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Li ni u

th gi lnh lamdethi.sty ti bin son mt s ton thi

Olympic, m cc hc tr ca ti lm bi tp khi hc tp LATEX.

ph v cc bn ham hc ton ti thu thp v gom li thnh cc

sch in t, cc bn c th tham kho. Mi tp ti s gom khong

51 bi vi li gii.

Rt nhiu bi ton dch khng c chun, nhiu im khng

hon ton chnh xc vy mong bn c t ngm ngh v tm hiu

ly. Nhng y l ngun ti liu ting Vit v ch ny, ti c

xem qua v ngi dch l chuyn v ngnh Ton ph thng. Bn c

th tham kho li trong [1].

Rt nhiu on v mi hc TeX nn cu trc v b tr cn xu, ti

khng c thi gian sa li, mong cc bn thng cm.

H Ni, ngy 2 thng 1 nm 2010

Nguyn Hu in

51GD-05

89/176-05 M s: 8I092M5

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Mc lc

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

Mc lc ..................................................... 4

Chng 1. thi olympic Russian.... .. .. .. .. .. .. .. .. .. .. .. . 5

Chng 2. thi olympic Nam Phi .. .. .. .. .. .. .. .. .. .. .. .. .. 9

Chng 3. thi olympic Ty Ban Nha.. .. .. .. .. .. .. .. .. .. 12

Chng 4. thi olympic i Loan .. .. .. .. .. .. .. .. .. .. .. .. 16

Chng 5. thi olympic Th Nh K .. .. .. .. .. .. .. .. .. .. . 23

Chng 6. thi olympic Ukraina .. .. .. .. .. .. .. .. .. .. .. .. . 27

Chng 7. thi olympic Anh... .. ... .. ... ... ... .. .. ... .. .. 34

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Chng 1

thi olympic Russian

1.1. Chng minh rng cc s t 1 n 16 c th vit c trn cng 1dng nhng khng vit c trn 1 ng trn, sao cho tng ca 2 s

bt k ng lin nhau l 1 s chnh phng.

Li gii: Nu cc s vit trn 1 ng trn th ng cnh s 16 ls x, y khi 16+1 16 + x, 16 + y 16+15, suy ra: 16 + x = 16 + y = 25mu thun. Cc s c th c sp xp trn 1 dng nh sau:

16, 9, 7, 2, 14, 11, 5, 4, 12, 13, 3, 6, 10, 15, 1, 8.

1.2. Trn cnh AB v BC ca tam gic u ABC ly im D v Ktrn cnhAC v ly imE vM sao choDA + AE = KC+ CM = AB.

Chng minh rng gc giaDM vKE bng

3.

Li gii: Ta c: CE = AC AE = AD.V tng t: CK = AM.

Xt php quay tm l tm ca tam gic ABC, gc quay2

3bin K

thnh M, bin E thnh D, t suy ra iu phi chng minh.

1.3. Mt cng ty c 50.000 cng nhn, vi mi cng nhn tng sngi cp trn trc tip v cp di trc tip ca anh ta l 7. Vo th

2 mi cng nhn a ra mt s ch dn v gi bn photo ca n cho

mi cp di trc tip ca anh ta (nu anh ta c). Mi ngy sau mi

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6 Nguyn Hu in, HKHTN H Ni

cng nhn gi tt c cc ch dn m anh ta nhn c vo ngy hm

trc v gi bn photo ca chng cho tt c cp di trc tip ca anh

ta nu anh ta c hoc anh ta phi t thc hin nu khng c cp di

trc tip. C nh th cho n th 6 khng cn ch dn no a ra.

Hay ch ra rng c t nht 97 cng nhn ko c cp trn trc tip.

Li gii: Gi s k l s cng nhn ko c cp trn trc tip, vo ngyth 2 s ch dn c a ra nhiu nht l7k, vo ngy th 3 nhiu

nht l 6.7k vo ngy th 4 nhiu nht l 36.7k vo ngy th 5 mi

cng nhn nhn c 1 ch dn ko c cp di trc tip, v vy mi

cng nhn c 7 cp trn trc tip, mi ngi a ra nhiu nht l 6

ch dn v c nhiu nht l216.7k/7 cng nhn nhn c ch dn.

Chng ta c:

50.000 k + 7k + 42k + 252k + 216k = 518k vk 97.

1.4. Cc cnh ca tam gic nhn ABC l cc ng cho ca hnhvungK1, K2, K3. Chng minh rng mim trong ca tam gicABC c

th c ph bi 3 hnh vung.

Li gii: Gi I l giao im ca 3 ng phn gic ca tam gicABC, v cc gc l nhn nn IAB. IBA < 45 v vy tam gic IAB cth c ph bi hnh vung m ng cho ca n lAB v tng

t i vi tam gic IBC v tam gic ICA.1.5. Cc s t 1 ti 37 c th c vit trn 1 dng sao cho mi s l

c ca tng tt c cc s ng trc n. Nu s u tin l 37 v s

th 2 l 1 th s th 3 l bao nhiu.

Li gii: Gi s cui cng l x, x phi l c ca tng tt c cc sl37x19, v x = 19 v s th 3 phi l c ca 38 khc 1 hoc 19, vy

s th 3 l 2.

1.6. Tm cc cp s nguyn tp, q sao chop3

q5 = (p + q)2.

Li gii: Ch c nghim duy nht l (7, 3), u tin gi s c p v qu khng bng 3. Nu chng ng d vi Module 3, v tri th chia

ht cho 3 nhng v phi th khng, nu chng khng cng ng d

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thi olympic Russian 7

module 3 th v phi chia ht cho 3 nhng v tri th khng v th

khng xy ra kh nng ny. Nu p = 3 ta c q5 < 27, khng c s

nguyn t no tha mn. V vy q = 3 v p3 243 = (p + 3)2 ch cnghim duy nht lp = 7.

1.7. (a) a. thnh ph Mehico hn ch giao thng mi xe otoring u phi ng k 2 ngy trong 1 tun vo 2 ngy oto

khng c lu thng trong thnh ph. Mt gia nh cn s dng

t nht 10 chic oto mi ngy. Hi h phi c t nht bao nhiu

chic oto nu h c th chn ngy hn ch cho mi chic oto.

(b) b. Lut c thay i cm mi oto ch 1 ngy trong 1 tun

nhng cnh st c quyn chn ngy cm . Mt gia nh hi

l cnh st gia nh c quyn chn 2 ngy lien tip khng

b cm cho mi xe v ngay lp tc cnh st cm xe oto vo 1 trong

nhng ngy khc. Hi gia nh cn t nht bao nhiu xe oto

nu h s dng 10 chic mi ngy.

Li gii:(a) Nu n xe oto c s dng, s ngy c s dng l 5n. M mi

gia nh s dng t nht l 10 xe nn 7x10 5n v th n 14.Trong thc t 14 xe tha mn yu cu ca u bi ton: 4 xe b

cm vo ngy th 2 v th 3, 4 xe b cm vo ngy th 4 v th

5, 2 xe b cm vo ngy th 6 v th 7, 2 xe b cm vo ngy

th 7 v ch nht, 2 xe b cm vo ngy ch nht v th 6.(b) 12 xe oto l s xe h cn, u tin chng ta ch ra rng n 11

xe khng tha mn. Khi c n ngy xe b cm, mi ngy nhiu

nht lbn

7xe c lu thng nhng n 11, bn

7< 10. i vi

n = 12, gia nh cn a ra 2 ngy lin tip cho mi xe trong

nhng ngy xe ko b cm lu thng.

1.8. Mt a gic u 1997 nh c chia bi cc ng cho ko ctnhau to thnh cc tam gic. Hy ch ra rng c t nht mt tam gic

nhn.

Li gii: ng trn ngoi tip a gic u 1997 nh cng l ngtrn ngoi tip mi tam gic. V tm ca cc ng trn khng nm

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trn bt k ng cho no nn n phi nm trong mt tam gic, v

th tam gic phi l tam gic nhn.

1.9. Vit cc s t 1 n 1000 trn bng, Hai ngi chi ln lt xai 1 s trong cc s , cuc chi kt thc khi cn li 2 s:

Ngi chi th 1 thng nu tng cc s cn li chia ht cho 3, cctrng hp cn li ngi chi th 2 thng. Ngi chi no c chin

thut chin thng.

Li gii: Ngi chi th 2 c chin thut chin thng, nu ngichi th 1 xa i s x, ngi chi th 2 xa i s 1001 x v th tngca 2 s cui cng l 1001.

1.10.C 300 qu to, khng c qu no nng hn 3 ln qu khc. Hych ra rng c th chia cc qu to ny thnh 4 nhm m khng c

nhm no c cn nng hn 112 ln nhm khc.

Li gii: Sp xp cc qu to tng dn theo trng lng, v ghptng i mt qu nh nht vi qu nng nht, sau li em 1 qu

nh nht tip theo vi 1 qu nng nht tip theo tip tc cho n

ht. Ch rng khng cp no nng hn 2 ln cp khc. Nu a, d v

b, c l 2 nhm vi

a b c d th a + d 4a 2b + 2c

b + c 3a + d 2a + 2d.

By gi cc cp nng nht v nh nht to thnh 4 nhm, khng

cn nng ca nhm no gp 2/3 ln nhm khc v

e f g h l cc cp th e + h 3e 32

(f + g)

f + g 2e + h 32

(e + h).

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Chng 2

thi olympic Nam Phi

2.11.Cho A0B0C0 v mt dy A1B1C1,A2B2C2 ... c xy dngnh sau:

Ak+1, Bk+1, Ck+1 l im tip xc ca ng trn ngoi tipAkBkCk vi

cc cnhBkCk,CkAk,AkBk theo th t.

(a) hy xc nhAk+1Bk+1Ck+1 tAkBkCk(b) chng minh: lim

kAkBkCk = 60

0

Li gii:(a) Ta c AkBk+1 = AkCk+1(v y l 2 tip tuyn xut pht t

mt im).V vy Ak+1Bk+1Ck+1 l tam gic cn vi AkBk+1Ck+1 =900 Ak2 .Tng t ta c: CkBk+1Ck+1 = 900 Ck2 . Hn naBk+1 =(Ak+Ck)

2= 900 Bk

2

(b) Ta c Bk+1 600 = 900 Bk2 600 = Bk600

2V Bk 600 = B0600(2)k Hin nhin limkBk = 60

0

2.12.Tm tt c cc s t nhin tho mn:khi chuyn ch s u tinxung cui,s mi bng 3,5 ln s ban u.

Li gii:Cc ch s nh th c dng sau: 153846153846153846. . . 153846

Hin nhin nhng s tho mn gi thit phi bt u bi 1 hoc 2.

Trng hp 1: S c dng: 10N + A viA < 10N.V 72(10N + A) =

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10A + 1 A = (7.10N2)13

Ta c 10N 1, 3, 4, 9, 10, 12(mod13).V th A s l mt s nguyn t nu:N 5(mod6).T ta c kt qu nh trn.Trng hp 2: S c dng 2.10N + A,A < 10N.Theo chng minh

trn A =

(14.10N4)13 .Nhng v A < 10

N

,tc 10N

< 4.V l.2.13.Tm tt c cc hm: f : Z Z sao cho:

f(m + f(n)) = f(m) + n, m, n Z

Li gii:R rng: f(n) = kn vi k = 1, k = 1.Ta s chng minh y l kt quduy nht.Cho n = 0 ta c f(m + f(0)) = f(m).Xt 2 trng hp:

Trng hp 1: f(0) = 0.cho m = 0 c f(f(n)) = n.Gn f(n) bi n

ta c: f(m + f(f(n))) = f(m + n) = f(m) + f(n) f(n) = nf(1) vn = f(f(n)) = n(f(1))2 f(1) = 1.y l kt qu trn.Trng hp 2: f(0) = 0. f(n) l hm tun hon v b chn .tf(M) f(n), n.nhng f(M + f(1)) = f(M) + 1.Mu thun gii thit.

2.14.Cho mt ng trn v mt im P pha trn ng trn trongmt phng to .Mt ht nh di chuyn dc theo mt ng thng t

P n im Q trn ng trn di nh hng ca trng lc.Khong

cch i t P trong thi gian t l: 12

gt2 sin vi g khng i v l gc

gia PQ vi mt phng nm ngang.Hy ch ra v tr im Q sao chothi gian di chuyn t P n Q l t nht.

Li gii:Cu hi l tm gi tr nh nht ca PQ

sinhoc gi tr ln nht ca sin

PQ.

Biu din mt php nghch o im P vi nh l qu o ca chnh

n.im cc i trn s c(gi l Q) s vch ra mt im Q

vi gi

tr ln nhtP Q

sin , vi cao khc nhau gia P vQ

.Nh vy P l

im pha trn ng trnQ

l im pha y ng trn. tm Q

hy ch rng P,Q,Q

cng thuc mt ng thng.Do cch tm

nh sau:

(a): Tm im pha y ng trn gi lQ

(b): Tm giao caP Q

vi ng trn , l im cn tm.

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thi olympic Nam Phi 11

2.15.C 6 im c ni vi nhau tng i mt bi nhng on mu hoc mu xanh..chng minh rng:C mt chu trnh 4 cnh cng

mu.

Li gii:

Gi cc im l A,B,C,D,E,F.D thy lun c mt chu trnh tam giccng mu(Tht vy:Xt mt s nh no ,s c 5 cnh t cc nh

y m t nht 3 trong s chng cng mu.gi s l mu v cc

cnh ny i n A,B,C.Nu mt s cnh gia A,B,C l mu ta c

iu phi chng minh,nu khng ta cng c iu phi chng minh)

Khng mt tnh tng qut,gi cc cnh AB,BC,CA l mu .Nu

mt trong cc nh ca cnh mu khc chy n A,B,C,ta c iu

cn tm.Nu 2 trong 3 im D,E,F ca cnh mu xanh chy n 2

trong s cc nh cng mu A,B,C,ta cng c iu cn tm.Trng

hp duy nht khng xy ra l nu mt trong cc im D,E,F ca

cnh mu to bi nhng im khc A,B,C;khng mt tnh tng

qut gi s AD,BE,CF l mu .Cc cnh khng theo l thuyt l

DE,EF,FD.Nu mt cnh trong s chng mu ta c dng hnh

trn(v d nu DE mu th DABE l )Nu mt trong s chng

mu xanh th DCEF mu xanh.Ta chng minh c bi ton.

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Chng 3

thi olympic Ty Ban Nha

3.16.Tnh tng bnh phng ca 100 s hng u tin ca mt cp scng, vi gi thit tng100 s hng bng1 v tng cc s hng thhai, th t, ..., th mt trm bng1

Li gii:Gi 100 s hng u tin ca cp s cng l x1, x2, x3, , x100 v d lcng sai ca cp s cng theo gi thit th nht ta c:

x1 + x2 + x3 + + x100 = 12

(x1 + x100) .100 = 1 x1 + x100 = 150

.

Theo gi thit th 2 ta c

x2 + x4 + + x100 = 12

((x1 + d) + x100) .50 = 1 x1 + x100 + d = 125

.

Suy rad = 350

v x1 + x100 = x1 + (x1 + 99d) = 150 hayx1 = 14950 . T ta tm c:

x21 + x22 + + x2100 = 100x21 + 2dx1 (1 + + 99) +

12 + + 992 .

Vy

x21 + x22 + + x2100 =

14999

50

3.17.A l mt tp gm16 im to thnh mt hnh vung trn mi cnhc 4 im. Tm s im ln nht ca tp A m khng c 3 im no

trong s cc im to thnh mt tam gic cn

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thi olympic Ty Ban Nha 13

Li gii: S im ln nht cn tm l 6 c c bng cch ly ccim hai cnh k nhau nhng b i im chung ca hai cnh .

u tin gi4 im bn trong khng c chn, nhng im cn li

to thnh 3 hnh vung, nn nhiu nht2 ng thng ng t mi

hnh vung c chn. Nh vy chng ta c th cho rng mt trong

s cc im trong c chn trong s sau l im O

D A1 A2 A3

C Z1 O Z2

E B1 B2 B2

C D C E

Khng c im no cng tn gi A,B,C,D,E c chn, vy nu ta

khng chn Z1, Z2 mt ln na nhiu nht6 im c th c chn.Nu chn Z1 nhng khng chn Z2 th A1, A2, B1, B2 cng khng c

chn, v c A3 v B3 cng khng c chn, v vy mt trong hai

im A v B phi b i, mt ln na s im ln nht l 6. Trng

hp chn Z2 nhng khng chn Z1 tng t. Cui cng nu Z1 vZ2c chn th c Ai v Bi u khng c chn, v vy s im ln

nht l6.

3.18.Vi mi Paraboly = x2 +px + q ct hai trc ta tai3 im phnbit, v mt ng trn i qua3 im . Chng minh rng tt c cc

ng trn u i qua mt im c nh

Li gii: Tt c cc ng trn u i qua im (0, 1). Gi s(0, q) , (r1, 0) , (r2, 0) l 3 im m Parabol qua, do r1 + r2 = p.Gi s (x a)2 + (y b)2 = r2 l ng trn lun i qua 3 im trndo a = p

2v

1

4p2 + (q b)2 =

r p

2

2+ b2 =

1

4(r1 r2)2 + b2

hayq2

2.qb =

q do b = q+12 , khi im i xng vi im (0, q)

qua ng knh nm ngang l im (0, 1)

3.19.Cho p l s nguyn t. Tm tt c k Z sao cho

k2 pk l snguyn dng.

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Li gii: Gi tr k cn tm lk = (p1)2

4vi p l s l ( tr trng hp

p =2 ). Trc ht xt p = 2, trong trng hp ny ta cn k2 2k =(k 1)2 1 l mt s chnh phng dng trng hp ny khng thxy ra v ch c duy nht hai s chnh phng lin tip l 0 v1

Gi s p l s l. u tin ta xt trng hp k chia ht cho p, hayk = np, khi k2 pk = p2n (n 1), n vn 1 l hai s nguyn t lintip. Do c hai khng th l s chnh phng.

Gi s k v p l hai s nguyn t cng nhau, khi k v k p cngl hai s nguyn t cng nhau. k2 pk l s chnh phng khi vch khi k v k p l cc s chnh phng, k = m2, k p = n2. Do

p = m2 n2 = (m + n) (m n). Suy ram + n = p, m n = 1 vk = (p+1)24 ,hoc m + n = 1, m n = p vk = (p1)24

3.20.Chng minh rng trong tt c cc t gic li c din tch bng1, th tng di cc cnh v cc ng cho ln hn hoc bng

2

2 +

2

Li gii: Thc t ta cn ch ra rng tng di cc cnh ca t gicli ln hn hoc bng 4 v tng di cc ng cho ca t gic li

ln hn hoc bng 2

2. i vi trng hp ng cho ta s dng

cng thc tnh din tch A = 12

d1d2 sin , vi l gc giu hai ng

cho. T gi thit cho din tch t gic bng 1 suy ra d1d2 2, pdng bt ng thc AG-GM suy ra d1 + d2

2

2, ng thc xy ra

khi v ch khi di hai ng cho bng nhau v vung gc vinhau.

i vi trng hp cnh ta s dng cng thc tnh din tch

A = (s a) (s b) (s c) (s d) abcd cos2 B + D2

,

vi s = a+b+c+d2

, B v D l hai gc i din nhau. T gi thit cho

din tch t gic bng 1 ta suy ra (s a) (s b) (s c) (s d) 1, lis dng bt ng thc AG-MG ta li suy ra

4 (s a) + (s b) + (s c) + (s d) = a + b + c + d

, ng thc xy ra khi v ch khi a = b = c = d.

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thi olympic Ty Ban Nha 15

T ta suy ra kt lun, c hai ng thc xy ra khi v ch khi

t gic li l hnh vung.

3.21.Lng gas chnh xc mt chic t hon thnh mt vng ngua c t trongn bnh gas t dc ng ua. Chng minh rng

c mt v tr m xe c th bt u vi mt bnh gas rng, c thhon thnh mt vng ng ua m khng s ht gas ( gi s xe c

th cha mt lng gas khng gii hn)

Li gii: Ta s dng phng php qui np theo n, trng hp n = 1d dng thy c. Cho n + 1 bnh cha phi c mt bnh chaA m

t t c th ti c bnh cha B m bnh khng c gas

cho mt vng ua. Nu chng ta dn bnh B vo bnh A v b bnh

B i, theo gi thit quy np c 1 im xut pht m xe c th hon

thnh vng ua, cng im xut pht nh th cho hon thnh vngua vi lng phn pht ban u ca bnh cha.

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Chng 4

thi olympic i Loan

4.22.Choa l mt s hu t, b,c,d l cc s thc vf : R [1;1] l 1hm tha mn:

f(x + a + b) f(x + b) = c[x + 2a + [x] 2[x + a] [b]] + d

vi mi x R. Chng minh rng hm f tun hon, tc l tn ti sp > 0 sao chof(x + p) = f(x) vi mix R.

Li gii: Vi mi s nguyn n ta c:f(x + n + a) f(x + n)

= c [x b + n] + 2a + [x b + n] 2 [x b + n + a] [b] + d= c [x b] + n + 2a + [x b] + n 2 [x b + a] + n [b] + d= c [x b] + 2a + [x b] 2 [x b + a] [b] + d= f(x + a)

f(x)

Ly s nguyn dng m sao cho am l mt s nguyn. Khi vi mi

s t nhin k ta c:

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thi olympic i Loan 17

f(x + kam) f(x)

=x

j=1

mi=1

(f(x + jam + ai) f(x + jam + a (i 1)))

= k

m

i=1

(f(x + ai) f(x + a (i 1)))

= k (f(x + am) f(x))

Do f(x) [1;1], f(x + kam) f(x) b chn nn f(x + kam) f(x) phibng 0. Suy raf(x + kam) = f(x), v vyf(x) l hm tun hon.

4.23.Cho on thngAB. Tm tt c cc imC trong mt phng saocho tam gicABC sao cho ng cao k t A v trung tuyn k t B

c di bng nhau.

Li gii: Gi D l chn ng cao k t A vE l chn ng trungtuyn k t B. Gi F l chn ng vung gc k t E xung BC.

Khi EF//AD v E l trung im ca AC, v vy EF = 1/2(AD) =

1/2(BE) vEBC = /6 (tt c cc gc u c hng tr khi c ).By gi, cho P l mt im sao cho B l trung im caAP. Khi

BE//PC, v th P CB = EBC v khng i. Qu tch tt c cc im

C sao cho P CB khng i l mt ng trn. Do , qu tch cc

im C bao gm hai ng trn bng nhau, ct nhau ti B, P (Mt

tng ng vi gc /6 v mt tng ng vi gc /6 ). Trong trnghp c bit, khi tam gic ABC cn ta thy rng mi ng trn u

c bn knh AB v tm sao cho ABQ = 2/3 (khng c hng).

4.24.Cho s nguynn 3, gi thit rng dy s thc dnga1, a2,...,antho mnai1 + ai+1 = kiai vi dyk1, k2,...,kn l dy s nguyn dng

bt k. (trong a0 = an van+1 = a1 ). Chng minh rng

2n k1 + k2 + ... + kn 3n

Li gii: Bt ng2n k1 + k2 + ... + kn

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c chng minh da vo AM-GM vi ch rng:

k1 + k2 + ... + kn =ni=1

aiai+1

+ai+1

ai

chng minh c bt ng k1 + k2 + ... + kn 3n, ta cn chngminh k1 + k2 + ... + kn 3n 2 vi n 2 , bng phng php quy nptheo n. Vi n = 2, nu a1 a2 th 2a2 = k1a1 , v th hoc a1 = a2 vk1 + k2 = 4 = 3.2 2, hoc a 1 = 2a2 vk1 + k2 = 4 = 3.2 2. Vi n > 2,ta c th gi thit tt c cc ai khng bng nhau, khi tn ti i sao

cho ai aa1, ai+1 m du bng khng xy ra t nht mt trong haitrng hp. Khi aiaa1 + ai+1 < 2ai v do ki = 1. Ta kt lun rng

dy m b i s hng ai cng tha mn iu kin cho vi ki1 v

ki+1 gim i 1 n v v b i s hng ki. Theo gi thit quy np, tng

caki s hng nh hn hoc bng 3(n 1) 2, tng ca cc s ki banu nh hn hoc bng 3n 2, do ta c iu phi chng minh.

4.25.Cho k = 22n + 1 vi n l s nguyn dng bt k. Chng minhrngk l mt s nguyn t khi v ch khik l mt c ca3(k1)/2 + 1.

Li gii: Gi s k l mt c ca3(k1)/2 + 1. iu ny tng ngvi 3(k1)/2 1( mod k). V vy3k1 1( mod k). Vi d sao cho 3d 3mod k. V vy, d khng l c ca (k 1)/2 nhng l c ca k 1,mt khc (k

1) l c cad nn d = k

1 (bi vi d phi nh hn k).

Do , k l s nguyn t.

Ngc li, nu k l s nguyn t3k

=k3

=23

= 1

3(k1)/2 3k

14.26.Cho t dinABCD. Chng minh rng:

(a) Nu AB = CD,AD = BC,AC = BDth cc tam gic

ABC,ACD,ABD,BCD l cc tam gic nhn.

(b) Nu cc tam gic ABC,ACD,ABD,BCD c cng din tch th

AB = CD,AD = BC,AC= BD

Li gii: (a) Theo gi thit 4 mt ca t din bng nhau, ta c gctam din mi nh c to bi ba gc khc nhau ca mt mt.

Gi M l trung im ca BC. Theo bt ng thc trong tam gic,

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AM + MD > AD = BC = 2MC. Cc tam gic ABC v DBC l bng

nhau, v th AM = DM. Do 2MD > 2MC; ngha l, MD ln hn

bn knh ca ng trn nm trn mt phng BC D vi ng kinh

BC. Do , D nm ngoi ng trn v gc BDC l gc nhn. Tng

t nh vy, ta chng minh c cc gc cn li (bi ton ny l bi

ton USAMO thng 2/1972, tham kho cc cch gii khc quyn

sch USAMO ca Klamkin).

(b) V AB v CD khng song song (v ABCD)l hnh t din), ta c

th chn hai mt phng song song l mt phng (P) chaAB v(Q)

chaCD. Gi khong cch gia mt phng (P) v(Q) ld. Gi A, B

ln lt l hnh chiu ca A, B trn (Q), gi C, D ln lt l hnh

chiu ca C, D trn (P). V cc tam gic ACD v BC D c cng din

tch v chung y CD, nn chng c cng chiu cao h. Ta v hnh

tr vi trc CD v bn knh h; r rng A, B thuc hnh tr ny. Haiim ny cng thuc mt phng (P) v mt phng (P) giao vi hnh

tr ti mt hoc hai ng thng song song vi CD.

A v B khng th cng nm trn mt trong hai ng thng ny,

v th hnh tr v mt phng (P) s giao nhau ti hai ng thng,

mt ng thng i quaA v mt ng thng i quaB. Hai ng

thng song song vi nhau v cch u ng thng CD mt khong

l(h2 d2)1/2. V vy, ng thng CD chia i on thng AB.Tng t, ta chng minh c ng thng AB chia i on thng

CD, qua php chiu, mt phng (Q) bin thnh mt phng (P),

ng thng AB chia i on thng CD. AB v CD ct nhau

ti trung im mi ng nn ACBD l hnh bnh hnh, do ta

c AC = BD (gi khong cch gia chng l x) v BC = AD(gi

khong cch gia chng ly). Vy ta c:

AC =

AC2 + CC21/2

=

x2 + d21/2

=

BD 2 + DD2

1/2

= BD

AC =

BC2 + CC21/2

=

y2 + d21/2

=

AD2 + DD21/2

= AD

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S dng hai cp cnh khc lc u, ta cng c AB = CD

4.27.Cho X l mt tp hp cc s nguyn cho bi cng thc:

a2k102k + a2k210

2k2 + ... + a2102 + a0

Vi k l mt s nguyn khng m v a2i {1, 2,..., 9} vi i = 0, 1,...,k.Chng minh rng, mi s nguyn c dng 2p3q , vi p, q l nhng snguyn khng m, chia ht mt phn t no ca X.

Li gii: Mi s nguyn khng chia ht cho 10 th u chia ht chomt s hng no caX. Ta nhn thy rng lun tn ti mt phn

t trong Xc2p 1 ch s l bi ca 4p, vi mi s nguyn p khngm. iu ny dn ti php quy np theo p: vi p = 0, 1 v nu x l

mt bi s vi p = k, th ta c th chn a2k sao cho x + a2k102k 0(mod 4k+1) v 102k

1( mod 4k)

By gi, ta chng minh rng bt k s nguyn n no khng chia

ht cho 10 th chia ht cho mt s hng no ca x. Gi n = 2pk

vi k l. Khi vi b bn trn ta c th tm ra bi s ca 2p

trong tp hp X. Gi m l bi s v d l s ch s ca m v f =

10d+1 1. Bng h qu Euler ca nh l Fermat, 10fk( mod f k) . Vvym

10(d+1)(fk) 1 / 10d+1 1 chia ht cho 2pk v thuc X

4.28.Xc nh tt c cc s nguyn dng k tn ti mt hm f :N

Z tha mn:

(a)f(1997) = 1998(b) Vi mia, b N, f(ab) = f(a) + f(b) + kf(gcd(a, b))Li gii: Hm f nh trn tn ti vi k = 0 v k = 1. Ta ly a = b,thay vo (b) ta c f(a2) = (k + 2)f(a). p dng hai ln ta c:

f(a4) = (k + 2)f(a2) = (k + 2)2f(a)

Mt khc ta c:

f(a4) = f(a) + f(a3) + kf(a) = (k + 1)f(a) + f(a3)

= (k + 1)f(a) + f(a) + f(a2) + kf(a)

= (2k + 2)f(a) + f(a2) = (3k + 4)f(a)

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Chn a = 1997 f(a) = 0 suy ra(k + 2)2 = 3k + 4 c nghim k = 0; k =1.Vi k = 0 ta c:

f(pe11 ...penn ) = e1g (p1) + ... + eng (pn)

vi m l c nguyn t ca 1997, g(m) = 1998 v g(p) = 0 vi mi s

nguyn t p = m.Vi k = 1 ta c

f(pe11 ....penn ) = g (p1) + ... + g (pn)

4.29.Cho tam gic ABC nhn vi O l tm ng trn ngoi tip vbn knhR. AO ct ng trn ngoi tip tam gic OBC D, BO ct

ng trn ngoi tip tam gicOCA E, vCO ct ng trn ngoi

tip tam gicOAB F. Chng minh rngOD.OE.OF 8R3

.Li gii: Gi D, E, F ln lt l giao im caAO vBC , BO vCA,CO vAB. Do chng l nh caD,E,F qua php ca ng trn

ngoi tip tam gic ABC, Do OD.OD = OE.OE = OF.OF = R2. V

vy, bt ng thc ca bi ton tng ng vi:

AO

ODBO

OECO

OF 8

Gi h1, h2, h3 ln lt l khong cch t O n AB,BC,CA. Khi

AO/OD = [AOB]/[BOD] = (ABh1)/(BD h2), tng t BO/OE =(BC h2)/(CEh3) vCO/OF = (CAh3)/(AFh1). Do ta c:

AB.BC.CA

AF.BD .CE=

(AF + FB) (BD + DC) (CE + EA)AF.BD .CE

8

AF.FB.BD.DC.CE.EA

AF.BD .CE

= 8

FB.D C.A

AF.BD .CE= 8

Du bng xy ra khi v ch khi AF = FB,BD = DC,CE = EA, hay

ABC l tam gic u.

4.30.ChoX = 1, 2,...,n vin k 3 vFk l mt tp con gmk phnt caX sao cho hai tpFk bt k c nhiu nhtk 2 phn t chung.

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Ch ra rng tn ti mt tp Mk ca X c t nht [log2 n] + 1 phn t

khng c cha trong mt tp con no ca Fk.

Li gii: Nu k log2 n th iu phi chng minh hin nhin ng,v th ta gi s k < log2 n. tm = [log2 n] + 1. Vi mi tp con k 1 s

hng ca tp Xthuc ti a mt tp con ca Fk v mi s hng caFk bao gm cc tp con k(k 1) phn t. Ta c:

(Fk) 1k

n

k 1

= 1

n k + 1

n

k

Mt khc, chn mt tp con m s hng bt k caX, s s hng ca

Fk l:

m

k

(Fk)

nk

1

n k + 1

m

k

iu c th chng minh rng con s sau nh hn 1, v th mt

tp con m phn t no khng c cha s hng no ca Fk.

Hin nhin

m

k

i

m

i

= 2m v ta cng c th chng minh

m

k

3.2

m3 vi m

k

3 bng phng php quy np theo m. V

vy ta c:

1

n k + 1

m

k

3n

4 (n k + 1) < 1

vi n 3. Ta c iu phi chng minh.

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Chng 5

thi olympic Th Nh K

5.31.Cho tam gic ABC vung ti A, gi H l chn ng cao k A.Chng minh rng tng bn knh cc ng trn ni tip cc tam gic

ABC, ABH, ACH bng AH.

Li gii: t a = BC,b = CA,c = AB v s = a+b+c2

. Cc tam gic ABH

v ACH ng dng vi tam gic ABC vi t s tng ng a/c vb/c

p dng cng thc din tch tam gic bng bn knh ng

trn ni tip nhn vi na chu vi, suy ra bn knh cn tm lab

a+b+c; ac

aba+b+c

; bc

aba+b+c

v tng ca chng l abc

= AH

5.32.Dy s{an}n=1 , {bn}n=1 c cho bi:a1 = , b1 = , an+1 = an bn, bn+1 = an + bn vi min 1

C bao nhiu b s thc(, ) tha mna1997 = b1 vb1997 = a1?

Li gii: 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

5.33.Trong mt hip hi bng , khi mt cu th chuyn t i X cx cu th sang iY cy cu th, lin on nhn cy x triu lat iY nu y x nhng phi tr lix y triu la cho i X nu

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x > y. Mt cu th c th di 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 16,16,21,22,22,23

cu th. Tng s tin ln nht m lin on c th kim c trong

sut ma chuyn nhng l bao nhiu?

Li gii: Chng ta tha nhn rng s tin ln nht kim c bikhng bao gi cho php mt cu 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 xtrc khi nhn mt cu th v s tin m lin on kim c bng

tng ca cc s . By gi ta xem xt cc s c ghi bi mt i

m kt thc c nhiu hn 20 cu th. Nu s lng cu th ti a

ca i trong sut qu trnh chuyn nhng l k > n th cc sk 1 vk xut hin lin tip v b i 2 s th tng s tng ln. Vth tng ca cc s trong i bng t nht l20 + 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 chnhxc l nhng con s c c bi vic lun chuyn cu th t i kt

thc t hn 20 cu th sang i kt thc c nhiu hn 20 cu th.

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

5.34.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 vab = cd = 1/4. TnhAC+ CE theoa,b,c.

Li gii: Nu gi 2, 2, 2, 2 l cc cung chn bi cc cnh a,b,c,dtng ng th:

AC = 2 sin ( + ) = a21 b2

4+ b

21 a2

4

Tng t vi CD.

Tng qut vi R l bn knh ng trn ngoi tip ng gic, th th

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thi olympic Th Nh K 25

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 cc s a,b,c,d.

5.35.Chng minh rng vi mi s nguyn t p 7, tn ti mt snguyn dng n v cc s nguynx1, x2,...,xn, y1, y2,...,ym khng chia

ht chop sao cho:

x21 + y21 x22 (modp)

x22 + y22 x23 (modp)

.....

x

2

n + y

2

n x2

1 (modp)

Li gii: Gi n l cp ca 5/3 mod p, v t xi = 3n1i5i1, yi =43n15i1 th mi ng d trn l tng ng tr h thc cui cng.

H thc c dng 52n 32n (modp) (ng).Vy ta c iu phi chngminh.

5.36.Cho cc s nguynn 2. Tm gi tr nh nht ca :

x51

x2 + x3 + ... + xn +

x52

x3 + x4 + ... + xn + x1 + ... +

x5n

x1 + x2 + ... + xn1

Vix1, x2,...,xn l cc s thc tha mnx21 + x22 + ... + x

2n = 1.

Li gii: tS = x1 + x2 + ... + xnS dng BT Chebyshevs cho hai dy xi

Sxi v x4i (c hai dy u l

dy tng). Ta c:

x5iS xi

x1

S x1 +x2

S x2 + ... +xn

S xn

x41 + x

42 + ... + x

4n

n

p dng bt ng thc hm li ta c:

x1S x1 +

x2S x2 + ... +

xnS xn

1

n 1

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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

1

n 1 .1

n2=

1

n (n 1)

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

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Chng 6

thi olympic Ukraina

6.37.Mt li hnh ch nht c t mu theo kiu bn c, v trongmi c mt s nguyn. Gi s rng tng cc s trong mi hng v

tng cc s trong mi ct l s chn. Chng minh rng tng tt c cc

s trong en l chn.

Li gii: Gi s cc mu t l v en, trong vung gc tritrn l mu . (V tng tt c cc s trong li hnh ch nht l

chn nn iu cn chng minh cng tng ng vi tng cc

mu l s chn.)

Tng cc hng th nht, th ba, . . . (t trn xung), v cc ct th

nht, th ba, . . . (t tri sang) bng tng cc s trong cc mu en

tr i hai ln tng tt c cc s trong cc mu . V tng ny l

s chn nn tng cc s trong cc en l chn.

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6.38.Tm tt c cc nghim thc ca h phng trnh sau: x1 + x2 + + x1997 = 1997x41 + x42 + + x41997 = x31 + x32 + + x31997.

Li gii: Ta s chng minh h trn ch c nghim x1 = x2 = =x1997 = 1.

tSn = xn1 + + xn1997. Theo bt ng thc lu tha trung bnh 1,S4

1997

1/4 S1

1997= 1

v S4

1997

1/4

S31997

1/3=

S4

1997

1/3

do S4

1997 1.

V vy bt ng thc lu tha trung bnh xy ra du bng, ngha l

x1 = = x1997 = 16.39.K hiud(n) l s l ln nht trong cc c s ca s t nhin n.

Ta xc nh hmf : N N sao cho

f(2n 1) = 2n v f(2n) = n + 2nd(n)

vi min N

Tm tt c cc s k sao chof(f(. . . f (1) . . .)) = 1997, f c lpkln.

Li gii: Ch c mt gi tr ca k l499499.Vi mi s chn (2a)b, b l s l va 1,

f((2a)b) = (2a1)b +(2a)b

b= (2a1)b + 2a = (2a1)(b + 2).

Do ly tha cao nht ca hai trong f((2a)b) bng mt na lu

tha cao nht ca hai trong (2a)b. V vy, ly lp a ln caf ti (2a)b

c mt s l.1"power mean inequality" (trong bn ting Anh), tm dch l "bt ng thc lu

tha trung bnh" (ND)

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thi olympic Ukraina 29

Ngoi ra, c l ln nht caf((2a)b) lb + 2, ln hn c l ln nht

ca(2a)b hai n v. V vy sau a ln lp f ti (2a)b, c l ln nht l

b + 2a. Do sau a ln lp f ta nhn c s l nn fa((2a)b) phi bng

b + 2a.

Sau mt ln lp f ti 2n

1, n

1, ta nhn c 2n. ta = n

vb = 1. Ta thy phi lp thm n ln na mi nhn c s l, s

l, b + 2a = 2n + 1. V vy phi lp n + 1 ln cho f ti 2n 1 nhnc s l tip theo, 2n + 1.

Ta d thy l khng c s l no xut hin hai ln khi lp f ti 1, bi

v dy cc s l nhn c t vic lp f ti bt k s nguyn dng

no cng l mt dy tng ngt. c bit, tn ti mt gi tr k sao cho

f(k)(1) = 1997, v s k l duy nht.

By gi ta chng minh bng quy np rng

fn(n+1)

21(1) = 2n 1.

D thy iu ng ti n = 1. Sau gi s n ng vi n = k,

f(k+1)(k+2)

21(1) = fk+1

f

k(k+1)2

1(1)

= fk+1(2k 1) = 2k + 1

V 1997 = 2(999) 1, ta c

f999.1000

21(1) = f499499(1) = 1997,

nn k = 499499 tho mn yu cu bi, v do lp lun trn, k l

duy nht.

6.40.Hai ng gic uABCDE vAEKPL trong khng gian sao choDAK = 60o. Chng minh rng hai mt phngACK vBAL vung gc.

A

B

C

D E

KP

L

T

O

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Li gii: Nu ta quayAEKPL quanh trc AE, bt u v tr trngnhau vi ABCDE, th gc DAK tng cho n khi AEKPL li nm

trn mt phng cha ABCDE. (iu ny c th nhn c bng

cch tnh ton thy tch v hng ca vect AD v AK tng trong

mt khong xc nh). V vy c mt gc duy nht gia cc mt

phng sao cho gc DAK = 60. Tht vy, iu ny xy ra khi cc

hnh ng gic l hai mt ca khi mi hai mt u 2 (do tnh cht

i xng, tam gic DAK l u trong trng hp ny). c bit, mt

phng BAL l mt phng cha mt th ba qua nh A ca khi mi

hai mt u, ta gi mt l BALTO.

S dng thm du phy sau tn im k hiu hnh chiu tng

ng ca cc im xung mt phng ABCDE. Khi K nm trn

ng phn gic ca gc DEA. Lp h trc ta sao cho ABCDE

nm trn mt phng Oxy theo chiu kim ng h, vKE cng chiuvi chiu dng ca trc Ox. Xt hai vect

EK v

EA. Vect th nht

khng c thnh phn y, v vect th hai khng c thnh phn z. Do

tch v hng ca hai vect bng tch cc thnh phn x ca

chng. V AE vKE to vi nhau gc 54,EA c thnh phn x c

tnh bi AE(cos 54) = cos54o.EK c thnh phn x l KE. Do

tch v hng ca chng bng KE(cos54). Mt khc EK .EA =EK.EA. cosKEA = (1).(1). cos(108) = cos72. Lp phng trnh thai kt qu trn ta nhn c KE = cos 72

cos 54

Cc tnh ton trn vect sau y s chng t KA vAC vung gc:

KA.

AC = (

KE+

EA).

AC

= KE ACcos54 CA ACcos72

=cos 72

cos 54cos 54 cos 72 = 0

V BO vung gc vi AC vBO song song vi AT, nn AT vung gc

vi AC. V vy cc im K, A , T thng hng. V K,A,T khng thng

hng nn mt phng AKT vung gc vi mt phng cha ABCDE.Quay xung quanh A, ta suy ra mt phng ACK vung gc vi mt

2khi thp nh din u (ND)

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thi olympic Ukraina 31

phng chaBALTO, t c iu phi chng minh.

6.41.Cho phng trnhax3 + bx2 + cx + d = 0 c ba nghim thc phnbit. Hi phng trnh sau c bao nhiu nghim thc:

4(ax3 + bx2 + cx + d)(3ax + b) = (3ax2 + 2bx + c)2

Li gii: tP(x) = ax3 + bx2 + cx + d vQ(x) = 2P(x)P(x) [P(x)]2,ta s m s nghim thc ca Q(x) = 0. Ta c th thc hin mt s

php rt gn m khng thay i s nghim thc caQ(x). Trc tin

ta chiaP cho mt h s t l sao cho a = 1. Sau ta tnh tin x sao

cho nghim ng gia caP l0, tc ld = 0 vc < 0. Khi :

Q(x) = 3x4 + 4bx3 + 6cx2

c2.

By gi ta p dng quy tc du Descartes m s nghim ca

Q(x). Du caQ(x) l +, s, , v ca Q(x) l +, s, , , s ldu cab. Trong trng hp no th mi dy trn cng ch c ng

mt ln i du. V vyQ c mt nghim m v mt nghim dng,

tc lQ c tt c hai nghim thc.

6.42.K hiuQ+ l tp tt c cc s hu t dng. Tm tt c cc hmsf : Q+ Q+ sao cho vi mix Q+:

(a) f(x + 1) = f(x) + 1(b) f(x2) = f(x)2.

Li gii: Ch c mt hm s tho mn l f(x) = x. T (a), f(x + n) =f(x) + n vi mi n nguyn dng. tx = pq vi p, q nguyn dng. Ta

c:

f

p + q2

q

2=

q + f

p

q

2= q2 + 2qf

p

q

+ f

p

q

2.

Mt khc,

fp + q2q2 = f(p + q2)2

q2

= f

q2 + 2p +

p2

q2

= q2 + 2p + f

p

q

2.

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T hai phng trnh trn suy ra

2qf

p

q

= 2p

v vyf(p

q) =

p

q.

6.43.Tm s nguynn nh nht sao cho vi min s nguyn tu , tnti18 s nguyn trong c tng chia ht cho18.

Li gii: Gi tr nh nht can ln = 35; tp hp 34 phn t gm 17s khng v17 s mt cho ta thy rng n 35. V vy ta ch cn phichng minh rng vi 35 s nguyn bt k lun tm c 18 s trong

c tng chia ht cho 18. Thc ra ta s chng minh rng vi mi n,

trong 2n 1 s nguyn lun tm c n s c tng chia ht cho n.Ta s chng minh khng nh trn bng quy np theo n. D thy

iu ng vi n = 1. Nu n l hp s, vitn = pq, ta c th ly ratp p s nguyn m tng chia ht cho p cho n khi cn li t nht

2p 1 s; ta c 2q 1 tp nh vy, v li theo gi thit quy np, c qtp trong s c tng (ca pq s) chia ht cho q.

By gi gi s n = p l s nguyn t. S x chia ht cho p nu v ch

nu xp1 1 (mod p). V vy nu khng nh trn sai th tng ccs dng (a1 + + ap)p1 trn mi tp con {a1, . . . , ap} ca cc s cho ng d vi Cp12p1 1 (mod p). Mt khc, tng ca cc s dngae11

aepp vi e

1+

+ e

p p

1 lun chia ht cho p: nu k

p

1 cc

s ei khc khng th mi tch c lp li Cpk2p1k ln, v cui cng

l mt bi ca p. S mu thun chng t khng nh trn l

ng trong trng hp ny. (Ch : chng minh bi ton cho,

cn phi chng minh trc tip cho trng hp p = 2, 3.)

6.44.Cc im K,L,M,N nm trn cc cnh AB,BC,CD,DAca hnh hp (khng cn l hnh hp ng) ABCDA1B1C1D1.

Chng minh rng cc tm mt cu ngoi tip ca cc t din

A1AKN,B1BKL,C1CLM,D1DM N l cc nh ca mt hnh bnh hnh.

Li gii: a vo h ta vi ABCD song song vi z =0. Gi E,F,G ,H l tm ng trn ngoi tip cc tam gic

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Chng 7

thi olympic Anh

7.45.Gi sM vN l hai s nguyn dng c 9 ch s c tnh cht lnu bt k ch s caM c thay bi ch s caN tng ng th ta

c mt bi ca 7. Chng minh rng vi bt k mt s t c bng

cch thay mt ch s caN tng ng bi mt ch s ca M cng l

mt bi ca 7.

Tm mt s nguynd > 9 sao cho kt qu trn vn cn ng khiM v

N l hai s nguyn dng cd ch s.

Li gii: Kt qu ng vi bt k d 2 ( mod 7). VitM =

k 10

k, N =

nk10k

, ymk, nk l cc ch s. Th vi bt k k, 10k

(nkmk) 0M (mod 7). Ly tng theo k, chng ta c M N dM 2M ( mod 7),v vyN M ( mod 7), c 10k(mk nk) N ( mod 7). Vy khithay bt k ch s trong N bi ch s tng ng trong M chng ta

t c mt s chia ht cho 7.

7.46.Trong tam gic nhnABC, CF l mt ng cao, vi F trnAB,vBM l mt trung tuyn, viM trnCA. ChoBM = CF vMBC =

F CA, chng minh rng tam gicABC l u.

Li gii: Gi s ACF = CBM = A, v gi s CM = AM = m. ThMB = CF = 2m cos A. Theo nh l hm s Sin,CM

sinCBM=

MB

sinMCB,

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thi olympic Anh 35

v v vysinMCB = 2cos A sin A = sin 2A.

iu ny a n hai kh nng. Nu MCB +2A = 180, th CMB =

A = MBC. Khi CB = MC v MB = 2MCsin A. Cng c MB =

CF = ACcos A = 2MCcos A. Do sin A = cos A v vy A 45 MCB

90, mu thun.

V vy chng ta c MCB = 2A, v vyACF = BC F. Do tam

gic ACF ng dng vi tam gic CBM, v vyCAF = BC M. Do

BC = AB, vy tam gic ABC l u.

7.47.Tm s cc a thc bc 5 vi cc h s khc nhau t tp{1, 2, . . . , 9} m chia ht chox2 x + 1.Li gii: Cho phng trnh bc 5 lax5 + bx4 + cx3 + dx2 + ex + f = 0.Cc nghim cax2 x + 1 khng phi l cc nghim thc cax3 + 1,

lei/3

ve5i/3

. Do a thc bc 5 l chia ht cho x2

x + 1 nuv ch nu

ae5i/3 + be4i/3 + cei + de2i/3 + eei/3 + f = 0.

Ni cc khc, v vy i sin 60(a b + d + e = 0), hoc a d = e b va/2 b/2 c d/2 + e/2 + f = 0, hay e + 2f + a = b + 2c + d hoc (va d = e b) a d = c f = e b. iu ny ko theo 1/12 ca a thcs c cc h s p + k,q ,r + k,p,q + k, r vi k > 0 vp q r.Vi k cho, c C39

k cc gi tr cap,q,r sao cho r +k

9. Tuy nhin,

cc h s phi khc nhau, nn chng ta phi tr i. C 9 2k cchla chn 2 s khc nhau theo k, v 7 k cch la chn cc s cnli. Tuy nhin, chng ta m c cc s dng x, x + d, x + 2d sinh

i, v c 9 2k s.Do , vi k cho trc, chng ta c

C39k (9 2k)(7 k) + 9 3k

a thc. Cng li, c (1+4+10+20+35+56)

(42+25+12+3)+(3+6) = 53

a thc dng trn, v53.12 = 636 cc a thc tt c.

7.48.TpS = {1/r : r = 1, 2, 3, . . .} cc s nguyn dng nghc o lpthnh cp s cng vi di ty . Chng hn, 1/20, 1/8, 1/5 l mt

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cp s nh vy, vi di l 3 vi phng sai l 3/40. Hn na, c

mt cp s cc i trongS vi di 3 v n khng th m rng thm

v bn phi v bn tri caS (1/40 v11/40 khng l phn t caS).(a) Tm mt cp s cc i trongS c di 1996.

(b) C hay khng mt cp s trongS c di 1997?

Li gii: C mt cp s cc i c di n, vi mi n > 1. Theo nhl Dirichlet ko theo rng c mt s nguyn t p c dng 1 + dn vi

mi s nguyn dng d. By gi xt cp s

1

(p 1)! ,1 + d

(p 1)! , . . . ,1 + (n 1)d

(p 1)! .

V cc mu s chia ht t s, nn mi phn thc l mt s nguyn

nghc o, nhng vi (1 + nd)/(p 1)! = p/(p 1)! khng phi v p ls nguyn t. Do c dy cp s cc i. ( gii (a), n gin ly

p = 1997.)

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