8/8/2019 VMO 2010 - Solution

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Li gii ngh VMO 2010Trn Nam Dng

Trng i hc Khoa hc t nhin thnh ph H Ch Minh

Bi 1. Gii h phng trnh x4 y4 = 240x3

2y3 = 3(x2

4y2)

4(x

8y)

.

Li gii.

Cch 1. Nhn phng trnh th hai vi 8 ri cng vi phng trnh th nht, ta c

x4 8x3 + 24x2 32x + 16 = y4 16y3 + 96y2 256y + 256,

hay(x 2)4 = (y 4)4.

T y, ta suy ra ngay x = y 2, hoc x = 6 y.

Trng hp x = y 2. Thay vo phng trnh u, ta c8y3 + 24y2 32y + 16 = 240,

y3 3y2 + 4y + 28 = 0,(y + 2)(y2 5y + 14) = 0.

T y ta tm c y = 2 v x = 4. Trng hp x = 6 y. Thay vo phng trnh th nht ca h cho, ta c

24y3 + 216y2 864y + 1296 = 240,

y3 9y2 + 36y 44 = 0,(y 2)(y2 7y + 22) = 0,

suy ra y = 2 v x = 4.

Vy h cho c hai nghim l (x, y) = (4, 2) v (x, y) = (4, 2).

Cch 2. t y = 2t thay vo phng trnh v vit li h di dngx4 + 16 = 16(t4 + 16) (1)

x3 3x2 + 4x = 16(t3 3t2 + 4t) (2)Ch y ch l p n tham kho, khng phi l p n chnh thc.

1

8/8/2019 VMO 2010 - Solution

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2 Trn Nam Dng

Nhn cho hai phng trnh ny, ta c

(x4 + 16)(t3 3t2 + 4t) = (t4 + 16)(x3 3x2 + 4x). (3)

D thy nu (x, t) l nghim ca h th xt = 0 nn ta chia hai v ca phng trnh trn chox2t2 th c x2 +

16

x2

t 3 + 4

t

=

t2 +

16

t2

t 3 + 4

t

.

T y nu t u = x +1

xv v = t +

4

tth ta c

(u2 8)(v 3) = (v2 8)(u 3),

u2v v2u 3(u2 v2) + 8(u v) = 0,(u v)[uv 3(u + v) + 8] = 0. (4)

T (1) ta suy ra rng x v t cng du. Do p dng bt ng thc AM-GM ta d dng suyra u, v hoc cng 4 hoc cng 4. Suy ra (u 3) v (v 3) lun ln hn hay bng 1 hoclun nh hn hay bng 7. Suy ra uv 3(u + 3) + 8 = (u 3)(v 3) 1 0. Du bng ch cth xy ra khi u = v = 4.

T l lun trn v t (2) ta suy ra u = v, t suy ra x = t hoc x =4

t.

Trng hp x = t. Thay vo phng trnh (1) ta c t4 + 16 = 16(t4 + 16), v nghim.

Trng hp x = 4t

. Thay vo phng trnh (1), ta c

256

t4+ 16 = 16(t4 + 16),

16(16 + t4)

t4= 16(t4 + 16),

t4 = 1,

suy ra t = 1. T ta c cc nghim (x, y) = (4, 2) v (x, y) = (4, 2).Nhn xt. Li gii 1 kh ngn gn nhng l mt tng khng d ngh ra. Nu nh tx = 2u, y = 2v v a v h phng trnh

u4 v4 = 152(u3 2v3) = 3(u2 4v2) 2(u 8v)

th c l s d nhn thy cc h s nh thc hn.

D sao th y l mt tng khng mi. N c s dng VMO 2004, bng B. Thm chxt v mt mt no th bi VMO 2004 cn kh hn bi nm nay.

C th bi VMO 2004 nh sau: Gii h phng trnh sau

x3 + 3xy2 = 49x2 8xy + y2 = 8y 17x .

8/8/2019 VMO 2010 - Solution

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Li gii ngh VMO 2010 3

Cch gii p n ca bi ny nh sau: t x + y = u, x y = v th x = u + v2

, y =u v

2v h

c th a v dng

u3 + v3 = 98 3u2 + 5v2 = 9u 25v .Sau nhn phng trnh th hai vi 3 ri cng vi phng trnh th nht th c

(u 3)3 + (v + 5)3 = 0.R rng cch gii ny tng ng vi cch gii th nht ca VMO 2010.

Tuy nhin, bi VMO 2004 cn c mt cch gii n gin hn l nhn phng trnh th hai (cah ban u) vi 3 ri cng vi phng trnh th nht v a v dng

(x + 1)[(x 1)2 + 3(y 4)2] = 0.

Bi 2. Cho dy s (an) xc nh bi

a1 = 5, an =n

an1n1 + 2

n1 + 2 3n1 vi mi n = 2, 3, 4, . . .

(a) Tm cng thc tng qut tnh an.

(b) Chng minh dy (an) gim.

Li gii.

(a) T cng thc truy hi ta suy ra

an

n= an1

n

1

+ 2n1 + 2

3n1.

Thay n bng n 1, n 2, . . . , 2, ta can1n1 = a

n2n2 + 2

n2 + 2 3n2,. . . ,

a22 = a11 + 2

1 + 2 31.

Cng cc ng thc trn v theo v ri gin c cc s hng bng nhau hai v, ta c

ann = a11 +

n

k=2(2k1 + 2 3k1) = 5 + 2n 2 + 3n 3 = 2n + 3n.

T suy ra an = n

2n + 3n.

(b) chng minh (an) l dy gim, ta vit

an+1n =n

2n + 3n(2n + 3n) > 3(2n + 3n) > 2n+1 + 3n+1. (Do n

2n + 3n > 3)

T y suy raan >

n+1

2n+1 + 3n+1 = an+1.

Nhn xt. Ging nh bi 2 nm ngoi, bi ny l bi cho im. Li gii phn (b) trnh by trny c th coi l gn gng nht. C th chng minh bt ng thc a

n> a

n+1 bng nhiu cch

khc na nhng u rm r hn.

8/8/2019 VMO 2010 - Solution

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4 Trn Nam Dng

Bi 3. Cho ng trn (O). Hai im B, C c nh trn ng trn, BC khng phi ng knh. LyA l mt im trn ng trn khng trng vi B, C. AD, AE l cc ng phn gic trong v ngoica gc BAC. I l trung im ca DE.Qua trc tm tam gic ABC k ng thng vung gc viAI ct AD, AE ti M, N.

(a) Chng minh rng M N lun i qua mt im c nh.

(b) Tm v tr im A sao cho din tch tam gic AM N ln nht.

Li gii.

Gi 2 l ln cung nh BC. Khi gc BAC bng hoc 180 .

(a) Gi X l im i xng ca O qua BC suy ra X c nh. Ta c

OX = 2d(O, BC) = 2R cos = AH,

v OX AH (v cng vung gc vi BC) nn AOXH l hnh bnh hnh. Suy ra AO HX (1).

Li c (CBDE) = 1 nn theo h thc Newton ta c ID2 = IB IC. M IA = ID (tam gicADE vung ti A) nn IA2 = IB IC. iu ny chng t IA tip xc vi (O), t y suy raIA OA (2).

T (1) v (2) suy ra XH AI, m M N i qua H v AI nn M, N, X thng hng. Vy M Ni qua X c nh.

(b) Ta c OAC =180 AOC

2= 90ABC = HAB v AD l phn gic BAC. Suy ra

AD cng l phn gic gc OAH. M AE AD suy ra AE l phn gic ngoi gc OAH, do (AO, AH, AD, AE) = 1. Li c AO M N, do vy H l trung im ca M N. T y ta c

SAMN = 2SAHN = HA HN sinAHN.

M tam gic AM N vung ti A nn HN = HA = 2R cos khng i. T suy ra SAMN tgi tr ln nht bng 4R2 cos2 khiAHN = 90. iu ny xy ra khi v ch khi AOX = 90.T y ta tm c hai v tr SAMN t gi tr ln nht l hai u mt ca ng knh vunggc vi OX (tc l song song vi BC).

Nhn xt. Li gii trn y dng n hng im iu ha v chm iu ha. Chng ta c thkhng dng n cc kin thc ny m ch s dng cc tnh ton v gc.

Bi 4. Chng minh rng vi mi n nguyn dng, phng trnh x2 + 15y2 = 4n c t nht n nghimt nhin.

Li gii.

Ta chng minh bng quy np theo n. Vi n = 1 ta c nghim (2, 0), vi n = 2 ta c nghim(4, 0) v (1, 1). D thy rng nu (x1, y1), (x2, y2), . . . , (xn, yn) l n nghim phn bit caphng trnh

x2 + 15y2 = 4n

th (2x1, 2y1), (2x2, 2y2), . . . , (2xn, 2yn) l n nghim phn bit ca phng trnh

x2 + 15y2 = 4n+1,

8/8/2019 VMO 2010 - Solution

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Li gii ngh VMO 2010 5

v cc nghim ny u c x, y cng chn. Do thc hin php chng minh quy np, tach cn chng minh b sau

B . Vi mi n

2, phng trnh

x2 + 15y2 = 4n

c nghim (x, y) m x, y cng l.

Chng minh. Vi n = 2 mnh ng (x = y = 1). Gi s (x, y) l nghim ca phng trnh

x2 + 15y2 = 4n,

vi x, y l. Xt cp cc s

x + 15y

2,

x y2

,

x 15y2 , x + y2

. D dng kim tra c

chng u l nghim t nhin ca phng trnh

x2 + 15y2 = 4n+1.

By gi, do x, y cng l nn ch c th xy ra hai trng hp.

x v y cng ng d vi nhau theo modulo 4 (ng d1 hoc 3). Khi cp nghim th hail cp nghim l.

Trong hai s x v y c mt s chia 4 d1 v mt s chia 4 d3. Khi cp nghim th nht lcp nghim l.

B c chng minh v bi ton c gii quyt hon ton.

Nhn xt. R rng tng nhn i nghim ca phng trnh vi tham s n c nghim

ca phng trnh vi tham s n + 1 l kh hin nhin. im mu cht ca li gii l tm thmmt nghim khc.

tng xt b nghim nh trong li gii n t hng ng thc Fibonacci quen thuc

(x2 + Dy2)(z2 + Dt2) = (xz + Dyt)2 + D(xt yz)2 = (xz Dyt)2 + D(xt + yz)2.Bi ton ny c l c xut x t bi Moscow MO nm 1985. C th nh sau: Chng minh rngmi s 2n vi n 3 u biu din c di dng 2n = 7x2 + y2 vi x, y l cc s nguyn l.

tng gii trong p n MMO ging nh b trn (hay ngc li th ng hn!): Xt cc

cp nghim x + y2

, 7x y2

, x y2 , 7x + y

2 .

Chnh bi ton ny sau ny cn c s dng ti vng 3 ca Bulgarian MO nm 1996.

Bi 5. Cho bng 3 3 v n l mt s nguyn dng cho trc. Tm s cc cch t mu khng nh nhaukhi t mi bi mt trong n mu. (Hai cch t mu gi l nh nhau nu mt cch nhn c t cchkia bi mt php quay quanh tm.)

Li gii.

Ta nh s cc vung nh hnh v bn di. Ta nhn thy ch c ba php quay bin hnhvung thnh hnh vung l cc php quay cc gc 90, 180 v 270 theo chiu kim ng h(php quay 360 gi nguyn hnh vung).

8/8/2019 VMO 2010 - Solution

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6 Trn Nam Dng

B1 A1 B1A4 C A2B4 A3 B3

Qua cc php quay ni trn th C lun khng i. Vi php quay 90

th A1 A2 A3 A4 A1 v B1 B2 B3 B4 B1. Tng t vi php quay 270 th A1 A4 A3 A2 A1 v B1 B4 B3 B2 B1. Do cc php quay ny s to ra mt php t gingnh c khi v ch khi cc A c t cng mu v cc B c t cng mu.

Vi php quay 180 th A1 A3, A2 A4, B1 B3, B2 B4 do php quay ny to ramt php t ging nh c khi v ch khi cc cp (A1, A3), (A2, A4), (B1, B3), (B2, B4) ct cng mu.

By gi ta bt u m s cch t. C n cch t mu C. Trong s cc cch t mu 8 xungquanh, ta chia lm ba loi.

Loi 1. Gm cc cch t m cc A c t cng mu v cc B c t cng mu. C ncch chn mu t cc A v n cch chn mu t cc B. Suy ra c n2 cch t loi1. Vi loi ny th cc php quay ni trn khng to ra cch t mi nn s cc cch t loi1 khng thu c t nhau qua cc php quay l n2.

Loi 2. Gm cc cch t m cc cp (A1, A3), (A2, A4), (B1, B3), (B2, B4) c t cngmu nhng cc cch t ny khng thuc loi 1. D thy C n4n2 cch t nh vy (chak n s trng nhau qua php quay). Vi loi ny th php quay cc gc 90 v 270 sto ra cc cch t mi (s phi loi i bt) nhng hai cch t ny trng nhau. Nh vy cn4 n2

2s cch t loi 2 khng thu c t nhau qua php quay.

Loi 3. Gm cc cch t m t nht hai trong 4 cp ni trn c t khc mu. D thyc n8 n4 cch t nh vy. Tuy nhin, mi mt cch t s tng ng thm vi ba ccht khc thu c qua php quay 90, 180 v 270. Nh vy s cch t loi 3 khng thu

c t nhau qua php quay ln8 n4

4.

Nh vy, s cch t cn tm l

N = n

n2 +

n4 n22

+n8 n4

4

=

n9 + n5 + 2n3

4.

Nhn xt. Bi ny c thi gian v bnh tnh th khng kh. Tuy nhin trong p lc phng thi

th phi rt bn lnh mi dm lm. V y cng thuc dng bi d sai p s.

V dng bi tng t th c mt s bi sau (tuy nhin d hn bi VMO ca mnh kh nhiu):

1. Hai ca hnh vung 7 7 c t bng mu vng. Cc cn li c t bng mu . Hai ccht c coi l tng ng nhau nu chng c th thu c t nhau bng mt php quay trnmt phng ca hnh vung. m s cc cch t mu khng tng ng.

(AIME 1996)

2. ng trn c chia thnh p cung bng nhau, trong p l s nguyn t. Hi c bao nhiu ccht cc cung bng a mu. Hai cch t c coi l ging nhau nu c th thu c t nhau bngmt php quay.

(Kvant, trong bi bo vit v nh l nh Fermat ca Senderov v Spivak)

8/8/2019 VMO 2010 - Solution

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Li gii ngh VMO 2010 7

p s bi AIME lC249 24

2+

24

2= 300 (bn c thy c nt ging?).

p s bi Kvant lap a

p

+ a. Ch t kt qu ny ta suy ra nh l nh Fermat.