bmo 2012 problem and solution

8/9/2019 BMO 2012 Problem and Solution

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A B C Γ O

ABC > 90◦ D AB

AC C D AO E

AC F Γ

D E

BF E CF D F

(x + y) (z + x)(z + y) ≥ 4(xy + yz + zx ),

x y z

(x + y) (z + x)(z + y) + ( y + z ) (x + y)(x + z ) + ( z + x) (y + z )(y + x).

n P n = {2n , 2n − 1 · 3, 2n − 2 · 32 , . . . , 3n }

X P n S X X S = 0 y 0 ≤ y ≤ 3n +1 − 2n +1

Y P n 0 ≤ y − S Y < 2n

Z +

f : Z + → Z +

f (n !) = f (n )! n

m − n f (m ) − f (n ) m n



∩ AO = {K } G Γ

A D,C,G E ADG

GE AD G E B

CDF = GDK = GAC = GF C F G

CF D F F BE = F BG = F AG = GF K = GF E F G

BF E F CF D BF E

F



(x + y) (z + x)(z + y) ≥ 2xy + yz + zx

x y z

(x + y)2 (z + x)(z + y) ≥ 4x2 y2 + y2 z 2 + z 2 x2 + 4 xy2 z + 4x2 yz + 2 xyz 2

x3 y + xy3 + z (x3 + y3 ) ≥ 2x2 y2 + xyz (x + y)

(xy + yz + zx)(x −y)2

≥ 0,

x,y,z

√ x + y, √ y + z, √ z + x K = √ xy + yz + zx/ 2

K = 12√ x + y√ z + x sin α,

α √ x + y √ z + x

cos α = x + y + z + x −y −z 2 (x + y)(z + x)

= x (x + y)(z + x)

sin α = √ 1 −cos2 α =√ xy + yz + zx

(x + y)(z + x).

√ x + y√ y + z √ z + xcyc

√ x + y ≥ 16K 2 .

√ x + y√ y + z √ z + x4K ≥ 2

K cyc √ x + y/ 2

R ≥ 2r



α = 3/ 2 1 + α > α 2

y Y Y = S = 0 i = 0 m

S Y + 2 i 3m − i ≤ y Y Y {2i 3m − i}.

Y P m S Y ≤ y

P m

2m , 2m α, 2m α 2 , . . . , 2m α m Y

Y

1 + α > α 2 Y

Y = P m y = 3m +1 −2m +1

Y

2m Y (S Y −2m ) + 2 m − 1 3 > y

y −S Y < 2m 2m

Y y −S Y < 2m

3m +1 − 2m +1 = (3 − 2)(3m + 3 m − 1 · 2 + · · · + 3 · 2m − 1 + 2 m ) = S P m

P m 2m

m a = 3/ 2 Qm = {1,a ,a 2 , . . . , a m }

x 0 ≤ x ≤ 1 + a + a2 + · · ·+ am X Qm

0 ≤ x −S X < 1

m

m = 1

S = 0

S {1 } = 1

S {a } = 3/ 2

S {1 ,a } = 5/ 2

m x 0 ≤x ≤ 1 + a + a2 + · · ·+ am +1 0 ≤ x ≤ 1 + a + a2 + · · ·+ am

X Qm Qm +1 0 ≤ x −S X < 1

am +1 −1a −1

= 1 + a + a2 + · · ·+ am < x x > a m +1

am +1

−1

a −1 = 2(am +1

−1) = am +1

+ ( am +1

−2) ≥ am +1

+ a2

−2 = am +1

+ 14 .

0 < (x −am +1 ) ≤ 1 + a + a2 + · · ·+ am

X Qm 0 ≤ (x −am +1 ) −S X < 1 0 ≤ x −S X < 1

X = X {am +1 } Qm +1



Z +

f

a ≥ 3

f (a) = a

a!, (a!)!, · · · f f

a1 < a 2 < ·· · < a k < ·· · n ak −n ak −f (n) =f (ak ) −f (n) k Z +

ak −n ak −n ak −f (n) −(ak −n) =n −f (n) f (n) = n f = Z +

f 2 p ≥ 5

( p − 2)! ≡ 1 p p ( p − 2)! − 1

( p−2)! −1 f (( p−2)!) −f (1) p f (( p−2)!) −f (1) = ( f ( p−2))! −f (1)

f (1) = 1 f (1) = 2 p ≥ 5 p (f ( p − 2))! − f (1) f ( p

− 2) < p p

( p − 1)! − 1 ( p − 1)! − 2 f ( p − 2) ≤ p − 2

p−3 = ( p−2) −1 f ( p−2) −f (1) ≤ ( p−2) −1 f ( p−2) = f (1) f ( p−2) = p −2 p−2 ≥ 3 f

f ( p−2) = f (1) p ≥ 5

n p−2 −n f ( p−2) −f (n) = f (1) −f (n) p ≥ 5 f (n) = f (1) f 1 2

f (n 0 ) = n0 m−n 0 |f (m)−m m Z + f (n 0 ) = n0

n 0 Z + f (m) −m f (m) = m

m Z + f (n 0 ) = n 0 n0 ≥ 3 f

(nk )∞k =0

nk = nk − 1 !

f (3) = 3

f (n) = n n Z +

f (3) = 3 f (1) = f (1)! f (2) = f (2)! f (1), f (2) {1, 2}

4 = 3!−2|f (3)!−f (2) f (3) = 3 f (3) {1, 2} n > 3

n!−3|f (n)!−f (3) 3 f (n)! f (n)! {1, 2} f (n) {1, 2} n Z +

f m, n {f (n), f (m)} = {1, 2}

k = 2 + max {m, n } f (k) = f (m) k − m|f (k) − f (m)

|f (k) −f (m)| = 1 k −m ≥ 2

f ≡ 1, f ≡ 2, f = Z +

bmo 2012 problem and solution

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