[PDF] IMO 2018 Solution Notes - Evan Chen




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[PDF] IMO 2018 Solution Notes - Evan Chen

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14273_6IMO_2018_notes.pdf Ѿ䁼㢱᜽ѿ dzǴ dzǴ dzR Ǵ Ȝ ABC DE

ABAC AD=AE BD

CE ABAC FG

DEFG n3 a1;a2;:::;an aiai+1+ 1 =ai+2 i= 1;2;:::;n n 1 10 4 26
571
83109
2018

11 + 2 ++ 2018

(x;y) x;y2 f1;:::;20g

400

p5 K K a1a2 N nN a1 a2+a2 a3++an1 an+an a1 (an) ABCDABCD=BCDA X ABCD \XAB=\XCD\XBC=\XDA: \BXA+\DXC= 180

Ȝ

Ȝ

ABC DE

ABAC AD=AE BD

CE ABAC FG

DEFG

ǥ AXFDAEGY

XY XFkAB M

ÔXFdAB NA

BCFMN G DEXY XF=AD=AE=Y G ÔXFdY G

ÔMFÔNG MNkFG

MNDE \A Ƕ

ǥ bcfga

da= 

2f+a+babf

2b  a=fab f ea=gac g AD=AE ea da 2 =c b)(g2ac)2 (f2ab)2=g2c f2b )bc(bg2cf2)a2=g2f4cf2g4b=f2g2(f2cg2b) )bca2= (fg)2)  fg a 2 =bc: fg a X AX?FG FG \A \A

ǥ ` \A DF

G0 FG0?`E0 AC GE0=GC ě AD=AE0 E=E0

FG0?` \FBD=1

2C+x\GCA=1

2B+x x FGkMNMN \FAB=1

2Cx\GAC=1

2Bx

R

BF= 2R

1 2Cx  :

BD= 2BF

1 2C+x  = 4R 1 2C+x  1 2Cx 

DA=ABBD= 2RC4R

1 2C+x  1 2Cx  = 2R  C2 1 2C+x  1 2Cx  = 2R[C(C2x)] = 2R2x: AE0= 2R2x

FG0kDE G=G0

ǥ FDGE J

K A BCDE FGJ K 4BFD 4JAD FB=FD AJ=AD

4CGE 4KAEGC=GE AK=AE

AK=AE=AD=AJ;

DEJK A

]KED=]KJD=]KJF=]KGF; Ƕ KJ GK ACK ABC AK=AD=AE(AB;AC)

K AB J AC K6=J:

Ȝ

n3 a1;a2;:::;an aiai+1+ 1 =ai+2 i= 1;2;:::;n n 3jn (1;1;2;1;1;2;:::)

ǥ aiai+1ai+2

aiai+1ai+2= [ai+21]ai+2=a2i+2ai+2 =ai[ai+31] =aiai+3ai: a2i+2ai+2=aiai+3ai X i a2i+2=X i aiai+3()X (aiai+3)2= 0: 3

1 x2+ 1 =x 3jn

ǥ an

+ >1 ++ a1= 0 a2= 0a3>0a4>0 ++ ++ a1=x <0 a2=y <0 a3= 1 +xy >1 a4= 1y(1 +xy)<0 a5= 1 + (1 +xy)(1y(1 +xy))<1: a6a5= (1 +a5a4)(1 +a3a4) =a4(a5a3)>0 a5>1> a3 ++ ++++::: a1a2+ 1 =a3<0< a4= 1 +a2a3)a1< a3 a2>0 a1< a3< a5< ::: + ++ +

Ȝ

1 10 4 26
571
83109
2018

11 + 2 ++ 2018

n= 2018N= 1 + 2 ++n d d d=ab d!a A 1 + 2 ++n

N= 1 + 2 ++n f1;2;:::;ng B

A BC DXY XY B B C C D bn/21c n+1n+2 1;2;:::;n A!B D (n+ 1) + (n+ 2) ++ (n+ (bn/21c))>1 + 2 ++n n2018

Ȝ

Ȝ

(x;y) x;y2 f1;:::;20g

400

p5 K K K= 100 100

200

1

2200 = 100

100
44 44 2 664

1 2 3 4

3 4 1 2

2 1 4 3

4 3 2 1

3 775
4 1/4

Ȝ

a1a2 N nN a1 a2+a2 a3++an1 an+an a1 (an)

S(n) =an+1an

a1+an an+1 n > N p p p n > N p(an)< p(an+1) p(an+1) =p(a1) p(an) =p(an+1) p(an)> p(an+1) p(an+1)p(a1) ěp p(aN+1)p(aN+2) p(aN+1)p(aN+2)::: K > N p(aK)< p(aK+1) =p(aK+2) ==p(a1) dzǴ p(a1) p(a1)>0 ν (a1)n > N pa1 p(an) pja1 an+1jan n p an2Zan2Zp p xn p(xn) xn xn n

Ƕ

an/a1 n=N+ 1;N+ 2;::: 1 1

Ȝ

ABCDABCD=BCDA X ABCD \XAB=\XCD\XBC=\XDA: \BXA+\DXC= 180

ǥ

ABCD=BCDA

ě X

ABCD \D=\A\A=\B

DABCABCD:

DABC ABCD X X X ABCD \BXA+\DXC= 180 XX

ǥ dz Ǵ

AB AD=CB CD BA BC=DA DC BD !ACAC A

C !BDBD

PABCD P 1 ABCD 1

A0B0=PA

ABPB=PC

BCPB=1

B0C0 A0B0=B0C0 B0C0=C0D0=D0A0 ]XAB=]XCD ]XAB=]XAP+]PAB=]PX0A0+]A0B0P ]XCD=]XCP+]PCD=]PX0C0+]C0D0P ]A0X0C0=]A0B0P+]PD0C0=](A0B0;B0P) +](PD0;C0D0) =](A0B0;B0P) +](PD0;A0B0) =]D0PB0:

A0B0kC0D0

]B0X0D0=]A0PC0: ]AXB=]AXP+]PXB=]PA0X0+]X0B0P ]CXD=]CXP+]PXD=]PC0X0+]X0DP0 ]AXB+]CXD= (]PA0X0+]X0B0P) + (]PC0X0+]X0D0P) =]A0X0P+]X0PA0+]PX0B0+]B0PX0 ]C0X0P+]X0PC0+]PX0D0+]D0PX0 =]PX0A0+]BX0P+]PX0C0+]D0X0P +]A0PX0+]X0PB0+]C0PX0+]X0PD0 =]A0PB0+]C0PD0+]B0X0C+]D0X0A ]A0PB0+]C0PD0+]B0X0C+]D0X0A= 0: A B CDX YQ P

Q P Y (BQC)

(AQD) ]AXC=]DPB=]BQD=]BQY+]Y QD=]BCY+]Y AD =](BC;CY) +](Y A;AD) =]Y CA=]AY C:

XACY XBDY

ěX6=Y AB < AC Q !BD T BA(QBC) QB=QD=QT=pQAQC \BAD <180 (QBCY)

ABCD ABCD

X ]DXA=]DXY+]Y XA=]DBY+]Y CA =](DB;Y B) +](CY;CA) =]CY B+ 90 =]CQB+ 90=]APB+ 90: ]BXC=]DPC+ 90: X X Y
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