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Phng php ta trong mt phng

Gio vin: Nguyn Trung Ngha - THPT chuyn Quc Hc Hu 1

MC LC

Trang

Tm tt kin thc 2

Cc bi ton v im v ng thng 4

Cc bi ton v tam gic 6

Cc bi ton v hnh ch nht 13

Cc bi ton v hnh thoi 16

Cc bi ton v hnh vung 17

Cc bi ton v hnh thang, hnh bnh hnh 19

Cc bi ton v ng trn 21

Cc bi ton v ba ng conic 31

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TM TT KIN THC

1. Phng trnh ng thng

ng thng i qua im ( );o oA x y v c VTCP ( );u a b= c PTTS l = += +

o

o

x x at

y y bt.

ng thng i qua im ( );o oA x y v c VTPT ( )=

;n a b c PTTQ l ( ) ( ) + = 0o oa x x b y y . ng thng i qua hai im ( );A AA x y v ( );B BB x y c phng trnh: =

A A

B A B A

x x y y

x x y y.

ng thng i qua hai im ( );0A a v ( )0;B b vi 0a v 0b c phng trnh: + =1x ya b

.

ng thng song song hoc trng vi Oy c phng trnh l ( )+ = 0 0ax c a . ng thng song song hoc trng vi Ox c phng trnh l ( )+ = 0 0by c b . ng thng i qua gc ta O c phng trnh l + = 0ax by ( )2 2 0a b+ . nu (d) vung gc vi + + =( ') : 0d ax by c th (d) c phng trnh l + = 0bx ay m . nu (d) song song vi + + =( ') : 0d ax by c th (d) c phng trnh l ( )+ + = 0 ax by m m c . ng thng c h s gc k c phng trnh l = +y kx b . ng thng i qua im ( );o oA x y v c h s gc k c phng trnh l ( ) = o oy y k x x . = +( ) :d y kx b vung gc vi = + = ( ') : ' ' . ' 1d y k x b k k . = +( ) :d y kx b song song vi = + =( ') : ' ' 'd y k x b k k .

2. Khong cch v gc

khong cch t ( );o oA x y n + + =( ) : 0ax by c tnh bi cng thc: ( ) + + =+2 2

,o oax by c

d Aa b

M, N cng pha i vi ng thng + + =( ) : 0ax by c ( )( ) + + + + > 0M M N Nax by c ax by c M, N khc pha i vi ng thng + + =( ) : 0ax by c ( )( ) + + + + < 0M M N Nax by c ax by c cho hai ng thng + + =( ) : 0ax by c v + + =( ') : ' ' ' 0a x b y c th:

phng trnh hai ng phn gic ca cc gc to bi v ' l + + + += + +2 2 2 2

' ' '

' '

ax by c a x b y c

a b a b

( ) + =+ +2 2 2 2

' 'cos ; '

. ' '

aa bb

a b a b

+ =' ' ' 0aa bb .

3. ng trn ng trn (C) tm ( );o oT x y , bn knh R c phng trnh l ( ) ( ) + =2 2 2o ox x y y R . phng trnh + + + + =2 2 2 2 0x y ax by c vi + >2 2 0a b c l phng trnh ca mt ng trn

vi tm ( ) ;T a b v bn knh = + 2 2R a b c . cho ng thng + + =( ) : 0ax by c v ng trn (C) c tm ( );o oT x y v bn knh R . Lc :

( ) tip xc (C) ( ) + + = =+2 2

;o oax by c

d T R Ra b

.

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4. ng elip

x

y

F2F 1 O

M

nh ngha: ( ) { }= + =1 2| 2E M MF MF a

Phng trnh chnh tc:

( ) ( )+ = <



CC BI TON V IM V NG THNG

B04: Cho hai im A(1; 1), B(4; 3). Tm im C thuc ng thng =2 1 0x y sao cho khong cch t C n ng thng AB bng 6.

S: C C1 243 27

(7;3), ;11 11

A06: Cho cc ng thng ln lt c phng trnh: + + = = =1 2 3: 3 0, : 4 0, : 2 0d x y d x y d x y . Tm to im M nm trn ng thng d3 sao cho khong cch t M n ng thng d1 bng hai ln khong cch t M n ng thng d2. S: M(22; 11), M(2; 1) B11: Cho hai ng thng : 4 0x y = v : 2 2 0d x y = . Tm ta im N thuc ng thng d sao cho ng thng ON ct ng thng ti im M tha mn . 8OMON = .

S: ( )0; 2N hoc 6 2;5 5

N

Ton hc & Tui tr: Cho ng thng : 2 2 0d x y = v hai im A(0 ; 1) v B(3 ; 4). Tm ta ca im M trn d sao cho 2 22MA MB+ nh nht.

S: M(2 ; 0) chuyn H Vinh: Cho hai im A(1 ; 2) v B(4 ; 3). Tm ta im M sao cho o135AMB = v khong cch t im M n ng thng AB bng 10

2.

S: ( )0;0M hoc ( )1;3M D10: Cho im A(0; 2) v l ng thng i qua O. Gi H l hnh chiu vung gc ca A trn . Vit phng trnh , bit khong cch t H n trc honh bng AH.

S: 2 ng : ( ) x y5 1 2 5 2 0 = B04(d b): Cho im I(2; 0) v hai ng thng d x y d x y1 2: 2 5 0, : 3 0 + = + = . Vit phng trnh ng thng d i qua im I v ct hai ng thng d1, d2 ln lt ti A, B sao cho IA IB2=

.

S: : 7 3 14 0d x y + + = Ton hc & Tui tr: Cho hai ng thng 1 2: 1 0; : 2 1 0d x y d x y+ + = = . Lp phng trnh ng thng d i qua ( )1; 1M v ct 1 2;d d ln lt ti A v B sao cho 2MB MA=

.

S: : 1d x = Ton hc & Tui tr: Cho hai im ( ) ( )2;5 , 5;1A B . Vit phng trnh ng thng d i qua A sao cho khong cch t B n d bng 3.

S: : 7 24 134 0d x y+ = Ton hc & Tui tr: Cho im ( )3;4M v hai ng thng 1 : 2 3 0d x y = v 2 : 0d x y = . Vit phng trnh ng thng d i qua M ct 1d ti A, ct 2d ti B sao cho 2MA MB= v im A c tung dng. chuyn Phan Bi Chu - Ngh An: Cho ba im A(1 ; 1), B(3 ; 2) v C(7 ; 10). Vit phng trnh ng thng i qua A sao cho tng khong cch t B v C n l ln nht.

S: : 4 5 9 0d x y+ = chuyn H Long - Qung Ninh: Cho tam gic ABC c nh A(0 ; 4), trng tm ( )4 / 3;2 / 3G v trc tm trng vi gc ta . Tm ta B, C bit B Cx x< .

S: ( ) ( )1; 1 , 5; 1B C

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ng Thc Ha - Ngh An - 2013: ( ) ( ) ( ) + =2 2: 1 2 10C x y c tm l I. Vit phng trnh ng thng d cch O mt khong bng 5 v ct (C) ti hai im phn bit A, B sao cho din tch tam gic IAB ln nht. S: =: 2 5 0d x y S GD&T Vnh Phc - 2014: Cho hai ng thng + =1 : 2 3 0d x y v =2 : 2 1 0d x y ct nhau ti. Vit phng trnh ng thng d i qua O v ct 1 2,d d ln lt ti A, B sao cho 2IA=IB. S: =: 3 4 0d x y hoc =: 0d x chuyn H Vinh - 2013: Cho hai ng thng = + =1 2: 2 0, : 2 2 0d x y d x y . Gi I l giao im ca 1 2,d d . Vit phng trnh ng thng i qua M(-1;1) ct 1 2,d d ln lt ti A, B sao cho AB = 3IA. S: + = 0x y hoc 7 6 0x y+ = chuyn Nguyn Quang Diu - ng Thp - 2014: Cho im A(0;2) v ng thng : 2 2 0.d x y + = Tm trn d 2 im M, N sao cho tam gic AMN vung ti A v AM=2AN, bit honh v tung ca N l nhng s nguyn. S: M(2;2), N(0;1) chuyn L T Trng - Cn Th - 2014: Cho im A(4;-7) v ng thng : 2 4 0x y + = . Tm im B trn sao cho c ng ba ng thng

1 2 3, ,d d d tha mn khong cch t A n

1 2 3, ,d d d u bng 4

v khong cch t B n 1 2 3, ,d d d u bng 6.

S: ( )2;1B hoc 6 13;5 5

B

*****

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CC BI TON V TAM GIC 1. Tam gic thng 1.1. Tm ta ca im A04: Cho hai im A(0; 2) v ( ) 3; 1B . Tm ta trc tm v ta tm ng trn ngoi tip ca tam gic OAB. S: ( ) ( )H I3; 1 , 3;1 B08: Hy xc nh to nh C ca tam gic ABC bit rng hnh chiu vung gc ca C trn ng thng AB l im H(1; 1), ng phn gic trong gc A c phng trnh + =2 0x y v ng cao k t B c phng trnh + =4 3 1 0x y .

S: C 10 3;3 4

D10: Cho tam gic ABC c nh A(3; 7), trc tm l H(3; 1), tm ng trn ngoi tip l I(2; 0). Xc nh to nh C, bit C c honh dng. S: ( )C 2 65;3 + B11: Cho tam gic ABC c nh 1 ;1

2B

. ng trn ni tip tam gic ABC tip xc vi cc cnh BC,

CA, AB tng ng ti cc im D, E, F. Cho D(3 ; 1) v ng thng EF c phng trnh 3 0y = . Tm ta nh A, bit A c tung dng.

S: 133;3

A

D11: Cho tam gic ABC c nh ( )4;1B , trng tm ( )1;1G v ng thng cha phn gic trong ca gc A c phng trnh 1 0x y = . Tm ta cc nh A v C. S: ( ) ( )4;3 , 3; 1A C B13: Cho tam gic ABC c chn ng cao h t nh A l 17 1;

5 5H

, chn ng phn gic trong ca

gc A l ( )5;3D v trung im ca cnh AB l ( )0;1M . Tm ta nh C. S: ( )9;11C D13: Cho tam gic ABC c im ( )9 / 2;3 / 2M l trung im ca cnh AB, im ( )2;4H v ( )1;1I ln lt l chn ng cao k t B v tm ng trn ngoi tip tam gic ABC. Tm ta nh C. S: ( )1;6C D03(d b): Cho tam gic ABC c nh A(1; 0) v hai ng thng ln lt cha cc ng cao v t B v C c phng trnh tng ng l: x y x y2 1 0, 3 1 0 + = + = . Tnh din tch tam gic ABC. S: B C( 5; 2), ( 1;4) S 14= D04(d b): Cho im A(2; 3) v hai ng thng d x y d x y1 2: 5 0, : 2 7 0+ + = + = . Tm to cc im B trn d1 v C trn d2 sao cho tam gic ABC c trng tm G(2; 0). S: ( ) ( )1; 4 , 5;1B C A06(d b): Cho tam gic ABC c nh A thuc ng thng x yd : 4 2 0 = , cnh BC song song vi d. Phng trnh ng cao BH: x y 3 0+ + = v trung im ca cnh AC l M(1; 1). Tm to cc nh A, B, C.

S: A B C2 2 8 8; , ( 4;1), ;3 3 3 3

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B06(d b): Cho tam gic ABC c nh A(2; 1), ng cao qua nh B c phng trnh x y3 7 0 = v ng trung tuyn qua nh C c phng trnh x y 1 0+ + = . Xc nh to cc nh B v C ca tam gic. S: B(2; 3), C(4; 5) A07(d b): Cho tam gic ABC c trng tm G(2; 0), phng trnh cc cnh AB: x y4 14 0+ + = , AC: x y2 5 2 0+ = . Tm to cc nh A, B, C.

S: A(4; 2), B(3; 2), C(1; 0) Ton hc & Tui tr: Cho tam gic ABC bit ba chn ng cao tng ng vi ba nh A, B, C ln lt l ( )' 1;1A , ( )' 2;3B v ( )' 2;4C . Vit phng trnh cnh BC.

S: 2 3 3 1 5 2 013 10 13 10 13 10

x

+ + + =

Ton hc & Tui tr: Cho tam gic ABC c : 5 2 7 0; : 2 1 0AB x y BC x y+ + = = . Phng trnh ng phn gic trong gc A l 1 0x y+ = . Tm ta im C.

S: 11 4;3 3

C

Ton hc & Tui tr: Cho tam gic ABC bit C(4 ; 3). ng phn gic trong v trung tuyn k t nh A ca tam gic ln lt c phng trnh 2 5 0x y+ = v 4 13 10x y+ . Tm ta im B.

S: ( )12;1B Ton hc & Tui tr: Cho tam gic ABC bit ( )1;1A , trc tm H(1 ; 3), trung im ca cnh BC l im M(5 ; 5). Xc nh ta cc nh B v C ca tam gic ABC. ng Thc Ha - Ngh An: Cho tam gic ABC c : 2 3 0d x y = l ng phn gic trong gc A. Bit ( ) ( )1 16;0 , 4;4B C ln lt l hnh chiu vung gc ca B, C ln cc ng thng AC, AB. Xc nh ta ca A, B, C.

S: ( ) 21 21 31 11; 1 , ; , ;4 4 4 4

A B C

L Hng Phong - Thanh Ha: 1. Cho tam gic ABC c A(5 ; 2). Phng trnh ng trung trc on BC l 6 0x y+ = , trung tuyn CC l 2 3 0x y + = . Tm ta cc nh B, C. 2. Cho tam gic ABC c A(1 ; 5). Phng trnh : 2 6 0BC x y = . Tm ng trn ni tip I(1;0). Tm ta cc nh B, C. S: 1. ( ) ( )23/ 5;55/ 3 , 28 / 3; 14 / 3C B 2. ( ) ( )4; 1 , 4; 5B C

chuyn H Vinh: Cho tam gic ABC c trng tm G(1 ; 1); : 2 1 0d x y + = l phng trnh ca ng cao k t nh A. Cc nh B, C thuc ng thng : 2 1 0x y + = . Tm ta cc im A, B, C bit tam gic ABC c din tch bng 6.

S: ( ) ( ) ( )1;3 , 3; 1 , 1;1A B C hoc ( ) ( ) ( )1;3 , 3; 1 , 1;1A C B L Thi T - Bc Ninh: Cho tam gic ABC bit ng cao k t nh B v ng phn gic trong gc A ln lt c phng trnh l 1 2: 3 4 10 0; : 1 0d x y d x y+ + = + = . im M(0 ; 2) thuc ng thng AB ng thi cch C mt khong bng 2 . Tm ta cc nh ca tam gic ABC.

S: ( ) ( ) ( )4;5 , 3; 1/ 4 , 1;1A B C hoc ( )31/ 25;33/ 25C THPT Cu Xe: Cho tam gic ABC bit ng cao k t nh C v ng trung trc on BC ln lt l 2 0;3 4 2 0x y x y + = + = . im ( )4; 2A . Tm ta cc nh B, C.

S: ( ) ( )1/ 4;9 / 4 , 7 / 4;1/ 4B C

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THPT Triu Sn 4: Cho tam gic ABC bit ng cao k t nh A v ng phn gic trong gc B ln lt c phng trnh l 2 2 0; 1 0x y x y = = . Tm ta cc nh ca tam gic ABC, bit M(0 ; 2) thuc ng thng AB v AB = 2BC.

S: ( ) ( ) ( )3;1/ 2 , 2;1 , 7 / 4;3/ 2A B C Qunh Lu 2 - Ngh An: Cho tam gic ABC c din tch bng 12 6 6+ , ( ) ( )2;0 , 4;0A B , bn knh ng trn ngoi tip bng 5. Tm ta im C bit tung ca C dng.

S: ( )0;4 2 6C + hoc ( )2;4 2 6C + chuyn Nguyn Quang Diu - ng Thp: Cho tam gic ABC c 5AB = , ( )1; 1C , ng thng

: 2 3 0AB x y+ = . Trng tm G thuc ng thng : 2 0d x y+ = . Tm ta ca A, B. S: ( ) ( )4; 1/ 2 , 6; 3 / 2A B hoc ( ) ( )4; 1/ 2 , 6; 3 / 2B A GSTT.VN - 2013: Cho tam gic ABC c M(0;-1) nm trn cnh AC. Bit AB=2AM, ng phn gic trong gc A l : 0d x y = , ng cao i qua nh C l ' : 2 3 0d x y+ + = . Tm ta cc nh ca tam gic ABC.

S: ( ) ( )

11;1 , 3; 1 , ; 2

2A B C

ng Thc Ha - Ngh An - 2013: Cho tam gic ABC c 135oBAC = , ng cao : 3 10 0BH x y+ + = ,

trung im ca cnh BC l 1 3;2 2

M

v trc tm H(0;-10). Bit tung ca im B m. Xc nh ta cc nh ca tam gic ABC. ng Thc Ha - Ngh An - 2013: Cho tam gic ABC c trc tm H, : 4 0BC x y + = , trung im ca cnh AC l M(0;3), ng cao AH ct ng trn ngoi tip tam gic ABC ti N(7;-1). Xc nh ta cc nh ca tam gic ABC v vit phng trnh ng trn ngoi tip tam gic HBC. chuyn L Qu n - Qung Tr - 2013: Cho tam gic ABC c trng tm G(1;2), im M(-2;1) nm trn ng cao k t A. ng thng BC c phng trnh 1 0x y = . Tm ta im B bit 0Bx > v din tch tam gic ABC bng 24. S: B(7;6) chuyn H Vinh - 2013: Cho tam gic ABC c A(-1;-3), B(5;1). im M nm trn on thng BC sao cho MC=2MB. Tm ta im C bit rng MA = AC = 5 v ng thng BC c h s gc l mt s nguyn. S: C(-4;1) Ton hc & Tui tr - 2014: Cho tam gic ABC c A(1;2), trng tm G(1;1) v trc tm 2 10;

3 3H

.

Tm ta hai nh B v C ca tam gic. S: B(-1;0) v C(3;1) Hng Quang - Hi Dng - 2014: Cho tam gic ABC c din tch bng 2. Phng trnh ca ng thng AB l 0x y = . im M(2;1) l trung im ca cnh BC. Tm ta trung im N ca cnh AC. S: B(3;2) v C(1;0) S GD&T Vnh Phc - 2014: Cho tam gic ABC c nh C(5;1), M l trung im ca BC, im B thuc ng thng : 6 0d x y+ + = . im N(0;1) l trung im ca AM, im D(-1;-7) khng nm trn ng thng AM v khc pha vi A so vi ng thng BC, ng thi khong cch t A v D ti ng thng BC bng nhau. Xc nh ta cc im A, B. S: B(-3;-3) v A(-1;3)

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chuyn Nguyn nh Chiu - ng Thp - 204: Cho tam gic ABC c ( ) ( ) ( )0;2 3 , 2;0 , 2;0A B C v BH l ng cao. Tm ta ca im M, N trn ng thng cha ng cao BH sao cho ba tam gic MBC, NBC v ABC c chu vi bng nhau.

S: 8 24 3 24 6 3 8 24 3 24 6 3; , ;13 13 13 13

M N

+ + +

chuyn H Vinh - 204: Cho tam gic ABC c phng trnh ng thng cha ng cao k t B l 3 18 0x y+ = , phng trnh ng thng trung trc ca BC l 3 19 279 0.x y+ = nh C thuc ng

thng : 2 5 0.d x y + = Tm ta nh A bit rng 135 .oBAC = S: A(4;8) chuyn L T Trng - Cn Th - 2014: Cho tam gic ABC c H(1;1) l chn ng cao k t nh A. im M(3;0) l trung im ca cnh BC v .BAH HAM MAC= = Tm ta cc im A, B, C.

S: ( ) ( ) ( )1 3;1 2 3 , 1;2 , 7; 2A B C HSP H Ni - 2014: Cho tam gic ABC c AC>AB, C(6;0) v hai ng thng : 3 10 0d x y = , : 3 3 16 0.x y + = Bit rng ng thng d cha ng phn gic trong ca gc A, ng thng

vung gc vi cnh AC v ba ng thng , d v trung trc ca cnh BC ng qui ti mt im.

S: 4 2;3 3

B

chuyn H Vinh - 204: Cho tam gic ABC c M(2;1) l trung im cnh AC, im H(0;-3) l chn ng cao k t A, im E(23;-2) thuc ng thng cha trung tuyn k t C. Tm ta im B bit im A thuc ng thng : 2 3 5 0d x y+ = v im C c honh dng. S: ( )3; 4B Nguoithay.vn - 2014: Cho tam gic ABC c A(1;5), im B nm trn ng thng

1: 2 1 0d x y+ + = v

chn ng cao h t nh B xung ng thng AC nm trn ng thng 2: 2 8 0d x y+ = . Bit

M(3;0) l trung im ca cnh BC. Tm ta ca cc im B v C. 1.2. Vit phng trnh ng thng D09: Cho tam gic ABC c M(2; 0) l trung im ca cnh AB. ng trung tuyn v ng cao qua nh A ln lt c phng trnh l = =7 2 3 0, 6 4 0x y x y . Vit phng trnh ng thng AC.

S: AC x y: 3 4 5 0 + = chuyn Phan Bi Chu - Ngh An: Cho tam gic ABC c trc tm ( )1;4H , tm ng trn ngoi tip l ( )3;0I v trung im ca cnh BC l ( )0; 3M . Vit phng trnh ng thng AB bit B c honh dng.

S: : 3 7 49 0AB x y+ = chuyn H Ni - Amsterdam: Cho tam gic ABC v im ( )0; 1M . Phng trnh ng phn gic trong ca gc A v ng cao k t C ln lt l 0; 2 3 0x y x y = + + = . ng thng AC i qua M v AB = 2AM. Vit phng trnh cnh BC.

S: : 2 5 11 0BC x y+ + =

Ton hc & Tui tr - 2013: Cho tam gic ABC c C(5;4), ng thng : 2 11 0d x y + = i qua A v song song vi BC, ng phn gic trong AD c phng trnh 3 9 0x y+ = . Vit phng trnh cc cnh cn li ca tam gic ABC.

S: + = + = + =: 2 13 0, : 2 3 0, : 2 4 0AC x y BC x y AB x y

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Ton hc & Tui tr - 2014: Cho tam gic ABC c A(-1;3), trng tm G(2;2). Bit im B, C ln lt l thuc cc ng thng : 3 3 0d x y+ = v ' : 1 0d x y = . Vit phng trnh ng thng i qua A c h s gc dng sao cho tng khong cch t B v C n l ln nht.

S: + =: 3 6 0x y chuyn Nguyn nh Chiu - ng Thp - 2014: Cho tam gic ABC c phng trnh ng cao AH l

3 3.x = Phng trnh ng phn gic trong gc ABC , ACB ln lt l 3x y , 3 6 3 0.x y+ = Bn knh ng trn ni tip tam gic ABC bng 3. Vit phng trnh cc cnh ca tam gic ABC, bit nh A c tung dng.

S: : 3 18 0, : 0, : 3 0AC y x BC y AB y x+ = = = 2. Tam gic cn 2.1. Tm ta ca im B03: Cho tam gic ABC c = ,AB AC = 90oBAC . Bit M(1; 1) l trung im cnh BC v ( )2/ 3; 0G l trng tm tam gic ABC. Tm ta cc nh A, B, C. S: A(0; 2), B(4; 0), C(2; 2) B09: Cho tam gic ABC cn ti A c nh A(1; 4) v cc nh B, C thuc ng thng : =4 0x y . Xc nh to cc im B v C, bit din tch tam gic ABC bng 18.

S: B C11 3 3 5; , ;2 2 2 2

hoc B C3 5 11 3; , ;

2 2 2 2

A10: Cho tam gic ABC cn ti A c nh A(6; 6); ng thng i qua trung im ca cc cnh AB v AC c phng trnh + =4 0x y . Tm to cc nh B v C, bit im E(1; 3) nm trn ng cao i qua nh C ca tam gic cho. S: B(0; 4), C(4; 0) hoc B(6; 2), C(2; 6) A05(d b): Cho tam gic ABC cn ti nh A c trng tm G 4 1;

3 3

, phng trnh ng thng BC l

x y2 4 0 = v phng trnh ng thng BG l x y7 4 8 0 = .Tm ta cc nh A, B, C. S: A(0; 3), B(0; 2), C(4; 0) chuyn L T Trng - Cn Th: Cho tam gic ABC cn ti B, c : 3 2 3 0AB x y = . Tm ng trn ngoi tip tam gic ABC l I(0 ; 2). im B thuc trc Ox. Tm ta im C.

S: ( )3 1;1 3C Qunh Lu 1 - Ngh An: Cho tam gic ABC cn ti A c : 2 2 0; : 2 1 0AB x y AC x y+ = + + = , im M(1 ; 2) thuc on BC. Tm ta im D sao cho .DB DC

nh nht.

S: D(0 ; 3) Nguyn c Mu - Ngh An: Cho tam gic ABC cn ti A, nh B thuc : 4 2 0d x y = , cnh AC song song vi d. ng cao k t nh A c phng trnh 3 0x y+ + = , im M(1 ; 1) nm trn AB. Tm ta cc nh ca tam gic ABC.

S: ( ) ( ) ( )0; 3 , 2 / 3; 1 / 3 , 8 / 3; 11 / 3A B C chuyn Phan Bi Chu - Ngh An - 2013: Cho tam gic ABC cn ti A. Gi D l trung im ca AB.

Bit rng 11 5;3 3

I

v 13 5;3 3

E ln lt l tm ng trn ngoi tip tam gic ABC, trng tm tam

gic ADC. Cc im M(3;-1), N(-3;0) ln lt thuc cc ng thng DC, AB. Tm ta cc im A, B, C bit A c tung dng.

S: ( ) ( ) ( ) 7;5 , 1;1 , 3; 3A B C

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chuyn H Vinh - 2013: Cho tam gic ABC cn ti A, c trc tm H(-3;2). Gi D, E l chn ng cao k t B v C. Bit rng im A thuc ng thng : 3 3 0d x y = , im F(-2;3) thuc ng thng DE v HD=2. Tm ta im A.

S: ( )3;0A Lng Th Vinh - H Ni - 2014: Cho tam gic ABC cn ti A. Gi N l trung im ca AB. Gi E v F ln lt l chn ng cao h t cc nh B, C ca tam gic ABC. Tm ta ca nh A bit rng

E(7;1), 11 13;5 5

F

v phng trnh ng thng CN l 2 13 0.x y+ =

S: ( )7;9A 2.2. Vit phng trnh ng thng B06(d b): Cho tam gic ABC cn ti B, vi A(1; 1), C(3; 5). im B nm trn ng thng d x y: 2 0 = . Vit phng trnh cc ng thng AB, BC. S: AB: x y23 24 0 = , BC: x y19 13 8 0 + = Ton hc & Tui tr: Cho hai ng thng 1 : 2 1 0d x y + = v 2 : 2 7 0d x y+ = . Lp phng trnh ng thng i qua gc ta O v to vi 1 2;d d mt tam gic cn c y thuc ng thng .

S: 118

3 8 0;5

x y S + = = hoc 232

3 6 0;5

x y S+ = =

Ton hc & Tui tr: Cho tam gic ABC cn ti A. Bit : 2 1 0; : 4 3 0AB x y BC x y+ = + + = . Vit phng trnh ng cao k t nh B ca tam gic ABC.

S: 31 22 9 0x y+ = Ton hc & Tui tr: Cho hai ng thng 1 2: 3 3 0; : 3 3 2 0d x y d x y = + = ct nhau ti A. Lp phng trnh ng thng d ct 1 2;d d ln lt ti B v C sao cho tam gic ABC u c din tch bng 3 3 . 3. Tam gic vung 3.1. Tm ta ca im

A02: Xt tam gic ABC vung ti A, phng trnh ng thng BC l =3 3 0x y , cc nh A v B thuc trc honh v bn knh ng trn ni tip bng 2. Tm ta trng tm G ca tam gic ABC.

S: G17 4 3 6 3

;3 3

+ +

, G24 3 1 6 2 3

;3 3

D04: Cho tam gic ABC c cc nh A(1; 0), B(4; 0), C(0; m) vi 0m . Tm ta trng tm G ca tam gic ABC theo m. Xc nh m tam gic GAB vung ti G.

S: mG m1; , 3 63

=

B07: Cho im A(2; 2) v cc ng thng: + = + =1 2: 2 0, : 8 0d x y d x y . Tm to cc im B v C ln lt thuc d1 v d2 sao cho tam gic ABC vung cn ti A. S: B(1; 3), C(3; 5) hoc B(3; 1), C(5; 3) D04(d b): Cho tam gic ABC vung A. Bit A(1; 4), B(1; 4), ng thng BC i qua im K 7 ;2

3

.

Tm to nh C. S: ( )3;5C D07(d b): Cho im A(2; 1). Trn trc Ox, ly im B c honh Bx 0 , trn trc Oy, ly im C c

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tung Cy 0 sao cho tam gic ABC vung ti A. Tm cc im B, C sao cho din tch tam gic ABC ln nht. S: B(0; 0), C(0; 5) D07(d b): Cho cc im A(0; 1), B(2; 1) v cc ng thng + + =1 : ( 1) ( 2) 2 0d m x m y m , + + =2 : (2 ) ( 1) 3 5 0d m x m y m Chng minh d1 v d2 lun ct nhau. Gi P l giao im ca d1 v d2. Tm m sao cho PA + PB ln nht. S: Ch : PA PB PA PB B2 2 2 2( ) 2( ) 2A 16+ + = = . Do max(PA+PB)=4 khi P l trung im ca

cung AB. Khi P(2; 1) hay P(0; 1) m = 1 hoc m = 2. Ton hc & Tui tr: Cho tam gic ABC vung ti A. ng thng : 4 3 4 0BC x y = . Cc nh A, B thuc trc honh v din tch tam gic ABC bng 6. Tm ta trng tm G ca tam gic ABC. Ton hc & Tui tr -2012: Cho tam gic ABC vung ti A, cc nh A, B thuc trc honh v din tch tam gic ABC bng 6. ng thng BC c phng trnh l 4 3 4 0x y = . Tm ta trng tm G ca tam gic ABC.

S:

4 43; , 1;3 3

G G

chuyn Nguyn Quang Diu - ng Thp: Cho ( )1;2A v ng thng : 2 3 0d x y + = . Tm trn d hai im B v C sao cho tam gic ABC vung ti C v AC = 3BC.

S: 3 6;5 5

C

v 13 16;15 15

B

hoc 1 4;3 3

B

chuyn H Ni - Amsterdam: Cho tam gic ABC vung cn ti A . ng thng : 7 31 0BC x y+ =

. im 51;2

N

thuc ng thng AC, im ( )2; 3M thuc ng thng AB. Xc nh ta cc nh ca tam gic ABC.

S: ( ) ( ) ( )1;1 , 4;5 , 3;4A B C Nguoithay.vn - 2014: Cho tam gic ABC vung cn ti A c I l trung im ca cnh BC. Gi M l

trung im ca IB v N l im nm trn on thng IC sao cho NC=2NI. Bit rng 11; 42

M

, phng

trnh ng thng AN l 2 0x y = v im A c honh m. Tm ta cc nh ca tam gic ABC. 3.2. Vit phng trnh ng thng B10: Cho tam gic ABC vung ti A, c nh C(4; 1), phn gic trong gc A c phng trnh

+ =5 0x y . Vit phng trnh ng thng BC, bit din tch tam gic ABC bng 24 v nh A c honh dng.

S: BC: x y3 4 16 0 + =

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CC BI TON V HNH CH NHT

1. Tm ta ca im

B02: Cho hnh ch nht ABCD c tm

1I ; 02

, phng trnh ng thng AB l x 2y + 2 = 0 v AB =

2AD. Tm ta cc nh A, B, C, D bit rng nh A c honh m. S: A(2; 0), B(2; 2), C(3; 0), D(1; 2) D12: Cho hnh ch nht ABCD. Cc ng thng AC v AD ln lt c phng trnh l 3 0x y+ = v

4 0x y + = . ng thng BD i qua im ( )1 / 3;1M . Tm ta cc nh ca hnh ch nht. S: ( ) ( ) ( ) ( )3;1 , 3; 1 , 1;3 , 1; 3A C D B A13: Cho hnh ch nht ABCD c im C thuc ng thng : 2 5 0d x y+ + = v ( )4;8A . Gi M l im i xng ca B qua C, N l hnh chiu vung gc ca B trn ng thng MD. Tm ta cc im B, C bit rng ( )5; 4N . S: ( ) ( ) 1; 7 , 4; 7C B Ton hc & Tui tr: Cho hnh ch nht ABCD bit : 2 1 0; : 7 14 0AB x y BD x y = + = . ng cho AC i qua im M(2 ; 1). Tm ta cc nh ca hnh ch nht. S: ( ) ( ) ( ) ( )1;0 , 7;3 , 6;5 , 0;2A B C D Lng 4 - Ngh An: Cho hnh ch nht ABCD c din tch bng 12, tm I thuc ng thng

: 3 0d x y = v 92

Ix = , trung im ca mt cnh l giao im ca d v trc Ox. Tm ta cc nh

ca hnh ch nht. S: ( ) ( ) ( ) ( )2;1 , 5;4 , 7;2 , 4; 1A B C D Nguyn c Mu - Ngh An: Cho hnh ch nht ABCD c din tch bng 16, phng trnh ng thng : 3 0AB x y + = , im I(1 ; 2) l giao im ca hai ng cho. Tm ta cc nh ca hnh ch nht.

S: ( ) ( ) ( ) ( )2;5 , 2;1 , 0; 1 , 4;3A B C D hoc ( ) ( ) ( ) ( )2;5 , 2;1 , 0; 1 , 4;3B A D C Lng Th Vinh - H Ni - 2012: Cho hnh ch nht ABCD c din tch bng 12, tm 9 3;

2 2I

v trung

im ca cnh AD l M(3;0). Tm ta cc nh ca hnh ch nht. S: ( ) ( ) ( ) ( )2;1 , 5;4 , 7;2 , 4; 1A B C D

ng Thc Ha - Ngh An - 2013: Cho hnh ch nht ABCD c cc cnh AB, DA tip xc vi ng

trn ( ) ( ) ( )2 2: 2 3 4C x y+ + = , ng cho AC ct (C) ti cc im 16 23;5 5

M v N thuc trc Oy.

Tm ta cc nh ca hnh ch nht ABCD, bit im A c honh m, im D c honh dng v din tch tam gic AND bng 10.

S: ( ) ( ) ( ) ( ) 4;5 , 4;0 , 6;0 , 6;5A B C D chuyn HKHTN H Ni - 2013: Cho hnh ch nht ABCD c din tch bng 12. Tm I ca hnh ch nht l giao im ca hai ng thng 1 : 3 0d x y = v 2 : 6 0d x y+ = . trung im ca mt cnh l giao im ca 1d vi trc honh. Tm ta cc nh ca hnh ch nht ABCD. chuyn Nguyn Quang Diu - ng Thp - 2014: Cho hnh ch nht ABCD c din tch bng 6, ng cho : 2 9 0AC x y+ = . im M(0;4) nm trn cnh BC, ng thng CD i qua im N(2;8). Tm ta cc nh ca hnh ch nht ABCD bit nh C c tung l mt s nguyn.

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S: ( ) ( ) ( ) ( )3;3 , 2;2 , 1;5 , 0;6A B C D chuyn Nguyn Quang Diu - ng Thp - 2014: Cho hnh ch nht ABCD c hai nh B, C thuc trc tung. ng cho : 3 4 16 0AC x y+ = . Bn knh ng trn ni tip tam gic ACD bng 1. Tm ta cc nh ca hnh ch nht.

S: ( ) ( ) ( ) ( )4;1 , 0;1 , 0;4 , 4;4A B C D hoc ( ) ( ) ( ) ( ) 4;7 , 0; 7 , 0;4 , 4;4A B C D chuyn Nguyn Quang Diu - ng Thp - 2014: Cho hnh ch nht ABCD c din tch bng 48, nh D(-3;2). ng phn gic ca gc BAD c phng trnh 7 0x y+ = . Tm ta nh B bit im A c honh dng.

S: ( )5;8B Hng Quang - Hi Dng - 2014: Cho hnh ch nht ABCD c nh C(3;-1). Gi M l trung im ca cnh BC, ng thng DM c phng trnh 1 0y = . Bit nh A thuc ng thng : 5 7 0d x y + = v im D c honh m. Tm ta cc nh A v D.

S: ( )2 ;5 , 2;15

A D

S GD&T Bc Ninh - 2014: Cho hnh ch nht ABCD c : 2 1 0AD x y+ = , im I(-3;2) thuc BD sao cho 2IB ID=

. Tm ta cc nh ca hnh ch nht bit 0Dx > v 2AD AB= . S: ( ) ( ) ( ) ( )5;11 , 11;8 , 5; 4 , 1; 1A B C D

S GD&T Bc Ninh - 2014: Cho hnh ch nht ABCD c AD = 2AB. Gi M, N ln lt l trung im ca AD, BC. Trn ng thng MN ly im K sao cho N l trung im ca ca MK. Tm ta cc nh ca hnh ch nht bit K(5;-1), : 2 3 0AC x y+ = v 0Ay > .

S: ( ) ( ) ( ) ( )1;1 , 3;1 , 3; 3 , 1; 3A B C D Can Lc - H Tnh - 2014: Cho hnh ch nht ABCD c AB = 2AD. Gi N l trung im ca cnh BC, M l im thuc cnh CD sao cho DC=4DM. Bit ta M(1;2), phng trnh ng thng AN l 4 5 0.x y + = Tm ta nh A bit 0,5Ax < .

S: ( )1;1A Ton hc & Tui tr - 2014: Cho hnh ch nht ABCD c B(1;1). Trng tm ca tam gic ABC nm trn ng thng : 3 2 0.d x y = im N(4;6) l trung im ca cnh CD. Tm ta nh A.

S: ( ) 9 571;3 , ;5 5

A A

Nguoi thay.vn - 2014: Cho hnh ch nht ABCD c hai im E, F ln lt nm trn cc cnh AB, AD sao cho EB=2EA, FA=3FD. Bit rng F(2;1), phng trnh ng thng CE l 3 9 0x y = , tam gic CEF vung ti F v im C c honh dng. Tm ta cc nh ca hnh ch nht ABCD. Nguoi thay.vn - 2014: Cho hnh ch nht ABCD c din tch bng 30 v nh B nm trn ng thng : 2 2 0d x y = . Trung im ca AB l M(4;3) v im N(1;-3) nm trn ng thng CD. Tm ta

cc nh ca hnh ch nht ABCD, bit im B c tung dng. Nguoi thay.vn - 2014: Cho hnh ch nht ABCD c din tch bng 30 v hai im M(1;4), N(-4;-1) ln lt nm trn cc ng thng AB, AD. Phng trnh ng cho AC l 7 4 13 0.x y+ = Tm ta cc nh ca hnh ch nht ABCD bit im A v D u c honh m. 2. Vit phng trnh ng thng A09: Cho hnh ch nht ABCD c im I(6; 2) l giao im ca hai ng cho AC v BD. im M(1; 5) thuc ng thng AB v trung im E ca cnh CD thuc ng thng : + =5 0x y . Vit phng trnh ng thng AB. S: y x y5 0, 4 19 0 = + =

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Ton hc & Tui tr: Cho hnh ch nht, hai ng cho ln lt c phng trnh l + =1 : 7 4 0d x y + =2; : 2 0d x y . Vit phng trnh ng thng cha cnh ca hnh ch nht bit n i qua im

( )3;5M . S: 3 12 0x y = hoc 3 14 0x y + =

Ton hc & Tui tr: Cho hnh ch nht ABCD c din tch bng 6, : 2 12 0BD x y+ = . ng thng AB i qua im M(5 ; 1), ng thng BC i qua N(9 ; 3). Vit phng trnh cc cnh ca hnh ch nht, bit im B c honh ln hn 5.

S: : 6 0; : 6 0; : 0; : 8 0AB x y BC x y AD x y CD x y+ = = = + = hoc : 6 0; : 6 0; : 12 0; : 4 0AB x y BC x y AD x y CD x y+ = = = + =

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CC BI TON V HNH THOI

1. Tm ta ca im Lng Ti 2 - Bc Ninh: Cho ABCD l hnh thoi vi AC = 2BD, tm I(2 ; 1). im ( )0;1/ 3M thuc ng thng AB, im N(0 ; 7) thuc ng thng CD. Tm ta nh B bit B c honh dng.

S: ( )1; 1B chuyn Quc Hc Hu: Trong mt phng vi h ta Oxy, cho hnh thoi ABCD c AC = 2BD. Bit ng thng AC c phng trnh 2 1 0x y = ; nh ( )3;5A v im B thuc ng thng

+ =: 1 0d x y . Tm ta cc nh B, C, D ca hnh thoi ABCD.

S: ( ) ( ) ( )1;2 , 3;0 , 1; 3B D C hoc ( ) 13 4 13 313; 2 , ; , ;5 5 5 5

B D C

Thun Thnh 3 - Bc Ninh - 2014: Cho hnh thoi ABCD c phng trnh cnh BD l 0x y = , ng thng AB i qua im ( )1; 3P , ng thng CD i qua ( )2; 2 3Q . Tm ta cc nh ca hnh thoi, bit AB AC= v im B c honh ln hn 1.

S: ( ) ( ) ( ) ( ) 1 3; 3 1 , 2;2 , 3 1; 1 3 , 4; 4A B C D Lng Giang 1 - Bc Giang: Cho hnh thoi ABCD c phng trnh cnh AC l 7 31 0x y+ = , hai nh B, D ln lt thuc cc ng thng + =1 : 8 0d x y v + =2 : 2 3 0d x y . Tm ta cc nh ca hnh thoi bit din tch ca hnh thoi bng 75 v nh A c honh m.

S: ( ) ( ) ( ) ( ) 10;3 , 0;8 , 11;6 , 1;1A B C D GSTT.VN - 2013: Cho hnh thoi ABCD bit : 3 1 0; : 5 0AB x y BD x y+ + = + = . ng thng AD i qua im M(1;2). Tm ta cc nh ca hnh thoi.

S: ( ) ( )4;1 , 0;5B D S GD&T Vnh Phc - 2014: Cho hnh thoi ABCD c : 1 0AC x y+ = . im E(9;4) nm trn ng thng AB, im F(-2;-5) nm trn ng thng CD v 2 2AC = . Xc nh ta A, B, C, D bit im C c honh m.

S: ( ) ( ) ( ) ( )0;1 , 3;0 , 2;3 , 1;4A B C D 2. Vit phng trnh ng thng

Cho hnh thoi ABCD c tm I(3;3) v AC = 2BD. im 42;3

M

thuc ng thng AB, im

133;3

N

thuc ng thng CD. Vit phng trnh ng thng BD bit 3Bx < .

S GD&T Bc Ninh - 2014: Cho hnh thoi ABCD c 60oABC = , ng trn (C) c tm I bn knh R=2 tip xc vi tt c cc cnh ca hnh thoi (tip xc vi AB v CD ln lt ti M v N, tung ca I dng). Bit phng trnh ng thng : 3 1 0MN x y+ = , ng thng AD khng vung gc vi trc tung v i qua im P(3;0). Vit phng trnh ng thng AB, AD.

S: : 3 4 5 3 0; : 3 3 3 0AB x y AD x y + = + =

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CC BI TON V HNH VUNG

1. Tm ta ca im A05: Cho hai ng thng =1 : 0d x y v + =2 : 2 1 0d x y . Tm to cc nh hnh vung ABCD bit rng nh A thuc d1, nh C thuc d2 v cc nh B, D thuc trc honh. S: A(1; 1), B(0; 0), C(1; 1), D(2; 0) hoc A(1; 1), B(2; 0), C(1; 1), D(0; 0) A12: Cho hnh vung ABCD. Gi M l trung im ca cnh BC, N l im trn cnh CD sao cho

CN = 2ND. Gi s 11 1;2 2

M

v ng thng : 2 3 0AN x y = . Tm ta im A.

S: ( ) ( )1; 1 , 4;5A A Ton hc & Tui tr: Cho ba ng thng 1 2: 3 4 4 0; : 6 0d x y d x y = + = v 3 : 3 0d x = . Tm ta cc nh ca hnh vung ABCD bit A, C thuc 3d , B thuc 1d v C thuc 2d .

S: ( ) ( ) ( ) ( )3;3 , 2;2 , 1;3 , 4;2A B C D hoc ( ) ( ) ( ) ( )1;3 , 2;2 , 3;3 , 4;2A B C D chuyn Vnh Phc: Cho hnh vung ABCD c M l trung im ca cnh BC, phng trnh ng thng

: 2 0DM x y = v ( )3; 3C . Bit nh A thuc ng thng : 3 2 0d x y+ = . Tm ta cc im A, B, D.

S: ( ) ( ) ( )1;5 , 3; 1 , 5;3A B D T K - Hi Dng: Cho hnh vung ABCD c ( )2;6A , nh B thuc : 2 6 0d x y + = . Gi M, N ln lt l hai im trn hai cnh BC, CD sao cho BM = CN. Bit AM ct BN ti 2 14;

5 5I

. Xc nh ta

im C. S: C(0 ; 0) hoc C(4 ; 8)

Lng 4 - Ngh An: Cho hnh vung ABCD c tm 3 1;2 2

I

. Cc ng thng AB, CD ln lt i

qua ( )4; 1M , ( )2; 4N . Tm ta cc nh ca hnh vung bit im B c honh m. S: ( ) ( ) ( ) ( )2;3 , 1;1 , 1; 2 , 4;0A B C D chuyn H Long - Qung Ninh: Cho hnh vung ABCD c nh C(1 ; 2). Gi M l trung im ca BC. ng thng DM c phng trnh 2 7 0x y+ = . nh A thuc ng thng : 5 0d x y+ = . Tm ta A, B, D.

S: ( ) 1 17 1 151;6 , ; , ;2 4 2 4

A B D

ng Thc Ha - Ngh An: Cho hnh vung ABCD c nh A thuc : 4 0d x y = . ng thng BC, CD ln lt i qua M(4 ; 0) v N(0 ; 2). Bit tam gic AMN cn ti A, xc nh ta cc nh ca hnh vung. S: ( ) ( ) ( ) ( )1; 5 , 2; 2 , 1; 1 , 2; 4A B C D hoc ( ) ( ) ( ) ( )1; 5 , 5; 3 , 3;3 , 3;1A B C D S GD&T Vnh Phc: Cho ( ) ( ) ( )2 2: 2 3 10C x y + = ni tip hnh vung ABCD. Xc nh ta cc nh ca hnh vung bit ng thng cha cnh AB i qua im M(-3;-2) v im A c honh dng.

S: ( ) ( ) ( ) ( ) 6;1 , 0; 1 , 2;5 , 4;7A B C D

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chuyn Quc Hc Hu - 2014: Trong mt phng vi h ta Oxy, cho hnh vung ABCD. Gi M, N ln lt l trung im ca cc cnh AB v CD. Bit rng 1 ;2

2M

v ng thng BN c phng trnh

2 9 34 0x y+ = . Tm ta cc im A v B bit rng im B c honh m. S: ( ) ( )1;4 , 0;0B A chuyn Nguyn Quang Diu - ng Thp - 2014: Cho hnh vung ABCD c A(2;-4), nh C thuc ng thng : 3 2 0d x y+ + = . ng thng : 2 0DM x y = vi M l trung im ca AB. Tm ta cc nh B, C, D ca hnh vung, bit im C c honh m. S: ( ) ( ) ( ) 4; 2 , 2;4 , 4;2B C D S GD&T Vnh Phc - 2014: Cho hnh vung ABCD c : 3 0BD x y+ = , im M(-1;2) thuc ng thng AB, im N(2;-2) thuc ng thng AD. Xc nh ta cc nh ca hnh vung bit 0

Bx > .

S: ( ) ( ) ( ) ( )2;2 , 1;2 , 1;1 , 2;1A B C D Tnh Gia 1 - Thanh Ha - 2014: Cho hnh vung ABCD c D(5;1). Gi M l trung im ca BC, N l im thuc ng cho AC sao cho AC=4AN. Tm ta im C bit phng trnh ng thng MN l 3 4 0x y = v M c tung dng. S: C(5;5) ng Thc Ha - Ngh An - 2014: Cho hnh vung ABCD. Gi E l trung im ca cnh AD,

11 2;

5 5H

l hnh chiu vung gc ca B ln CE v 3 6;5 5

H

l trung im ca on BH. Xc nh

ta ca cc nh ca hnh vung ABCD bit im A c honh m. S: ( ) ( ) ( ) ( )1;2 , 1; 2 , 3; 2 , 3;2A B C D chuyn Lng Th Vinh - ng Nai - 2014: Cho hnh vung ABCD c A(1;1), AB=4. Gi M l trung im cnh BC, im 9 3;

5 5H

l hnh chiu vung gc ca D ln AM. Tm ta cc nh cn li ca

hnh vung bit 2.Bx <

S: ( ) ( ) ( )1; 3 , 5; 3 , 5;1B C D Nguoithay.vn - 2014: Cho hnh vung ABCD c M(2;2) l trung im ca cnh AB, ng thng i qua nh C v trung im ca cnh AD c phng trnh l 7 46 0.x y+ = Xc nh ta cc nh ca hnh vung ABCD bit im C tung m. 2. Vit phng trnh ng thng

Cho hnh vung ABCD bit cc im ( ) ( ) ( ) ( )2;1 , 4; 2 , 2;0 , 1;2M N P Q ln lt thuc cc cnh AB, BC, CD, DA. Vit phng trnh cc cnh ca hnh vung ABCD.

S GD&T Vnh Phc - 2014: Cho hnh vung ABCD c nh A thuc ng thng : 4 0d x y = , ng thng BC i qua im M(4;0), ng thng CD i qua im N(0;2) v tam gic AMN cn ti A. Vit phng trnh ng thng BC. S: : 3 4 0BC x y = hoc : 3 12 0BC x y+ =

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CC BI TON V HNH THANG, HNH BNH HNH

1. Tm ta ca im B13: Cho hnh thang cn ABCD c hai ng cho vung gc vi nhau v AD = 3BC. ng thng BD c phng trnh 2 6 0x y+ = v tam gic ABD c trc tm ( )3;2H . Tm ta cc nh C v D. S: ( )1;6C v ( )4;1D hoc ( )8;7D chuyn Vnh Phc: Cho hnh bnh hnh ABCD c din tch bng 4. Bit ( ) ( )2;0 , 3;0A B v giao im I ca hai ng cho AC v BD nm trn ng thng :d y x= . Tm ta ca C v D.

S: ( ) ( )3;4 , 2;4C D hoc ( ) ( )5; 4 , 6; 4C D Yn Kh - Ph Th: Cho hnh bnh hnh ABCD c A(1 ; 2), : 2 1 0BD x y+ + = . Gi M l mt im nm trn ng thng AD sao cho A nm gia M v D, AM = AC. ng thng : 1 0MC x y+ = . Tm ta cc nh cn li ca hnh bnh hnh.

S: ( ) ( ) ( )1/ 2; 2 , 7;8 , 13/ 2;12B C D GSTT.VN - 2013: Cho hnh bnh hnh ABCD c A(1;5). im H(1;3) l hnh chiu vung gc ca B trn AC v ng trung trc ca BC c phng trnh 4 5 0x y+ = . Tm ta cc im B, C, D.

S: ( ) ( ) ( ) 2; 6 , 4; 2 , 1; 3B C D chuyn Nguyn Quang Diu - ng Thp - 2013: cho hnh thang ABCD vi hai y l AB v CD, bit B(3;3), C(5;-3). Giao im I ca hai ng cho nm trn ng thng : 2 3 0d x y+ = v CI = 2BI. Xc nh ta ca im A v im D bit tam gic ACB c din tch bng 12, 0; 0A Ix x< > .

S: ( ) ( ) 1;3 , 3; 3A D Ton hc & Tui tr - 2014: Cho hnh thang vung ABCD vung ti A v D c AB AD CD= <

( ), 1;2B , ng thng BD c phng trnh 2 0y = . Bit ng thng : 7 25 0d x y = ct cc on thng AD, CD ln lt ti hai im M, N sao cho BM vung gc vi BC v tia BN l tia phn gic ca gc MBC . Tm ta im D bit D c honh dng. S GD&T Bc Ninh - 2014: Cho hnh thang vung ABCD vung ti A(1;1) v B. Trn cnh AB ly im M sao cho BM = 2AM, im N(1;4) l hnh chiu vung gc ca M trn ng thng CD. Tm ta cc nh B, C, D bit CM vung gc vi DM, im B thuc ng thng : 2 0d x y+ = .

S: ( ) ( ) ( )2;4 , 1;5 , 3;3B C D S GD&T Vnh Phc - 2013: Cho hnh thang cn ABCD c AB=2CD. Phng trnh cc ng thng AC l 4 0x y+ = v ng thng BD l 2 0x y = . Tm ta cc nh ca hnh bnh hnh bit honh ca A v B dng v din tch ca hnh bnh hnh bng 36.

S: A(7; 3), B(7; 5), C(1; 3), D(1; 1) chuyn L T Trng - Cn Th - 2014: Cho hnh bnh hnh ABCD c A(4;0), phng trnh ng thng cha trung tuyn k t B ca tam gic ABC l 7 4 5 0.x y+ = Phng trnh ng trung trc ca on BC l 2 8 5 0.x y+ = Tm ta cc im B, C, D.

S: ( ) ( ) ( )1; 3 , 2; 1 , 3; 4B C D 2. Vit phng trnh ng thng o Duy T - Thanh Ha: Cho hnh thang cn ABCD c din tch bng 18, : 2 0CD x y + = . Hai ng cho AC v BD vung gc nhau v ct nhau ti I(3 ; 1). Vit phng trnh ng thng BC, bit C c hong m. S: : 2 1 0BC x y+ =

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chuyn Quc Hc - Hu - 2013: Cho ABCD l hnh thang vung ti A v B, c din tch bng 50, nh C(2;-5), AD = 3BC. Bit rng ng thng AB i qua im 1 ;0

2M

, ng thng AD i qua N(-3;5). Vit phng trnh ng thng AB bit ng thng AB khng song song vi cc trc ta . S: + =: 4 3 2 0AB x y hoc + + =: 6 8 3 0AB x y

*****

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CC BI TON V NG TRN

1. Vit phng trnh ng trn D03: Cho ng trn (C): + =2 2( 1) ( 2) 4x y v ng thng d: x y 1 = 0. Vit phng trnh ng trn (C) i xng vi ng trn (C) qua ng thng d. Tm ta cc giao im ca (C) v (C). S: C x y2 2( ) : ( 3) 4 + = , A(1; 0), B(3; 2) B04: Cho hai im A(2; 0), B(6; 4). Vit phng trnh ng trn (C) tip xc vi trc honh ti im A v khong cch t tm ca (C) n im B bng 5. S: C x y C x y2 2 2 21 2( ) : ( 2) ( 1) 1, ( ) : ( 2) ( 7) 49 + = + = A07: Cho tam gic ABC c A(0; 2), B(2; 2), C(4; 2). Gi H l chn ng cao k t B; M v N ln lt l trung im ca cc cnh AB v BC. Vit phng trnh ng trn i qua cc im H, M, N. S: H(1; 1), x y x y2 2 2 0+ + = D07: Cho ng trn + + =2 2( ) : ( 1) ( 2) 9C x y v ng thng + =: 3 4 0d x y m . Tm m trn d c duy nht mt im P m t c th k c hai tip tuyn PA, PB ti (C) (A, B l cc tip im) sao cho tam gic PAB u. S: m = 19, m = 41 A09: Cho ng trn + + + + =2 2( ) : 4 4 6 0C x y x y v ng thng : + + =2 3 0x my m , vi m l tham s thc. Gi I l tm ca ng trn (C). Tm m ct (C) ti hai im phn bit A, B sao cho din tch IAB ln nht. S: m= 0 hoc = 8/15m . A10: Cho hai ng thng + =1 : 3 0d x y v =2 : 3 0d x y . Gi (T) l ng trn tip xc vi d1 ti A, ct d2 ti hai im B, C sao cho tam gic ABC vung ti B. Vit phng trnh ca (T), bit tam gic ABC c din tch bng 3

2 v im A c honh dng.

S: T x y2 2

1 3( ) : 1

22 3

+ + + =

B10: Cho im ( )2; 3A v elip (E): + =2 2 13 2x y

. Gi F1 v F2 l cc tiu im ca (E) (F1 c honh m); M l giao im c tung dng ca ng thng AF1 vi (E); N l im i xng ca F2 qua M. Vit phng trnh ng trn ngoi tip tam gic ABF2.

S: x y2

2 2 3 4( 1)3 3

+ =

B12: Cho hai ng trn 2 21( ) : 4C x y+ = v 2 2

2( ) : 12 18 0C x y x+ + = v ng thng : 4 0d x y = . Vit phng trnh ng trn (C) c tm thuc ( )2C , tip xc vi d v ct ( )1C ti hai

im phn bit A, B sao cho AB vung gc vi d. S: 2 2( ) : ( 2) ( 2) 8C x y + =

D12: Cho ng thng : 2 3 0d x y + = . Vit phng trnh ng trn (C) c tm thuc d, ct trc Ox ti A v B, ct trc Oy ti C v D sao cho AB = CD =2.

S: 2 2( ) : ( 3) ( 3) 10C x y+ + + = A13: Cho ng thng =: 0x y . ng trn (C) c bn knh 10R = ct ti hai im A v B sao

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cho 4 2AB = . Tip tuyn ca (C) ti A v B ct nhau ti mt im thuc tia Oy. Vit phng trnh ng trn (C).

S: + =2 2( ) : ( 5) ( 3) 10C x y

B09: Cho ng trn (C): + =2 2 4( 2)5

x y v hai ng thng = =1 2: 0, : 7 0x y x y . Xc nh

to tm K v tnh bn knh ca ng trn (C1); bit ng trn (C1) tip xc vi cc ng thng 1, 2 v tm K (C) S: K R8 4 2 5; ,

5 5 5

=

D02(d b): Cho hai ng trn: C x y x C x y x y2 2 2 21 2( ) : 10 0, ( ) : 4 2 20 0+ = + + = . Vit phng trnh ng trn i qua cc giao im ca (C1), (C2) v c tm nm trn ng thng d: x y6 6 0+ = . S: x y2 2( 12) ( 1) 125 + + = B03(d b): Cho ng thng d x y: 7 10 0 + = . Vit phng trnh ng trn c tm thuc ng thng : x y2 0+ = v tip xc vi ng thng d ti im A(4; 2). S: x y2 2( 6) ( 12) 200 + + = A04(d b): Cho im A(1; 1) v ng thng d x y: 1 2 0 + = . Vit phng trnh ng trn i qua A, qua gc to O v tip xc vi ng thng d. S: 2 2( 1) 1x y+ = hoc 2 2( 1) 1x y+ + = A05(d b): Cho ng trn (C): x y x y2 2 12 4 36 0+ + = . Vit phng trnh ng trn (C1) tip xc vi hai trc ta Ox, Oy ng thi tip xc ngoi vi ng trn (C). S: C x y C x y C x y2 2 2 2 2 21 2 3( ) : ( 2) ( 2) 4, ( ) : ( 18) ( 18) 18, ( ) : ( 6) ( 6) 36 + = + = + + = D05(d b): Cho 2 im A(0;5), B(2; 3) . Vit phng trnh ng trn i qua hai im A, B v c bn knh R = 10 . S: x y x y2 2 2 2( 1) ( 2) 10, ( 3) ( 6) 10+ + = + = D06(d b): Cho im A(1; 1) v ng thng d x y: 1 2 0 + = . Vit phng trnh ng trn (C) i qua im A, gc to O v tip xc vi ng thng d. S: C x y y C x y x2 2 2 21 2( ) : 2 0, ( ) : 2 0+ = + + =

B07(d b): Cho ng trn (C) c phng trnh x y x y2 2 2 4 2 0+ + + = . Vit phng trnh ng trn (C) c tm M(5; 1) v (C) ct (C) ti cc im A, B sao cho AB 3= . S: C x y C x y' 2 2 ' 2 21 2( ) : ( 5) ( 1) 13, ( ) : ( 5) ( 1) 43 + = + = . chuyn Nguyn Quang Diu - ng Thp: Cho tam gic ABC vung cn ti A(1; 2). Vit phng trnh ng trn (T) ngoi tip tam gic ABC bit tip tuyn ca (T) ti B l ng thng : 1 0d x y = . S: ( ) ( )22: 1 2T x y+ = hoc ( ) ( ) ( )2 2: 2 3 2T x y + = chuyn H Long - Qung Ninh: Cho im M(2 ; 1) v ng thng : 1 0d x y + = . Vit phng trnh ng trn i qua M v ct d ti hai im A, B sao cho tam gic ABM vung ti M v c din tch bng 2. S: ( ) ( )2 21 2 8x y + = Lng Giang 2 -Bc Giang: Cho ( ) 2 2: 4 3 4 0C x y x+ + = . Tia Oy ct (C) ti im A. Lp phng trnh ng trn (C) c bn knh bng 2 v tip xc ngoi vi (C) ti A.

S: ( ) ( ) ( )2 2' : 3 3 4C x y + =

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Nguyn ng o - Bc Ninh: Cho 1 2: 2 6 0; : 2 0d x y d x y+ = + = v 3 : 3 2 0d x y = . Vit phng trnh ng trn (C) c tm thuc 3d , ct 1d ti A v B, 2d ti C v D sao cho t gic ABCD l hnh vung.

S: ( ) ( ) ( )2 2: 1 1 18 / 5C x y + = H Vinh: Cho ng trn ( ) 2 2: 2 4 20 0C x y x y+ + = v im ( )5; 6A . T A v tc tip tuyn AB, AC ca ng trn (C) vi B, C l cc tip im. Vit phng trnh ng trn ni tip tam gic ABC.

S: ( ) ( )2 2 252 24

x y + + =

Ton hc & Tui tr: Vit phng trnh ng trn c bn knh bng 2, c tm I nm trn ng thng 1 : 3 0d x y+ = v ng trn ct ng thng 2 : 3 4 6 0d x y+ = ti A, B sao cho

o120AIB = .

Ton hc & Tui tr: Cho im ( )2; 1M v ng trn 2 2( ) : 9C x y+ = . Vit phng trnh ng trn ( )1C c bn knh bng 4 v ct (C) theo mt dy cung qua M c di nh nht.

S: ( )2 2

1

4 3 2 3: 2 1 16

5 5C x y

+ + + =

; ( )

2 2

1

4 3 2 3: 2 1 16

5 5C x y

+ + + =

Ton hc & Tui tr: Cho tam gic ABC c A(1 ; 0), ng cao k t B v C ln lt c phng trnh 2 1 0x y + = v 3 1 0x y+ = . Vit phng trnh ng trn ngoi tip tam gic ABC.

S: 2 2 36 10 43( ) : 07 7 7

C x y x y+ + =

S GD&T Vnh Phc - 2013: Cho tam gic ABC vung cn ti A(1;2). Vit phng trnh ng trn (C) ngoi tip tam gic ABC bit ng thng : 1 0d x y = tip xc vi ng trn (C) ti im B.

S: ( ) ( ) ( ) + =2 2: 2 3 2C x y hoc ( ) ( )+ =22: 1 2C x y GSTT.VN - 2013: Cho A(1;5) v + + =2 2( ) : 2 6 0C x y x y . Vit phng trnh ng trn (C') c tm nm trn : 2 0d x y+ + = , i qua A v ct (C) ti 2 im phn bit M, N sao cho 2 2MN = .

S: ( ) + + =

2 223 15 377

:4 4 8

C x y hoc ( ) + + + =

2 25 3 305

:4 4 8

C x y

Hng Vng - Bnh Phc - 2014: Cho hnh vung ABCD, A(-1;2). Gi M, N ln lt l trung im ca AD v DC, E l giao im ca BN vi CM . Vit phng trnh ng trn ngoi tip tam gic BME bit : 2 8 0BN x y+ = v 2Bx > .

S: ( ) ( ) ( ) + =2 2: 1 3 5C x y Ton hc & Tui tr - 2014: Cho hai im A(1;2), B(3;4) v ng thng : 3 0d y = . Vit phng trnh ng trn (C) i qua hai im A, B v ct d ti hai im phn bit M, N sao cho o60MAN = .

S: ( ) ( ) ( )2 2: 3 2 4C x y + = Ton hc & Tui tr - 2014: Cho im A(1;2) v ng trn ( ) + + + =2 2: 2 4 1 0C x y x y . Vit phng trnh ng trn (C') c tm A v ct (C) ti hai im phn bit M v N sao cho din tch tam gic AMN t gi tr ln nht.

S: ( ) ( ) ( )2 2' : 1 2 12C x y + = Ton hc & Tui tr - 2014: Cho im A(-1;2) v ng thng : 3 4 7 0d x y + = . Vit phng trnh ng trn (C) c bn knh R = 1, i qua A v ct d theo dy cung BC sao cho tam gic ABC c din tch bng 4 / 5 .

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S: ( ) ( ) ( )+ + =2 2: 1 1 1C x y hoc ( ) + + =

2 21 43

: 125 25

C x y

Ton hc & Tui tr - 2014: Cho hai ng thng 1 2: 1 0; : 1 0d x y d x y+ = + = . Lp phng trnh ng trn (C) ct d1 ti A v d2 ln lt ti hai im B, C sao cho tam gic ABC l tam gic u c din tch bng 24 3 .

S: ( ) ( ) ( ) + + =2 2: 2 1 32C x y hoc ( ) ( ) ( )+ + =2 2: 2 3 32C x y Ton hc & Tui tr - 2014: Cho ( )1 3 1 1 3 4; , ; , ; , 2;0 .

2 2 5 52 2A B C D

Vit phng trnh

ng trn (T) c tm l im D v ct ng trn ngoi tip tam gic ABC theo mt dy cung c di bng 2. chuyn Trn i Ngha - HCM - 2014: Cho hai ng thng 1 2: 4 3 8 0; : 4 3 2 0 + = + + =d x y d x y v ng trn ( ) 2 2: 20 2 20 0.C x y x y+ + = Vit phng trnh ng trn (C') tip xc vi (C) v ng thi tip xc vi ng thng d1 v d2.

S: ( ) ( )22: 1 1C x y+ = hoc ( ) ( ) ( )2 2: 100 1 6561C x y + = Tnh Gia 1 - Thanh Ha - 2014: Cho tam gic nhn ABC ni tip ng trn tm I(1;2), bn knh R=5. Chn ng cao k t B v C ln lt l H(3;3) v K(0;-1). Vit phng trnh ng trn ngoi tip t gic BCHK, bit A c tung dng.

S: ( )2 2

7 1 25:

2 2 2C x y

+ + =

chuyn H Vinh - 2014: Cho hai im A(1;2), B(4;1) v ng thng : 3 4 5 0.x y + = Vit phng trnh ng trn i qua A, B v ct ti C, D sao cho CD=6.

S: ( ) ( ) ( )2 2: 1 3 25C x y + + = ; ( )2 2

43 51 1525:

13 13 169C x y

+ =

2. Tm ta ca im D06: Cho ng trn (C): + + =2 2 2 2 1 0x y x y v ng thng + =: 3 0d x y . Tm to im M nm trn d sao cho ng trn tm M, c bn knh gp i bn knh ng trn (C), tip xc ngoi vi ng trn (C). S: M(1; 4), M(2; 1) A11: Cho ng trn 2 2( ) : 4 2 0C x y x y+ = v ng thng : 2 0x y + + = . Gi I l tm ca (C), M l im thuc . Qua M k cc tip tuyn MA v MB n (C) (A v B l cc tip im). Tm ta im M, bit t gic MAIB c din tch bng 10. S: ( ) ( )2; 4 , 3;1M M D13: Cho ng trn + =2 2( ) : ( 1) ( 1) 4C x y v ng thng =: 3 0y . tam gic MNP c trc tm trng vi tm ca (C), cc nh N v P thuc , nh M v trung im ca cnh MN thuc (C). Tm ta im P. S: ( ) ( )1;3 , 3;3P P A02(d b): Cho ng thng d x y: 1 0 + = v ng trn (C): x y x y2 2 2 4 0+ + = . Tm to im M thuc d m qua ta k c hai ng thng tip xc vi (C) ti A v B sao cho AMB 060= . S: M M1 2(3;4), ( 3; 2)

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D05(d b): Cho ng trn (C) c phng trnh: C x y x y2 2( ) : 4 6 12 0+ = . Tm ta im M thuc ng thng d c phng trnh: x y2 3 0 + = sao cho MI = 2R, trong I l tm v R l bn knh ca ng trn (C). S: M M 24 63( 4; 5), ;

5 5

B07(d b): Cho ng trn (C): x y x y2 2 8 6 21 0+ + + = v ng thng d x y: 1 0+ = . Xc nh to cc nh hnh vung ABCD ngoi tip ng trn (C), bit A nm trn d. S: A(2; 1), B(2; 5), C(6; 5), D(6; 1) hoc A(6; 5), B(6; 1), C(2; 1), D(2; 5) Ton hc & Tui tr: Cho ng trn 2 2 3( ) :

2C x y+ = v parabol ( ) 2:P y x= . Tm trn (P) cc im

M t k c hai tip tuyn n (C) v gc gia hai tip tuyn bng 60o. S: ( )2; 2M hoc ( )2; 2M

Ton hc & Tui tr: Cho : 3 4 5 0d x y + = v 2 2( ) : 2 6 9 0C x y x y+ + + = . Tm ta im M thuc (C) v im N thuc d sao cho MN nh nht.

S: 2 11 1 7; , ;5 5 5 5

M N

Ton hc & Tui tr: Cho ng trn 2 2( ) : ( 1) ( 3) 1C x y+ + = v im 1 7;5 5

M

. Tm trn (C) nhng im N sao cho MN nh nht.

S: ( )8 / 5;19 / 5N Trung Gi - H Ni: Cho tam gic ABC vung cn ti A ngoi tip ng trn ( ) 2 2: 2C x y+ = . Tm ta ba nh ca tam gic ABC bit A thuc tia Ox.

S: ( ) ( ) ( )2;0 , 2,2 2 , 2, 2 2A B C + chuyn Vnh Phc: Cho ( ) ( )2 2: 4 4C x y + = , im E(4 ; 1). Tm ta im M trn trc tung sao cho t k c hai tip tuyn MA, MB n (C) vi A, B l tip im v ng thng AB i qua E.

S: ( )0;4M S GD&T Vnh Phc - 2013: Cho ( ) + =2 2: 25C x y , im M(1;-2). ng trn (C') c bn knh bng 2 10 . Tm ta tm ca (C') sao cho (C') ct (C) theo mt dy cung qua M c di nh nht.

S: ( )1;2 hoc (3;6) chuyn Vnh Phc - 2013: Cho ( ) + =2 2: 2 4 4 0C x y x y . Tm ta cc nh ca tam gic u ABC ngoi tip (C) bit A thuc ng thng : 1d y = v 0Ax > . S: A(6; 1), B(-4; -1), C(1; 8) chuyn Phan Bi Chu - Ngh An - 2013: Cho im A(2;0) v ( ) ( ) + + =2 2( ) : 1 2 5C x y . Tm ta hai im B, C thuc (C) sao cho tam gic ABC vung ti B v c din tch bng 4. S: ( ) ( )16 8 6 122; 4 , ; , 0;0 , ;

5 5 5 5B B B B

, C(0; -4)

chuyn Nguyn Tri - Hi Dng - 2013: Cho ( )+ =22( ) : 1 1C x y . Tm ta im M thuc ng thng : 3 0d y = sao cho cc tip tuyn ca (C) k t M ct trc honh ti hai im phn bit A, B v bn knh ng trn ngoi tip tam gic MAB bng 4. S: M(2;3) hoc M(-2;3)

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Nguyn Hu - Ph Yn: Cho tam gic ABC c ng trn ngoi tip ( )2 2( ) : 4 10C x y + = , A(1 ; 1), trng tm 11 1;

3 3G

. Tm ta ca B v C ( )0Cy > . S: ( ) ( )3; 3 , 7;1B C o Duy T - Thanh Ha: Cho ( ) 2 2: 2 24 0C x y x+ = c tm I ; ng thng : 3 4 28 0d x y+ = . Chng minh d tip xc vi (C). Tm ta im A trn (C), im B v C trn d sao cho tam gic ABC nhn I lm trc tm v trung im cnh AC thuc (C), bit im C c honh dng. S: ( ) ( ) ( )2; 4 , 0;7 , 12; 2A B C D09: Cho ng trn + =2 2( ) : ( 1) 1C x y . Gi I l tm ca (C). Xc nh to im M thuc (C) sao cho = oO 30IM . S: ( )3/ 2; 3 / 2M HSP H Ni - 2014: Cho ng trn ( ) 2 2: 2 6 15 0C x y x y+ = ngoi tip tam gic ABC c A(4;7). Tm ta cc nh B v C bit H(4;5) l trc tm ca tam gic ABC. S: ( ) ( ) +1 2 6;2 , 1 2 6;2B C hoc ( ) ( ) +1 2 6;2 , 1 2 6;2C B H Ni -Amsterdam - 2014: Cho tam gic ABC c nh A(1;5). Tm ng trn ni tip v ngoi tip ca tam gic ABC ln lt l I(2;2) v 5 ;3

2K

. Tm ta cc nh B v C.

S: ( ) ( )1;1 , 4;1B C hoc ( ) ( )1;1 , 4;1C B Ng Gia T - Vnh Phc - 2014: Cho tam gic ABC c trung tuyn v phn gic trong nh B c phng trnh ln lt l 2 3 0, 2 0x y x y+ = + = . im M(2;1) nm trn ng thng cha cnh AB; ng trn ngoi tip tam gic ABC c bn knh bng 5 . Bit nh A c honh dng, hy xc nh ta cc nh ca tam gic ABC . S: ( ) ( ) ( )1;1 , 3;1 , 1; 3B A C c Th - H Tnh - 2014: Cho ng trn ( ) 2 2: 9C x y+ = , ng thng : 3 3y x = + v im A(3;0). Gi M l mt im di ng trn (C) v B l im sao cho t gic ABMO l hnh bnh hnh. Tm ta trng tm G ca tam gic ABM, bit G thuc v G c tung dng. S: ( )3; 3G Ton hc & Tui tr - 2013: Cho ng trn ( ) 2 2: 4 2 4 0C x y x y+ = c tm l I v ng thng

: 1 0d x y + = . Tm ta im M thuc d t M c th k c hai ng thng tip xc vi (C) ti A, B sao cho t gic IMAB l hnh vung. S: ( )1 2 2;2 2 2M hoc ( )1 2 2;2 2 2M + + Ton hc & Tui tr - 2014: Cho tam gic ABC nhn. Gi E, F ln lt l chn ng cao h t B, C. nh A(3;-7), trung im ca BC l im M(-2;3) v ng trn ngoi tip tam gic AEF c phng trnh ( ) ( ) ( )2 2: 3 4 9C x y + + = . Xc nh ta cc im B v C. Ton hc & Tui tr - 2014: Cho ( ) ( ) ( )2 2: 1 2 5C x y + = l phng trnh ng trn ni tip tam gic u ABC. ng thng BC i qua im 7 ;2

2M

. Xc nh ta im A.

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Ton hc & Tui tr - 2014: Cho ng trn ( ) 2 2: 2 2 2 0C x y x y+ + = v + + =: 2 10 0d x y . T mt im M bt k trn d k cc tip tuyn MA v MB n (C) (A, B l cc tip im). Xc nh ta im M sao cho khong cch t O n ng thng AB t gi tr ln nht.

S: 14 58;3 3

M

Ton hc & Tui tr - 2014: Cho ng trn ( ) ( ) ( )2 2: 1 2 2C x y + + = v hai im A(3;5) v B(5;3). Xc nh ta im M trn (C) sao cho din tch tam gic MAB c gi tr ln nht. S: ( )0; 3M Ton hc & Tui tr - 2014: Cho ng trn ( ) 2 2: 5+ =C x y v ng thng : 3 2 0.x y = Tm ta im A, B trn tam gic OAB c 10

5OA = v c cnh OB ct ng trn (C) ti M sao

cho MA=MB (vi O l gc ta ). S: ( ) 4 222;4 , ;

5 5

B B

Ton hc & Tui tr - 2014: Cho tam gic ABC c trc tm H(5;5), phng trnh ng thng cha cnh BC l 8 0.x y+ = Bit ng trn ngoi tip tam gic ABC i qua hai im M(7;3) v N(4;2). Tnh din tch tam gic ABC. Phan Chu Trinh - Nng - 2014: Cho ng thng + =: 3 0.d x y Qua im A thuc d k hai ng thng tip xc vi ng trn ( ) ( ) ( )2 2: 2 1 4 + =C x y ti B v C. Gi G l trng tm ca tam gic ABC. Tm ta ca im A, bit AG=2. S: ( ) ( )2;5 , 2;1A A chuyn H Vinh - 2014: Cho tam gic ABC c nh A(3;3), tm ng trn ngoi tip I(2;1), phng trnh ng phn gic trong gc BAC l 0x y = . Tm ta cc nh B, C bit rng 8 5

5BC = v gc

BAC nhn.

S: ( ) 8 60;2 , ;5 5

B C hoc ngc li

chuyn Nguyn Quang Diu - ng Thp - 2014: Cho tam gic ABC c trc tm H(-1;3), tm ng trn ngoi tip I(3;-3) v chn ng cao k t nh A l K(-1;1). Tm ta cc nh A, B, C. S: ( ) ( ) ( )1; 5 , 5;1 , 1;1A B C hoc ( ) ( ) ( )1; 5 , 1;1 , 5;1A B C chuyn L T Trng - Cn Th - 2014: Cho tam gic ABC vung ti A(-1;1) v c tm ng trn ni tip l I(1;5). ng thng vung gc vi IA ti A ct ng trn ngoi tip tam gic AIC ti im th hai l D(-7;4). Tm ta im B. S: ( )17;7B H Huy Tp - Ngh An - 2014: Cho ng trn ( ) + =2 2: 25C x y ngoi tip tam gic nhn ABC c ta cc chn ng cao h t B, C ln lt l M(-1;-3), N(2;-3). Hy tm ta cc nh A, B, C bit rng im A c tung m. S: ( ) ( ) ( )0; 5 , 5;0 , 4;3A B C H Huy Tp - Ngh An - 2014: Cho tam gic ABC cn ti A(0;3) v hai im B, C thuc ng trn ( ) + =2 2: 9.C x y Tm ta ca B, C bit rng tam gic ABC c din tch ln nht v im B c honh dng.

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S: 27 3 27 3; , ;2 2 2 2

B C

chuyn Lam Sn - Thanh Ha - 2014: Cho im A(1;-3) v ng trn ( ) + + =2 2: ( 2) ( 6) 50C x y co tm l im I. Tm ta im M thuc (C) sao cho s o ca gc AMI ln nht. S: ( ) ( )7; 1 , 5; 5 M M ng Thc Ha - Ngh An - 2014: Cho tam gic ABC vung ti A. Gi M l im trn cnh AC sao cho AB=3AM. ng trn tm I(1;-1) ng knh CM ct BM ti D. Xc nh ta cc nh ca tam gic ABC bit ng thng BC i qua im 4 ;0

3N

, phng trnh ng thng CD l 3 6 0x y = v

im C c honh dng. S: ( ) ( ) ( )2; 1 , 2;2 , 3; 1A B C Nguoithay.vn - 2014: Cho tam gic ABC c ng cao AH, H thuc cnh BC sao cho BC=4BH. ng trn ngoi tip tam gic ABH c phng trnh l + + =2 2 2 4 20 0x y x y . im A nm trn ng thng : 2 3 7 0d x y = v din tch tam gic ABC bng 60. Tm ta cc nh ca tam gic ABC, bit im A v C c honh m. Nguoithay.vn - 2014: Cho ng trn ( ) + + =2 2: ( 1) ( 1) 20C x y v ng thng : 3 4 8 0.d x y = Vit phng trnh ng trn (T) c tm nm trn d v ct (C) ti hai im A, B sao cho 2 5AB = , bit ng thng AB to vi ng thng d mt gc vi 1cos .

10 =

3. Vit phng trnh ng thng B06: Cho ng trn (C): + + =2 2 2 6 6 0x y x y v im M(3; 1). Gi T1 v T2 l cc tip im ca cc tip tuyn k t M n (C). Vit phng trnh ng thng T1T2. S: Chng t to x y0 0( ; ) ca T1, T2 tho phng trnh x y2 3 0+ = .

D11: Cho im ( )1;0A v ng trn 2 2( ) : 2 4 5 0C x y x y+ + = . Vit phng trnh ng thng ct (C) ti hai im M v N sao cho tam gic AMN vung cn ti A. S: : 1y = hoc : 3y =

Ton hc & Tui tr: Cho im M(2 ; 1) v ng trn ( ) ( ) ( )2 2: 1 2 5C x y + = . Vit phng trnh ng thng i qua M ct (C) ti hai im phn bit A v B sao cho AB nh nht. B02(d b): Cho hai ng trn: (C1): x y y2 2 4 5 0+ = v (C2): x y x y2 2 6 8 16 0+ + + = . Vit phng trnh tip tuyn chung ca hai ng trn (C1) v (C2). S: 4 tip tuyn chung: x y y y x42 3 5 2 0; 1; 3

3+ = = =

D02(d b): Cho hai ng trn: C x y x C x y x y2 2 2 21 2( ) : 10 0, ( ) : 4 2 20 0+ = + + = . Vit phng trnh tip tuyn chung ca cc ng trn (C1), (C2). S: x y7 5 25 2 0+ = B05(d b): Cho 2 ng trn 2 21C x y( ) : 9+ = v C x y x y2 22( ) : 2 2 23 0+ = . Vit phng trnh trc ng phng d ca 2 ng trn (C1) v (C2). Chng minh rng nu K thuc d th khong cch t K n tm ca (C1) nh hn khong cch t K n tm ca (C2). S: d x y: 7 0+ + = , xt OK IK2 2 16 0 = < OK < IK

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A07(d d): Cho ng trn (C): x y2 2 1+ = . ng trn (C) tm I(2; 2) ct (C) ti cc im A, B sao cho AB 2= . Vit phng trnh ng thng AB. S: Ch AB OI. Phng trnh AB: y x 1= Ton hc & Tui tr: Cho ng trn 2 2( ) : 6 2 1 0C x y x y+ + = . Vit phng trnh ng thng d song song vi ng thng : 2 4 0x y = v ct (C) theo mt dy cung c di bng 4. S: 1 : 2 4 0d x y + = hoc 2 : 2 6 0d x y = Phc Bnh - Bnh Phc: Cho hai ng trn ( ) 2 21 : ( 1) 1/ 2C x y + = , ( ) 2 22 : ( 2) ( 2) 4C x y + = . Vit phng trnh ng thng d tip xc vi ( )1C v ct ( )2C ti hai im phn bit AB sao cho

2 2AB = .

S: 2 0; 2 0; 7 6 0;7 2 0x y x y x y x y+ = = + = = ng Hng H - Thi Bnh: Cho ( ) 2 21 : ( 6) 25C x y + = v ( ) 2 22 : 13C x y+ = ct nhau ti A(2 ; 3). Vit phng trnh ng thng d i qua A v ct ( )1C , ( )2C theo hai dy cung c di bng nhau.

S: : 2 0d x = hoc : 3 7 0d x y + = H Vinh: Cho ng trn ( ) 2 2: 4 2 15 0C x y x y+ + = . Gi I l tm ng trn (C). ng thng d i qua im ( )1; 3M ct (C) ti hai im AB. Vit phng trnh ca d bit tam gic IAB c din tch bng 8 v AB l cnh ln nht.

S: : 3 0d y + = hoc : 4 3 5 0d x y+ + =

THPT L Xoay: Cho ( ) ( ) ( )2 21 : 1 2 4C x y + = v ( ) ( ) ( )2 22 : 1 3 2C x y + = . Vit phng trnh ng thng d i qua im A(1 ; 4) ct ( )1C ti M, ( )2C ti N sao cho AM = 2AN.

S: : 1 0d x = hoc : 2 7 0d x y + = chuyn i hc quc gia H Ni: Cho ng trn ( ) 2 2: 2 2 23 0C x y x y+ + = . Vit phng trnh ng thng i qua im A(7 ; 3) v ct (C) ti B v C sao cho 3AB AC= .

S: 3 0y = hoc 12 5 69 0x y = chuyn Nguyn Quang Diu - ng Thp: Cho ( ) 2 2: 8 9 0C x y x+ = v im ( )1; 1M . Vit phng trnh ng thng i qua M ct (C) ti hai im A, B sao cho MA = 3MB. S: 2 3 0x y = hoc 2 1 0x y+ + = chuyn Nguyn Hu - H Ni: Cho ( ) 2 2: 2 4 0C x y x y+ = v im M(6 ; 2). Vit phng trnh ng thng d i qua M ct (C) ti hai im A v B sao cho 2 2 50MA MB+ = . S: 3 12 0x y+ = hoc 3 0x y = ng Thc Ha - Ngh An: Cho ( ) 2 2: 10 10 30 0C x y x y+ + = . Vit phng trnh ng thng d tip xc vi (C) bit d ct tia Ox ti A, tia Oy ti B sao cho

2 2

1 1 1

5OA OB+ = .

S: : 2 5 0d x y+ = hoc : 2 5 0d x y+ =

i hc s phm H Ni: Cho im M(0 ; 2) v ( ) 2 2: 14

xH y = . Lp phng trnh ng thng d i

qua im M ct (H) ti hai im phn bit A, B sao cho 53

MA MB=

.

S: : 2d y x= + hoc : 2d y x= +

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S GD&T Vnh Phc - 2013: Cho ( ) + + =2 2: 4 6 12 0C x y x y v im ( )2;4 3M . Vit phng trnh ng thng d ct (C) ti hai im A, B sao cho tam gic MAB u. S: 0y = hoc 4 3 9

2y =

chuyn Vnh Phc - 2013: Cho tam gic ABC cn ti A(4;-13) v ( ) + + =2 2: 2 4 20 0C x y x y l phng trnh ng trn ni tip tam gic ABC. Vit phng trnh ng thng BC.

S: : 3 7 5 10 0BC x y + + = on Thng - Hi Dng - 2014: Cho tam gic ABC c nh A(-3;4), ng phn gic trong ca gc A c phng trnh 1 0x y+ = v tm ng trn ngoi tip tam gic ABC l I(1;7). Vit phng trnh cnh BC, bit din tch tam gic ABC bng 4 ln din tch tam gic IBC.

S: + =:15 20 131 0BC x y hoc + =: 9 12 114 0BC x y

Ton hc & Tui tr - 2012: Cho M(2;1) v ng trn ( ) ( ) ( ) + =2 2: 1 2 5C x y . Vit phng trnh ng thng d qua M ct (C) ti hai im phn bit A, B sao cho di on thng AB nh nht.

S: =: 1 0d x y Ton hc & Tui tr - 2014: Cho hai ng trn ( ) ( )22: 1 4C x y+ + = v ( ) ( )2 2' : 1 2C x y + = . Vit phng trnh ng thng d tip xc vi (C) v ct (C') ti hai im phn bit A, B sao cho AB = 2.

S: =: 1 0d y hoc =: 2 0d x Qunh Lu 1 - Ngh An - 2014: Cho tam gic ABC ni tip ng trn (T) c tm ( )3 / 2;0I v (T) tip xc vi ng thng : 4 2 19 0x y + = . ng phn gic trong ca gc A c phng trnh l

1 0.x y = Vit phng trnh ng thng BC, bit din tch tam gic ABC bng ba ln din tch tam gic IBC v im A c tung m.

S: : 2 2 0BC x y+ = hoc : 4 2 11 0BC x y+ + = chuyn H Long - Qung Ninh - 2014: Cho ng trn ( ) 2 2: 9 18 0C x y x y+ + = v hai im A(4;1), B(3;-1). Cc im C, D thuc (C) sao cho ABCD l hnh bnh hnh. Vit phng trnh ng thng CD.

S: : 2 6 0CD x y + = hoc : 2 1 0CD x y + = Nguoithay.vn - 2014: Cho im M(3;1) v ng trn ( ) ( )22: ( 2) 2 10. + =C x y Vit phng trnh ng thng d i qua M, ct (C) ti hai im A, B sao cho khong cch t giao im ca hai tip tuyn vi (C) ti A v B n trc honh bng 3.

*****

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CC BI TON V BA NG CONIC

1. Tm ta ca im D08: Cho parabol (P): =2 16y x v im A(1; 4). Hai im phn bit B, C (B v C khc A) di ng trn (P) sao cho gc = 090BAC . Chng minh rng ng thng BC lun i qua mt im c nh. S: Vit PT ng thng BC BC i qua im c nh I(17; 4) A10: Cho elip

2 2

( ) : 14 1

x yE + = . Tm ta cc im A v B thuc (E), c honh dng sao cho tam

gic OAB cn ti O v c din tch ln nht.

S: 2 22; , 2;2 2

A B

hoc 2 22; , 2;2 2

A B

A03(d b): Cho parabol y x2 = v im I(0; 2). Tm to hai im M, N thuc (P) sao cho IM IN4=

.

S: M N(4; 2), (1;1) hoc M N(36;6), (9;3)

D05: Cho im C(2; 0) v elip (E): x y2 2

14 1

+ = . Tm to cc im A, B thuc (E), bit rng hai im A, B i xng vi nhau qua trc honh v tam gic ABC l tam gic u.

S: A B2 4 3 2 4 3; , ;7 7 7 7

hoc A B2 4 3 2 4 3; , ;

7 7 7 7

Ton hc & Tui tr: Cho A(3 ; 0) v ( ) 2 2: 19

xE y+ = . Tm ta cc im B, C thuc (E) sao cho

tam gic ABC vung cn ti A. Ton hc & Tui tr: Cho ( ) 2:P y x= . Tm ta im B v C trn (P) sao cho tam gic OBC u. S: ( ) ( )6;2 3 , 6; 3B C hoc ( ) ( )6;2 3 , 6; 3C B Ton hc & Tui tr: Cho ( ) 2 2: 1

16 4

x yE + = v im A(0 ; 2). Tm ta im B v C trn (E) sao cho

tam gic ABC u.

S: 16 3 22 16 3 22; , ;13 13 13 13

B C

hoc 16 3 22 16 3 22; , ;13 13 13 13

C B

Ton hc & Tui tr: Cho ( ) 2 2: 125 16

x yE + = v mt tiu im 1( 3;0)F . Tm ta im A trn (E) sao

cho 1AF nh nht. S: ( )5;0A v 1 2AF =

Chu Vn An - H Ni - 2014: Cho ( ) 2 2: 19 4

x yE + = c hai tiu im 1F v 2F vi

10

Fx < . Tm ta

im M trn (E) sao cho 2 21 22MF MF+ nh nht. Tm gi tr nh nht .

S: 3 4;5 5

M

v gi tr nh nht l 36.

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Nguoithay.vn - 2014: Cho ( ) 2 2: 116 12

x yE + = c hai tiu im 1F v 2F vi

10

Fx < . Tm ta im M

trn (E) sao cho bn knh ng trn ni tip tam gic 1 2MF F bng 2.3

2. Vit phng trnh ba ng conic

A08: Vit phng trnh chnh tc ca elip (E) bit rng (E) c tm sai bng 53

v hnh ch nht c s

ca (E) c chu vi bng 20. S: x y

2 21

9 4+ =

A12: Cho ng trn 2 2( ) : 8C x y+ = . Vit phng trnh chnh tc ca elip (E) c di trc ln bng 8 v (E) ct (C) ti 4 im to thnh bn nh ca mt hnh vung. S:

2 2

( ) : 11616

3

x yE + =

B12: Cho hnh thoi ABCD c AC = 2BD v ng trn tip xc vi cc cnh ca hnh thoi c phng trnh 2 2 4x y+ = . Vit phng trnh chnh tc ca elip (E) i qua cc nh ca hnh thoi bit A thuc Ox.

S: 2 2

( ) : 120 5

x yE + =

A06(d b): Cho elip (E): x y2 2

112 2

+ = . Vit phng trnh hypebol (H) c hai ng tim cn l y x2= v c hai tiu im l hai tiu im ca elip (E). S: (H): x y

2 21

2 8 =

D06(d b): Lp phng trnh chnh tc ca elip (E) c di trc ln bng 4 2 , cc nh trn trc nh v cc tiu im ca (E) cng nm trn mt ng trn. S: (E): x y

2 21

8 4+ =

Ton hc & Tui tr: Cho elip (E) i qua im ( )2; 3M v c phng trnh ng chuNn l 8 0x + = . Vit phng trnh chnh tc ca elip (E).

S: ( ) 2 2: 116 12

x yE + = hoc ( ) 2 2: 1

52 39

x yE + =

Ton hc & Tui tr: Cho parabol ( ) 2:P y x= v im ( )1; 1M . Gi s A, B l hai im phn bit khc M, thay i trn (P) sao cho MA MB . Chng minh rng ng thng AB lun i qua mt im c nh. chuyn Nguyn Hu - H Ni: Cho ng trn ( ) 2 2: 16C x y+ = . Vit phng trnh chnh tc ca elip (E) c tm sai 1/ 2e = bit elip ct (C) ti 4 im A, B, C, D sao cho AB song song vi trc honh v AB = 2CD.

S: ( ) 2 2: 1256 64

15 5

x yE + =

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chuyn HSP H Ni - 2013: Cho parabol ( ) 2: 4P y x= . ng thng d i qua im 5 ;12

M

ct (P) ti hai im E v F sao cho ME=MF. Tnh di on EF. S: ( ) ( ) =4;4 , 1; 2 , 3 5E F EF o Duy T - Thanh Ha: Cho ng trn ( ) 2 2: 10 16 0C x y x+ + + = v im T(1 ; 0). Vit phng trnh chnh tc ca hipebol (H) bit (H) nhn tm ca (C) lm mt tiu im v c hai tim cn ln lt song song vi hai tip tuyn k t im T n (C).

S: ( ) 2 2: 175 25

4 4

x yH =

chuyn H Vinh: Cho parabol ( ) 2: 4P y x= c tiu im F. Gi M l im tha mn iu kin 3FM FO=

; d l ng thng bt k i qua M ct (P) ti hai im phn bit A v B. Chng minh tam

gic OAB l tam gic vung. chuyn Vnh Phc: Vit phng trnh chnh tc ca elip (E) bit rng c mt nh v hai tiu im ca (E) to thnh mt tam gic u v chu vi hnh ch nht c s ca (E) bng 24 12 3+ .

S: ( ) 2 2: 136 27

x yE + =

chuyn Quc Hc Hu - 2013: Cho elip (E) c hai tiu im 1F v 2F vi ( )1 3;0F . Vit phng trnh chnh tc ca elip (E) bit rng tn ti mt im M thuc elip (E) sao cho tam gic 1 2F MF c din tch bng 1 v vung ti M.

S: ( ) + =2 2: 14 1

x yE

chuyn Phan Bi Chu - Ngh An - 2013: Vit phng trnh chnh tc ca hypebol (H), bit hnh ch nht c s ca (H) c din tch bng 48 v mt ng chuNn ca (H) c phng trnh 5 16 0x + = .

S: ( ) =2 2: 116 9

x yH

chuyn Phan Bi Chu - Ngh An - 2013: Vit phng trnh chnh tc ca elip (E) bit rng khi M thay i trn (E) th di nh nht ca OM bng 4 v di ln nht ca 1MF bng 8 vi 1F l tiu im c honh m.

S: ( ) + =2 2: 125 16

x yE

H Huy Tp - Ngh An - 2014: Vit phng trnh chnh tc ca elip (E) bit rng (E) c tm sai bng 45

v ng trn ngoi tip hnh ch nht c s ca (E) c phng trnh 2 2 34x y+ = . Tm ta im M trn (E) sao cho M nhn hai tiu im di mt gc vung v M c honh dng.

S: + =2 2

125 9x y ;

5 7 9;

4 4M

*****

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