imo shortlist

Bi ging THSGQG thng 12/2014

Trn Xun Bang-Trng THPT Chuyn V Nguyn Gip 1

Mt s bi ton HH t ImoShortlist 2012. G1. Cho tam gic ABC, J l tm ng trn bng tip gc A. ng trn (J) tip xc BC, CA, AB ln lt ti M, L, K. cc ng thng LM v BJ ct nhau ti F, v cc ng thng KM v CJ ct nhau ti G. Gi S l giao im ca cc ng thng AF v BC, T l giao im ca cc ng thng AG v BC. Chng minh rng M l trung im ca ST. Li gii. t , ,CAB ABC BCA . ng thng AJ l phn gic gc CAB nn

2JAK JAL

. V 090AKJ ALJ nn K, L thuc ng trn ng

knh AJ. Tam gic KBM cn nh B do BM = BK. BJ l phn gic KBM nn

0902

MBJ

v 2

BMK

. Tng t 0902

MCJ

v 2

CML

nn

BMF CML = 2

. Suy ra LFJ MBJ BMF = 0902

-

2

=2

LAJ .

Hn na, F v A cng v mt pha i vi JL nn F thuc ng trn ng knh AJ. Hon ton tng t, G cng thuc ng trn ng knh AJ. Do

090AFJ AGJ . D thy cc ng thng AK v BC i xng nhau qua ng thng BF. Mt khc ,AF BF KM BF nn SM = AK. Tng t TM = AL. Nhng AK = AL. Suy ra SM = TM. Bnh lun: Sau khi pht hin A, F, K, J, L, G cng thuc mt ng trn, c rt nhiu cch khc gii tip bi ton. V d, t cc t gic JMFS v JMGT ni

tip ta c th tm thy 2

TSJ STJ

.

S

F

T

G

J

L

K

M C

B

A



Mt kh nng khc l s dng thc t l cc ng AS v GM l song song

(c hai u l vung gc vi phn gic BJ), do 1MS AGMT GT

G2. Cho ABCD l mt t gic ni tip c ng cho AC v BD gp nhau ti E. Cc tia AD v BC ct nhau ti F. G l im sao cho ECGD l mt hnh bnh hnh. H l im i xng ca E qua AD. Chng minh rng DHFG l t gic ni tip. Li gii. EAB EDC v DEC AEB (i nh) DCE DCA DBA EBA

Suy ra DG CE CDEB EB AB

(1)

Mt khc CD FDFCD FABAB FB

(2)

T (1) v (2) suy ra DG FDEB FB

(3)

Ta li c CDA ABF (Do ABCD ni tip) GDC DCA (Do CGDE l hnh bnh hnh) DCA DBA suy ra GDF GDA GDC DCA ABF DBA DBF EBF

GDF EBF (4) T (3) v (4) suy ra DGF BEF FGD BEF V H i xng E qua FD nn 0 0180 180FHD FED BEF FGD Suy ra pcm. G3. Cho tam gic ABC nhn vi cc ng cao AD, BE, CF. Tm ng trn ni tip cc tam gic AEF v BDF l I1 v I2 tng ng. Tm ng trn ngoi

G

H

F

E

D

C

B



tip cc tam gic tam gic ACI1 v BCI2 l O1 v O2 tng ng. Chng minh rng I1I2//O1O2. Li gii. t , , .CAB ABC BCA Ta thy chng minh ABI2I1 l t gic ni tip. Tht vy, gi I l giao ca cc tia AI1 v BI2. E, F l cc giao im khc B, C ca ng trn ng knh BC vi cc cnh AC, AB. Do

,AEF ABC AFE ACB AEF ABC vi t s ng dng cosAEAB

.

Mt khc 1 1 cosI A rIA r

(t s ng dng)

Suy ra 21 1 1cos (1 cos ) 2 sin 2I A IA I I IA I A IA IA

Tng t 22 2 sin 2I I IB

Suy ra 2 2

1 2 . 2 2 .2 2I I IA IAsin IBsin I I IB

Vy ABI2I1 ni tip. Nh th, I c cng phng tch i vi hai ng trn (O1) v (O2). C l mt giao im ca (O1) v (O2). Suy ra ng thng CI l trc ng phng ca (O1) v (O2). Do 1 2CI OO (1) By gi ta chng minh 1 2.CI I I (2) Gi Q l giao im ca ng thng CI vi I1I2. Ta c:

1 1 1 1 2( )II Q I IQ II Q ACI CAI II I ACI CAI .

I

F

E

D

I2

I1

O2

O1

C

BA



rng 1 2 2II I

(Do ABI2I1 ni tip) v

2ACI

,

2CAI

, suy ra

01 1 902 2 2

II Q I IQ

. (2) c chng minh.

T (1) & (2) suy ra I1I2//O1O2. Cch khc. Trc tin ta s chng minh A, I1, I2, B cng thuc mt ng trn. Gi l tm ni tip tam gic ABC, ta cn chng minh I I1.IA = I I2.IB Gi P, Q, R ln lt l tip im ca (I) vi AC, AB, BC. Khi I , I1 i xng vi nhau qua PQ. Tht vy, d thaays EFBC l t gic ni tip nn suy ra 1AEI ABI .

Suy ra 1AEI ABI 1AIAE

AB AI 1 osAI c A

AI

Gi '1I l im i xng vi qua PQ. Ta c:

' ' '' 0 01 1 11

sinsin sin(90 2 ) sin(90 ) cos

sin

AI QI AQIAQI IQP A A

AI QI AQI .

Do '1 1I I . Hon ton tng t I v I2 i xng vi nhau qua QR. V th 21 2 1 2. . 2II IA II IB r AI I B l t gic ni tip. Mt khc ta li c 1 2CI I I (1) Tht vy, gi X l giao im ca 1 2,CI I I



Ta c:

0 0 01 1

1 1180 90 90

2 2XI I XII ABI AIC ABC ABC

Suy ra 1 2CI I I Do cn chng minh 1 2CI OO Gi Y l giao im th hai ca (O0) v (O2). Khi 1 2CY OO (2) Mt khc, v l tm ng phng ca 3 ng trn (O1 ), (AI1I2B) v (O2) nn C, Y, I thng hng. (3) T (1), (2) v (3) ta c 1 2 1 2/ / .OO I I G4. Cho ABC l mt tam gic vi AB AC v ng trn (O) ngoi tip. Phn gic BAC ct BC ti D v E l im i xng ca D qua trung im M ca BC. Cc ng thng qua D v E vung gc vi BC ct cc ng AO v AD ln lt ti X v Y. Chng minh rng t gic BXCY ni tip c trong mt ng trn. Li gii. Phn gic BAC v ng trung trc ca BC gp nhau ti P l trung im ca cung nh BC. OP vung gc BC ti M. K hiu l Y' l im i xng ca Y qua ng thng OP. Khi BYC = BY'C t BXCY' l t gic ni tip suy ra BXCY l t gic ni tip.

Ta cXAP =OPA =EYP ( do OA = OP v EY // OP). Nhng {Y, Y'} v {E, D} l cp im i xng qua OP nn EYP = DY'P v do XAP = DY'P = XY'P. Suy ra XAY'P l t gic ni tip. Ta c: XDDY' = ADDP = BDDC.



Suy ra BXCY' l t gic ni tip. G5. Cho tam gic ABC c BCA = 900, D l hnh chiu ca C trn AB. Mt im X thuc on thng CD v K, L l cc im trn cc on AX, BX theo th t sao cho BK = BC v AL = AC. Gi M l giao im ca AL v BK. Chng minh MK = ML Li gii. Gi Q, N ln lt l cc giao im ca cc tia AX, BX vi ng trn ngoi tip tam gic ABC khc A, B. Thy ngay 090ANB AQB . Gi P l giao

ME

P

X

Y'Y

D

O

CB

A

XN

Q

P

LK

M

E

D

C

BA



im ca cc ng thng AN, BQ. Khi X l trc tm tam gic PAB nn P thuc ng thng CD. rng t gic PNDB ni tip nn AN.AP = AD.AB = AC2 = AL2. Do tam gic ALP vung L hay PL tip xc ng trn (A, AC). Tng t PK tip xc ng trn (B, BC). Mt khc, gi E l giao im ca cc ng trn (A, AC) v (B, BC) th E i xng C qua AB nn E thuc ng thng CD. Ta c PL2 = PC.PE = PK2 suy ra PL = PK, suy ra hai tam gic vung PML PMK . V th MK = ML.

Cch 2. D thy, nu gi E l giao ca cc ng trn (A, AC) v (B, BC) khc C th E i xng vi C qua AB. Ta cng c AC v AE cng tip xc vi ng trn (B, BC). Gi Q l giao ca ng thng AK v (B, BC) khc K. Do AC, AE cng tip xc (B, BK) nn t gic CQEK l t gic iu ha. Do P l giao ca hai tip tuyn ca (B, BC) ti K v Q thuc ng thng CE. Ta c (PXCE) = - 1. Tng t gi N l giao ca ng thng BL vi ng trn (A, AC) th P' l giao im ca cc tip tuyn ca (A, AC) ti L v N thuc ng thng CE v (P'XCE) = - 1. Do P' trng P. Khi ng thng CE l trc ng phng ca hai ng trn (A, AC) v (B, BC) nn PL = PK. Hai tam gic vung PML PMK nn MK = ML.

Cch 3. Gi U l giao im ca ng thng CD vi vi ng trn i qua ba im A, D, L. Do AC = AL nn AD. AB = AC2 = AL2. Do

ALD ABL AUD ALD DBL . Do UD BDUAD BXDAD DX

AB

C

D

E

M

K L

P

Q

N

X



Hai tam gic UDB v ADX u vung ti D nn ng dng. Suy ra

( DKB KAB)DUB DAX DKB do . Suy ra D, K, U, B cng thuc mt ng trn. Mt khc 090ULA UDA v 090UKB UDB nn AU AL v UK BK . p dng nh l Carnot cho tam gic MAB vi UL, UK, UD ng quy ta c 2 2 2 2 2 2( ) ( ) ( ) 0KM KB DB DA LA LM . Hn na 2 2 2 2 2 2 2 2 2 2, , ,BK BC AL AC BD CB CD AD AC CD . T suy ra

2 2LM KM hay LM = KM. G6. Cho ABC l mt tam gic c O l tm ng trn ngoi tip v I l tm ng trn ni tip. Cc im D, E v F ln lt thuc cc cnh BC, CA v AB sao cho BD + BF = CA v CD + CE = AB. Cc ng trn ngoi tip cc tam gic BFD v CDE ct nhau ti P D. Chng minh rng OP = OI. Li gii. Theo nh l Miquel cc vng trn (AEF) = (CA), (BFD) = (CB) v (CDE) = (CC) c mt im chung. V vy (CA) i qua im chung P D l im chung ca (CB) v (CC). Gi A', B', C' ln lt l giao im ca cc tia AI, BI, CI vi cc ng trn (CA), (CB) v (CC). B . Cho XOY . ng trn (XAY) = (C) ct phn gic gc XAY ti L. Khi 2 cos

2AX AY AL

.

Tht vy, rng L l trung im cung XY khng cha A. t XL = YL = u, XY = v.

U

A B

C

D

KL

X



Theo nh l Ptolemy AX.Y L + AY. XL = AL. XY (AX + AY) u = AL.v. (1)

Ta c LXY = 2

v = 2cos2

u. (2)

T (1) v (2) suy r b c chng minh. p dng b choBAC = v ng trn (C) l (CA), ct AI ti A' cho

ta 2AA'cos2

= AE + AF = BC.

Tng t cho BB' v CC'. p dng b vi BAC = v (C) l ng trn ng knh AI. Sau X, Y l nhng tip im ca (I) vi AB v AC, do

AX = AY = 12

(AB + AC - BC). V vy, ta c 2AIcos2

= AB + AC - BC

Tip theo, nu (C) l ng trn (O) ca BAC v tia AI ct (C) ti M A ta

c 2AM cos2

= AB + AC.

Tm li ta c:

2AA'cos2

= BC, 2AIcos2

= AB + AC - BC, 2AM cos2

= AB + AC. (*)

Suy ra AA'+ AI = AM. Nh th cho ta AI = MA' nn O thuc trung trc ca on IA'. Do OI = OA'. Tng t, ta c OB' = OC' = OI. Vy I, A', B', C' cng cng cch u O. chng minh OP = OI, ta ch cn chng minh rng I, A ', B', C' v P cng thuc mt ng trn. Ta c th gi nh P khc vi I, A', B', C'. Gi s A', B', P, I l phn bit; sau cng thuc mt ng trn l (A'P, B'P) = (A'I, B'I). Do A, F, P, A' cng thuc ng trn (CA) nn: (A'P, FP) = (A'A, FA) = (A'I, AB). Tng t nh vy (B'P, FP) = (B'I, AB). V th (A'P, B'P) = (A'P, FP) + (FP, B'P)

v u

u

LO

Y

X

A



= (A'I, AB) - (B'I, AB) = (A'I, B'I). y chng ta gi nh rng P F. Nu P = F th P D, E. Ta c (A'F, B'F) = (A'F, EF) + (EF, DF) + (DF, B'F). G7. Cho ABCD l mt t gic li c cc cnh BC v AD khng song song. rng c mt im E trn BC sao cho t gic ABED v AECD l ngoi tip c. Chng minh rng c mt im F trn AD cc t gic ABCF v BCDF ngoi tip c khi v ch khi AB song song vi CD. Li gii. Gi (C1) v (C2) l ng trn c tm O1 v O2 ngoi tip cc t gic ABED v AECD tng ng. Mt im F ch tn ti nh th nu (C1) v (C2) cng cng l cc ng trn ngoi tip ca cc t gic ABCF v BCDF tng ng. Gi giao im ca cc tip tuyn t k B n (C2) v t C n (C1) (khc vi BC) ct AD F1 v F2 tng ng. Ta cn phi chng minh rng F1 F2 khi v ch khi AB // CD.

(CC)

(CB)

(CA)

(C)

F

E

D

O

P

I

M

C'

B'

A'

CB

A

O2

O1

OF2F1

E

D

C

B

A



B . Cc vng trn (C1) v (C2) c tm O1 v O2 cng tip xc vi hai cnh ca gc xOy. Cc im P, S cng thuc mt cnh ca gc v cc im Q, R cng cnh kia ca gc sao cho RS, PQ l cc tip tuyn ca cc ng trn (C2), (C1) , mt khc (C1) ni tip tam gic OPQ v (C2) l ng trn bng tip gc O ca tam gic ORS. K hiu p = OO1.OO2. Ta c mt trong cc kt qu: OP.OR < p < OQ.OS, OP.OR > p > OQ.OS, OP.OR = p = OQ.OS. Tht vy, K hiu OPO1 = u, OQO1 = v, OO2R = x, y = OO2S, POQ = 2. V PO1, QO1 v RO2, SO2 l cc phn gic trong tam gic PQO v phn gic ngoi tam gic RSO, nn u + v = x + y (= 90 - ). (1) Theo nh l sin trong cc tam gic OO1P v OO2Q ta c:

2

1

sin( ) sin( ),

sin sin

OOOP u x

OO u OR x

Do u, x v l s o cc gc nhn, nn : 2

1

. sin sin( ) sin sin( ) sin( ) 0OOOP

OPOR p x u u x x u x uOO OR

Nh vy OP.OR p l tng ng vi x u, OP.OR = p khi v ch khi x = u. Tng t, p OQ.OS l tng ng v y, OQ.OS = p khi v ch khi v = y. Mt khc x u v v y l tng ng bi (1), vi x = u nu v ch nu v = y. B c chng minh. Tr li nh l : Vi cc b bn im {B, E, D, F1}, {A, B, C, D} v {A, E, C, F2}. Gi s OE.OF1 > p, chng ta c c

v

u

yx

O2

O1

S

R

Q

P

O



OE.OF1 > p OB.OD < p OA.OC > p OE.OF2 < p. Ni cch khc, OE.OF1 > p ta c: OB.OD < p < OA.OC v OE.OF1 > p > OE.OF2. Tng t nh vy, OE.OF1 < p ta c: OB.OD > p > OA.OC v OE.OF1 < p < OE.OF2. Trong nhng trng hp ny F1 F2 v OB.OD OA.OC, v vy cc ng thng AB v CD khng song song vi nhau . Trng hp OE.OF1 = p. B dn n OB.OD = p = OA OC v OE.OF1 = p = OE.OF2. Do F1 F2 v AB // CD. Bnh lun. Kt lun cng ng nu BC v AD l song song. Ngi ta c th chng minh mt trng hp gii hn b cho cc cu hnh hin th trong hnh bn di, vi O'r1 v O''r2 l tia song song, O'O'' r1r2 v O l trung im ca O'O''. Hai ng trn tm O1 v O2 tip xc O'r1 v O''r2. Cc ng thng PQ v RS l tip tuyn vi cc ng trn sao cho P, S thuc O'r1, v Q, R thuc O''r2, v vy m O, O1 v cng mt bn i vi PQ ; O, O2 v hai pha ca ca RS. t s = OO1+ OO2. Ta c mt trong nhng kt qu sau: O'P + O''R < s s > O''Q + O'S, O'P + O''R = s = O''Q + O'S.

S

RQ

P

O2 O1

r2

r1

O''

O'

O



2011. G1. Cho tam gic ABC nhn, ni tip ng trn tm O. (C) l mt ng trn m tm L ca n thuc cnh BC. Gi s (C) tip xc vi AB ti B' v AC ti C' v O thuc cung nh ' 'B C ca (C). Chng minh rng ng trn tm O ngoi tip tam gic ABC v (C) ct nhau ti hai im. Li gii. B', C' l hnh chiu ca L ln lt trn AB, AC v O nm trong tam gic AB'C', do COB



G3. Cho ABCD l mt t gic li vi cc cnh AD v BC khng song song nhau. Gi s rng ng trn ng knh AB v ng trn ng knh CD ct nhau ti E v F trong t gic. ng trn (CE) i qua hnh chiu vung gc ca E trn cc ng thng AB, BC, v CD. ng trn (CF) i qua hnh chiu vung gc ca F trn cc ng thng CD, DA v AB. Chng minh rng trung im ca EF thuc ng thng i qua hai im giao im ca hai ng trn (CE) v (CF). Li gii. K hiu l P, Q, R, S v cc hnh chiu ca E trn dng DA, AB, BC, v CD tng ng. Cc im P v Q nm trn ng trn ng knh AE, v vy QPE = QAE; Tng t QRE = QBE. V vy QPE + QRE = QAE + QBE = 900. Tng t SPE + SRE = 900, do ta c QPS + QRS = 900 + 900 = 1800. Suy ra t gic PQRS ni tip trong (CE). Tng t, c bn hnh chiu ca F ln c ng thng cha cnh t gic ABCD cng thuc ng trn (CF). K hiu l K giao im ca c ng thng AD v BC. Khng mt tnh tng qut trong gi nh rng A nm trn on DK. Gi s rng CKD 900. Cc im E, F khng th nm bn trong t gic ny. Tri gi thit. Do , cc ng thng EP v BC ct nhau ti im P', cc ng thng ER v AD ct nhau ti im R'. Cc im P' v R' cng thuc (CE). Tht vy, do cc im R, E, Q, B cng thuc mt ng trn nn QRK = QEB = 900 - QBE = QAE = QPE. V vy QRK = QPP', suy ra P' nm trn (CE). Tng t cho R'.

P'

R'

R

Q

P

F

E

N'

N

M'

MK

S

D

C

B

A



Gi M v N l hnh chiu ca ca F trn cc ng thng AD v BC tng ng, M' = FM BC, N' = FN AD. Bi nhng lp lun tng t trn, cc im M'v N' cng thuc (CE). Xem hnh di y: U = NN' PP', V = MM' RR'. Do cc im N, N ', P, P' cng thuc mt ng trn, do UN.UN' = UP.UP'. Suy ra U thuc trc ng hng ca (CE) v (CF). Tng t cho V. Cui cng, v EUFV l mt hnh bnh hnh, nn UV i qua trung im ca EF.

G4. Cho ABC l mt tam gic nhn. B0 l trung im ca AC v C0 l trung im ca AB. Gi D l hnh chiu A v G l trng tm ca tam gic ABC, (C0) l ng trn i qua B0 v C0 v tip xc vi ng trn (C) ngoi tip tam gic ABC ti X khc A. Chng minh rng cc im D, G, X thng hng.

Li gii 1. Nu AB = AC th li gii l tm thng. V vy m khng mt tnh tng qut, gi s AB < AC.

Php v t tm A t s 12

bin ng trn (C) thnh ng trn (C1) i qua B0

v C0 v tip xc vi ng trn (C) ngoi tip tam gic ABC ti X khc A. Tip tuyn ti A ca (C) v (C1), tip tuyn ti A ca (C) v (C0), ng thng B0C0 l ba trc ng phng nn ng quy ti W. Cc im A v D l i xng i qua ng thng B0C0 vi; do WX = WA = WD. Suy ra W l tm ca ng trn ngoi tip ca tam gic ADX. Hn na, chng ti c WAO = WXO = 900 do AWX + AOX = 1800. K hiu l T l giao im th hai ca (C) v ng thng DX. Lu rng O thuc (C1) .

V

U

A

B

C

D

S

KM

M'

N

N'

E

F

P

Q

R

R'

P'



Ta c DAT = ADX - ATD = 12

(3600 -AWX) - 12AOX =

1800 - 12

(AWX + AOX) = 900. V vy, AD AT, v do AT // BC.

Nh vy, ATCB l mt hnh thang cn. Gi A0 l trung im ca BC, v xt nh ca ATCB qua php v t h tm G

v t s - 12

, ta c h (A) = A0, h (B) = B0, v h (C) = C0 .

T tnh cht i xng qua B0C0, ta c TCB = CBA = B0C0A = DC0B0. Mt khc AT // DA0 suy ra h (T) = D. Do cc im D, G, v T thng hng, v X nm trn cng ng thng . Li gii 2. Gi cc im A0, O, v W nh trong Li gii 1 v cng gi s AB < AC. Gi Q l hnh chiu vung gc ca A0 trn B0C0. Do WAO = WQO = OXW = 900 nn nm im A, W, X, O, v Q nm trn mt ng trn. Ta c WQD = AQW = AXW = WAX = WQX. Do ba im Q, D, v X nm trn mt ng thng d.

O

(C)

(C0)

(C1)

X

C0B0

A0

G

T

W

D CB

A



Cui cung, php v t tm G t s - 12

bin tam gic ABC thnh tam gic

A0B0C0 v do bin D thnh Q. Do , cc im D, G, v Q thng hng v X nm trn ng thng . Bnh lun. C rt nhiu cch khc chng minh Q, D, v X thng hng . V d, gi J l giao ca AW v BC. Ta chng minh c A, D, X, v J cng cch u W v vy ADXJ t gic ni tip. Kt hp vi AWXQ l t gic ni tip ta c JDX = JAX = WAX = WQX. V BC // WQ, nn Q, D, v X l thc s thng hng. G5. Cho ABC l mt tam gic vi ng trn ni tip tm I v ng trn ngoi tip (C). Cho D v E l giao im th hai ca (C) vi cc tia AI v BI tng ng. DE ct AC ti im F, v ct BC ti mt im G. Cho P l giao im ca ng thng i qua F song song vi AD v ng thng qua G song song vi BE. Gi s rng cc tip tuyn ca (C) ti A v B gp nhau ti im K. Chng minh rng ba ng thng AE, BD, v KP l song song hoc ng quy. Li gii 1. Thy ngay IAF = DAC = BAD = BED = IEF suy ra AIFE l t gic ni tip. K hiu ng trn ngoi tip AIFE l (C1). Tng t nh vy, t gic BDGI l ni tip; k hiu ng trn ngoi tip BDGI l (C2). ng thng l trc ng phng ca (C) v (C1), ng thng BD l trc ng phng ca (C) v (C2). Gi t l trcng phng ca (C1) v (C2). Ba trc ng phng ny ng quy ti tm ng phng ca chng hoc l song song vi nhau. Ta s chng minh rng t ng thng PK. Gi L l giao im th hai ca (C1) v (C2). Khi t = IL. (Nu hai ng trn l tip xc vi nhau th L = I v t l tip tuyn chung) Gi K' L v P' L l giao im ca t vi cc ng trn ngoi tip cc tam

O

J

A

B CD

W Q

G

A0

B0C0

X

(C1)

(C0)

(C)



gic ln lt l ABL v BDL. Ta c: (AB, BK') = (AL, LK') = (AL, LI) = (AE, EI) = (AE, EB) = (AB, BK). Suy ra BK' // BK. Tng t nh vy chng ta c AK'k//AK, v do K' K. Tip theo, chng ta c: (P'F, FG) = (P'L, LG) = (IL, LG) = (ID, DG) = (AD, DE) = (PF, FG). Do P'F // PF v tng t P'G // PG. Do P' = P. iu ny c ngha rng t i qua K v P. Chng minh c kt thc. Solution 2. Gi M l giao im ca cc tip tuyn vi (C) ti D v E, gi T l giao im ca cc ng thng AE v BD. Nu AE v BD song song, cc im K v M nm trn ng TI (nh l mt h qu ca nh l Pascal, p dng cho cc hnh lc gic thoi ha ghi AADBBE v ADDBEE). Cc dng AD v BE l phn gic ca cc gc CAB v ABC, tng ng, v vy D v E l trung im ca vng cung BC v CA ca ng trn (C). Do , DM song song vi BC v EM l song song vi AC. p dng nh l Pascal cho lc gic suy bin CADDEB. Theo nh l, cc im F = CA DE, I = EB cho ta FI // BC // DM. Tng t, cc ng thng GI //AC // EM.

Xt php v t tm H t s FGED

bin E thnh G v bin D thnh F. V tam

gic EDM v tam gic GFI c cc cnh tng ng song song, nn php v t ny bin M thnh I. Tng t nh vy, do cc tam gic DEI v FGP c cnh tng ng song song, nn php v t ny bin I thnh P. Do , cc im M, I, P v H thng hng. Do , im P cng thng hng vi T, K, I, M.

(C2)

(C1)

(C)

D

IL

F

E

P'P

K'

K

W

CB

A



Bnh lun. Ngi ta c th chng minh rng IF//BC v IG//AC mt cch c bn hn. Do ADE = EDC v DEB = CED nn cc im I v C l i xng qua DE. Hn na, cung EA EC v DB DC ta c CFG = FGC, v vy tam gic CFG cn. Do , t gic IFCG l mt hnh thoi. G6. Cho ABC l mt tam gic cn AB = AC, v D l trung im ca AC. Phn gic ca BAC ct ng trn qua D, B, v C ti im E bn trong tam gic ABC. ng thng BD ct ng trn qua A, E, v B hai im B v F. Cc ng thng AF v BE ct nhau ti im I, cc ng thng CI v BD ct nhau ti im K. Chng minh rng I l tm ng trn ni tip ca tam gic KAB. Solution 1. Gi D' l trung im ca on AB, v M l trung im ca BC. V DD'BC l hnh thang cn nn ni tip c trong ng trn (C). V cung D'E v cung ED ca ng trn (C) bng nhau nn ta c ABI = D'BE = EBD = IBK, v vy I thuc phn gic gc ABK. Ta cn chng minh I thuc phn gic gc BAK.

T DFA = 1800 - BFA = 1800 - BEA = MEB = 12CEB = 1

2CDB

suy ra DFA = DAF nn tam gic AFD cn vi DA = DF. p dng nh l Menelaus cho tam gic ADF i vi ng thng CIK, cng vi cc ng dng nh l phn gic trong tam gic ABF, chng ta suy ra:

1 . . 2 . 2 . .2

AC DK FI DK BF DK BF DK BF

CD KF IA KF AB KF AD KF AD v do :

1 1BD BF FD BF KF DF AD

AD AD AD DK DK DK

Suy ra cc tam gic ADK v BDA l ng dng nn do DAK = ABD. Cui cng IAB = AFD - ABD = DAF - DAK = KAI suy ra im K thuc phn gic gc BAK.

A

(C)

PH

G

FI

K

D

E

B

A



Solution 2. Ta chng minh rng KI chia i AKB.

F

(C2)

(C1)

E

K

M

ID' D

CB

A

O3

O1

F

A

B C

DD' I

M

K

E

(C1)

(C3)



K hiu ng trn ngoi tip ca tam gic BCD v tm ca n bng (C1) v O1, tng ng. T t gic ABFE ni tip c, ta c DFE = BAE = DAE. Tng t ta c EAF = EBF = ABE = AFE. V vy DAF = DFA, v do DF = DA = DC. Suy ra, tam gic AFC c ghi trong ng trn (C2) vi tm D. K hiu trn ngoi tip tam gic ABD bng (C3), v tm ca n l O3. V trong (C1) ta c EB EC v cc tam gic ADE v FDE l ng dng, nn AO1B = 2BDE = BDA. V vy O1 nm trn (C3). Do O3O1D = O3DO1. ng thng BD l trc ng phng ca (C1) v (C3). im C thuc trc ng phng ca (C1) v (C2) suy ra AI.IF = BI.IE. Do K = BDCI l tm ng phng ca (C1), (C2) v (C3). ng thng AK l trc ng phng ca (C2) v (C3). Cc ng thng AK, BK v IK vung gc vi ng thng i qua hai tm O3D, O3O1 v O1D, tng ng. V O3O1D = O3DO1, ta nhn c KI l phn gic gc AKB. Solution 3. Mt ln na, gi M l trung im ca BC. Nh trong cc cch gii trc, ta c th suy ra ABI = IBK. Chng ta thy rng im I nm trn ng phn gic gc KAB. Cho G l giao im ca ng trn AFC v BCD, khc C. Cc ng thng CG, AF v BE l trc ng phng ca ba vng trn AGFC, CDB, v ABFE. V vy I = AF BE l tm ng phng ca ba vng v CG cng i qua I. Gc gia ng thng DE v tip tuyn ca ng trn (BCD) ti E bng EBD =EAF = ABE = AFE. Tip tuyn ca ng trn (BCD) ti E

B'

K

G

F

A

B C

DD'

I

M

E

(C1)



vung gc vi AM, ng thng DE vung gc vi AF. Tam gic AFE cn, do DE l ng trung trc ca on AF v do AD = DF. Suy ra, im D l tm ca ng trn (AFC), v ng trn ny i qua M v AMC = 900. Gi B' l im i xng ca B qua D, v vy ABCB' l mt hnh bnh hnh. V DC = DG chng ti c GCD = DBC = KB'A. Do , AKCB l t gic ni tip v do CAK = CB'K = ABD = 2MAI.

Cui cng l IAB = MAB - MAI = 12CAB - 1

2CAK = 1

2KAB

v do AI l phn gic gc KAB.

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