Nguyễn Hữu Điển

8/14/2019 Nguyn Hu in

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Nguyn Hu in

PHNG PHP GII TON

NH XUT BN GIO DC


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51GD-05

89/176-05 M s: 8I092M5


3/27

Li ni u

Sau khi mt lot cun sch v phng php gii ton c bn c n nhn

[]-[], nhng cun sch ny lin tc c ti bn v nhiu bn c khen hn l ch.iu ng vin ti thc hin bin tp cun sch ny, ng nh tn ca cun sch

l tuyn tp cc phng php v cc chuyn gii ton ch khng phi tuyn tpcc bi ton hay. Ta bit rt nhiu phng php hay c ti bin tp trongcc cun []-[], sau mt thi gian tm hiu k hn na th ti thy cc phng php

ny gii c rt nhiu dng bi ton khc nhau, trong tay ti c rt nhiu ti

liu m nhng cun sch trc khng c c. Ti bin tp cun sch ny cngc cc phng php gii ton m cc cun sch trc th hin v a thm mt

s phng php khc, cch nhn khc v vic gii ton. c ti liu ny cc bn

s thy tuy l phng php gii ton khc nhau nhng n c mt t tng thngnht l suy lun c l. S bi tp hay dng cc phng php gii khc nhau l v

cng nhiu, nn tt c nhng bi ton trong cc cun trc y ti khng a voy. Ti c gng chn nhng bi ton hay, mi vo tuyn tp ny. Nu c nhng biton trng vi cc tp sch trc th s c mt cch gii hon ton mi, bn c c

th so snh vi nhng cch gii c. Cun sch c chia lm hai phn ln:

Phn I. Cc phng php gii ton.

1. Phng php chng minh bng phn chng.

2. Phng php dng v d, phn v d v xy dng li gii.

3. Phng php nguyn l irichle

4. Phng php quy np ton hc

5. Phng php dng i lng bt bin6. Phng php dng i lng cc bin

7. Phng php t mu


4/27

4 Li ni u

8. Cc phng php khc

Phn II. Nhng chuyn c bn

1. T hp ri rc

2. L thuyt s

3. Bt ng thc4. Dy s

5. a thc

6. Phng trnh hm

7. Hnh hc

8. Thut ton v tr chi

Mi phn trn u c trin khai t d n kh v mt lgic c l cao. Cc bitp v v d c gii cn thn v d hiu nht. Bn c c th tm thy nhng li

gii khc hay hn, ngn hn nhng nhm mc ch m t phng php gii ton

nn y c th di hn. Phn cui ca mi chng l li gii ngay cc bi tp trongchng , nh s cc v d, bi tp l ln lt cng nhau cho n ht chng.

Cun sch dnh cho hc sinh ph thng yu ton, hc sinh kh gii mn ton,

cc thy c gio, sinh vin i hc ngnh ton, ngnh tin hc v nhng ngi yuthch ton hc ph thng. Trong bin son khng th trnh khi sai st v nhm

ln mong bn c cho kin. Mi gp gi v a ch: Ban bin tp sch Ton,

Nh xut bn Gio dc, 187b Ging V, H Ni.

Tc gi cm n ban bin tp Ton - Nh xut bn Gio dc H Ni ht sc

gip cun sch c in ra.

H Ni, ngy 2 thng 11 nm 2006

Nguyn Hu in


5/27

Nhng k hiu

Trong cun sch ny ta dng nhng k hiu vi cc ngha xc nh trong bng

di y:

N tp hp s t nhinN tp hp s t nhin khc 0Z tp hp s nguynQ tp hp s hu tR tp hp s thcC tp hp s phc du ng d dng v cng (tng ng vi +) m v cng tp hp rngCkm

t hp chp k ca m phn t... php chia ht

... khng chia htUCLN c s chung ln nhtBCNN bi s chung nh nhtdeg bc ca a thcIMO International Mathematics OlympiadAPMO Asian Pacific Mathematics Olympiad


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Mc lc

Li ni u . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

Nhng k hiu . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

Mc lc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

Chng I. thi olympic irland . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

I.1. Gii thiu Irland . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

I.2. Bi tp . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

Li gii bi tp chng 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

Chng II. Bi ton t Hn Quc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

II.1. Gii thiu . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12II.2. Bi tp . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

II.3. Li gii bi tp chng 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

Chng III. Cc thi olympic Canada . . . . . . . . . . . . . . . . . . . . . . . 17

III.1. Gii thiu . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

III.2. Bi tp . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

III.3. Li gii bi tp chng 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

Chng IV. Cc bi ton Rumania . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

IV.1. Gii thiu . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

IV.2. Bi tp . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20


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MC LC 7

IV.3. Li gii bi tp chng 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

Chng V. Cc bi ton Th Nh K . . . . . . . . . . . . . . . . . . . . . . . . . . 24

V.1. Gii thiu phng php . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

V.2. Bi tp . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

V.3. Li gii bi tp chng 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

Ti liu tham kho . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27


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Chng I

thi olympic irland

I.1. Gii thiu Irland . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

I.2. Bi tp . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

Li gii bi tp chng 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

I.1. Gii thiu Irland

I.2. Bi tp

Bi tp I.1. Tm tt c cc cp s nguyn (x, y) sao cho 1 + 1996x + 1998y = xy

Bi tp I.2. Cho ABC, M l im trong tam gic. Goi D,E,F ln lt l hnh

chiu ca M xung BC, CA, AD. Tm tp hp tt c cc im M tha mnF DE = 2

Bi tp I.3. Tm tt c cc a thc P(x) sao cho i vi mi x ta c :

(x 16) P(2x) = 16 (x 1) P(x)

Bi tp I.4. Cho a,b,c l cc s thc khng m sao cho a + b + c abc. Chngminh rng a2 + b2 + c2 abc


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Li gii bi tp chng 1 9

Bi tp I.5. Cho tp hp S = {3, 5, 7,...}. Vi mi x S ta t (x) l xc nhmt s nguyn duy nht sao cho: 2(x) < x < 2(x)+1

i vi a, b S ta nh ngha php ton

a

b = 2(a)1 (b

3) + a

a, Chng minh rng nu a, b S th a b Sb, Chng minh rng nu a,b,c S th (a b) c = a (b c)

Bi tp I.6. Cho t gic li ABCD c mt ng trn ni tip. Nu

A = B =2

3, D =

2, BC = 1

Tm di AD

Bi tp I.7. Gi A l tp con ca

{0, 1, 2,..., 1997

}gm hn 1000 phn t. Chng

minh rng A ch gm nhng ly tha ca 2 hoc hai phn t phn bit c tng lly tha ca 2

Bi tp I.8. Xc nh s t nhin n tha mn nhng iu kin sau:

a, Khai trin thp phn ca n gm 1000 sb, Tt c cc s trong khai trin l s l.c, Hai phn t bt k lin nhau trong khai trin ca n hn km nhau 2 n v

Li gii bi tp chng 1

I.1 Ta c: (x 1998)(y 1996) = xy 1998y 1996x + 1996.1998 = 19972Do 1997 l s nguyn t, nn ta c: x 1998 = 1; 1997; 19972. Vy c 6 gi tr (x, y)tha mn l

(x, y) =

1999, 19972 + 1996

,

1997, 19972 + 1996 ,(3995, 3993) , (1, 1) 19972 + 1998, 1997 , 19972 + 1998, 195

I.2 T cc t gic ni tip MDBF v MDCE ta c MDE = MCE v M DF = M BE do F DE = 2 M CB + MBC = 6 hay BM C = 56 M nm trn cung trn i qua Bv C

I.3 Goi d = degP v a l h s ca x trong P(x) vi s m ln nht. Khi h s ca xm ln nht bn tri l 2da phi bng 16a do d = 4Do v phi lc ny chia ht cho (x 1), nhng trong trng hp v phi li chia htcho (x 2), tng t l chia ht cho (x 4) v (x 8). Vy a thc P(x) l bi ca(x 1)(x 2)(x 4)(x 8) l tt c cc a thc tha mn.


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10 thi olympic irland

I.4 Gi s phn chng rng vi a,b,c > 0 m a2 + b2 + c2 < abc do abc > a2 a < bc.Lm tng t ta cng c b < ca,c < ab. Do abc a2 + b2 + c2 ab + bc + ca. Theo btng thc AM-GM v ab + bc + ca > a + b + c suy ra abc > a + b + c. Tri vi gi thit. Vybi ton c chng minh.

I.5 a, Hin nhin

b, Nu 2m

< a < 2m+1

, 2n

< b < 2n+1

tha b = 2m1 (b 3) + a 2m1 (2n 2) + 2m + 1 = 2n+m1 + 1

v a b 2m1 2n+1 4 + 2m+1 1 = 2m+n 1. V vy (a b) = m + n 1Nu 2p < c < 2p+1 th

(a b) c = 2m1 (b 3) + a c = 2m+n2 (c 3) + 2m1(b 3) + aV

a (b c) = a 2m1 (c 3) + b = 2m1 2n1(c 3) + b 3+ a = (a b) cI.6 Goi I l tm ng trn ni tip . Do ABC l tam gic u, BI C = 1050, IC B =

150,AID = 750, IDA = 450 nn

AD =BI

BC

AD

AI=

sin150

sin 1050sin750

sin450=

2sin150

I.7 Gi s tp A khng tha mn bi ton. Khi A s bao gm hn na s nguyn t51 ti 1997 m chng c chia thnh tng cp c tng l 2048 (V D : 51 + 1997 = 2048...).Tng t nh vy, A bao gm nhiu nht na s nguyn t 14 ti 50, gm nhiu nht nas nguyn t 3 ti 13, v c th c s 0, do A c tng cng 937 + 18 + 5 + 1 = 997 snguyn, tri vi gi thit A gm hn 1000 s nguyn t tp {0, 1, 2, ..., 1997}

I.8 t an, bn, cn, dn, en l s trong khai trin ca n, l nhng s l v hai s lin tipkhc nhau 2 n v do tn cng theo th t l 1, 3, 5, 7, 9 do

0 1 0 0 01 0 1 0 00 1 0 1 00 0 1 0 10 0 0 0 1

anbncndnen

=

an+1bn+1cn+1dn+1en+1

Gi A l ma trn vung trong biu thc . Ta tm gi tr ring ca ca A, gi s Av = vvi v = (v1, v2, v3, v4, v5). Do

v2 = v1v3 = v2 v1 =

2 1 v1

v4 = v3 v2 =

3 2 v1v5 = v4

v3 = 4 3

2 + 1 v1v v4 = v5, do 5 33 + = 3 2. Gii pt ny ta c = 0, = 1, =

3

tng ng ta c cc vect ring x1, x2, x3, x4, x5 l

(1, 0, 1, 0, 1) , (1, 1, 0, 1, 1) , (1, 1, 0, 1, 1) ,

1,

3, 2,

3, 1


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Li gii bi tp chng 1 11

v

(1, 1, 1, 1, 1) =1

3x1

2 +

3

6x4 +

2 36

x5

V vy

(a1000, b1000, c1000, d1000, e1000) = 3999

2

2 +

3

61,

3, 2,

3, 1

2 36

1,

3, 2,

3, 1

=

3499, 2.3499, 2.3499, 2.3499, 3499

V th kt qu ca bi ton l 8.3499


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Chng II

Bi ton t Hn Quc

II.1. Gii thiu . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

II.2. Bi tp . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

I I.3. Li gii bi tp chng 2 . . . . . . . . . . . . . . . . . . . . . . . . . 14

II.1. Gii thiu

II.2. Bi tp

Bi tp II.1. Ch ra rng vi mi s nguyn t cho trc p th tn ti nhng s t

nhin x,y,z, tho mn x2 + y2 + z2- .p=0 v 0


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II.2 Bi tp 13

P,Q,R,S cng nm trn mt ng trn.

A

B

C

D

P

Bi tp II.4. Cho p l mt s nguyn t sao cho p 1 (mod4). Hy tnhp1k=1

2k2

p

2k2

p

Bi tp II.5. Xt nhng hnh L sau y, mi hnh c to bi bn hnh vungn v ghp li.

Cho m v n l cc s t nhin ln hn 1. Chng minh rng mt hnh ch nht

kch thc mxn s c xp bi cc hnh cho khi v ch khi m.n l bi s ca 8.

Bi tp II.6. Cho nhng s thc a,b,c,x,y,z tho mn abc > 0 v xyz > 0.Chng minh rng:

a2x2

(by+cz)(bz+cy)+ b

2y2

(cz+ax)(cx+az)+ c

2z2

(ax+by)(ay+bx) 34


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14 Bi ton t Hn Quc

II.3. Li gii bi tp chng 2

II.1 Vi trng hp p = 2, ta c th ly x = 0, y = z = = 1.

By gi ta xt trng hp p > 2. Trc tin ta xt trng hp 1 l ng d bnhphng modun p, khi tn ti mt s t nhin a, 0 < a < p 1 sao cho a2 1(modp).

B (x, y, z) = (0, 1, a). V x2

+ y

2

+ z

2

= a

2

+ 1 chia ht cho p nhng 1 + (p 1)2

< p

2

nntn ti {1, 2,...,p 1} sao cho x2 + y2 + z2 .p = 0 .Tip theo, gi s (1) khng l ng d bnh phng modunp. Ta phi tm mt s k

no c k v p k 1 u l ng d bnh phng. Nu p12 l ng d bnh phngth chn k=p12 . Nu ngc li, th mi ng d trong s

p12 cc ng d bnh phng

khc khng s ri vo trong cc cp {1, p 2}, {2, p 3},...,p32 , p+12 . Theo nguyn lPigeonhole Principle s c mt cp (k, p k 1) m c hai s k v (p k 1) u l ngd bnh phng nh ta nh tm.V vy, ta c th chn x, y 0, 1, ..., p12 sao cho x2 k(modp) v y2 p k 1(modp).Cho z = 1, ta c x2 + y2 + z2 chia ht cho p v x2 + y2 + z2 < p2. Gi tr s c xcnh nh trng hp trc.

II.2 Cho x = y, ta c f(0) = 0.Cho x = 1, y = 0 ta c f(1) = f(1).Cho x = a, y = 1, sau cho x = a, y = 1 ta c:

f(a2 1) = (a 1) [f(a) + f(1)]f(a2 1) = (a + 1) [f(a) f(1)]

Cho cc v phi ca cc phng trnh bng nhau v gii phng trnh i vi f(a) tac f(a) = f(1).a vi mi a.Nh vy, mi hm s no tho mn rng buc cho phi c dng f(x) = kx vi hng s kno . Ngc li, bt k hm s no c dng f(x) = kx vi hng s k no r rng utho mn yu cu bi ton.

II.3 Cc gc xt n u l cc gc nh hng ngoi tr cc trng hp ni khc i.Gi s chng ta c mt tam gic tu XY Z vi tm ng trn ni tip l im I v tm

ng trn bng tip IX i din vi gc X. Suy ra X , I , I X thng hng. Ta cIY IX=

2 =IZ IX v vy t gic IY IXZ l ni tip c v XIXY=IIXY=IZ Y hay Y IXX=Y ZI.

X

Y Z

I X

I

Gi I1, I2 ln lt l tm ng trn ni tip cc tam gic ABD v tam gic ACD.T gi thit ta suy ra P, Q l cc tm ng trn bng tip ca tam gic ABD i din


15/27

II.3 Li gii bi tp chng 2 15

vi gc A v gc D, tng t R, S l cc tm ng trn bng tip ca tam gic ACD idin vi gc A v gc D.p dng kt qu ca phn trn vi (X, Y, Z, I X) l (A , D, B, P), (D, A , B, Q), (A,D,C,R)v (D,A,C,S) ta c, AP D=I1BD, AQD=ABI1, ARD=I2CD , v ASD=ACI2.Khi coi cc gc sau l khng nh hng, ta thy I1BD, ABI1, I2CD v ACI2 u bngAQD

2=ACD

2.

Hn na, cc gc trn u cng mt hng, nn nu coi chng l nhng gc nh hng,chng s bng nhau. Nh vy (tr li vi nhng gc nh hng) ta c: AP D = AQD =ARD = ASD v bn im P,Q,R,S cng nm trn cung trn trng bi A, D.

II.4 Vi mi s thc x, t{x}=x [x] [0, 1) .

Ta c 2k2

p

=2k

2

p-2k2

p

vk2

p

=k

2

p-k2

p

Ta c

2k2

p

-2k2

p

=2k2

p

-2k2

p

Nu

{x}

< 1

2

th 2{

x} {

2x}

={

x}

2{

x}

=0Nu {x} 12 th 2 {x} {2x}=2{x} (2 {x} 1)=1Nh vy, tng cn tnh trong bi ra s bng l s cc phn t k trong [1, p 1] sao chok2

p

1

2 , hay bng vi s ng d k khc khng m k2 l ng d m un p vi mt s

no trongp+12

, p 1.V p l s nguyn t ng d vi 1 m un p, ta bit 1 d2 (modp), vi d l mt sno . Phn chia cc ng d m un p khc khng thnh p12 cp dng {a,da} sao choa2 (da)2 (modp).V vy, c ng mt ng d trong mi cp m bnh phng ca n ng d vi mt sno trong

p12 , p 1

, v c tt c p12 ng d nh th.

T suy ra tng cho bng p12 .

II.5 Trc tin ta chng minh rng nu 8 \ mn, th hnh ch nht mxn c th c xpbi cc hnh cho.Trng hp 1: C m v n u l s chn. Khng mt tnh tng qut ta gi s rng 4 \ m,2 \ n. Hai hnh cho c th ghp c mt hnh ch nht kch thc 4x2, v m.n/8 hnhch nht nh vy s ghp thnh mt hnh ch nht kch thc mxn (gm n/2 hng v m/4ct).Trng hp 2: Hoc m hoc n l. Khng mt tnh tng qut, ta gi s rng m l s l. Khi 8 \ n. V m > 1 nn m 3. Ta c th ghp c mt hnh kch thc 3x8 nh hnh v sau:


16/27

16 Bi ton t Hn Quc

Nhng hnh 3x8 nh vy c th ghp thnh hnh ch nht (3xn).Nu m = 3, ta ghp xong. Trong trg hp ngc li, m > 3, th phn cn li (m 3)xnc th ghp nh trong trng hp 1 v 2\(m 3).By gi ta s chng minh rng: nu hnh ch nht c kch thc (mxn) c ghp bi cchnh trn th 8\m.n. V mi mt hnh L c din tch l 4 nn 4\(m.n). Khng mt tnh tngqut, gi s 2\n, v t m hng trong hnh ch nht mxn thnh hai mu en trng cnhnhau. Mi mnh hnh ch L trong hnh ch nht c ghp s gm mt s l en hnhvung. V c tt c 1 s chn (nx

m2

) vung mu en, nn hnh ch nht c ghp cha

1 s chn cc hnh ch L, m ta t s l 2k. Nh vy m.n = 8k, hay 8\mn.

II.6 t v tri ca bt ng thc l S. V abc>0 v xyz>0 nn ta c bz +cyby +czsuy ra

(by + cz)(bz + cy) (by + cz)2 2 (by)2 + (cz)2t = (ax)2; = (by)2; = (cz)2, khi ta c:

a2x2

(by+cz)(bz+cy) a2x2

2[(by)2+(cz)2] =

+

p dng tng t cho hai bt ng thc, ta cS 1

2 (

+ +

+ +

+ )

S dng bt ng thc Cauchy-Schwarz, ta c( + + + + + )(( + ) + ( + ) + ( + ))( + + )2

m v phi bng12

( )2 + ( )2 + ( )2 + 3( + + ) 32 (2 + 2 + 2)

Do ,S 12 ( + + + + + ) 12 (++)

2

(2+2+2) 34

Vy bi ton c chng minh.


17/27

Chng III

Cc thi olympic Canada

III.1. Gii thiu . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

III.2. Bi tp . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

I II.3. Li gii bi tp chng 3 . . . . . . . . . . . . . . . . . . . . . . . . 18

III.1. Gii thiu

III.2. Bi tp

Bi tp III.1. C bao nhiu cp s (x; y) nguyn dng vi x y tho mngcd(x, y) = 5! v lcm(x, y) = 50! ?

Bi tp III.2. Cho trc mt s hu hn cc khong ng c di bng 1 sao

cho hp ca chng l khong ng [0, 50], chng minh rng tn ti mt tp con ca

cc khong khng giao vi tt c cc khong khc.

Bi tp III.3. Chng minh rng:1

1999 a3 > a4 > a5 > ksao cho:

n = C3a1 C3a2 C3a3 C3a4 C3a5

a

3

=

a(a 1)(a 2)6

Bi tp IV.5. Cho P1P2 Pn l mt a gic li trong mt phng. Gi s rngvi cp nh Pi, Pj, tn ti nh V ca a gic sao cho PiV Pj = 3 . Chng minhrng n = 3

IV.3. Li gii bi tp chng 4

IV.1 Trc ht ta gi s rng tt c cc nghim ca P l bng nhau, khi ta vit cdi dng:P(x) = a(z z0)n vi a, z0 C v n N. Nu A1, A2, , An l cc im ri rc trong R2sao cho f(A1) = f(A2) = = f(An) th A1, A2, , An nm trn ng trn vi tm l(Re(z0), Im(z0)) v bn knh l n

|f(A1)| , suy ra cc im l cc nh ca mt a gic

li.Ngc li, ta gi s rng khng phi tt c cc nghim ca P l bng nhau, khi P(x) cdng:

P(x) = (z z1)(z z2)Q(z) vi z1 v z2 l 2 nghim phn bit ca P(x) sao cho |z1 z2|l nh nht. Gi l l ng thng i qua hai im Z1 v Z2 vi Z1 = (Re(z1), Im(z2)),Z2 =(Re(z2), Im(z2)), v t z3 = 12 (z1 + z2) sao cho Z3 = (Re(z3), Im(z3)) l trung im caZ1Z2. K hiu s1, s2 ln lt l cc tia Z3Z1, Z3Z2, v r = f(Z3) 0. Ta phi c r 0, biv nu ngc li ta c z3 l mt nghim ca P sao cho:|z1 z3| |z1 z2|, iu ny l mu thu vi |z1 z2| l nh nht.Do

limZZ3

Zs1

f(Z) = limZZ3

Zs1

f(Z) = +.

v f lin tc, tn ti Z4 s1 v Z5 s2 sao cho f(Z4) = f(Z5) = r. Do vy f(Z3) =f(Z4) = f(Z5) v Z3, Z4, Z5 khng phi l cc nh ca a gic li. Do vy, f khng phil olympic.

IV.2 Ta c n 2 bt k v tm s cc hm tng ng. Nu f : {1, 2, , n} {1, 2, 3, 4, 5}phi tha mn cho th f(n) = 3 bi nu ngc li th f(n 1) 0 hoc f(n 1) 6,v l. K hiu an, bn, dn, en l s cc hm f : {1, 2, , n} {1, 2, 3, 4, 5} tha mn tnhcht cho sao cho f(n) tng ng bng 1, 2, 4, 5. Khi a2 = e2 = 2 v b1 = d2 = 1, v


22/27

22 Cc bi ton Rumania

do vy vi n 2 :

an+1 = en + dn, bn+1 = en

en+1 = an + bn, dn+1 = an

Ta cn tm an + bn + dn + en vi

n

2. Ta c: a2 = e2 v b2 = d2; bng quy np ta c

an = en v bn = dn n 2. Do vy vi n, ta c:

an+2 = en+1 + dn+1 = an+1 + bn+1 = an+1 + en = an+1 + an

do vy, {an}n2 tha mn nh dy Fibonaci {Fn}n0, vi cc ch s c chn sao cho:F1 = 0 v F1 = 1. Bi v a2 = 2 = F2 v a3 = e2 + d2 = 3 = F3, vy suy ra an = Fn vi n.Do ,an + bn + dn + en = 2(an + bn) = 2en+1 = 2an+1 = 2Fn+1 vi n 2 v 2Fn+1 thamn tnh cht cho.

IV.3 Nu s cc s xk m ln hn s cc s xk dng th (a1, , an) l mt hon v ca(x1, , xn)(tng ng l (x1, , xn)) sao cho a1, , an l mt dy khng gim. Docch xy dng, lc lng P cc s dng ak khng nhiu phn t hn lc lng N cc s

mak, v do vy |P| n

1

2 . V N l khc rng v a1, , an l khng gim, cc phn tca P l ak0+1 < ak0+2 < < ak0+l vi k0 > 0

Gi s rng 1 i n 1. Trong dy x1, , xn phi c hai phn t k nhau xjv xk sao cho xj ai v xk ai+1 suy ra 0 ai+1 ai xk xj 1. Do vy,ak0+1 ak0 + 1 1,ak0+2 ak0+1 + 1 2.K hiu P v N ln lt l tng ca cc s trong P v N. Mt khc ta c:

|P N| |P N| |2P|

2(1 + 2 + +

n 12

)

n 12 .( n 12 + 1) = n2

14

IV.4 Ta thy rng: n + C3m > 2m + 1 vi m ln hn gi tr N, bi v v tri l bc 3 vih s cao nht dng trong khi v phi l tuyn tnh vi m.

Nu m 0(mod4), th C3m = m(m1)(m2)6 l chn bi v t s chia ht cho 4 cn mus th khng.Nu m 3(mod4) th C3m = m(m1)(m2)6 l l bi v c t s v mu s uchia ht cho 2 nhng khng chia ht cho 4. Do vy ta chn m > maxk, N sao cho n + C3ml s l.

Ta vit: 2a + 1 = n + C3m > 2m + 1. Ta thy rng:(C3a+3 C3a+2) (C3a+1 C3a) =

C2a+2

C2a = 2a + 1. Do vy

n = (2a + 1)

m

3

=

a + 3

3

a + 2

3

a + 1

3

+

a

3

=

m

3

tha mn yu cu bi ton v a + 3 > a + 2 > a + 1 > a > m > k


23/27

IV.3 Li gii bi tp chng 4 23

IV.5 Trong li gii ny ta s dng kt qu sau:Cho tam gic XYZ sao cho XY Z 3 th tam gic l u hoc max{Y X , Y Z } > X Z.Tng t nu XY Z 3 th tam gic XYZ u hoc min{Y X , Y Z } < X ZChng ta ch ra rng tn ti cc nh A , B, C v A1, B1, C1 sao cho:(i) tam gic ABC vA1B1C1 l tam gic u v (ii) AB(tng ng l A1B1) l khong cch nh nht (ln nht)khc 0 gia 2 nh. Hn na, A, B l 2 nh phn bit sao cho AB c di nh nht, v

C l nh sao cho ACB =3 . Khi max{AC,CB} AC, tam gic ABC phi l tam

gic u. Tung t, ta chn A1, B1 sao cho A1B1 c di ln nht, v nh C1 sao choA1C1B1 =

3 , khi tam gic A1B1C1 l tam gic u.

Ta ch ra rng ABC = A1B1C1. Cc ng thng AB,BC, CA chia mt phng thnh7 phn. Gi DA gm cc phn do tam gic chia m nhn BC lm bin v cc phn cto ra ti phn to ra cc nh B v C. Tng t ta nh ngha cho DB v DC. Bi v agic cho l li, nn mi hoc nm trong 1 phn hoc trng vi A, B, C.

Nu 2 im bt k trong A1, B1, C1 , gi s l A1, B1 nm trong min DX , thA1XB1 A1B1, mu thun vi A1B1 l ln nht.

Hn na, khng c hai im trong A1, B1, C1 trong cng 1 phn. By gi ta gi srng mt trong cc im A1, B1, C1 ( gi s l A1) nm trn 1 phn(gi s l DA). Biv min

{A1B, A1C

} BC, ta c BA1C

3 . Ta c B1 khng nm trong DA. Bi v a

gic cho l li, B khng nm trong tam gic AA1B1, v tng t C khng nm trongtam gic AA1B1. T c B1nm trn min ng c bin l cc tia A1B v A1C. Tngt, vi C1. Hn na, 3 = B1A1C1 BA1C = 3 , du bng xy ra khi B1 v C1 ln ltnm trn tia A1B v A1C . Bi v a gic cho l li,nn iu ny ch xy ra khi B1 vC1 ln lt bng B v C -trong trng hp BC = B1C1, ta c tam gic ABC v A1B1C1l bng nhau.

Mt khc, khng c im no trong A1, B1, C1 nm trn DA DB DC, do chngln lt trng vi A,B,C. Trong trng hp ny, tam gic ABC v A1B1C1 l trng nhau.


24/27

Chng V

Cc bi ton Th Nh K

V.1. Gii thiu phng php . . . . . . . . . . . . . . . . . . . . . . . . . . 24

V.2. Bi tp . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

V.3. Li gii bi tp chng 5 . . . . . . . . . . . . . . . . . . . . . . . . . 25

V.1. Gii thiu phng php

V.2. Bi tp

Bi tp V.1. Cho tam gic ABC vung ti A, gi H l chn ng cao k A. Chngminh rng tng bn knh cc ng trn ni tip cc tam gic ABC, ABH, ACH

bng AH.

Bi tp V.2. Dy s{

an}

n=1 ,

{bn

}

n=1 c cho bi:

a1 = , b1 = , an+1 = an bn, bn+1 = an + bn vi mi n 1

C bao nhiu b s thc (, ) tha mn a1997 = b1 v b1997 = a1?


25/27

V.3 Li gii bi tp chng 5 25

Bi tp V.3. Trong mt hip hi bng , khi mt cu th chuyn t i X c xcu th sang i Y c y cu th, lin on nhn c y x triu la t i Y nuy x nhng phi tr li x y triu la cho i X nu x > y. Mt cu th c thdi chuyn ty thch trong sut ma chuyn nhng. Hip hi bao gm 18 i, tt

c cc i u bt u ma chuyn nhng vi 20 cu th. Kt thc ma chuyn

nhng, 12 i kt thc vi 20 cu th, 6 i cn li kt thc vi 16,16,21,22,22,23cu th. Tng s tin ln nht m lin on c th kim c trong sut ma chuyn

nhng l bao nhiu?

Bi tp V.4. Ng gic ABCDE li c cc nh nm trn ng trn n v, cnh

AE i qua tm ng trn . Gi s AB = a,BC = b,CD = c,DE = d v

ab = cd = 1/4. Tnh AC+ CE theo a,b,c.

Bi tp V.5. Chng minh rng vi mi s nguyn t p 7, tn ti mt s nguyndng n v cc s nguyn x1, x2,...,xn, y1, y2,...,ym khng chia ht cho p sao cho:

x21 + y21 x22 (modp)x22 + y

22 x23 (modp)

.....x2n + y

2n x21 (modp)

Bi tp V.6. Cho cc s nguyn n 2. Tm gi tr nh nht ca :x51

x2 + x3 + ... + xn+

x52x3 + x4 + ... + xn + x1

+ ... +x5n

x1 + x2 + ... + xn1

Vi x1, x2,...,xn l cc s thc tha mn x21 + x22 + ... + x

2n = 1.

V.3. Li gii bi tp chng 5

V.1 t a = BC,b = CA,c = AB v s = a+b+c2 . Cc tam gic ABH v ACH ng dngvi tam gic ABC vi t s tng ng a/c v b/c

p dng cng thc din tch tam gic bng bn knh ng trn ni tip nhn vi na chuvi, suy ra bn knh cn tm l ab

a+b+c ;ac

aba+b+c ;

bc

aba+b+c v tng ca chng l

abc

= AH

V.2 Lu rng a2n+1+b2n+1 =

2 + 2

a2

n + b2

n

, tr = = 0. Chng ta cn 2+2 =

1. V vy c th t = cos , = sin , t bng phng php quy np ta ch ra =cos n, = sin n. T c 1998 b s: (0; 0) v (cos ;sin ) vi = k3998 , k = 1, 3, ..., 3997

V.3 Chng ta tha nhn rng s tin ln nht kim c bi khng bao gi cho php mtcu th chuyn n i nh hn. Chng ta cng c th gi k lc bng mt cch khc.Mt i bng c x cu th th c ghi l x trc khi giao dch mt cu th hoc x trckhi nhn mt cu th v s tin m lin on kim c bng tng ca cc s . By gi


26/27

26 Cc bi ton Th Nh K

ta xem xt cc s c ghi bi mt i m kt thc c nhiu hn 20 cu th. Nu s lngcu th ti a ca i trong sut qu trnh chuyn nhng l k > n th cc s k 1 vk xut hin lin tip v b i 2 s th tng s tng ln. V th tng ca cc s trongi bng t nht l 20 + 21 + ... + (n 1).Tng t nh vy, tng ca cc s trong i bng kt thc c n < 20 cu th t nht l20 19 ... (n + 1). V nhng con s ny chnh xc l nhng con s c c bi viclun chuyn cu th t i kt thc t hn 20 cu th sang i kt thc c nhiu hn 20 cuth. S sp xp dn n s tin kim c l ln nht. Trong trng hp , tng l:

(20 + 20 + 21 + 20 + 21 + 20 + 21 + 22) 2(20 + 19 + 18 + 17) = 17

V.4 Nu gi 2, 2, 2, 2 l cc cung chn bi cc cnh a,b,c,d tng ng th:

AC = 2sin ( + ) =a

2

1 b

2

4+

b

2

1 a

2

4

Tng t vi CD.Tng qut vi R l bn knh ng trn ngoi tip ng gic, th th AC2 + BD2 = 1 m

AC = a

R2

b2

+ b

R2

a2

Khi , dn n biu thc cha R2 di du cn v ta gii phng trnh i vi R theo ccs a,b,c,d.

V.5 Gi n l cp ca 5/3 mod p, v t xi = 3n1i5i1, yi = 43n15i1 th mi ng dtrn l tng ng tr h thc cui cng. H thc c dng 52n 32n (modp) (ng).Vyta c iu phi chng minh.

V.6 t S = x1 + x2 + ... + xnS dng BT Chebyshevs cho hai dy xi

Sxi v x4i (c hai dy u l dy tng). Ta c:

x5i

S xi x1

S x1+

x2

S x2+ ... +

xn

S xnx41 + x

42 + ... + x

4n

n

p dng bt ng thc hm li ta c:

x1S x1 +

x2S x2 + ... +

xnS xn

1

n 1

p dng bt ng thc gi tr trung bnh ta c:

x4in

1/2 x2i

n=

1

n

Ta c kt lun: x5iS xi n

1n 1 .

1n2

= 1n (n 1)

ng thc xy ra khi x1 = x2 = ... = xn = 1n .


27/27

Ti liu tham kho

[1] H. Rademacher, Higher Mathematics from an Elementary point of view,Birkhauser, 1983.

[2] Martin Aigner and Gunter M. Ziegler, Proofs from the book, Springer, 1999.[3] T. Andreescu, R. Gelca, Mathematical olympiad challenges, Birkhauser, 2002.

[4] Loren C. Larson, Problem-Solving through problems, Springer-Verlag, 1983.

[5] L. E. Dickson, New first course in the theory of equations John Wiley & Sons,1946.

[6] Judita Cofman, What to solve? Problems and Suggestions for Young Mathe-maticians, Clarendon Press-Oxford, 1989.

[7] Nguyn Hu in, Phng php irichle v ng dng, NXBKHKT, 1999.

[8] Nguyn Hu in, Phng php Quy np ton hc, NXBGD, 2000.[9] Nguyn Hu in, Nhng phng php in hnh trong gii ton ph thng,

NXBGD, 2001.

[10] Nguyn Hu in, Nhng phng php gii bi ton cc tr trong hnh hc,NXBKHKT, 2001.

[11] Nguyn Hu in, Sng to trong gii ton ph thng, NXBGD, 2002.

[12] Nguyn Hu in, a thc v ng dng, NXBGD, 2003.

[13] Nguyn Hu in, Gii phng trnh v nh nghim nguyn, NXBHQG,2004.

[14] Nguyn Hu in, Gii ton bng phng php i lng bt bin, NXBGD,2004.

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