

De thay M K = 2 M N T u do suy ra ket luan ciia bai toan

1.41 Cho tam giac A B C a) Xac djnh diem M sao cho A M = A C + 2AB

b) Cho N la diem thoa man B N = k B C Xac dinh k sao cho A, M, N thang hang

Jiuang ddn gidi

a) Dung diem D sao cho A D = 2 A B Tiep theo dung hinh binh hanh A C M D Khi do A M = A C + 2 A B M la diem can xac dinh

Boi dumig HSG Hinh hoc 10

b) Cho tam giac ABC c6 cac canh la a = BC, b = A C , c = A B va trong tarn G

C h i i n g m i n h rang neu a.GA + b.GB + c.GC = 0 t h i tam giac A B C la tam

Vay tam giac ABC la tam giac deu

1.43 Cho tam giac ABC T i m tap hop nhung diem M sao cho

o BA, M C la hai vecto cung p h u o n g A? ;

M nam tren d u o n g thang qua C va song song v o i AB

b) G<?i I ia t r u n g d i e m A C K h i do M A + M C = 2 M i T u do ta c6

M A + k M B + M C = 0<=> 2 M i + k M B = 0 o M I = — M B <=> M I , M B

2 „, cimg p h u o n g hay M , B, I thang hang

Vay tap hop cac diem M la d u o n g thang BI

V o i G la t r p n g t a m tam giac ABC

Vay tap h g p cac d i e m M la d u o n g thang qua G va song song v o i A B d) Gpi I la t r u n g d i e m A B va J la t r u n g diem CJ K h i do

i s ^ + M B + 2 M C - kBC o 2 M I + 2 M C = kBC o 4MJ Vay tap h g p cac d i e m M la d u o n g thang qua J va song song v o i BC 1.44 Cho hai d i e m A , B co'dinh T i m tap h g p n h i m g d i e m M sao cho

a) c)

c) 3 M A + M B - 2 M C = 3 M B - 2 M A - M C b) Ggi I va J la hai d i e m thoa m a n 2 i A + 3IB = 0, 4 j A - JB = 0 Cho M la d i e m thoa m a n yeu cau bai toan K h i do ta c6

2 M A + 3MB = 5 M I , 4 M A + M B = 5MJ Suy ra S M I 5 M J o M I = MJ

Vay tap h g p cac d i e m M la d u o n g t r u n g tryc doan IJ

33

Boi ditmg HSG lihih hoc 10

A-TOMTATLI'THUYET

1 True toa do

• Toa do cua vecto u la so' a dugc xac djnh boi u = ai i

Toa do ciia diem M la so'm dugc xac djnh boi O M = m i

Dp dai dai so cua vecto A B tren mot true la mot so dupe ki hieu la AB va

dupe xac djnh boi he thue AB = AB.i (T la vecto don vi cua true)

* H§ thuc Sa-lo: A B + B C = A C

voi mpi diem A, B, C tren true Ox

2 Jie true toa dd

• Trong he true toa dp Oxy

b cimg phuong vai a^O <=> x' = kx,y' = ky vai k E R

• Neu I la trung diem doan AB thi Xj = ^ ^ A ± ^ ; = ^^ + 76

• Neu G la trpng tam tam giac A B C thi

_ X A + X B +

B MOTSODANGTOAN

^ang l.Mot sohai todn ve toa do cua diem tren mot true

^di 28 Tren tryc x 'Ox cho 3 diem A, B, C Ian lupt eo toa dp la -3, 2, 5

a) Tinh toa dp cac vecto A B , B C , C A , - 3 A B + 4BC + 2 C A

b) Tim toa dp diem M tren true sao cho - 3 A M + B M + M C = 0

G i a i a) Toa dp ciia vecto A B la 2 - (-3) = 5

To?i dp cua vecto BC la 5 - 2 = 3 , ; >

Toa dp cua vecto C A la - 3 - 5 = -8

Toa dp cua vecta - 3 A B + 4BC + 2 C A la -3.5 + 4.3 + 2(-8) = -19

b) Gpi X la toa dp ciia M Khi do toa dp ciia cac vecto A M , B M , M C Ian lupt la

X + 3, X - 2, 5 - x Theo gia thiet

- 3 A M + B M + M C = 0 » - 3 ( x + 3) + X - 2 + 5 - X = 0 <=> X =-2

^dj 29 Tren mot true cho 4 diem A, B, C, D Chung minh rang

AB.CD + AC.DB + AD.BC = 0 (He thuc Euler)

Giai, Gpi toa dp eiia 4 diem A, B, C, D Ian lupt la a, b, e, d

Khi do vetrai cua dang thiic can chung minh la (b - a)(d - c) + (c - a)(b - d) + (d - a)(e - b) " ' Thuc hien khai trien va riit gpn ta eo dieu phai chung minh

(I

i)Qng 2 Tinh todn ve toa do diem - vecto tren mat phang

• Sic dung cac cong thUc ve toa do diem - vecto

• Su dung dieu kien cua hai vecto ciing phuong

•,{! :ai;;.| -f ',r j'iil; ia

,.-: A ; • ••••• f - ^ >•

®di30.Choeaevecta a = (-3;l), b = (4;3), c - ( - 2 ; 6 ) ,ay:r^,^tifM

a) Xac djnh toa dp cua eae veeta 2a - 3b , - 3a + 4b + 2e - I » >;

b) Xac dinh hai so x, y sao cho c = xa + y b it <[

^ ''L ; G i a i a) + Ta CO 2a = (-6; 2), 3b = (12; 9) - ' - ^ ' ; r '

Boi duong HSG Hinh hgc 10

^ang 3 Ba diem A, B, C thang hang, khong thang hang

• Tinh toa do cac vecta AB, AC ' " ' > ? ; r^r

Bang tqa dg chting minh hai vecta A B , A C khong cung phuang

Suy ra A, B, C khong thang hang

5>t-—=>AB, A C khong cung phuong | ^, - , t w ,

Vay ba diem A , B, C khong thSng hang lA' i ' > ' •

^ d i 32 Cho A ( - l ; 2 ) , B(3;l)

a) A B cat true Oy tai N T i m toa do diem N ^ ' ' ^

b) T i m tpa dp diem M tren ( d ) : y = 1 de A, B, M thang hang

t)ang 4 Xdc dinh tga do diem M thod dim ki^n cho truac

• Tim dac tncng hinh hoc cua diem M * •: » - - < »• *

» Chuyen cac dac tntng hinh hgc sang toa do

^di 33 Trpng mat phang Oxy cho ba diem A ( l ; - 2 ) , B(3;4), C(0;5)

a) Tinh toa dp cac vecta A B , A C Suy ra A , B, C la ba d i n h cua m p t tam giac

T i m toa dp trpng tam G cua tam giac do

b) T i m to? dp diem D sao cho A B C D la hinh binh hanh '

do I

U ' 2

Bai 34 Cho tam giac A B C v a i B(l;-1), C(6;-3) Biet trpng tam ciia tam giac nam tren true hoanh va trung diem M cua canh A B each deu hai tryc tpa dp Xac djnh tpa dp diem A

Gia su A(a; b) K h i do trpng tam G ciia tam giac c6 tpa dp

Boi dumig HSG Hinh hoc 10

CLUYENTAP

1.45 Tren true x' Ox cho hai d i e m A va B c6 toa do Ian l u a t la 2 va 8

a) T i m toa do d i e m M biet rang M B + 2 M A = 0

b) T i m toa do d i e m N biet rang N B - 3 N A - 0 , ' , < n i f M o

c) T i m toa dp d i e m P d oi x i i n g v o i B qua A * A ' • i *

' Jiuang dan gidi

a) Tpa dp M i a 4 b) Tpa dp N l a - 1 c) Tpa d p P l a - 4 '^"^^^^^

AC CB 1.46 Tren true x'Qx cho 3 d i e m A , B, C phan biet thoa m a n dieu k i ^ n = —

Gpi I la diem thoa m a n 3IA + IB - 2iC = 0

• T i n h toan ve tpa dp diem - vecta tren mat phang

1.47 Cho hai vecta a = (2;5),b = (2; 3) K h i do vecto v = 2a - 3b cung p h u o n g

1.48 T r o n g mat phang tpa dp, m o i menh de sau day d u n g hay sai ?

a) H a i vecto d o i nhau k h i va chi k h i chiing c6 hoanh dp d o i nhau va t u n g dp

b) N e u vecta a 6 cung h u a n g v d i vecta I t h i no cd hoanh d p d u o n g

Cty TNHU MIV DVVH Khang Vic

Neu vecta a 0 ngupc h u d n g vdi vecto j thi no cd tung dp am

J) Neu vecta a khong cung p h u o n g vdi ca hai vecto i va j thi ca hai toa do cua no deu khac 0

e) Hai vecto a = (-3; V3) va b = (V3; -1) la hai vecta cung h u d n g

Jiuang d&n gidi

a) D u n g b) D i i n g V I i = (1;0) va neu a^O cung h u o n g v d i i thi a = k i = (k;0) v d i k > 0 c) D u n g (tuong t u cau b)

d) D u n g Ta luon vie't dupe a = (x;y) = xi + y j ^ i''' *

Do gia thiet nen ca x va y deu khac 0 v j n / - j, „ « 1 ^,

e) Sai Ta cd a = - V 3 b = > a , b ngupc h u d n g <\i ' i uv h im '

1.49 Cho cac vecto u = — i + 3 i , v = 2i + xi

a) Xaedjnh xsao cho hai vecta u , v ciing phuong

b) Xae d i n h x sao cho vecto 2u - 3 v cung p h u o n g v d i vecto i • y,,^:^

•:r • ' • : J i ^ g d^ri gidi , , ^t^j ejjp yV-,f

a) Hai vecto u , v cung p h u o n g k h i va chi k h i = — <z> x = -12

2 b) Ta cd 2u = -T + 6], 3'v = 61 + 3x] => 2u - 3v = -7T + (6 - 3x)]

Vecto 2u - 3v ciing p h u o n g v d i vecto i k h i 6 - 3x = 0 <::> x = 2 1.50 Cho cac vecto a = (-3; 4), b = (-1;3), c - (2;0) ?

a) T i m toa dp cac vecto 2a + b, - a + 3b - 2c .'^'^ " b) Xae d i n h k, 1 sao cho ka + lb + e = 6

c) Xae d j n h cac so k, 1 sao cho vecta c ciing p h u o n g vdi vecto v = ka + lb vi

I V1= 5

a) T i m toa dp cac vecto 2a + b = ( - 8 ; ] ! ) , - a + 3b - 2 c = ( - 2 ; 5 ) ' "' b) T a c d ka = ( - 3 k ; 4 k ) , l b = ( - l ; 3 1 ) : ^ k a + lb = ( - 3 k - l ; 4 k + 3l) ^

Bdi duang HSG Hinh hpc 10

c) Ta thay hai vecto cung phuang khi 4k + 31 = 0<:>l = -4k

Khi do dp dai vecta v la j-3k -1| =

5 k

-3k + - k

3 _5|k

Theo gia thiet = 5 o k = ±3

Vay ta c6 hai dap so (k;l) = (3;-4) hoac (k;l) = (-3;4) ^ f ^

1.51 Trong cac m^nh de sau, m^nh de nao diing ?

a) Neu O H = 1, OK = 2 voi H , K Ian lugt la hinh chieu ciia A len Ox, Oy thi

diem A c6 toa dp la A ( l ; 2). j , - * • > > , - - v , r (•H / n / ? r., ••^-vi^-.- '

h) Neu A(x;y) va C B = x i - y j vai y^O thi A va B doi xiing nhau qua true Oy

c) Neu hai diem A va B c6 hoanh dp doi nhau, tung dp doi nhau thi trung

diem doan AB la go'c toa dp O i

d) Neu A(a; 1) va B(-a; 3) thi trung diem doan AB nam tren true hoanh

JIudng dan gidi

a) Sai (phai sua lai la O H = 1, OK = 2,

b) Sai (doi xiing qua Ox), • ;' ^ - ^, '^^wVl,

c) Diing, d) Sai (nSm tren Oy) ^>no-nui:'fnh v - , u • v : k > l l i,

• Xac dinh tpa dp diem thoa man dieu kifn cho tmoc

a) Chung minh rang A, B, C la ba dinh ciia mpt tam giac Tim toa dp trpng tam

cua tam giac do

b) Tim diem D tren true hoanh sao cho A, B, D thang hang • >5 < f i f i

c) Tim diem E tren true tung sao cho ABCE la hinh thang vai hai day la AB, CE

a) Taco A B - ( 4 ; - 3 ) , A C = (12;-l)

Hai vecta AB,AC khong cimg phuang nen ba diem A, B, C khong thang

hang Vay A,B,C la ba dinh cua mpt tam giac

^10 n\

Gpi G la trpng tam tam giac ABC De c6 G

3 3 j b) Lay D(x;0) la diem nam tren true hoanh Khi do A D = (x + 2;-5)

Ba diem A, B, D thang hang <=> AB, A D cung phuang

o = — < » 3 x + 6 = 2 0 o x = — Vgy D

I 3

4 n

Cty TNHH MTV DWH Khang Viet

Lay E(0; y) la diem nam tren true tung Khi do C E = (-10; y - 4)

A B C E la hinh thang <=> A B , C E ciing phuang

, d £ = ^ c > 4 y - 1 6 = 30<=>y = — V a y E

2 )

153 Trong mat phang t<?a dp cho ba diem A(2;-l), B(x;2) va C(-3;y)

a) Xae dinh x, y sao cho B la trung diem eiia doan AC

b) Xac dinh x, y sao cho tam giac ABC nhan go'c toa dp O lam trpng tam c) Tim h? thiic lien h? giija x, y sao cho A, B, C la ba diem thang hang

b) Tim X, y sao cho ba diem A, B, C tao thanh mpt tam giac nhan D lam trpng tam

Huong ddn gidi

a) Taco AB = (0;-2),AC = ( x - 2 ; - 4 ) , DC = (x + 2 ; - 3 - y ) ABCD la hinh binh hanh <:>AB = DC v a A , B , C khong thang hang

D l thay la ba diem A, B, C khong thang hang Do do :

^ B C nhan D lam trpng tam khi va chi khi

Boi dumg HSG Hinh hoc 10

1.55 Cho giac A B C biet A(3; 7), trong tam G - ; 3 , diem B riclm tren tia Oy vo

> Goi C(xc ; 0) va B(0 ; yB) Ian lirgt nam tren cac t r u o O x va Oy sao cho AA\]Q

' ^ f3 + 0 + x c = 7 ^ j X c = 4 * , , ,

Vay B(0 ; 2) va C(4 ; 0)

1.56 Trong mat phSng toa do cho ba diem M ( 2 ; - 3 ) , N ( - 1 ;2), P(3;-2)

a) Xac djnh toa do diem Q sao cho M P + M N - 2 M Q = 0

b) Xac djnh toa do ba dinh cua tam giac ABC sao cho M , N , P Ian luot la trung

diem cua BC, CA, AB

Jlacmg ddn gial

a) T a c o M N = (-3;5),MP = (1;1) => M N + M P = ( 2;6)

Giasir Q ( x ; y ) K h i d 6 M Q = (x - 2;y + 3) => 2 M Q = (2x - 4;2y + 6)

Cty TNHH MTV DWH Khang Vict

b) Hai diem M va N Ian lugt tren hai d u o n g thang A B va A D Gia su

A M = x A B , A N = y A D T i m dieu k i f n cua x va y sao cho ba diem M , N , B thSnghang ' 0,-'CTT 'ir! 4 /A ' b

Boi duang HSG Hinh hgc 10

• phuong Nhu the ta phai c6

^ y Z ^ - Z ± ^ 2 z 2 i2 2 2 2 = o « x y - x 2 - ( 2 y - x y +^ \ y 2 x - x 2 ) = 0 ' '

.'i* Hay x y - x ^ - 2 y + xy-2x + x^ = O o x y = x + y

Vay dieu kien de M, N, C thang hang la xy = x + y ' '

MNP CO cung trong tam. • ,, ,,»,f >/r e/-v ^ < O u i cisi : x > i j - t i

= (GA + GB+GC) + (AM+BN + CP) = O

1.61 Chung minh rMng dieu ki^n can va du de hai hinh binh hanh ABCD va

Tu do suy ra dieu phai chung minh

doi xung ciia M qua C

a) Tinh cac vecto AM, AN theo hai vecto AB = a, AC = b

debadi&nB, I, Jthanghang t ; I d r ' *^ ; fc"' H'

Jiucng ddn gidi

a) Dithay BM = J B C , B N = JBC Tvr do

4 5:

Boi diam^ liSC Uiiili hoc 10

Cac vecto A M , A N ciing phuong Suy ra

b) Chiing minh O A + OB + O C + O D = 40I voi O la diem bat ki •

c) Xac dinh diem M deM M A + MB + M C + M D I c6 gia trj be nhat

d) Tim tap hop nhirng diem N sao cho A B cung phuong vai vecto

v = N A + N B + N C + N D

Jiwmg dan gidi

a) Goi M, N Ian lugt la trung diem cua AB va C D

T a c o rA + iB = 2 i M , i C + ID = 2iN

Nhuthe: I A + IB + I C + ID = 0 ^ 2 I M + 2IN = 0

<=> I l a trung diem doan M N

b) Ap dung quy tac ve hif u hai vecta, ta c6: *"

D o d o l M A + MB + M C + M D I dat gia trj be nhat o MI = 0

V a y M s Ithi I M A + MB + M C + M D I dat gia tri be nhat

d) Ap dung ke't qua cau b), ta c6: v = N A + r^ + N C + N D = 4 N I

N h u vay, v ciing phuang vai A B o NI ciing phuang voi AB <=> NI song

song vai AB

N h u the tap hgp nhiing diem N sao cho v ciing phuang vai A B la duang

thang qua I va song song vai AB , w :

ft,, fNlUl MVV DVVn Khaug Viet

J 65- 8'^^ A B C Cho d, d' la hai duong thang phan biet thay do!

nhung luon song song vol BC Biet d cat hai canh AB, A C Ian lugt tai M, N ; d' cat hai canh AB, A C Ian lugt tai P, Q

Chung minh vecto v = MQ + PN c6 huang khong doi

= (k + 1)AC - ( k + 1)AB = (k + 1 ) ( A C - A B ) = (k + I)BC '

- ; 4 , trung diem canh C D la P(2;2) Hay xac dinh tga do cac dinh A ,

Vay V luon cung huang vai B C 1.66, Cho hinh binh hanh A B C D c6 trung diem AB la M(l;5), trung diem B C

I la trung diem A C nen

I cung la trung diem MP nen I

_ + xc _ 7 + 2x

(3 7_\

[I'l]

Boi duang HSG Hinh hoc 10

b) Gia sit M, N , P thoa m a n M A = - M B , N B = - 2 N C , PC = 2PA , ^

H a y xac d i n h tpa dp cac d i e m A, B, C , j , j ,

Jiucfngdangiai vj ^ ^^^^^ ^ ^ ,^ Q , , , , ; ; ^ a) C h u n g m i n h hai vecto M N , M P khong cung p h u a n g /'I

Jiu&ng dan gidi

a) Gia su A ( a ; 0 ) , B ( 0 ; b ) V i M la t r u n g d i e m cua dogin A B nen

N h u t h e M C = 2 M D o - ^

c = - 2

2 Vay hai d i e m can t i m la C ( - 2 ; 0 ) , D 0 ; -

- ^ - ; - l J • Biet tam a i a h i n h b i n h hanh n a m tren tr\ic hoanh Hay qua d i e m M

Jiu&ng ddn gidi

Gpi I la tam h i n h b i n h hanh K h i do l(a;0) Gpi N la d i e m d o i x u n g cua M

qua I Ta c6 N thuoc canh A D va N 2a - - ; 1

Ta tinh dug-c cac vecto A D = (4;-l), A N =

V i hai vecto A N , A D cung p h u a n g nen 4(-2) =

giac A B C de | M A + M B + M C | C6 gia trj be nhat

Jiurnig dan giai

G(?i G la trong tam cua AABC, ta c6 : M A + M B + M C = 3 M G

wA

Boi dumg HSG Hinh ht?c 10

Voi moi goc a (0" < a < 180"), ta xac dinh

mot diem M(x ; y) tren nua duong tron don

visaocho MOx = a

K h i d o

sin a = y , cosa = x ,

tana = - (x^O), cota = - ( y ^ O )

Cac so sin a , cos a , tan a , cot a duac goi la cac gia tri lugng giac ciia goc a

• Tinh chat: Vai hai goc bii nhau la a va 180°-a ta c6 :

{Wfci sin(180° - a) = sin a ; cos(180° - a) = -cosa;

tan(180°-a) = - t a n a (a ^ 90°) ;

• cot(180°-a) = -cota (O" < a < 180")

2 Gid tri lumg gid£ cua mdt s6goc ddc biet

B MOT SO DANG TOAN

'Dang 1 Tinh cac gid tri luong gidc

• Su dung dinh nghta gid tri luang gidc

• Sti dung moi quan hegiita hai goc phu nhau, bii nhau

^ d i 2 Tinh gia tri cac bieu thuc sau day :

S = cosl2° + cos 36° + cos 60° + cos 84° + cos 96° + cos 120° + cos 144° + cos 168''

P = tan 10° tan 20° tan 30° tan 40° tan 50° tan 60° tan 70° tan 80°

Giai

kSu dung quan he giira cac gia tri lugng giac ciia hai goc bu nhau

|cosl2° + cosl68° = cos36° + cosl44° = cos 60° + cosl20° ,

= cos 84° + cos 96° =0 ,,,,, iDo do S = 0

Taco: tan 80° = cot 10°, tan 70° = cot 20°,

tan 60° = cot 30°, tan 50° = cot 40°

iTvrdo:

P = ( tan 10° cot 10°) (tan 20° cot 20°) (tan 30° cot 30°) (tan 40° cot 40°) = 1

ft,-

'Dang2 Chung minh he thuc lien quan den gid tri lucmggidc

• Sic dung dinh nghia hinh hoc cua cdcgid tri lucmggidc ,

• Ap dung cdc hang dang thuc

^ d j 3 Chung minh cac he thuc sau day : a) sin^a + cos^a = l ;

b) cosa + sinatana = —-— voi a 9^90° ;

cosa

Boi duang HSG Hinh hoc 10

c) cot^x cos^x = cot^x - cos^x ;

Giai

a) Ta da biet sin^ a + cos^a = 1, vai a la goc nhpn

Six d y n g m o i lien giiia hai goc b u nhau ta de dang suy ra ket qua

b) Ta CO cosa + s i n a t a n a = cosa + sina sin a cos^ a + sin^ a 1

COS^ X ? •'' ' ' '

c) cot^x cos^x = cot2x(l - sin^x) = cot^x ^y—.sin x = cot^x - cos^x

s i n X

+ Cach 1 : Ta CO sin^x + cos^x = 1 => (sin^x + cosZx)^ = 1

+ Cach 2 : A p dung hang dang thiic a^ + b^ = (a + b)^ - 2ab, ta c6

sin^x + cos''x = (sin^x)2 + (cos^x)^ = (sin^x + cos^x)^ - 2sin2x.cos2x

= 1 -2sin2x.cos2x ;vv j ^'^^••!••.••:-.•.••

C LUYENTAP - • " ' " ^j{&f^«feft:;f'^-;if;V;

• Tinh cac gia trj lugng giac '' • •'

2.1 T i n h gia t r i cua cac bieu thiic

2.2 Tinh cac bieu thiic sau day:

a) A = cos^ 22° + cos^ 23° + cos^ 41° + cos^ 49° + cos^ 68° + cos^ 67°;

b) B = t a n l ° tan2° tan88° tan89°

J^udrng (Mn gidi

a) Ta C O cos 49° = sin 41°, cos 67° = sin 23°, cos 68° = sin 22°

D o do: A = cos222° + sin^ 22° + cos223° + sin^ 23° + cos241° + sin^ 41° = 3

CtyTNHHMTV I) VVI! Khang Vie,

I,) Taco: B = ( t a n l ° tan89'').(tan2° tan88°) (tan44° tan46'').tan45° = 1

2.3

3 a) Cho cosx = T i n h sinx, tanx, cotx b) Cho tanx = 4 ( 0 ° < x < 9 0 ° ) Tinh sinx, cosx " "

c) Cho cotx = - 2 (90° < x < 1 8 0 ° ) T i n h sinx, cosx

c) sinx = - ^ , cosx = — • ' • • ^ - i o u f ' v n ; fti :.) c; •

2.4 a) Cho tan x + cot X = m Tinh tan^ x + cot^ x, tan^ x + cot^ x , , „ 3sinx+2cosx ,

5sinx - cosx c) C = 2sin''x + 4cos^x, biet Ssin 'tx + 3cos''x = 2

JIuong ddn gidi

a)» =(tanx + cotx)^ =tan^x + c o t ^ x - 2 t a n x c o t x = tan^ x + cot^ x - 2

T u do tan^ x + cot^ x = - 2

• tan^ X + cot^ X = (tan x + c o t x ) ( t a n ^ x - t a n x cot x + c o t ^ x ) = m ( m ^ - 2 - l )

hay tan^ X + cot^ x = m ( m ^ - 3)

D) Ta CO : tanx = 4 => cosx 0 nen :

3 s i n x + 2cosx _ cosx(3tanx+ 2) 3tanx + 2 _ 3.4 + 2 _ 14 5sinx - cosx cosx(5tanx - 1 ) 5tanx - 1 5 4 - 1 19

^) E)at t = sin^x, ta CO cos^x = 1 - 1 Do do : " Ssin^x + Scos^x = 2 5t2 + 3(1 -1)^ = 2 0 8t2 - 6t + 1 = 0 <=> t = - hoac t = - •

Boi duang HSG Hinh hqc 10

• t = sin^x = i => cos^x = 1 - — = i , t a c 6 : A = 2sin''x + 4cos''x = 1

• t = sin^x = - => cos^x = - , ta CO C = 2sin''x + 4cos"x = —-

• Chung minh cac h^ thuc lugng giac don gian

2.5 Riit gon cac bieu thuc sau day

a) A = sin^ x + sin^ x cos^ x + cos^ x;

b) B = sinxcosx(tanx + cotx) ;

c) C = 2(sin^ x + cos^ x) - 3(sin'' x + cos"* x)

\ , Jluxmgclangidi

a) Ap dung h? thuc sin x + cos x = l t a c o ,^ , t

A = sin^ X + sin^ X cos^ X + cos^ X

= sin^ x{sin^ x + cos^ x) + cos^ x = sin^ x + cos^ x = 1

^ sin X cos X ^ = sin x + cos X = 1

b) B = sin X cos x( tanx + cotx) = sin x cos x

Vcosx sinx,

c) A p d u n g a 2 + b2 = (a + b)2-2abvaa3 + b3 = (a + b)3 -3ab (a + b)

=> sin^x + cos*x = 1 - 3sin2xcos2x va sin^x + cos-*x = 1 - 2sin2cos2x

Thay vao C ta suy dugc C =-1

2.6 Chung minh cac dSng thuc (voi dieu kien dang thuc c6 nghia)

cos X cos^ X cos^ X

2 , cos^ X sin^ X + cos^ x

cosa = (l + cosa)(l+tana) (a?^90°

c) —X— 1 + cota 1 + cosa 1 + cosa = 1 sin a sin a sin^ a 1-cos a 1-cosa

§2 TICH V 6 Hl/QNG CUA HAI VECTQ

A T6MTATUTHUYET

/ Goc giCca hai vecta

• Cho hai vecto a va b khac vecto 0

3

Tir diem O ba't ki, ta ve cac vecta OA = a va OB = b

Khi do AOB duoc goi la goc giiia hai vecto a va b, kihi?ula(a,b) v;

• ( a , b ) = 90° a l b

'^inh nghla tich v6 hu&ngcua hai vecta

Tich v6 huong cua hai vecto a va b la mot so,

ki hi^u la a.b, dugc xac dinh boi cong thuc a.b = lallblcos(a, b)

Tinh chat cua tkh vd huong: '

Voi mpi vecta a, b, c va mpi so thuc k, ta c6:

1) a.b = b.a (Tinh chat giao hoan);

2) (ki)b = k(ab) ;

"Boi duang MSG Hitth hgc 10

3) a.(b + c) = a.b + a.c (Tinh cha't phan pho'i d o i v 6 i phep c o n g ) ;

a (b - c) = a.b - a.c (Tinh chat phan pho'i d o i v d i phep t r u ) ;

4) a.b = 0 < = > a l b

5) Binh phuang v6 huang: Binh p h u o n g v6 h u o n g cua m p t vecto bang binh

p h i r o n g d p dai cua vecto do : = I a 1^

• Cac hang dang thuc ve b i n h p h u o n g v6 h u o n g :

(a + b ) 2 = a ' + 2ab + b ' ; ( a - b ) ^ = a ' - 2 a b + b

a ^ - b ^ = ( a - b ) ( a + b )

Cho O A = a, OB = b Tich v6 h u o n g cua hai vecto

a va b bang tich v6 h u o n g cua a v o i OB' = b ' la

a.b = a.b' hay OA.OB = O A O B ' .r

• Chii y: Cho ducmg tron (O) va mpt diem M

D u n g cat tuyen M A B v o i (O), ta d i n h nghia:

Phuong tich cua d i e m M d o i v o i d u o n g tron (O),

k i hi|u la P M / ( 0 ) la so' dugc xac d i n h boi bieu thiic:

^ M / ( 0 ) = M A M B = d ^ - R ^ (d = M O ) ;

Neu M nam ngoai d u o n g tron (O) va M T la

tiep tuyen ciia (O) t h i PM/(O) = MT^ =

4 $ieu thiic tga dO cua tich v6 huang

Trong mat phang Oxy, cho a = (x; y) v a b = (x'; y ' ) K h i do : *

Cty TNHH MTV DWH Khang Viet

i)ang 1 Tinh tich v6 huang cua hai vecto

• Ap dung dinh nghia

I • Ap dung cong thiic hinh chieii

$di 4- Cho h i n h thoi A B C D canh a va goc A = 60° T i n h cac tich v 6 h u o n g sau day: A B A D , AB.BC, A D C A , A C A B , AC.BD

Goi O la giao diem ciia A C va BD

Ta CO tam giac A B D can va A = 60° nen no la tam giac deu

^di 5 Cho hai vecto a va b Cho bie't

H a y t i n h cac tich v6 h u o n g a(2a - b), (Sa + 4 b ) ( - 2 a + 3 b )

57

Boi ihiang HSG Hinh hgc 10

'Bdi 6 Cho tarn giac deu ABC c6 canh bang a Tinh gia tri ciia bieu thiic

Tudo F = a 2 + ^ + a 2 - 3 a 2 = ^ "')/^ ,T-u:,'V' firs

'£>Qng 2 Chung minh he thiic lien quan den tich vd huang

• Su: dung cdc hang dang thiic ve tich vd huong

• Chii y den he thiic a =

^di 7 Cho tarn giac ABC Chung minh rang :

a) AB.AC = ^ ( A B ^ + AC^ - BC^); b) cosA=

Giii

AB? + AC^ - BC?

2AB.AC

a) Tac6BC^=BC = ( A C - A B ) = A C ^ - 2AB.AC + AB

Tu do ta C O AB.AC = |(AB^ + AC^ - BC^)

b) Tu ket qua tren, ta c6

AB.AC = AB.AC cos A - -(AB? + AC? - BC^) => cosA= AB^+AC -BC^

2AB.AC

Chu^

1 Cdch viel BC^ - BC^ hay dugc sit dung trong viec xu li cdc blnh phimng do dai

2 Tit bai todn tren ta thdy rang ne'u biel do dai ba cqnh cua tarn gidc ABC thi ta hodn

iodn tinh dugc cdc tich vd huang TiBAC,BC.BA,CA^CB vd tie do suy ra dugc do

Ian cua ba gdc

<6di 8 Cho hinh binh hanh ABCD

Chung minh rang AC^ + BD^ = 2(AB? + AD^) ; , f ; ' ;' ^? ' •'0/

Giai

Taco: AC = AB + AD AC^ = AB^ + AD^ + 2 A B A D ;

BD = AD - AB BD^ = AB^ + AD^ - 2 A B A D

Cpng hai dang thuc tren vetheo ve, ta dugc dang thuc can chung minh

Cty TNHH M i'V I) V VII Kliang Viel

^ang 3 Chiing minh hai vecto vudnggdc

Dua vdo tilth chat cua tich vd huong A./; = 0 <=> A ± 5

$0.19- S'^'^ '^'^"^ ^ a) Chiing minh rang MA.BC + MB.CA + MC.AB = 0 t,) Dira vao ke't qua tren, chung minh rang trong mot tarn giac thi ba duong cao dong quy

Giai • ' • Ji-r

) Ap dung quy tac ve hieu vecto, ta c6:

MA.BC = MA.(MC - MB) = MA.MC - MA.MB MB.CA = MB.(MA - MC) = MB.MA - MB.MC MC.AB = MC.(MB - MA) = MC.MB - MC.MA Cgng (1), (2), (3) ve theo ve, ta dugc MA.BC +MB.CA+ MC.AB = 0 (*) b) Cho hai duong cao AA' va BB' cua AABC cat nhau tai H

Theo ket qua (*) cau a) ta c6

0 )

(2) (3)

HA.BC + HB.CA + H C A B = 0

Vi A H 1 BC va BH 1 AC nen HA.BC = 0, HB.CA = 0

Tu (4) va (5) ta thu dugc

(4) ; (5)

HC.AB = 0 hayCH 1 AB

Noi each khac ba duong cao ciia AABC dong quy tai H

Sdi 10 Cho tarn giac ABC c6 duong cao A H va AB^ - BH^ = CH.BH Chung minh rang tam giac ABC vuong tai A

at" 11 Cho hinh vuong ABCD c6 canh bang 1 Dat AB = a , AD = b Goi M ,

N la hai diem tren hai canh AB, AD sao cho A M = D N = ^ '

^) Tinh C M , BN theo cac vecto a,b

Tinh C M B N Suy ra CM 1 B N

59

Boi duang HSG Hlnh hgc 10

Giai a) BM = -|AB=^BM = - - a , N D = i b

Tie day suy ra CM 1 B N

<8di J2 Cho tarn giac A D D ' c6 D A D ' > 90° Dung ngoai tarn giac A D D ' hai h i n h

vuong ABCD va AB'C'D' Ggi M la trung diem DD' Chtmg minh rang

Til- gia thiet AD.AB = AD'.AB^ = 0 •

Cho nen 2 A M - B B " = (AD + AD^)(AB' - A B ) = AD.AB' - AD'.AB - 0

•Vay A M I B B ' '

b) Ta CO BAB- + DAD' = 180° ^ cosBAB' = -cosDAD'

Suy ra A B A F =-AD.AD^ Do do :

4|AM|^ = 4AM^ = ( A D + A D ' ) ^ = |AD|^ + 2 A D A D ' + I A D "

= |AB|^ -2AB.AB' + \ABf = ( A B - A F ) ^ = BB'^

Cty TNUll M'fV nVVIi Khang Vi$t

Vay tap hop nhung diem M la duong tron duang kinh A I Gpi G la trpng tam ciia tam giac A B C Ta c6 MA+Nffi + M C = 3iv^

Taco ( M A - M C ) ( M A + M B + M C ) = 0 C > C A 3 M G = 0 C > C A 1 M G

V^y tap hop cac diem M la duong thang qua G va vuong goc vol A C

$di 1'^- ^^'^ diem A , B va k la mpt so'khong doi Tim tap h^p nhirng diem

ivl trong moi truong truong hop sau :

Neu k = — ^ thi M triing vol diem O

Neu k < thi tap hgp M la tap rong

b) Gpi H la hinh chie'u cua M tren AB, O la trung diem ciia AB Ta c6

M A ^ - M B ^ =MA^-Sffi^ = ( M A + M B ) ( M A - M B ) = k

o 2 M O B A = 2AB.OH = k (*)

Vi A, B, O CO djnh nen (*) cho ta H co dinh

Vay tap h^p cac diem M can tim la duong thang A vuong goc vai AB tai H

Chu y d cau b) ta da ap dyng mpt tinh chat ciia tich v6 huang: Vdi hai

; vecto AB va CD cung nam tren tryc (O; i ) thi

T2

AB.CD = (AB.i).(CD.i) = AB.CD.i = AB.CD

61

Boi duaiigHSG Hinh hgc 10

^ang 5 Bie'u thiic toa do ciia tick v6 hu6itg

Su dung cdc cong thtic vetoa do diem, vecto, do ddi vecta, goc ^iita hai vectt

'Bdi 15 Cho cac vecto a = (3; 2), b = (-4; 7)

a) Xac djnh toa do va do dai cua cac vecto 2a + 3b -'-i '#5 > >

b) Tinh cosin ciia goc giira hai vecto 2a va 3b ,! ' / v r ? ! v •.{',>•{''

b) Xac djnh cac so x va y sao cho vecto c = xa + yb c6 do dai bang 1 va c

vuong goc voi vecto a - b

Giai

a) Ta CO 13 = (4a - 3b)^ = 16a^ + 9h^ - 24ab =^ ab = i

Ma ab= a b cos(a,b)=>cos(a,b) = ^=:>(a,b) = 60°

b) Ta CO c^ = (xa + yb)^ = x^a^ + y^b^ +2xya.b = x^ + y^ + xy

Mat khac, do c vuong goc voi vecto a - b nen

c(a - b) = 0 (xa + yb)(a - b) = 0 <:i> xa^ - yb^ + (y - x)ab = 0

<idi 17 Cho tarn giac ABC c6 A(-3;2), B(4;-l),C(-2;5)

a) Tim toa dp cac vecto AB, AC va tinh cosA cua tarn giac ABC Suy ra dien

tich tam giac ABC ;

b) Tim toa do vecto v sao cho I v 1= 1 va v vuong goc voi u sau :

Suy ra u = 2 AB - 3BC + 4CA = (28; -36)

Giasu v = (x;y)

Vi = 1 nen ^x^ + y^ = 1 o x^ + y^ = 1 Mat khac u 1 v o u.v = 0 o 28x - 36y = 0 <=> x = - y 'vrdo: X + y = ! < = > — y + y =!<::> 130^^2 y = l < = > y = ±

\hu y : O cau a) trong bai tren ta tinh duac dien

ich AABC theo cong thitc:

'a AABH vuong tai H => BH = ABsinA nen

SAABC = - AB AC sin A

^ang 6 Tim tga dg diem thoa man dim ki^n cho tneoc

• Xdc dinh dac trung hinh hgc cua diem can tim Sau do chuyen sang cdc rdngbugcvetga dg

• Sit dung dieu kien ding phuang cua hai vecto, cong thiic tinh goc

• Chu y cdc tinh chat mot so diem dac bi^t trong tam giac ABC

^ _ • 0 4 + 06 + 0 0 x.+Xn+Xr yA+VB+Vc^

+ Trgngtam G: OG = =>G ^ - " ^ ^

Bdi dumtg HSG Hinh hgc 10

+ True tam H: AH.BC = 0

BH.CA = 0 + Tam duimg iron ngoai Hep I: IA = IB = IC

^di iS.Cho A ( - l ; 2), B(3;l) ^ , ,

a) Tim diem M tren true hoanh sao cho MA = MB '

b) Tim diem N tren true tung sao cho A, B, N thang hang ,

^di jaCho hai diem A(7;26) va B(12;12) Tim diem C sao cho tam giac ABC

vuong can tai C ,

Giai

Giasu C(x;y).Khid6 CA = (7-x;26-y),CB = (12-x;12-y)

Tam giac ABC vuong can t^i C khi va chi khi

Giai h? phuang trinh ta dupe hai nghif m (x; y) =

I V^y CO hai diem C thoa man yeu cau bai toan la

Cty TNHH MTV DWH Khang Viet

P^i 20.Cho A(-4 ; 2), B(2 ; 6), C(0 ; -2) Xac dinh toa dp trpng tam G, tryc tam

H, tam duong tron ngoai tie'p I cua tam giac ABC

^ang 7 Gidi bdi todn hinh hgc bang phuong phdp tga dp

^ 21 Cho duong tron (O) c6 hai day cung AB va CD vuong goc va cat nhau

Im I Chung minh rang duong thang qua I va vuong goc voi AC di qua

|trung diem BD

IChpn h^ true toa dp Ixy voi tr\ic Ix chua AB va true ly chua DC (nhu hinh

|ve) Khi do 1(0; 0) D^t A(a; 0), B(b; 0), C(0; c), D(0; d) >•;:

|Tac6:Pi /(0)= iA.iB = i C i D o a b = cd (1)

Boi dumig nSG Hhth hoc 10

b a

Goi M la trung diem B D , suy ra M —; —

Ta Chung minh IM ± AC ,

2 ' 2 Taco I M = , AC = (-a;c)

b Dodo: lM.AC = -a.^+c.4 = ^ ^ ( 2 )

2 2 2 Thay (1) vao (2), ta dugc IM.AC = 0 =^ I M 1 A C

Do do I M ± A C , noi each khac duong thSng

qua I va vuong goc AC di qua trung diem BD

'Bdi 22 Cho tarn giac ABC vai cac duong trung tuyen AD, BE, CF Goi P, Q, R

tuong ung la trung diem AD, BE, CF Tinh gia trj

AQ^ + AR^ + BP^ + BR^ + CP^ + CQ^

F =

AB^ + BC^ + CA^

Giai

Xet trong he true toa do nhan A lam goc va true hoanh la tia A B , true tung

la duang thang qua A vuong goc A B Khi do A(0 ; 0) Dat B(4x ; 0), C(4y ;

4z) Suy ra : D(2x + 2y; 2z), E(2y; 2z), F(2x; 0), tu do ta duge :

Q(2x + y;z), R(x + 2y;2z), P(x+y;z)

Taco : ir^-'

A Q ^ = (2x+y)^ + = 4x^ + 4xy + y^ + z^;

AR^ =(x+2y)^ +4z^ =x^ +4xy + 4y^ +4z^;

BP^ = (3x - y )^ + z^ = 9x^ - 6xy + y^ + z^;

BR2 =(3x-2yf +4z2 ^9y^ -6xy+4y2 +42^;

CP2 = (x - 3y)^ + 9z2 = x^ - 6xy + 9y2 + 9z2;

CR^ =(2x-3yf+9zi^ =4x^-12)^+9/+9z2

Tu do A Q ^ + A R ^ + Bp2 + BR^ + CP^ + CQ^ = 28(x2 - xy + y^ + z^)

Tuong tu nhu tren ^ ^

A B ^ =16x2,BC2 =16x2 _32xy+16y2 +16z^CA2 =16y2 +16z2

Va A B ^ + B C ^ + C A ^ =32(x2-xy+ y2+z2)

Vay F = -

Cty TNHH MTV DWH Khang Viet

<^(ing 8 Bdi todn lien quan Aen phuang tich mot diem dot vai duang tron

Ap dung PM/(O) = M A - M B = - ( v a i d = M O , R Id ban kinh ciia (O))

(goi 23 Cho tarn giac ABC vuong tai A c6 duang cao AA' Duong tron duang

ki'nh A A ' cat AB va AC Ian lugt tai M va N

a) Chung minh AM.AB = AN.AC, tii ket qua nay hay chung to rang tu giac

BMNC noi tiep dugc duong tron

b) Chiing minh BM BA^

CN CA"

Gidi

a) Cac tarn giac AA'B va AA'C vuong tai A'

CO M A ' va N A ' la cac duong cao nen

A'A^ = AM.AB; A'A^ = AN.AC

Do do : AM.AB = AN.AC

Vi A M cung huang voi AB, A N cung huong vol A C do do ta c6

Dung duang tron di qua 3 diem B, M , C, ki hi^u duang tron nay la (BMC)

Duong tron (BMC) cat AC tai N ' Ta c6:

3 •

Chu y: 6 hai tren ta dd chung minh dugc ket qua quan trong sau ; ' ''"

Neu AM.AB = AN.AC thi cdc diem B, M, N,C cung nam tren duang tron

67

Boi duong HSG Htnh hoc 10

Ngoai ra, neu MD.ME = MT thi MT la Hep tuyen cua duang trdn ngoqi tiep tarn

giac TDE

Day la hai each de chung minh mot tugiac noi tiep dime duang trdn

i>ang 9 ling dung tick v6 huang dechting minh hat dang thiic

Dua vdo hieu thiic toa do cua tick v6 huang

^ d i 24 a) Chung m i n h rang: I ax + by I < ^(a^ + b^)(x2 + y^) (i)

voi a, b, X, y la bon so'bat ki Dau " = " xay ra k h i nao ?

{(1) goi la hat dang thiee Bunhiacopxki, hay bat dang thuc Svacxa)

b) A p dung: Cho tam giac ABC Tren canh BC lay mot diem M bat k i Qua M

d u n g d//AB va dV/AC, d cat AC tai E va cat AB tai F

cos(u; v) = ±l <^ u , V ciang phuang <=> bx - ay = 0

b) A p dung dinh If Talet, ta c6:

• Tich v6 hwang, goc, dp dai cua vecta ,

2.7 Cho tam giac ABC can tai A va goc A =120° Hay xac dinh goc cua cac cap

= 3,

= 1,

= V2, ( i , b ) = 30° ; b) |i| = V3, |b| = 76, (a,b) = 45°; b| = 273, ( i , b ) = 60°; d) |a| = l , |b| = ^ , (a,b) = 90° ; b| = 73, (a,b) = 120° ; ^ g) |a| = 75, |b| = 75, (a,b) = 135°;

2.9 Cho hai vecto a va b Cac khang dinh sau day diing hay sai ?

b) Neu i.b = l thi l a l l b l = l ; c) Neu a.b = ac va a 0 thi b = c;

d) Neu a = b thi a = b hoac a = - b

Jivcang ddn giai

a) Dung.b)sai

Neu a ^ 0 thi c6 the xay ra truong hop a 1 (b - c) va h^c

d) Sai V I CO the xay ra truong hop a, b khong cimg phuong n h u n g

b <=>a = b

Boi duaiig IISG Hinh hoc 10

Jixjcang dan gidi

Chieu vecto B D len gia aia vecto A B , ta dugc vecto B A

Tu do A B B D = A B B A = - A B ^ = - a ^

Goi E la trung diem D C Khi do A B E D la hinh vuong

D B D C = D B D E = |DB||DE|COS(DB,DE) = 2a.a.cos0° = 2a.^

Ggi F la hinh chieu vuong goc ciia C len A B

B A C B = A B B C = a^

-a) T m h t i s o — b) Chiing minh rang AI vuong goc voi IC ^ ^

Jiuang dan gidi

2.15 Cho tarn giac ABC vuong tgi A voi AB = 2, AC = 3 Tia phan giac cua goc

BAC cat canh BC tai D Tinh dp dai doan AD | :i ffr r • y i j

J-Iic6ng d&n gidi I

2.16 Cho doan thang AB = a H la diem tren AB sao cho 2HA+HB - 0 Goi d

la duang thSng qua H va vuong goc voi AB M la mpt diem tren d Chung

Mat khac: HA.HB = -HA.HB = = - —

3 3 9 Vay MA.MB = M H ^ - 2a^

b) Khi M di dong tren (d), ap dung cong thiic hinh chieu ta c6

AM.AB = AB.AH = AB.AH = — (khong doi) t,

3

72

Cty TNHH MTV DWH Khang Vi?t

2,17 Cho hinh vuong ABCD M la diem tren canh AB Chung minh rang DM.DC + CM.CD khong doi khi M di dong tren c^nh AB

Jiudng dan gidi ^

Do AM, DC cung huong, ta c6 DM.DC = (DA + A M) DC = AM.DC = AM.DC

a4 CD = (CB + B M ) C D = BM.CD = BM.CD ' *

(vi BM,CD cimg huong) Do do DM.DC + CM.CD = AM.DC + BM.CD = CD(AM + BM) = a^ ' 2.18 Cho hinh chir nhat ABCD c6 canh bang AB = 2 va AD = 4 Goi M la trung diem canh AB va N la diem tren c^nh AD sao cho AN = kAD Xac dinh k

Vay voi k = - thi C M 1 B N

2.19 Cho hiiJi chii nhat ABCD M la diem tuy y tren mat phang Chung minh rang:

a) M A M C = M B M D ;

M A ^ + M C ^ = M B 2 + M D 2 ;

M A2 +M C ^ + 2 M B M D = 4 M O ^

Jiuang dan gidi

^) Gpi O la giao diem hai duong cheo ciia hinh chu nhat Ta c6

K I A M C = ( M O + O A ) ( M O + O C ) = ( M O + O A ) ( ! ^ ,

= M O ^ - O A ^ = M O ^ - O A ^ h no-ci j ^ v i r : , : ,

73

Boi dumig HSG Hinh hoc 10

Tvrong ty ta cung c6 MB.MD = MO^ - OB''

2.20 Cho hinh vuong ABCD c6 canh bang 1 Hai diem M, N thay doi tren hai

canh AB, AD sao cho AM = x (0 < x < 1), DN = y (0 < y < 1)

Tim dieu ki^n cua x, y sao cho CM 1BN

Chii y rang do CB, AN ngugc huong va BM, BA cimg huong

nen CB.AN = -(a - y)a va BM.BA = a(a - x)

Tu do ta CO CM 1 BN <=>-(a - y)a + a(a - x) = 0 <=> X = y

2.21, Cho tam giac ABC Tim tap hop nhiing diem M sao cho :

a) MA.MB = 0; b)MA(MC-MB) = 0;

c) (M\+ivffi)(lv^+iv®+ivC)=0 ; d)MA.MB = -MA.MB

a) Gia su M la diem thoa man MA.MB = 0

Ta CO MA.MB = 0 <=> MA 1 MB <=> M nam tren duong tron duang kinh AB

b) Ta CO M A ( M C - M B ) = 0 < » M A B C = 0 < » M A 1 B C < = > M nam tren duong

thang qua A va vuong goc voi BC ,

c) Goi, I la trung diem AB, G la trong tam tam giac ABC

Khi do + MB = 2MI va MA + 5 ^ + MC = 3MG

' Ta CO (MA + MB)(MA + MB + ivC) = 0 « 2 M I 3 M ^

nam tren duong tron duong kinh IG

74

Cty TNHH MTV DWHKhang Viet

Gia su MA.MB = -MA.MB <=> |MA||MB|COS(MA,MB) = -MA.MB

<r> cos(MA,Kffi) = -1«(MA,Nffi) = 180° M nam ben trong doan thang AB 2.22 Cho hai diem A, B va k la mpt so'khong doi Tim tap hop nhifng diem M ' thoa dieu kien: MA2 + MB2 = k2

JIudng ddn gidi '

Vai O la trung diem AB Ta c6 : ^ * ' ' ^

MA^ = MA^ = (MO + OAf = MO^ + 2MO.OA + OA^

MB^ = MB^ = (MO + OBf = MO^ + 2MO.OB + OB^

=> MA^ + MB^ = 2MO^ + 2MO.(OA + OB) +OA^ + OB^ (1)

Vi O la trung diem AB nen OA + OB = 6 va OA = OB, do do (1) trd thanh

ii) k2 = m2thi MO^ = 0<=> M = O : (L) = { O }

iii) k2 > m2 <=> MO = /^(k^ - ) : (L) la duong tron tam O c6 ban kinh la

• Bieu thuc tpa dp ciia tich v6 Huong

2.23 Cho hai vecto a = (1; 1), b (1; 0) Cac khang dinh sau day diing hay sai:

^) Xac dinh toa dg cac vecto a + 2b , 2a - 3b, 4a + 3b'

^) Tinh a + 2b,2a-3b

75

Boi duang HSG Hinh hoc 10

c) Tinh cac tich v6 huong a.b, 2a(4a + 3b), (2a - 3b)(4a + 3b)

Jiu&ng dan gidi

a) a + 2b = ( l ; 1 3 ) , 2 i - 3 b = (9 ; - 2 ) , 4 i + 3b = (9;32)

b) |a + 2b| = ^/T70, |2a-3b| = ^/85

c) a.b = 17, 2a.(4a + 3b) = 374, (2a-3b)(4a + 3b) = 17

2.25 Tinh goc giita cac cap vecto sau day:

Gpi a la goc giira hai vecto Khi do cosa=-j=^j~=

kli-AVi 2.26 Cho hai veco a = (3;6), b = (-2;5) [ ^'"^ ^^'^ '

a) Tim vecto u saocho u.a = 0 va u.b = -4 ,

b) Tim vecto v sao cho I v 1= 1 va v vuong goc voi vecto 3a - 2b

b) Tmoc het d l dang tinh dugc 3a - 2b = (13; 8)

Gia su v = (x; y ) Theo gia thiet ta c6

Giai h§ tren dugc hai nghi$m: (x; y) =

Cty TNHHMIV I)VVHKhang Viet

J,) Tim toa dp vecto v sao cho v 1 a va

a) Gia su c = xa + yb Khi do ta c6 he •

Jiuang ddn gidi

3 x - y = 4 2x + 5y = 7

X = 27

y = b) Gia su V = (x;y) Khi do V 1 a <=> v.a = 0 <=> 3x + 2y = 0

x2+y2=2 <=> <

y = x2 + r-3x^2 = 2

3x

y = ^ 13x2 = 8

m Giai hf, ta dugc hai nghi^m la: ^2726 3726^

2-28 Cho hai diem A(-3; 4), B(l; 5), C(0; 4)

3) Tinh toa dp ciia AB, BC + 2AB

b)Tinh IABI,IBC + 2ABI ' ' ' ' '

^)Tinh so do cac goc cua tam giac ABC

KM tUn'mv Uf-c: Itir.h hoc 10

Jludrng ddn gidi

Co hai vecto don vj ciing phuong voi AB = (1;1) la

2 ' 2 va V2 = r ^/2 V2 2 ' 2 ^ 2.30 Cho hai diem A(3; 4) va B ( - l ; 5) >

a) Tun diem M tren tryc hoanh sao cho khoang each tir do de'n hai diem A va

B bang nhau

b) Tim diem N tren tryc tung sao cho NAB = 45° '

JJucrng ddn gidi a) Gia six M tren trvc hoanh Khi do M(a ; 0), A(3; 4), B ( - l ; 5)

a) Tim diem M tren tryc hoarJi sao cho tam giac MAB vuong tai A

b) Tim diem N sao cho tam giac NAB nh^in goc tpa dp O lam tryc tam

Giai he ta duoc nghiem x = 6,y = 2 Vay c6 diem N(6;2)

a) Xac dinh toa do eac diem A va C sao cho OABC la hinh vuong nhan OB lam duong cheo

b) Tim hai diem M tren tryc hoanh va N tren tryc tung sao cho hirJi chieu cua

O len M N la diem B ':^:p: 'y v_ ^ •••^; , r ; ^

x 2+ ( 2 x - 5) 2= 1 0 [5x2-20x+15=0 Giai h$ tren ta dupe nghiem (x; y) = (3; 1), (x; y) = (1 ;-3)

xet: Dung dieu kien OB.BM = 0, OB.BN = 0 ta ciing tim dmc M, N