loi giai vmo 2011 (day 1 and 2)

7/25/2019 Loi Giai VMO 2011 (Day 1 and 2)

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BAN BIN TP DIN N TON HC MATH.VN

LI GII

THI HC SINH GII QUC GIA

NM HC 2010 2011

d

THNG 01 2011


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d


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LI NI U

Din n ton hc Math.vn cha y hai tui, nhng nhng nggp ca cc thnh vin trn din n vi cng ng dn ckhng nh Din n l ni trao i hu ch ca cc thy c gio,ca cc em hc sinh v nhng bn yu ton ... c nhiu biging hay, c nhng li gii p cho nhng bi ton kh, lni gp g trao i nhiu tng c o cho nhng vn tngchng n gin . . . Nhn li hn mt nm hot ng, chng ti thy c nhng du n:

T chc tng thut trc tip i hi Ton hc th gii n, ni ti nng v con ngi Vit Nam c khng nh bnggii thng Fields ca Gio s Ng Bo Chu. Nhng thngtin ca Math.vn c nhiu trang web trch ng.

T chc thi th nm 2010 vi 24 cht lng c a phnhc sinh v thy c nh gi cao.

T chc gii thi i hc khi A, B, D mn Ton c nhiu ligii hay c cng b sm nht.

Pht huy tinh thn , nhn k thi chn hc sinh gii Quc gia VitNam 2011, din n t chc gii VMO 2011. Chng ti tin tngy l mt ti liu tt cho cc bn hc sinh ang v s tham gia cccuc thi chn hc sinh gii tham kho.

BAN BIN TP DIN N MATH.VN


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4 DIN NTON HCMATH.VN

d


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MC LC

Li ni u . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3Li gii bi 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

Li gii bi 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

Li gii bi 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

Li gii bi 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

Li gii bi 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

Li gii bi 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

Li gii bi 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29


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d


8/31

BI S 1: BT NG THC

Bi 1. Cho xl s thc dng v nl s nguyn dng. Chng minhbt ng thcxn(xn+1+1)

xn+1

x+1

2

2n+1.

ng thc xy ra khi no?

Li gii 1. Ta s dng phng php quy np theon.Vi n = 1,btng thc ca ta tr thnh

x(x2+1)

x+1

x+1

2

3

.

Theo bt ng thc AM-GM, ta c

x(x2+1)=1

2 (2x) (x2+1)

1

2

(2x)+ (x2+1)

2

2=

(x+1)4

8.

T suy rax(x2+1)

x+1

(x+1)48

x+1=

x+1

2

3

.

V nh vy, bt ng thc cho ng vin= 1.

Tip theo, ta s chng minh rng nu bt ng thc ng cho n=k(k N)th n cng s ng vi n=k+1.Tht vy, theo gi thit quynp, ta c

x+12

2k+1

xk(xk+1+1)

xk+1 ,suy ra

x+1

2

2(k+1)+1=

x+1

2

2 x+12

2k+1

x+1

2

2xk(xk+1+1)

xk+1.

S dng nh gi ny, ta thy rng vic chng minh c th c av chng minh kt qu sau

x+1

2

2 xk(xk

+1+1)

xk+1 xk

+1(xk

+2+1)

xk+1+1.

Bt ng thc ny tng ng vi

(x+1)2

4x

(xk+2+1)(xk+1)

(xk+1+1)2 ,


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hay l(x+1)2

4x1

(xk+2+1)(xk+1)

(xk+1+1)2 1.

Do (x+1)24x= (x1)2 v(xk+2+1)(xk+1) (xk+1+1)2 =xk(x1)2 nnta c th thu gn bt ng thc li thnh

(x1)2

4x

xk(x1)2

(xk+1+1)2,

tng ng(x1)2 (x

k+1+1)24xk+1 0.

Bt ng thc ny ng v theo AM-GM, ta c (xk+1+1)2 4xk+1.

Nh vy, ta chng minh c nu khng nh bi ton ng chon=k (k N)th n cng ng cho n=k+1.T y, kt hp vi vic xc lp c tnh ng n ca bt ng thc cn chng minhchon= 1,ta suy ra n ng vi mi s nguyn dngn(theo nguynl quy np). Ngoi ra, c th thy c trong sut qu trnh chngminh, du ng thc ch xy ra ti mt im duy nhtx= 1.

Li gii 2. Ta s chng minh kt qu tng qut hn: Vi mi a,b> 0,th

a+b

2

2n+1

anbn(an+1+bn+1)

an+bn . (1)

Kt qu bi ton cho l trng hp ring khi a=xv b= 1.

D thy (1) l mt bt ng thc thun nht cho hai bin a, b,v vy

khng mt tnh tng qut ta c th chun ha cho a+ b = 2.Khi (1) c th vit li di dng fn(a, b) 0,trong

fn(a, b)= an+bnanbn(an+1+bn+1).

S dng bt ng thc AM-GM, ta c

ab(an1+bn1)(an+1+bn+1)

ab(an1+bn1)+ (an+1+bn+1)

2

2

=(a+b)2(an+bn)2

4= (an+bn)2,

t suy ra

an+1+bn+1 (an+bn)2

ab(an1+bn1).


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LI GIIVMO 2011 9

S dng nh gi ny, ta thu c

fn(a, b) an+bn

an1bn1(an+bn)2

an1+bn1

=an+bn

an1+bn1

an1+bn1an1bn1(an+bn)

=an+bn

an1+bn1fn1(a, b). (2)

T (2), thc hin cc nh gi lin tip, ta c

fn(a, b)

an+bn

an1+bn1fn1(a, b) (1)

an+bn

an1+bn1an1+bn1

an2+bn2fn2(a, b)

an+bn

an1+bn1an1+bn1

an2+bn2

a2+b2

a1+b1f1(a, b)

=an+bn

a+bf1(a, b). (3)

Mt khc, cng theo bt ng thc AM-GM th

f1(a, b)= a+bab(a2+b2)= a+b

1

2 (2ab) (a2+b2)

a+b1

2

(2ab)+ (a2+b2)

2

2= a+b

(a+b)4

8= 0.

Do , kt hp vi (3), ta suy ra fn(a, b) 0vi mi n N.

Li gii 3. Ta chng minh b sauB .Cho a, bl hai s thc dng. Khi , vi mi n 1,ta c

(ab)n(n1)

2 (an+bn) 2

a+b

2

n2. (4)

Chng minh.Khng mt tnh tng qut, ta gi s a+ b = 2v ta= 1+x, b= 1xvi0x< 1.Bt ng thc (4) c th vit li thnh

(1+x)(1x)

n(n1)2

(1+x)n+ (1x)n

2,

hay tng ng

g(x)= (1+x)n(n+1)

2 (1x)n(n1)

2 + (1+x)n(n1)

2 (1x)n(n+1)

2 2.


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Ta c

(1+x)n(n+1)

2 (1x)n(n1)

2

=

=n(n+1)

2(1+x)

n(n+1)2 1(1x)

n(n1)2

n(n1)

2(1+x)

n(n+1)2 (1x)

n(n1)2 1

=n

2(1+x)

n(n+1)2

1(1x)n(n1)

2 1

(n+1)(1x) (n1)(1+x)

= n(1+x)n(n+1)

2 1(1x)

n(n1)2 1(1nx)

v

(1+x)

n(n1)

2 (1x)n(n+1)

2=

=n(n1)

2(1+x)

n(n1)2 1(1x)

n(n+1)2

n(n+1)

2(1+x)

n(n1)2 (1x)

n(n+1)2 1

=n

2(1+x)

n(n1)2 1(1x)

n(n+1)2 1

(n1)(1x) (n+1)(1+x)

=n(1+x)n(n1)

2 1(1x)

n(n+1)2

1(1+nx),

do

g(x)= n(1+x)n(n1)

2 1(1x)

n(n1)12

(1+x)n(1nx) (1x)n(1+nx)

= n(1x2)n(n1)

2 1(1+x)n(1+nx)

1nx

1+nx

(1x)n

(1+x)n

.

T y ta thy g(x)c cng du vi h(x)= 1nx1+nx

(1x)n

(1+x)n.Tnh o hm

ca h(x),ta c

h(x)= 2n(1x)

n1

(1+x)n+1 2n

(1+nx)2= 2n(1

x

2

)

n1

(1+x)2n 2n

(1+nx)2

2n

(1+x)2n

1

(1+nx)2= 2n

1

(1+x)n

1

1+nx

1

(1+x)n+

1

1+nx

0

do theo bt ng thc Bernoulli th(1+x)n 1+nx(ch rng n 1).

Nh vy, h(x)l hm nghch bin trn [0, 1).Suy ra h(x) h(0) = 0,x [0, 1).M g(x)c cng du vi h(x)nn ta cng c g(x) 0vi

mi x [0, 1).Do vy g(x)l hm nghch bin trn [0, 1).T l lunny, ta suy ra g(x) g(0)= 2, x [0, 1).B c chng minh.

Quay tr li bi ton. Theo (4), ta c

(ab)k(k1)

2 (ak+bk) 2

a+b

2

k2, a, b> 0, k 1, (5)


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LI GIIVMO 2011 11

suy ra

a+b 2(ab)k1

2k

ak+bk

2

1k2

. (6)

Trong (6), cho a=xn, b= 1vk= n+1n > 1,ta c

xn+1 2xn

2(n+1)

xn+1+1

2

n2(n+1)2

.

T suy ra

xn(xn+1+1)xn+1

xn(xn+1+1)

2xn

2(n+1)

xn+1+1

2

n2(n+1)2

=xn(2n+1)2(n+1)

xn+1+1

2

2n+1(n+1)2

.

Nh th, php chng minh s hon tt nu ta ch ra c rng

xn(2n+1)2(n+1)

xn+1+1

2

2n+1(n+1)2

x+1

2

2n+1.

Bt ng thc ny tng ng vi

xn(n+1)

2 (xn+1+1) 2

x+1

2

(n+1)2.

y chnh l kt qu ca bt ng thc (5) p dng cho a = x, b = 1vk= n+1.Bi ton c chng minh xong.

Nhn xt. C th thy tng t nhin nht khi gii bi ny chnhl s dng php quy np. Li gii 3 tuy di v phc tp nhng ncng c ngha ring ca n. Tht vy, qua li gii ny ta c th thyc bt ng thc cho vn ng cho trng hpnl s thc ty khng nh hn1.Kt qu ny khng th suy ra c t hai li giibng quy np 1 v 2.


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d


14/31

BI S 2: GII HN CA DY S

Bi 2. Cho dy {xn}c xc nh bi

x1 = 1v xn =2n

(n1)2

n1i=1

xivi mi n 2.

Chng minh rng dy yn =xn+1xnc gii hn hu hn khi n+.

Li gii 1. T gi thit, ta suy ra vi min 1,th

n1i=1

xi =(n1)2

2nxn.

Do

xn+1 =2(n+1)

n2

ni=1

xi =2(n+1)

n2

xn+

n1i=1

xi

= 2(n+1)n2

xn+ (n1)

2

2nxn

= (n+1)(n

2

+1)n3

xn. (1)

S cng thc truy hi va tm c ny kt hp vi php quy np, tas chng minh

xn 4(n1), n 2. (2)

Do x2 = 22(21)2x1 = 4nn d thy (2) ng vin= 2.Gi s (2) ng vin=k 2,khi ta c

xk+1 =(k+1)(k2+1)

k3 xk

(k+1)(k2+1)

k3 4(k1)

=4(k21)(k2+1)

k3 =

4(k41)

k3 = 4k

4

k3< 4k,

suy ra (2) cng ng vi n =k+1.T y, p dng nguyn l quynp, ta c (2) ng vi mi n 2.

By gi, ta s i chng minh bi ton cho, c th ta s ch ra rngynl dy tng v b chn trn.

Chng minh yntng.Theo (1), ta c

yn =xn+1xn =(n+1)(n2+1)

n3 xnxn =

n2+n+1

n3 xn.


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Do

yn+1yn =(n+1)2+ (n+1)+1

(n+1)3

xn+1n2+n+1

n3

xn

=n2+3n+3

(n+1)3

(n+1)(n2+1)

n3 xn

n2+n+1

n3 xn

=xn

n3

(n2+3n+3)(n2+1)

(n+1)2 (n2+n+1)

=xn

n3

1+

n+2

(n+1)2

(n2+1) (n2+n+1)

=

xn

n3(n+2)(n2+1)

(n+1)2 n=

2xn

n3(n+1)2 > 0.

iu ny chng t ynl dy tng.

Chng minh ynb chn trn.S dng (2), vi mi n 2,ta c

yn =n2+n+1

n3 xn

n2+n+1

n3 4(n1)=

4(n31)

n3 < 4.

Do y1 < y2 < < yn < 4,hay ni cch khc, dy ynb chntrn bi4.

T hai kt qu va chng minh trn, ta d dng suy ra kt qu cnchng minh.

Nhn xt.Khi lm bi ton ny, c l cc bn hc sinh u khngkh tm ra cc tnh cht

xn+1 = (n+1)(n2+1)n3 xn.

yn =n2+n+1

n3 xn.

ynl dy tng.

V khi , cng vic cn li s ch l tm ra mt chn trn cho yn.Cth ni y chnh l yu t quan trng nht ca bi ton. Vic tm ra

nh gi (2) c th c gii thch nh sau: Ta bit rng hm phnthc f(x) = g(x)h(x) vi g(x), h(x)l cc a thc ng bc v h(x) > 0thb chn trn bi mt hng s. M quan st cng thc truy hi cayn,ta thy rng n

2+n+1n3

l hm phn thc theonvi n2+n+1l athc bc 2v n3 l a thc bc 3.T y ta cht c mt tng lnh gixnvi hm a thc bc nht theo n(chiu ) v khi yns


16/31

LI GIIVMO 2011 15

b chn trn bi mt hm phn thc vi t l a thc bc 3v mucng l a thc bc3,v nh th theo tnh cht va nhc li trn,ta bit chc rng yns b chn trn bi mt hng s.

Vi tng nh vy, ta mong mun c mt nh gi dngxn an+b.Ngoi ra, t cng thc truy hi trn ca xn,ta cng ngh n victhit lp nh gi ny bng quy np (v nh th l n gin hn c).Nh th, ta cn phi chn cc s thc a, bsao cho

(n+1)(n2+1)

n3 (an+b) a(n+1)+b.

Thc hin php khai trin, ta vit c bt ng thc ny li thnh

(a+b)n(n+1)b 0.

iu ny ng vi min 1,ta cn c a+b 0v2(a+b)+b 0.Vtt nhin, n gin, ta chn ngay a+b = 0v b < 0,tc a=b> 0.Khi , ta thu c bt ng thc dng

xna(n

1).

Ta thy rng nu c asao cho nh gi ny ng vi mt s n0no th nh gi cng s ng vi min n0(do l lun trn). V nhth, ta ch cn xt mt vi gi tr nnh v chn asao cho bt ngthc ng vi cc gi tr l c. Ngoi ra, ta thy bt ng thcs khng c tha mn vin= 1,nn ta s xt vi n= 2.Khi , btng thc tr thnh 4 a.V rt n gin, ta ngh ngay n vicchn a= 4.y chnh l ngun gc ca vic thit lp (2).

Li gii 2. p dng gi thit cho bi, ta c

n2xn+1

2(n+1)=

nk=1

xk =xn+n1k=1

xk =xn+(n1)2xn

2n=

(n+1)(n2+1)

n3 xn. (3)

T suy ra

xn+1xn = n2

+n+1n2

unvi un = xnn

. (4)

Do (3) nn ta c

un+1 =n2+1

n2 un =

1+

1

n2

un.


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Mt khc, d thy xn > 0, n N nn ta cng c un > 0, n N.Vvy, ta c th tlnun = vn.Khi

vn+1 = vn+ ln

1+ 1n2

> vn, n N.

Vy {vn}nN l dy tng. By gi, s dng bt ng thc c bnln(1+x) 0,ta thu c

vn+1 < vn+1

n2.

T y ta c

vn < v1+1+(n1)2k=2

1

k2< v1+1+

(n1)2k=2

1

k(k1)= v1+2

1

n1, n 2.

iu ny chng t {vn}nNb chn, hn na do {vn}l dy tng nnta suy ra c{vn}hi t. V un =evn v hm f(x)=ex l hm lin tcnn{un}nNcng hi t. T l lun ny kt hp vilimn

2+n+1n2

= 1v

(4), ta suy ra iu phi chng minh.


18/31

BI S 3: HNH HC PHNG

Bi 3.Cho ng trn (O),ng knh AB. P l mt im trn tiptuyn ca (O)ti B (P B).ng thng AP ct (O)ln th hai tiC. Dl im i xng ca Cqua O.ng thng DPct (O)ln thhai ti E.

(a) Chng minh rng AE, BC, POng quy ti M.

(b) Tm v tr ca im P din tch tam gic AMBl ln nht.Tnh din tch ln nht theo Rl bn knh ca (O).

Li gii. Trc ht xin nhc li khng chng minh b sau

B .Cho hnh thang ABCD, AB CD.Gi sACct BD ti OvADct BC ti I.Khi , OIi qua trung im AB v CD.

O

A B

P

C

D

E

F

M

Quay tr li bi ton:(a)Gi Fl giao im ca AEv BP .T tnh cht gc ni tip vng cao ca tam gic vung ta d thy AEC=ABC=BPC,vy t gic CPFEni tip. T suy ra

CPE=CFE, PCE=EFB.


19/31


Cng cc ng thc gc vi ch CEP = 90,ta suy ra

90 =CPE+PCE =CFE+EFB=CFB,

hayCF PB,v do CF AB.Gi Ml giao im ca CB v AE.p dng b cho hnh thangABFC,ta c MPi qua trung im AB hay MPi qua O.Vy AE,BC, OP ng quy ti M, l iu phi chng minh.

(b)p dng nh l Menelaus cho tam gic APOvi C, M, Bthnghng, ta d thy

OM

OP = CA

CA+2CP .T ta c

SMAB

SPAB= OMOP

= CACA+2CP .

Suy ra

SMAB =SPAB CA

CA

+2CP

SPAB CA

2CA

2CP

=SPAB CA

2

2BC

= BC PA2

CA2

2BC= 4R

2

4

2= R

2

2.

ng thc xy khi PB=

2R.


20/31

BI S 4: TON RI RC

Bi 4.Cho ng gic li ABCDEc cc cnh v hai ng cho AC,AD c di khng vt qu

3.Trong ng gic li ly 2011im

phn bit bt k. Chng minh rng tn ti mt hnh trn n v ctm nm trn cnh ca ng gic li ABCDE v cha t nht 403im trong s 2011im cho.

Li gii. Trc ht ta chng minh b sau

B . Cho im Inm trong tam gic XY Zc di cc cnh nhhn3.Khi ,

min{IX, IY, IZ}< 1.Chng minh.Tht vy, v X IY+Y IZ+ZIX=360nn trong bagc X IY, Y IZ, ZIXphi c mt gc khng nh hn120.Gi sX IY 120th trong tam gicIXY,theo nh l cosin ta c

3 XY2 = IX2+IY22IX IYcosX IY

IX2

+IY2

+IX IY 3min{IX2

, IY2

}.

T y a nmin{IX, IY} 1.B c chng minh. Quay tr li bi ton. Theo gi thit th cc tam gic ABC, ACD,ADEu c c ba cnh nh hn

3,m mi im trong2011im

gieo trong ng gic ABCDEu thuc min trong ca mt trong batam gic ny nn theo b , mi im phi cch mt nh no ca ng gic mt khong khng ln hn 1.Theo nguyn l Dirichlet,

c mt nh ca ng gic c khong cch khng ln hn1n t nht20115

= 403im. T ta c iu phi chng minh.


21/31


d


22/31

BI S 5: DY S C TNH CHT S HC

Bi 5. Cho dy s nguyn {an}xc nh bi

a0 = 1, a1 =1v an = 6an1+5an2vi mi n 2.

Chng minh rng a20122010chia ht cho 2011.

Li gii 1. Xt dy{bn}c xc nh nh sau

b0 =

1, b1 =

1

vbn =

6bn1+2016bn2vi mi

n

2.

Dy ny c phng trnh c trng

x26x2016= 0

c hai nghim l x= 42v x= 48.T y, s dng kin thc vphng trnh sai phn, ta tm c cng thc tng qut ca dy l

bn = 41 48n

+49 (42)n

90, n N.

Ngoi ra, ta cng d dng chng minh bng quy np rng

an bn (mod 2011), n N.

Theo , ta ch cn chng minh b2012+1 0 (mod 2011)na l xong.Ta c

b2012+1= 41 482012

+49 (42)2012

+9090

.

Do 2011l s nguyn t, v 2011, 90l hai s nguyn t cng nhaunn ta ch cn chng minh

41 482012+49 (42)2012+90 0 (mod 2011). (1)

M theo nh l Fermat nh, ta c

41 482012+49 (42)2012+90 41 482+49 422+90 (mod 2011)= 90b2+90= 90

6 (1)+2016 1

+90

= 90 2010+90= 90 2011 0 (mod 2011).

V vy, (1) ng. Bi ton c chng minh xong.


23/31


Li gii 2. Phng trnh c trng ca dy cho l x26x5= 0c hai nghim l 3

14v 3+

14,do ta d dng tm c cng

thc s hng tng qut ca dy l

an =

7214

3+14n+ 7+214314n

14

=7+

14

3+

14

n1 7+14314n114

=un+2vn,

trong

un =

3+

14n1+ 314n1

2, vn =

3+

14

n1 314n12

14.

S dng cng thc khai trin nh thc Newton, ta c

u2012 =1005

k=0C

2k20113

20112k14k = 32011+1005

k=1C

2k20113

20112k14k.

Do 1< 2k< 2011vi1 k 1005v2011l s nguyn t nn

C2k2011 = 2011

C

2k12010

2k

... 2011.

Mt khc, theo nh l Fermat nh th

32011

3 (mod 2011).

Do vy, kt hp cc lp lun li vi nhau, ta c

u2012 3 (mod 2011). (2)

Tng t vi vn,ta cng s dng khai trin Newton v thu c

v2012=

1006

k=1

C2k12011

320122k14k1=

141005

+

1005

k=1

C2k12011

320122k14k1.

n y, cng bng cch s dng tnh nguyn t ca 2011,ta thy

C2k12011= 2011

C

2k22010

2k1

... 2011


24/31

LI GIIVMO 2011 23

vi k {1, 2, ..., 1005}.V vy

v2012 141005 (mod 2011).

Do14 = 20252011= 4522011 452 (mod 2011)nn p dng nh lFermat nh, ta c

141005 452010 1 (mod 2011).

Suy rav2012 1 (mod 2011). (3)

T (2) v (3), ta ca201220103+2 12010 0 (mod 2011).

Bi ton c chng minh xong.


25/31


d


26/31

BI S 6: HNH HC PHNG

Bi6. Cho tam gic ABCkhng cn ti Av c cc gc ABC, ACB lcc gc nhn. Xt mt im Ddi ng trn cnhBCsao cho Dkhngtrng vi B, Cv hnh chiu vung gc ca Atrn BC.ng thngdvung gc vi BCti Dct ng thng AB, AC tng ng ti Ev

F.Gi M, Nv P ln lt l tm ng trn ni tip cc tam gicAEF, BD E v CDF. Chng minh rng bn im A, M, N, P cngnm trn mt ng trn khi v ch khi ng thng d i qua tmng trn ni tip tam gic ABC.

Li gii. Ta thy bi ton cho chnh l s kt hp c hc ca haikt qu sauB 1. Cho tam gic ABCkhng cn ti Av c cc gc ABC, AC Bl cc gc nhn. Xt mt im D di ng trn cnh BC sao cho D

khng trng vi B, Cv hnh chiu vung gc ca Atrn BC.ngthng dvung gc vi BC ti Dct ng thng AB, AC tng ngti E, F. (N), (P)ln lt l ng trn ni tip tam gic BD E, CDF.

Khi , di qua tm ni tip tam gic ABCkhi v ch khi tip tuynchung khc dca (N)v (P)i qua A.

B 2. Cho tam gic ABCkhng cn ti Av c cc gc ABC, AC Bl cc gc nhn. Xt im D di ng trn cnh BC sao cho D khngtrng vi B, Cv hnh chiu vung gc ca Atrn BC.ng thngdvung gc vi BC ti Dct ng thng AB, ACtng ng ti E,

F.Gi M, Nv P ln lt l tm ng trn ni tip cc tam gicAEF, BD Ev CDF.Khi , bn im A, M, N, Pcng nm trn mt

ng trn khi v ch khi tip tuyn chung khc dca (N)v (P)iqua A.

Nh vy, ta ch cn chng minh c hai b ny th bi ton cngc gii quyt xong.Chng minh b 1.Ta chng minh hai chiu.

Gi s tip tuyn khc dca(N)v(P)i qua A.Gi giao imca tip tuyn v d l T. D thy cc t gic TABDv T A C D

ngoi tip, do theo tnh cht c bn ca t gic ngoi tip,ta cAB +T D = AT+BDv AC +T D = AT+DC.

T hai ng thc ny, ta d thy

DBDC = AB AC.


27/31


Li c DB+DC =BCnn

DB =B A+BC AC

2, DC =

C A+CB AB

2.

VyDchnh l tip im ca ng trn ni tip tam gic ABCvi BC,hay di qua tm ni tip tam gic ABC.

d

A

B C

I

D

E

F

NPT

Gi s di qua tm ni tip ca tam gic ABC,khi ta c

ngay ng thcDBDC =

B A+BC AC

2

C A+CB AB

2= AB AC.

Gi giao im ca tip tuyn qua A(khc AB) ca (N)v dlT.T gic TABDngoi tip nn ta c

AB +T D = AT+BD.

Kt hp ng thc trn, ta suy raAC+T D = AT+DC.

iu ny chng t t gic T A D Cngoi tip. Vy ATtip xc(P),hay ni cch khc, ATl tip tuyn chung khc dca (N)v(P)i qua A.


28/31

LI GIIVMO 2011 27

Chng minh b 2.Ta chng minh hai chiu.

Gi s tip tuyn chung ca(N)v(P)i qua A,ta phi chng

minh bn im A, M, N, Pcng nm trn mt ng trn. rng E, M, Nthng hng v F, M, Pthng hng, do vy

N MP = 180EMF= 180

90+

E AF

2

= 90180B AC

2=B AC

2.

Mt khc, v tip tuyn chung khc dca (N)v (P)cng i

qua AnnN AP =

B AC

2.

Kt hp hai ng thc trn, ta suy ra t gic A M N Pni tip.

d x

A

B C

I

D

E

F

N PT

M

Gis A, M, N, Pcng nm trn mt ng trn, ta phi chngminh tip tuyn chung khc dca(N)v(P)i qua A.Cng tlp lun trn, ta c

N MP =B AC

2.


29/31


Do A, M, N, Pcng nm trn mt ng trn nn

N AP =N MP.

Kt hp vi trn, ta suy ra

N AP =B AC

2.

Qua Av tip tuyn A xca (N),ta c

N A x =B A x

2

.

Do

P A x=N AP N A x =B AC

2B A x

2=C A x

2=P AC.

T y suy ra Axi xng ACqua APm ACtip xc(P).VyAxtip xc(P).

Nhn xt.Phn thun ca b 1 l bi thi v ch Nga nm 2009(phn dnh cho lp 10).


30/31

BI S 7: A THC BT KH QUY TRN R

Bi 7. Cho nl s nguyn dng. Chng minh rng a thc

P(x, y)=xn+xy+yn

khng th vit di dng P(x, y)=G(x, y) H(x, y),trong G(x, y)vH(x, y)l cc a thc vi h s thc, khc a thc hng.

Li gii. Gi s tn ti cc a thc G(x, y) v H(x, y) tha mndegG 1, degH 1sao cho

P(x, y)=G(x, y) H(x, y).

Khi d thydegH+degG = n.T gi thit ta cG(x, 0)H(x, 0)=xn,suy ra tn ti k N, a1a2 = 1sao cho

H(x, 0)= a1xk, G(x, 0)= a2x

nk.

DoH(x, y)

vG(x, y)

l cc a thc nn ta c th vit c chngdi dng

H(x, y)= a1xk+yH1(x, y), G(x, y)= a2x

nk+yG1(x, y),

trong G1(x, y), H1(x, y)l cc a thc. Thay voP(x, y)v rt gn,ta c

a1xkyG1(x, y)+a2x

nkyH1(x, y)+y2G1(x, y) H1(x, y)=xy+y

n,

hay

a1xkG1(x, y)+a2x

nkH1(x, y)x= yn1

yG1(x, y) H1(x, y).

Cho y= 0,ta c

x= a1xkG1(x, 0)+a2x

nkH1(x, 0).

Gi s kv nku ln hn1.Khi ta c hai kh nng xy ra. C hai a thc G1(x, 0)v H1(x, 0)u ng nht 0,suy ra x

ng nht0(v l).

C mt trong hai a thc trn c bc 1,gi s l G1(x, 0).Khibccahngtcaonhtcaathcvphilk+degG1 > 1.


31/31


Do vy, trong hai sk, nkphi c mt s b hn2.Khng mt tnhtng qut, ta c th gi s k 1.

Nuk= 0th H1 0.Khi H(x, y) a1(loi). Nuk= 1th H1 b = 0.Khi H(x, y)= a1x+byhay

xn+xy+yn = 0, x=by

a1.

M iu ny cng khng th k c khi n= 2.

Vy bi ton c chng minh.

loi giai vmo 2011 (day 1 and 2)

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