Tr121-150

234 Cdc phuang phdp gidi todn qua cdc ky thi Olympic

Mac khac ta lai c6

1

1 1 ul^ uj + 2ik-l)

1 <

u 1 1

\ul + 2k-3 ul + 2k-lj 1 1

• .jjt; •• n - 1 ^ 1 /

fc=i ^ 2 ^ + 2A; - 1 + 2/c + 1

1 \

< 2Vwi + l u? + 2 n - i y 2 K + 1)

Ap dung bat dang thiJc Bunyakovski, ta c6

V/c ^ 1.

^ 1 . ^ ^ 1 / ( n - 1 ) / n - 1 ^

Do d6

ul<ul + 2{n-l) + n-1 2 (a2 + 1)

\

Tom lai

v/a2 + 2(n -l)Ûn ^ 2(a2 + l ) ^

• Bai toan 2. Cho day so :

1 til = 5; M„+i=u„H V n = : l , 2 , ...

Tim phdn nguyen [MIOOI]-

Day s6 ii„+i = u„ + 235

Lbi gidi. Ap dung ket qua cua bai todn 1 ta c6

V2025 < uiooi < y 2025 +

=^ 45 < uiooi < 45,5 => [wiooi] = 45.

Bai toan 3. Cho day so :

1

1000 48

til. 'DfU:, v;r: .

•

ui = 1; Un+i = t i n + — Vn = 1,2,

Tim gidi han l im n—•4-0O

I/O'/ gidi. Ap dung ket qua cua bai toan 1 ta c6

V2n - 1 ^ ^ ' 2 n - l +

n - 1

\ / n

n y/n n 2 V n n^

Chuyen qua gidi han khi n -> +oo va theo nguyen ly kep suy ra

u. lim , ^ Ti->+oo \ i / n

•

Nhdn xet: Vdi bai todn nay ta c6 the lam each khac nhu" sau: De thay lim ii„ = +oo, suy ra

" — • + 0 0

lim u: 2 _ ,,,2 ' n+ l " n = lim

ui n-v+oo 2 (n + 1) - 2n n - + o o 2

= 1.

ut Theo dinh ly Stolz thi lim — = 1 + 0 0 2n

236 Cdc phucfng phap guii loan qua cac ky tin Olympic

1.2. Trrfofng hgfp a = - 2

Day cung m trifdng hdp chiing ta hay gSp. Chung ta cung xem x6t cac bai todn sau

B a i toan 4. Cho day so :

u\ a {a> 0); tt„+i = u„ + — Vn = 1,2,...

CMng minh rang + 3 (n - 1) ^ u„ • " ^

< / 1 \ ^ 1

3 + 3 ( ^ ^ _ l + _ j + _ + ^ - + - ^ _ V n ^ l

Lflff gtai. Ta cung c6 nhan xet nhif tren

Un > 0, Vn

(tt„) la day tang thiTc siT w„ ^ u i , Vn

V d i mpi A; 1, ta c6

« t 1 = « l + 3 + 4- + ^ = ^ < i > «E + 3 = E < i > E ( 1 + 3) W . Uj^ k~l k=\

Suy ra « ^ > txf + 3 (n - 1) =^ u„ > + 3 (n - 1), Vn ^ 2 =^ u„ ^ </a3 + 3 ( n - l ) , Vn 1.

La i c6

< + 3 + + — -o 9 ( i t - l ) ^

n- l n-1

| : . 3 , , < g ( . 3 + 3 ^ ^ ^ ^ )

= .? + 3(n-l ) + -| + l + X: + jEp '^"^3. ^1 « i

Day so u^+i ^ Wn + (w„) 237

Suy ra

ti^ < + 3 n - 1 + — 1 " 1 1 " 1

a'' k 9 ^ k^ « =1 fc=l r*.-, ,:•

Tom la i ta c6: â^ + 3 (n - 1) ^ M„

' . .X'

< \ 1 \

0^ + 3 n - l + — a 3 y

1 ^ 1 1 , ^ 1 + - 6 + ) 7 + 7 ? V n ^ l .

fc=l /c=l

• Chung ta de ket qua danh gia ve phai ciia bat dang thiJc tren

nhu" vay de tuy thuoc vao hoan canh cu the ma su" dung ngat danh gia cho hdp ly.

B a i toan 5. Cho day so :

u i = a (a > 0); u„+i = w„ + — , Vn = 1,2,...

a) Chiing minh rang ton tai p,q > 0 sao cho l im (— n-.+oo \ J

Hay tim gidi han do.

A 1 b) Chiing minh rang ton tai r > 0 sao cho < r, Vn ^ 1

Lcfi gidi. a) Ap dung ket qua bai toan 4 ta c6

a3 + 3 ( n - l ) ûj

n n

+ 3(n - 1) 3 1 1 ^ 1 1 A 1

n na-^ na^ n ^ k 9n f—' k=\

+ 3 (n - 1) 3 1 \/2n 2

n no? na^ n 9n

238 Cdc phuang phdp giai todn qua cdc ky thi Olympic

n

ET:< k=l

Chuyen qua gidi han khi n —> +00 theo nguyen ly kep ta c6

l i m n—>+oo b) Ta c6

= 3. Vay p = q = 3.

1 1 Uk+i - Uk

UkUk+1

f 1 1

1 1 - >

\ 1

Uk Uk+\

n—1 ^ n—1

fc=i

Chpn r = - ta c6 dieu phai chu'ng minh. •

Nhqn xet: Phan a) ta c6 the dung dinh ly trung binh Cesaro nhif sau:

p = q = Z.

3. Do do

1.3. Trtfofng hijfp a =

Bai toan 6. C/io ây j'o (w„) : 1

u\= a {a> 0 ) ; = Wn H — 7 = Vn ^ 1.

Chiing minh rang Vn = 1 , 2 , . . .

1 1\4a /E 80,3 9

9/

Day so u^+i = + ( u n ) " 239

Lcfi gidi. Tuf cong thtfc truy hoi cua day ta c6 j

Uk + 1 \

^/u'k)

= + 3w| + 3 +

3 Q / 3 3\

2 ;

fc=i fc=i \ 3 3 3 3

=4> Wn > + - (n - 1) = av/a + - (n - 1) .

Mac khdc la i c6

/ ^ 1 \U k + 1

2 3

1

4 u 1 / ^ 1 \

2 u k J

u

2 u k 3 3 3

n - l

4ul +

n - l /

jk=i

\

1

k=\t /c=l

De y r^ng ta c6 cdc ddnh gid sau

1 n - l ^ n - l

E 4 < E - „ k=iul k=\ -{k-l)

J_ 2 : ^ 1

ai/a 3

240 Cdc phuang phdp gidi todn qua cdc Icy thi Olympic

n—1 ^ n—1

fc=l k=l -\

o-y/a^- -{k - 1)

1 :S fc=2

Suy ra

^ 9 ( f c - l ) 2

a3 + 9 2^ fc2 fc=i

J_ 1 9 - 1 ^

3 / 2 1 \ 1 1 ul < a ^ + - ( n - l ) + - - ^ 2 ( 7 1 - 1 ) + ^ ' o 3 ' n

3 ( n - 1) + A / 2 ( n - 1) ^ 3 1 1 2 4av/a Sa^ 9

T 6 m l a i , V n = 1 ,2 , . . .

' 3 \

3 / / 3 ( n - 1) + v ' S l n - 1) ^ 3 1 1

• N ftan xet: Ddnh gia tren chiTa that sir chSt do ta da lam troi qua mot so' btfdc trung gian, tuy nhien chi can nhiT vay cung du tfng dung trong cac bai toan gidi han cua day so. Chang han nhiT bai toan sau

B a i toan 7 (Chon dpi tuyen Vie t Nam du" I M O 1993). Cho day so {un) :

U l = 1; = u„ + Vn ^ 1.

Tim so thuc p sao cho day so

khdc khong.

, n ^ I CO gidi han hUu han

Day SOUr^+l =Un + [UnT 241

Lcfi gidi. A p dung ket qua cua bai toan 6 ta c6 3

n n ^ 3 J _ 1

^ 3 ( n - l ) + V 2 ( n - 1 ) ^ + 4av/H Sa^ 9

2n

V d i a = 1 thi - - — 2 2n

n

/ I _ J_\ +

143 72n' 2 2n • Y 2

Chuyen qua gidi han khi n -> +oo va theo nguyen ly kep ta

3 diTdc l i m n—»+oo I 77,

Neu p > - thi = 2 n n

3

3

- . U n + 0 0 . Neu p < - thi — ^ 0. 2 n

• P = 9 1^ g i ^ t r i can t im.

Nhqn xet: Sau khi chi ra dufdc l i m Un — +oo cd diTdc v—-n-»+oo • 2

(nhd xap x i u\^^ = < 1.4 V ul)

1 + 4 thi ta cd the lam theo mot trong hai each sau ma khong sijf dung danh gia nhiT tren.

Thuf nhat: ^ l i r n ^ {ul^^ - ul^

= l i m f W u„ + - (u„ + + + + M „ )

= l i m n — ' + 0 0

= l i m n — • + O 0 / 1 ^ Uny/Un V U^yJUn'

A / 1 + F = + l

3 2'

242 Cdc phucfng phdp gidi todn qua cdc ky thi Olympic

Thuf hai:

- 1

l_ 3

1 0 ( n -> +oo).

Kh i do ap dung quy t i c L'Hopital ta c6

/ 3 3 \ l im I , 1 -ul] = l im I—•O

Cuoi cung, dp dung dinh ly trung binh Cesaro thi

l im n—•+00 yn J

3 2"

1.4. TrrfoTng hcfp a =

B a i toan 8. Cho day so [un) :

1 •ui = 1; w„+i = t in + - — : , n = 1,2,...

77m tat cd cdc so thuc a sao cho day i—j,n = 1,2,... hoi tu vd

tim gidi han khdc khong cua no.

( 4\IM gidi. Ta C O u„ > 0 Vn ^ 1 va < > + - , Vn ^ 2

= ^ y < > y < I ^ + ^ , V n ^ 2 ^ ^ > ^ ( n - l ) , V n ^ 2 . (1)

Mat khac

+ 1 \

,VA; ^ 2.

Day .so = U n + 243

Suy ra

3 / . , 4 1

4 2 4 1 + : + . VA; > 2.

3 3 ^ ; ^ : ; 2 7 ^ s i^ti

Do do vdi moi n ^ 4, ta c6

^ < I + | ( ^ - I ) + | E - 7 ^

1 i 1 1

Dya vao (1) bat dang thiJc Bunyakovski ta c6

(2)

k=2 ^/ul_y^

<

<

' 3/, ,4

V n

1 11

< 1 + ^ V 2 ( n - 2 ) (3)

244 Cdc phucmg phdp gidi todn qua cdc ky thi Olympic

k=2

9 A 1 + 1 6 Z . ( f c - 2 ) 2

, 9 17

2 7 ^ 1

^ ^ 6 4 ^ ( ^ - 2 ) 3

1 2 7 ^ 1 64 ^ (A: - 2Y

, 27 59

< ^ + 32 = 32

TO (2), (3), (4) va (5) suy ra

4 1 /— 35 1 59

TO (1), (6) ta difcJc

(4)

(5)

(6)

4 / 1 \ 4 1 r-, r 35 1 59 ,, ,

Chuyen qua gidi han baft dang thiJc (7) khi n —> +oo ta du'dc

l im n—•+00

4

3'

La i do l im u„ = +oo nen suy ra n—>+oo

l im Ur. n—i-+cx)

3 = < + 0 0 khi cv > -

0 khi a < -

Day SOUn+l = Un + 245

ma

Do dd

n

l im n-^+oo n

' I A 4

Un a-- .Un

n

+ 0 0 khi Of > -

0 khi a < -o

Vay a - la gia tr i duy nha't de o \ J

l im n—»+oo

,n— 1,2,... hoi tu va

4

3'

•

1.5. Trrfofng hc^p a = 2

B a i toan 9. Cho day so (un) : u i G (0; 1 ) ; u„+i = - Vn ^ 1.

77/n gidi han l im —. n->+oo In n

Z ^ ' gidi. Ta chiJng minh l im u„ = 0. n—•+00

That vay, ta cd

0 < U2 = U l (1 — Ul) ^ ^ j = 4 = > « 2 G ( 0 ; 1 ) .

Bang quy nap ta du'dc u„ £ (0; 1) Vn. De thay day da cho giam nen suy ra ton tai l im M „ = a.

Chuyen qua gidi han bieu thiJc truy hoi da cho ta cd

a = a — o^<â = 0.

VSy l im u„ = 0.

246 Cd( [>'iiif(fin' plidp gidi todn qua cdc ky thi Olympic

Ap dung dinh ly trung binh Cesaro, ta c6

1 / 1 1

l im = l im — = l im n — • + 0 0 nUn n ->+oo n n ->+oo

1 \

\Un+l UnJ

Un - Un+l ,. Un - (u„ - ul)

l im = l im —. ^ — — -

l im — ^ = 1.

Suy ra l im nii„ = 1.

Ap dung dinh ly Stolz ta c6

n (1 — nu„) l im

n—*+oo I n n = l im

nUn (1 - nw„) n — + 0 0 Un In n

l - n u n hm l im •

I — + 0 O Un m n

1 — n

= 1. l im n - . + o o In n

- (n + 1) + n = l im Un+l Mi:

n->+oo l n ( n + l ) — I n n

- 1 - 1 l im

Un (1 - Un) Un

In

= l im n — • + 0 0

n + 1

n nUn

{ r ( l - w „ ) l n 1 + -

1

( 1 - 0 ) Ine = 1.

• B a i toan 10. C/io a > 0; a G (0; 1). Xet day (un) :

, aa Ul > 0; Un+i = { l - a ) U n +

Chiing minh rang day da cho c6 gidi han vd tim gidi han do.

Day so Un+i = + 247

Ufi gidi. X6t h^m s6' aa

f{u) = (1 - Q)U + - ^ r r , w > 0, a > 0, a G (0; 1) U "

c6 /'(u) = (1 - a) ( 1 - a u - i ) suy ra /'(w) = 0

Lap bang bien thien ta c6 / {u) ^ a", V « > 0. De thay w„ > OVn nen u„ ^ a"Vn ^ 2.

X6t hieu Uk - U k - i = ( a - ^ 0, V/c ^ 2.

Vay (ii„) bat dau giam tijf W2 bi chan difdi bdi a'* nen ton tai gidi han l im M „ = /.

Chuyen qua gidi han he thiJc truy hoi da cho ta c6 aa

l^{l-a)l + - ^ ^ l = a\

Vay l im Un = a°'. n — • + 0 0

•

2. Bmh luan

Neu a < 0 thi ta xay dtfng cac bat dang thufc dang y l i - ^ ^ 1 LS U

Un ^ - B i ^ , d d6 l im — = l im — = L nen l im — ^ n — • + 0 0 n n—»+oo77, n—>+oo n

B l-a

= L.

• Neu a > 0 thi ta xay dUng cac bai toan theo chieu hu'dng khdc bang each lap day tong hoac sijf dung cac dinh ly trung binh Cesaro, Stolz chang han nhU:

1) Cho [xn) •.xx = l\i = x„ + a; , n = 1, 2,.... T im gidi

4 han l im \ ... H

1

3

2) Cho {xn) : a;i = - ; x „+ i = X n - x l , n = 1,2,.... T im gidi

han l im (nx„) . NhU da nhan xet trong cdc bai toan n—>+oo

U trdn thi viec t im gidi han cua — c6 the dUdc thiTc hien

nP b^ng cdch suf dung dinh ly Stolz hoac dinh ly trung binh Cesaro. Tuy nhien vdi each danh gia va suf dung nguyen ly kep thi Idi giai bai toan UX nhien hdn va cung sd cap hdn.

Cdc phuang phdp gidi todn qua cdc ky thi Olympic

3. Bai t$p

1./ Cho day (tt„) : = 1; u„+i = w„ H Vn = 1,2,... ChiJng

1 7 minh rang V] — < - V/c ^ 1.

2. / Cho day : wi = 1; Wn+i = U n + — Vn = 1, 2,... ChiJng

minh rang E < 7 V/c ^ 1

3. / Cho day : l i i = 1 ; u„+i = u„ + - — Vn = 1,2,.. . Ztt-n.

Chufng minh rang y/nun < n + -• ^

1 + - + ... + -2 nj

1 / a 4./ Cho day (u„) : ui > 0; = - u„ +

2 V UnJ

Chufng minh day hoi tu va tim gidi han.

Vn ^ 1 (a > 0).

5./ Cho day : ui = a {a > 0); = tx„ + — Vn = 1,2,....

Tim gidi han lim —

1 / a \ 6./ Cho day (w„) : U i > 0; w„+i = - 2u„ + —- , Vn = 1,2,...

(a > 0). Tim gidi han cua day so' do.

7. / Cho day (w„) : = 1; u„+i = u„ + —p= Vn ^ 1. Tim m de

. . . . . ,. U1+U2 + ...+Un ton tai gicJi han hm .

n—.+00 m

8. / Cho day : u i G (0; 1 ) ; u^+i - w„ - n^ Vn ^ 1. Tim gidi han lim y/nun.

n—>+oo

9./ Cho day (ti„) : i t i = ti„+i = Un- u^Wn ^ 1. T m p G

sao cho ton tai lim n ' ' (nw„ — 1). n—>4-oo

P/idn hoach nguyen vd phdii lioach tdp hap 249

10./ Cho day (u„) : i t i = 1; Wn+i = Un^\n = 1, 2,... (p G N * ) .

•Un

Chufng minh rang lim —— = p + 1.

W.I Cho day (u„) : u i = 1; w„+i = u„ H ^ Vn = 1,2,...

1 1 + -

(p G N * , p > 1). Chtfng minh lim —— = 1 + - .

12./ Cho day (u„) : u i = 1; w„+i = + uf^^ Vn ^ 1. Tim gidi

PHAN HOACH NGUYEN V A PHAN HOACH TAP H O P

L6 Phuc Lfft S •

Phan hoach tSp hdp phan hoach nguyen 1 cdc chu de kha thu vi the hien mdi Hen he sau sac giiJa to hdp va so hoc, no cung xuat hien c' nhieu ky thi Olympic c^c cap. Nhffng bai toan trong phan nay rat phong phii, da dang mot so bai doi hoi cac lap luan kho. Nhieu bai c6 the giai bang ly thuye't ham sinh mot cdch rat turdng minh nhiTng de cho tinh thuan tuy cua chu de, ta chi tap trung xem xet cac each tiep can bang lap luan so' hoc, to hdp thong thudng. DiTdi day, chiing ta se tim hieu mot so cong cu chinh va cac kien thuTc sd lifdc de ap dung xijf ly cac van de. Ben canh cac phan dUdc khai thac chi tiet thi cung c6 vai phan khac chi mang tinh chat gdi md va rat can c6 sir chu dong cua ngudi doc trong viec tim hieu sau them nh^m bd sung, ho^n chinh day du kien thuTc cho minh trong noi dung thu vi nay.

1. So' Stirling loai mOt va loai hai 1.1. Chu trinh cua mpt hoan vi Trong cac bai to^n ve phan hoach tap hdp, ta khong the nao khong nhac den cdc sd Stirling. Day la cd sd de xay difng cac bai todn de'm mot cdch kh^ vffng chac, dong thdi ket hdp vdi cac cong cu giai tich va dai so' khac se giup xay difng dvfdc nhieu. ly thuyet khac.

Chu trinh trong mot hoan vi a i , 0 2 , 03,...,a„ cua cac sd 1, 2, 3 , . . . , n la mot tap con ciia cAc sd nay ma chung ddi vi tri cho nhau tao thanh mot vong. Ta xet hoan vi sau:

123456789-^ 142357698. / tSinh vien trUdng Dai hoc FPT thanh pho Ho Chi Minh.

250

rnan noch Ji nyiiyen vd phan hoach tap hap 251 Chang han trong hoan vi tren: / ( 2 ) = 4, / ( 3 ) = 2, / ( 4 ) = 3 nen CO the noi (2, 3, 4) tao thanh mot chu trinh. Cac sd khong ddi qua phep hoan vi diTdc coi la mot chu trinh rieng biet va do do, trong hoan vi tren c6 tat ca 5 chu trinh: (1), (423), (5), (76), (98).

1.2. SoT Stirling loai I ttf h t u

Sd hoin vi cua n phan tuf ma c6 diing A; chu trinh, dat la s (n , k).

Cong thu-c truy hoi s{n + 1, A:) - s{n, k - 1) + n • s{n, k). CMng minh. Ta se chia sd hoan vj cua n + 1 ma c6 k chu trinh thanh hai loai:

• Loai 1 gom cac hoan vi ma phan tuT {n + 1} la mot chu trinh rieng biet, con n phan tuT con lai hoan vi tao thanh A; - 1 chu trinh, c6 s(n, A; - 1) triTdng hdp.

• Loai 2 gom cac hoan vi c6 phan tuT ( n + 1} nam trong cac chu trinh da c6 cua n phan tvt dau tien, do c6 xet den M t\i nen phan tu- nay c6 n each dat vao day hoan vi da c6, nhu" the CO n • s{n, k) each.

Theo nguyen ly dem b^ng hai cdch, ta c6 cong thu-c da cho. Mot s d k e t qua cd ban

s{n. 0) ^ 0, . ( n , 1) = 2

s{n, n ) = l , s{n, n - I) =''^'^ ~ ^\

•

1.3. So Stirling loai II

Sd each chia tap hdp n phan tuf thanh k tap hdp con, dat la S{n, k).

Cong thu-c truy hoi S{n + 1, A;) = S{n, k - 1) + k • S{n , k).

252 Cdc phuong phdp gidi todn qua cdc ky thi Olympic

Chiirng minh. Chtfng minh cong thiJc nay tifdng tiT tren, ta thiTc hien bang each xet viec c6 xuat hien hoac khong cua tap con { n + 1} trong cdch chia (chu y rang each chia d day khong tinh thuT ttf).

M o t so k e t qua c d ban 5 ( n , n - 1) = Cl 5 ( n , 2) = 2"-^ - 1.

So' Stirling loai II nay cung l ien quan den bai toan ve so ham toan anh tif tap h d p eo k phan tuf sang tap h d p c6 k phan tur. •

1.4. C a c bai toan uTng dung

V i du 1. Cho bang 6 vuong (aij) vdi i = 1, n, j = 1, n vd dien cdc so ti( 1 den theo tM tvC tii trdi sang phdi, tit tren xuong dadi. NgUdi ta vie't ghep cdc hang cua bang nay theo thvC tu tvC trdi sang phdi dupe mot ddy X. Tiep tuc ghep cdc cot cua bang nay thdnh dang hang ngang cung tif trdi sang phdi dUcfc ddy Y. Mot phep bien doi cho phep doi cho hai so cho nhau. Hoi can it nhdt bao nhieu phep bien doi de diCa X ve Y ?

Hrfdfng d i n . Chu y rang de bien doi cac so trong hai chu trinh khac nhau cua mot bo k so, ta can it nhat k-l each. Do do, ta xay diTng day X, Y trong tru'dng h d p nho va dif doan so chu trinh rdi nhau, tinh todn so phep bien doi roi tdng quat len. •

Vi du 2. Cho A Id tap hap tat cd cdc hodn vi a= {ai, 0 2 , . . . , 02003) cila 2003 so nguyen duong ddu tien vd mot hodn vi thoa man dieu kien: Khong co tap con S ndo cda A md {uk \ E S} = S.

Vdi moi a = {a-[, 0 2 , . . . , 02003) G A, ta ki hieu

2003

d{a) - ^ ( a , - k)\ k=l

1. Tim gid tri nho nhdt cHa d{a), gpi gid tri nho nhdt do Id do.

2. Tim tdt cd cdc hodn vi a e A thoa man d{a) = do.

Hrfdfng dan. Dieu kien de bai noi len rang hoan v i da cho c6 dung 1 chu trinh, vdi moi hoan vi bat ky thuoe A, ta xet anh xa / : z —> a,. Ta eo:

{ 1 . / ( I ) , f{f{l)),...,r'Hl)} = {l, 2, 3 , . . . , 2003}.

Phan hoqch nguyen vd phan hogch tap hap 253

Dat Sk = If''-'{I) - f\l)\,k=l, 2003 va m < 2003 la so m nhien thoa man / ' " ( I ) = 2003.

Ta CO danh gia .

va

2003 = | r ( l ) - l | ^ 5 i + 52 + --- + 6'^ + l ,

2003 = i r (1) - 1| ^ 5 ^ ^ i + 5,„+2 + • • • + .2003 + 1. 2003

Tu" day suy ra £ Sfc ^ 2 • 2003 - 2. Nhif the fc=]

2003

^ ( « ) = E i / ' ~ ' ( i ) - / ' ( i ) r ^

2003 . :

2003

/2003

^ k

^ (2 • 2003 - 2? > 2003 - > 4 - 2 0 Q 3 - 8 .

Do d{a) la so chan nen d[d) ^ 4 • 2003 - 4 = 8006. •

Vi du 3. Cd bao nhieu each chia 15 do vat doi mot khdc nhau cho 5 ngudi. sao cho trong so do cd dung 2 ngudi khong nhdn duac do vat ndo ?

HiTdng d§n. Suf dung so Stirling loai II . Ket qua bai toan ro rang 1^ C'h • 5(15, 3). •

2. Phan hoach nguy§n

2.1. P h a n hoach nguyen

Viec phan chia mot so nguyen diTdng n thanh tdng cua cac so nguyen diTdng diTdc goi la phan hoach nguyen cua so nguyen dtfdng do.

2.2. P h a n hoach c6 tinh thuT trf

Chang han, c6 tat ca 7 phan hoach c6 thi? tiT cua 4 la )( '

4 = 1+1 + 1+1 - 2 + 1 + 1 = 1 + 2 + 1 = 1 + 1+2 = 2+2 - 3 + 1 = 1+3.


Be dem so phan hoach nay, ta suT dung cong thtfc truy hoi va noi chung bai loan dat ra d day thufdng de tiep can. •

So phan hoach cua n thanh 1, 2 cho bdi cong thufc ' '

r 5 ( l ) = l , 5(2) = 2, Iw'-^^'-'^-, [ S{n) = S{n - 1) + Sin - 2), n > 2

Cong thiJc tinh phan hoach c6 thuf tiT tdng qudt cua n Ik

n - l Sit) \

5(n) = ^ 5 ( z ) . i= i

2.3. Phan hoach khong tinh thit ttf (dang nay pho bien htfn)

Dem bang ham phan hoach pk{n) la so each phan hoach n thanh k thanh phan.

K i hieu them p{n) la so phan hoach n th^nh mot so thanh phan n

tuy y. Ta c6 p{n) = Yl Pk{n). Ta k i hieu A = ( A i , A2, A 3 , . . . , \k)

vdi Ai ^ A2 ^ A 3 ^ • • • ^ Afc va A i + A2 + A 3 + • • • + Afc = n la mot phan hoach cua n. Ta de dang tinh du'dc ,

p ( l ) - 1, p(2) = 2, p(3) = 3, p(4) - 5, p(5) = 7 , . . .

2.4. Bieu do Ferrers

M o i thanh phan trong phan hoach cua mot so nguyen dUdng bat ky CO the bieu dien bang cac dau cham tao thanh mot bieu do. Bieu do nay du'dc sap xep c6 chieu cao tang dan tuf trai qua phai hoac^ chieu dai giam dan tijf tren xuong duTdi.

Neu chuyen bieu do tCr ngang sang doc, ta nhan diTdc bieu do doi ngau cua no. Bieu do doi ngau vd i chinh no difdc goi la bieu do tif doi ngau.

L ien quan den cong cu de'm va chufng minh tinh chat cua phan hoach nguyen, ta con cd Young Tableux hay Durfee square cung rat thii v i .

, Phan hoach nguyen vd phan hoach tap hop 255

2.5. C a c bai toan uTng dung

V i du 4. Goi S{n) Id so each phan hoach cd tinh thvc tu cua n thanh cdc so 1, 3, 4. Mng minh rang S(2n) la so chinh phUcfng.

Hrfdng d i n . Ta tinh diTdc cong thu:c truy hoi cua S{n) la

{S{1) = \, 5(2) = ! , 5(3) = 2, 5(4) = 4, |; \ 5 ( n ) = 5 ( n - l ) + 5 ( n - 3 ) + 5 ( n - 4 ) , n > 4 ^'

Tinh Ian lu-dt cac so hang cua day, ta thay

5(2) = \) = 2\) = 3 , 5(8) = 5^,...

Do do, ta dy dodn va chu-ng minh 5(2n) = F„%i (vdi F„ la so Fibonacci thu" n) . Q

V i du 5. Til vice so sdnh so phan hoach cua n thanh cdc so I, 2 chiing minh rang

x+2y=n

Uxidng dSn. Ta suf dung phu-dng phap dem bang hai each.

Trirdc het, de tinh so phan hoach cua n thanh cac so 1, 2 thi ta CO cong thiJc

5(1) = 1, 5(2) = 2, S{n) = S{n-l) + S{n-2), n>2

hay 5 (n ) = F„+i.

Tiep theo, ta x6t so nghiem cua phUdng tfmh x + 2y = n. M o i nghiem nay se cd dang

( M , 1 , . . . . , 1 , 2, 2 , 2 , . . . , 2 ) . X y

M o i hoan v i cua cac bo so nay tao thanh mot phan hoach thoa man

de bai. Ta tinh diTcfc so hoan v i Ih iillH = Cf^,,. So sanh hai x\y\

ket qua, ta cd dieu phai chiing minh. ' ' •

256 Cdc phuang phdp gidi tudn qua cdc ky thi Olympic

3. Dinh ly Beatty va mof rong

3.1. NOi dung dinh ly

N e u a, h la hai so v6 t i diTdng thoa man i + ^ = 1 va xet cac tap

hdp A = {[ma\ m = 1, 2, 3 , . . . } , 5 = {[nh\ n = 1, 2, 3 , . . . } thi A n B = 0 , ^ U B = N * .

ChuTng minh. Ta se chi^ng minh bo de qua 2 phan sau:

(1) Khong ton tai so nguyen di/dng nao xuat hien d ca hai tap hdp. That vay, gia suf ngi/dc la i , ton tai so nguyen difdng fc, m , n thoa man A: = [ma\ \nb\ khi do

k <ma<k + \, k <nh <k + l

(do cac so ma, nh la so v6 t i ) . Do do

k k+\ k+\ - <m < , y < n < — - — , a a h 0

cpng cac bat dang thuTc nay ve theo ve, ta dufdc

i + -"^ <m + n<ik + l) \a bj - + 7 \a bJ

tUdng diTdng vdi k<m + n<k + l.

D i e u nay v6 ly do m + n la mot so nguyen difdng. D i e u gia sijf la sai nen (1) diing.

(2) M o i so nguyen dtfdng deu xuat hien d mot trong hai tap hdp. That vay, gia suT ngifdc la i , so nguyen difdng k khong xuat hien d ca hai tap hdp. Khi do, ton tai m , n sao cho

[ma\ <[{m + l)a\ [nb\ k < [{n+ l)a\

suy ra ma <k, k + 1< {m + l)a wa nb < k, k + 1 < {n + l)b. Do k k ^ , J , I t i I u ».

d6 m < - , n < va a 0

k+l k+l m + l> , n + l > — — .

a 0

Phdn hoach nguyen vd phdn hoach tap hap 257

c6ng cac ba't dang thtfc n^y ve theo ve, ta drfdc

m-\-n<k (- -

Wbj' {k+l) - + I

a h

< m + n + 2

tUdng diTdng v d i

m + n<k<m + n + l.

Day cung la mot dieu v6 ly nen ta thay (2) dung.

Do do, bd de du-dc chtfng minh. ' •

3.2. Tong quat cua dinh ly Beatty (dinh ly Lambek-Moser)

Xet ham so / : N * ^ N khong giam va khong bi chan. V d i moi ham so f* : N * N thoa man dieu kien

/ ( / * ( x ) ) < n ^ / ( / * ( n ) + l ) .

Khi do { / ( n ) + n I n G N } va {g{n) + n | n G N } la phan hoach cua tap so nguyen dufdng. Chang han neu xet f{n) = va / * (n) =

- l ] , thi hai day so tiTdng ting la }

{n^ + n \ e N*} , {[Vn - 1 + n] \,

chinh la phan hoach cua tap so nguyen dtfdng.

3.3. Cac bai toan itng dung

V i du 6. Cho so nguyen duang k. Day so {xn), n = 1, 2, 3 , . . . dugc xdc dinh nhu sau:

(i) xi = l. -

(ii) Vdi mSi so nguyen duang n ^ 1 thi Xn+i Id so nguyen duang be nhat khong thuoc tap

{xi, X2\; Xn] xi + k\ + 2k; x^ + Zk; Xn + nk]

258 Cac phuang phdp gidi todn qua cdc ky thi Olympic

Chitng minh rang ton tai so thuc a sac cho = \na\ moi

HtfdTng d i n . X e t da thtfc f{x) = x"^+ {k-2)-k\h. g o i a la ngh i em

dufdng cua no . Ta thay a G ( 1 , 2) va no la so v6 t i .

D 5 t 6 = a + A; th i - + 7 = 1 n e n theo d inh l y Beat ty th i c^c gia

t r i cua cac h a m so

/ ( n ) = [no] , g{n) = [nb] + kn t a i 1, 2, 3 , . . . ..

tao thanh phan hoach cua N* . B a n g quy nap, ta chufng m i n h du-dc

Xn = f{n). D

V i du 7. Xet day cdc so 3, 6, 7, 9, 10, 11 , 1 2 , . . . la co the bieu

dien thanh long cua it nhdt hai so nguyen ducfng lien tie'p. Got f{n)

Id so hang thvc n cua day. ChvCng minh rang

• f[n) = n+ [\og^{n + l o g2 (n ) ) .

Hrfdng dan. Ch i i y rang so non-pol i te thtif n ch lnh la 2 " " ^ T i r do

chufng m i n h bang quy nap b i e u thiJc cua day so t r en . •

4. C a c dinh ly cof ban ve phan hoach nguyen

D i r d i day, ta se l i e t ke va neu ra cac h i fdng chiJng m i n h sd lu'dc

cho cac cong thiJc t i n h hoSc danh gia so phan hoach nguyen .

(1) V d i n ^ A; t M pfc(n) = Pk-\{n)+Pk{n - k).

Chufng minh. Ch ia thanh cac tn fdng hdp theo cong thufc t ruy ho i . •

(2) So phan hoach cua n thanh d i ing k phan bang so phan hoach

cua n ma phan tuf I dn nhat la A;.

Chufng minh. Sur dung triTc t i ep bo de Ferrers. •

(3) So phan hoach cua n thanh k phan phan b i e t bang so phan /c(/c + 1)

hoach cua n -——- thanh k h o n g qua k phan.

I'hdn hoach nguyen vd phdn hoach tap hap 259

Chufng minh. X e p cac thanh phan cua phan hoach theo ch i eu tang dan r o i trfif vao m 6 i thanh phan cua phan hoach Ian lu-dt cac so

1, 2, 3 , . . . , fc. De thay td'ng cac so nhan diTcJc se la n - ^

va CO k h o n g qua k thanh phan ( thanh phan cuo i c6 the bang 0) . •

(4) G p i s{n) la so phan hoach cua n ma m o i phan deu Idn hdn 1. ChiJng m i n h rang

s{n) = p{n) - p{n - I).

Chufng minh. Ta chia phan hoach p{n) thanh ha i l o a i la phan hoach

CO thanh p h a n nho nha't Idn hdn 1 va phan hoach c6 thanh phan

nho nhat bang 1, d l thay cac so phan hoach t i fdng ufng la s (n ) ,

pin-I). •

(5) V d i m o i n t h i p{n + 2) + p{n) ^ 2p{n + 1).

Chufng minh. A p dung (7) v d i chu y rang s (n ) tang. V i e t l a i (8) thanh

p{n + 2) - p{n + 1) ^ p{n + 1) - p{n).

^•i •

(6) Chu'ng m i n h rang p{2n) ^ p{n) +p{n-\) + - +p{2) +p{l).

Chufng minh. Ch i i y rang v d i mS i phan hoach c6 dang p{r) c6 the du"dc t h e m bang m o t so hang 2n - r tao thanh phan hoach ci ia 2 n . •

(7) So phan hoach cua n thanh cac thanh phan le bang so phan

hoach cua n thanh cac thanh phan phan b ie t .

Chufng minh. G ia s\i lax + 2a2 + 3a3 H h kak t i fdng tfng v d i phan

hoach cua n thanh cac so 1, 2 , 3 , . . . , A; t rong do G {0 ; 1 } . V i e t

cac so A: thanh dang 2 " t v d i u , t G N va i l e . K h i do, ta cd the

b i6u d i e n tdng t r en thanh dang ^

n = l (2 °a i + 2^02 + 2^04 + •••) + 3(2°a3 + 2â6 + 2^012 + •••) + •••

Do do, so p h a n hoach t h ^ n h cac t h ^ n h phan phan b i e t k h o n g v i fdt

qua so phan hoach thanh cac thanh phan le . Ta l a i v i e t n =

2 6 0 Cdc phuang phdp gidi todn qua cdc ky thi Olympic

ta Viet Cfc = 2''" + 2'"=2 + 2'''= + thi l a i + 3a3 + 5a5 + • n = 1(2*" + 2*12 + 2''i3 + • • • ) + 3{2''3i + 2''32 + ...) + ... Cac thanh phan deu phan biet nen suy ra so phan hoach thanh cac thanh phan phan biet khong nho hdn so phan hoach thanh cac thanh phan le. Tijf day suy ra dieu phai chiJng minh. •

(8) Xet a, 6, c G N * va a > max {6, c } . K h i do, so phan hoach cua a - c thanh dung h - I phan khong viTdt qua c bang so phan hoach cua a-b thanh dung c - 1 phan khong vUdt qua b.

Chd'ng minh.

^6

= c - l

h-\{ b{ < 6 c - 1

Ta chiing minh bang each diing bieu do Ferrers, dau tien xet phan hoach CO 6 - 1 phan ma m6i phan khong qua c. Ta them mot dong CO dp dai c vao bieu do, ta c6 phan hoach thanh b phan cua a - 6 + c . Tiep theo, ta xoa cpt dau tien, difdc mot phan hoach khong qua b phan cua a-c, chuyen thanh bieu do doi n g l u , ta dtfpc phan hoach c6 c - 1 phan ma moi phan khong qua b. De thay cac bufdc nay cho thay tUPng ufng 1 - 1 giffa cac phan hoach nen ta c6 dpcm. •

(9) So phan hoach cua n thanh cac thanh phan 1, 2 la n + 1

(10) So phan hoach cua n thanh cac thanh phan 1, 2, 3 la so

nguyen gan vdi gia tr i J^^ nhat.

Chufng minh. C^c phan hoach nay tUPng iJng v d i so' nghiem cua cdc phiTdng trinh

x + 2y = n, a; + 2y + 3z = n .

Phan hoach nguyen vd phan hoach tap hap 261

Giai cac phufOng trinh Diophante nay, ta thu dtfpc cdc danh gia

Hay thu" tu" chiJug minh tiep cac dinh ly sau: r

(11) So phan hoach cua n thanh cac phan c h f n bang v d i so phan hoach ma mSi so' dufOc xuat hien so c h f n Ian.

(12) So phan hoach cua n thanh 3 phan bang so' phan hoach cua 2n thanh 3 phan c6 gia tri be hdn.

(13) ChuTng minh rang so phan hoach c6 bieu do Ferrers tu" doi n g l u cua bang vdi so' phan hoach ciia n thanh cac thanh phan le phan biet.

(14) ChiJng minh rang so phan hoach cua mot so' thanh cac phan vdi moi phan c h f n khong vifdt qua 1 bang vdi so phan hoach cua mot so vdi mot phan xuat hien khong qua 3 Ian.

(15) So phan hoach cua n thanh cac thanh phan giong nhau bang v d i so iTdc cua n.

5. P h a n hoach tap hgfp ^

5.1. Bai toan chia kô Euler

Ben canh cac bai toan ve so Stirling thi bai toan chia keo Euler, l ien quan den so to hdp lap hay dem so multiset, la dang toan kha cd ban va quen thupc.

Dang 1: S P each chia n v ien kep chp k em be ma em nap cung phai CP keo la C ^z j .

That vay, ta xep n vien keo da cho len mot diTdng thang va tinh so' each dat k — 1 ngan vao giita n — 1 khoang trong git?a hai vien keo de chia chung ra thanh k phan. Tong cpng c6 C^Zl each.

Dang 2: So each chia n vien keo cho k em be ma c6 the cd em khong cd keo la C^+fc_i.

That vay, ta "miTdn" them k vien keo v^ cho trifdc mSi em be, sau dd mdi chia n v ien keo cho ehiing thi difa ve bai toan dem so each chia n + k vien keo cho k em be ma em nao cung phai c6 keo. Tong cpng cd Cj^!^^_j each.

262 Cdc phuong phdp gidi todn qua cdc ky thi Olympic

5.2. Tong hgfp 12 dang cua phan hoach tap htfp i

Cdc b ^ i t o a n ve chia tap h d p hay c6 the h i n h dung la p h a n chia do v a t r a t g a n g u i t rong dcJi song nhu'ng t rong n h i e u t r i f d n g h d p , ta thvTcJng r a t d e b i n h a m iSn d a n d e n d e m thufa hoac t h i e u , t h a m chi k h o n g b i e t p h a i t i e p can theo phiTdng phap nao. T r o n g T o a n hoc, ngiTcfi ta d a t d n g h d p cac v a n d e do l a i va p h a n t h a n h 12 d ang v d i t e n g p i n o i t i e n g la " T w e l v e f o l d W a y " . D t f d i d a y , ta se x e t ba i t o a n C h u n g l a : X a c d i n h ta t ca so each cho n qua b o n g v ^ o k h o p .

Cdc k i e u p h a n hoach duTdc d a c t n f n g b a n g m o t d n h xa d i tijf cac

qua b o n g d e n cac h o p diTdc l i e t k e day d u d i f d i d a y , chu y r a n g ;

• A n h xa b a t k y : v iec p h a n chia la t u y y , k h o n g c6 r a n g buoc g i (c6 the m o t hop c6 n h i e u b o n g m a c i i n g c6 the c6 hop k h o n g CO qua b o n g nao, nhi fng tat n h i e n la m o t qua b o n g t h i c h i CO the chia vao m o t h o p ) . ' >

D d n a n h : m o i h o p ch i chuTa n h i e u nhat m o t qua b o n g .

T o a n a n h : m o i h o p chuTa i t nhat m o t qua b o n g . 6 d a y k h o n g

x e t song a n h v i ro r a n g k h i d o , m 6 i h o p p h a i chuTa d u n g m o t

qua b o n g va b a i t o a n k h o n g c o n n h i e u y nghia niJa. Q u y tfdc

= 0 v d i < a.

Trtfcfng hgfp 1. Cdc qua bong doi mot khdc nhau va cdc hop doi

mot khdc nhau.

1. A n h xa b a t k y : M o i qua b o n g d e u c6 the dufdc x 6 p v a o m o t

h o p t u y y . K e t qua la fc".

2 . D d n a n h : N e u n> k t h i k h o n g c6 each x e p thoa m a n v i theo n g u y e n l i D i r i c h l e t , se c6 1 h o p chufa n h i e u h d n 1 qua b o n g , ke't qua la 0. Ne'u n ^ A;, ta thay qua b o n g d a u t i e n c6 the d a t v a o k h o p , qua b o n g thuT hai c6 the dat v a o A; - 1 h o p con l a i v ^ ciif the'. K e t qua la k{k - l){k - 2)...{k - n).

3. T o d n d n h : V a n d e t i fdng iî nhif v iec p h a n hoach tap h d p c6 n p h a n tuT thanh k tap con khdc r 6 n g , tuy n h i e n chu y rang can quan tarn d e n thuf tiT cua cac tap con tiTdng ufng. Ke't qua Id k\S{n, k) v d i 5 ( n , k) la so S t i r l i n g l o a i I I .

Phan hoach nguyen va phan hoach tap hap 263

Trifofng hgfp 2. Cdc qua bong giong nhau va cdc hop dot mot khdc nhau. . , , , • ...

1. A n h xa ba't k y : D o cdc qua b o n g la g i o n g nhau n e n ta c h i x e t CO bao n h i e u qua dUdc dUa vao m o i h o p , tu'dng iJng v d i b a i toan chia k e o Euler t rong trufdng h d p k h o n g nhat t h i e t e m nao c u n g c6 k e o . K e t qua la C'^'^X-v ,

2. D d n d n h : T i f d n g tiT t r e n , ne'u n > k t h i k e t qua la 0. Ne'u n ^ k t h i ta t i n h so each chon n h o p t rong k hop . Ke't qua la

3. T o a n a n h : So trufdng h d p tu'dng tJug ch inh la ba i toan chia k e o E u l e r t rong trUdng h d p e m nao cung p h a i c6 k e o . Ke't q u d la Ctl

Trtfcfng hcfp 3. Cdc qua bong doi mot khdc nhau va cdc hop deu giong nhau.

1. A n h xa bat k y : D o cdc qua b o n g la khdc nhau n e n ta c6 the d d n h so c h i i n g bang cac so tiif 1 d e n n , d a n d e n b a i t o d n phan hoach tap h d p { 1 , 2, 3, . . . , n } thanh k h o n g qud k tap con. K e t qua la S{n, 1) + S{n, 2) + . . . -f- S{n, k).

2. D d n d n h : T i f d n g tvt t r e n , ne'u n > k t h i k e t qua la 0. Ne'u

k t h i ro r a n g ch i c6 1 each x e p , ke't qua la 1.

3. T o a n d n h : H o p nao c u n g p h a i c6 b o n g n e n no d u n g v d i b a i t o d n t i n h so S t i r l i n g l o a i I I . K e t qua la S{n, k).

Trufofng hgfp 4. Cdc qud bong deu giong nhau va cdc hop deu giong nhau.

1. A n h xa ba t k y : Tu'dng tog v d i b a i todn p h a n hoach so n thanh

k h o n g qud A; t h a n h p h a n . Ke't qua la p i ( n ) + p 2 ( n ) + -•

2. D d n d n h : TiTdng ttf t r e n , n e u n > k t h i ke't qua la 0. N e u n ^ k t h i ke't qua la 1.

3. T o a n d n h : Tu'dng iJng v d i ba i todn phan hoach so n thanh d u n g k t h a n h p h a n . Ke't qua cua ba i todn la Pk{n). : fiiv)

264 Cdc phuang phdp gidi todn qua cdc Icy thi Olympic i

\

5.3. Cac bai toan uTng dung V?

Vi du 8. Xdc dinh so nghiem tu nhien cua cdc phucfng trinh sau

(a) x i + + + • • • + X20 = 2013 vdi Xi ^ i, i = 1, n.

(b) xi + X2 + X3-\ X51 = 2013 vdi Xi, i = T~n Id cdc so le.

(c) XI + X2 + X3 + --- + X2o\ 2013 vdi Xi ^ 2, X2 ^ 3.

HifdTng dSn. Suf dung cong thiJc ve bai toan chia keo Euler ket hdp vdi ly cac dieu kien di kem. Chang han d cau a, ta can dat Xi = Ui + i d i diTa ve

?/i + ?/2 + ?/3 + • • • + ?/2o = 1803, vdi ^ 0.

•

Vi du 9. Trong mat phdng, cho tam gidc ABC. Mot diem P nam trong tam gidc duac goi la diem tot neu ta c6 the tim duac dung 27 tia chung goc P cat cdc canh cua tam gidc ABC vd chia tam gidc nay thanh 27 tam gidc con c6 dien tich bang nhau. Xdc dinh so tat cd cdc diem tot cua tam gidc ABC.

Hrfdng dSn. TrUdc het ta thay rang PA, P B , PC phai la 3_ trong 27 tia neu trong de bai. Gia suT SABC = 27 thi dien tich moi tam giac nho deu bang 1.

Dat SpBc = xi, SpcA = X2, SpAB = X3 vdi x i , X2, 23 Z+ v^ thoa man Xi + X2 + X3 = 27. (*)

tfng vdi mot bo { x i , X2, X3) thoa man dieu kien tren, ta c6

SpBcPA + SPCAPB + SPABPC = ~0.

Nen ton tai dung mot diem P thoa man de bai. Suy ra ta can dem so nghiem nguyen dUdng cua phiTdng trinh (*). Day lai la bai toan chia keo Euler quen thuoc, ket qua la Cfe = 325. •

Phdn hoach nguyen vd phdn hogch tap hap 265

6. Cac bai toan chon loc khac H

Bai 1. Chiing minh rang neu goif{r,n) Id so phdn hogch cua n thanh dang bo + h + b2 + ... + bs vdi 0 ^ i ^ s - I vd bi > r6,+i, r la mot so nguyen duang nao do thl ton tai mot so nguyen duang n ma f{r, n) khdc so phdn hoach cua n thanh cdc phdn tvt mot tap hap cdc so nguyen duang bat ky, tri( khi r = 1.

Lofi giai. Ta se chuTng minh dinh ly nay bang phan chufng.

Gia suf ton tai so r ^ 2 sao cho c6 mot tap hdp Sr cac so nguyen dUdng va neu goi so' phan hoach cua n thanh cac phan lay tu: tap hdp Sr la g{r, n) thi / ( r , n) — g{r, n) vdi moi n.

Ta cd / ( r , 1) = 1 nen 1 G Sr vi neu khong thi gir. 1) = 0. Tiep theo / ( r , 2) = 1 nen 2 ^ Sr vi neu khong thi g{r, 2) = 2.

Do / ( r , 3) = / ( r , 4) = • • • = / ( r , r + 1) = 1 nen ta cd diTdc 3, 4, 5 , . . . , r + 1 ^ 5..

Ta cd / ( r , r + 2) = 2 do phan hoach cua r + 2 thoa man bi>r- bi+i m r + 2 v^ (r + 1) + 1. Suy ra r + 2 G Sr vi neu khong thi gir, r + 2) = 1).

Ta cd bang the hien cac gia tri nhiT sau •{•.,.

n f{n) g{r, n) neu n ^ Sr

g{r, n) neu n ^ Sr

Ket luan

1 1 0 1 leS -r 2 1 1 2 2 i S - r 3 1 1 2 3^S-r

r + 1 1 1 2 r + l ^ S - r r + 2 2 1 2 r + 2 ^ 5 - r r + 3 2 2 3 r + 3 ^ 5 - r

2r + 1 2 2 3 2r + 2 ^ 5 - r 2r + 2 3 2 3 2r + 3 ^ 5 - r 2 r + 3 3 4 5 Mau thuan

266 Cdc phUcfng phdp gidi todn qua cdc ky thi Olympic

Tiep theo, ta se xac dinh gia tr i nho nhat cua n de /(r , n) = 3. Dieu nay xay ra khi n c6 dang n = 60 + 61 + 1 vdi 60 > rhi va h] > r hoSc n = 60 + 2 v d i 60 > 2r. 1 ,1 |

Gid tr i n trong trUdng hdp thu" nhat nho nhat la n = (r^ + r + 1) + (r + 1) + 1 = + 2r + 3 va trong trUdng hcJp thtf hai la n = (2r + 1) + 2 = 2r + 3.

Do 2r + 3 < + 2r + 3 vdi moi r dufdng nen n = 2r + 3 1^ gia tr i nho nhat cua n ma /(r , n) = 3. Trong trUdng hdp nay, cac phan hoach cua n la 2r + 3, (2r + 2) + 1, (2r + 1) + 2.

Do chi CO 1, r + 2 e Sr nen dan den c6 2 phan hoach cua 2r + 3 thanh cac phan chi c6

(r + 2) + l + l + . - - + l , 1 + 1 + 1 + +

Suy ra ta can c6 2r + 3 G 5^ de c6 g{r, 2r + 3) = 3.

Tiep theo, ta thay /(r , 2r + 4) = 3 do cac phan hoach cua 2r + 4 ma b, > rbi+x chi bao gom 2r + 4, (2r + 3) + 1, (2r 4- 2) + 2 (chu y rang r ^ 2).

M a t khac, cac phan hoach cua 2r + 4 thanh cac phan trong tap hdp { 1 , r + 2, 2 r + 3} la

( 2 r + 3 ) + l , ( r + 2 ) + ( r + 2 ) , ( r + 2 ) + l + l + -•- + 1, 1 + 1 + 1-f • • • + ! ,

nen g{r, 2r + 3) = 4 neu 2r + 4 ^ 5^ va g{r, 2r + 3) = 5 neu 2r + AeSr, mau thuan.

Tijf day ta c6 dieu phai chting minh. •

B a i 2. Goi f{n), g{n) idn luat la so phan hoach cua n thanh so chdn phan vd so le phan. Chiing minh rang f{n) — g{n) = (—1)"*

neu n = --m - (3m ± 1) vd bang f{n) = g{n) neu ngUcfc lai (so c6

1 dang n = - • m - (3m ± 1) dUOc goi la pentagonal number).

Lcfi giai. Goi Pe{n), Po{n) Ian lu'dt la so phan hoach cua n thanh so' chan phan va so le phan. Ta se thiet lap mot tiTdng iJng 1 - 1

giffa pe{n) va po{n) trong triTdng hdp n 7 - • m • (3m ± 1 ) .

Phan hoach nguyen vd phan hoach tap hap 267

G i d sur phan hoach A = (Ai , A 2 , . . . , A^) cua n c6 thanh phan nh6 nhat la s(A) = A^ va a{\) la so cac so nguyen du-dng lien tiep giam dan bat dau tiJf A i .

Ta xet hai phan hoach sau de hieu ro van de: A = (7, 6, 4, 3, 2) va A = (8, 7, 6, 5). Ta x6t cac triTdng hdp. • - '

• Neu s(A) ^ cr(A), ta them 1 vao moi phan trong s{X) thanh phan Idn nhat cua phan hoach A va bo di thanh phan nho nhat. (Chang han A = (7, 6, 4, 3, 2) A' = (8, 7, 4, 3)).

• Neu s(A) > cr(A), ta triif 1 vao moi phan cua cr(A) phan Idn nhat cua A va them mot thanh phan nho nhat c6 kich thu'dc a(A). (Chang han A = (8, 7, 4, 3) A' = (7, 6, 4, 3, 2)).

Cac phep bien doi trong hai triTdng hdp thay ddi tinh chan le cua so' phan cua phan hoach nen ta thay rang ton tai mot tu'dng iJng 1 - 1 g i a a pe{n) va Poiji).

Tuy nhien, trtfdng hdp 2 d tren se khong con dung neu nhu" 0-(A) = r, s(A) = r + 1 va gia tri cua n khi do la

n = (r + 1) + (r + 2) + • • • + 2r = i • r • (3r + 1). '

Con neu cr(A) = r, s(A) = r thi

n = r + (r + 1) + • • • + (2r - 1) = ^ • r • (3r - 1), • -

nen tuy theo tinh chan le cua r ma c6 su" chenh lech giiâ Pe(n) va Vo{n).

Po{n) ne'u n 7 - • r • (3r ± 1)

Po{n) + {-ly neu n = ]--r- (3r ± 1) 2 ,,,„,„

Chu-ng minh ho^n tat. •


B a i 3. Goi n Id so nguyen ducfng sao cho 1, 2, 3 , . . . , 3n cd the chia thdnh 3 tap hap Id A ,h,. 4 , j j , ,

{ C i , 0 2 , . . . , a „ } , { 6 1 , 62, • • • , { C l , C2 , . . . , C „ } ,

sao cho tti + bi = Ci, i = 1, n. Chiing minh rang neu n thoa thi 3n + 1, 4n, 4n + 1 ciing thoa.

Lcfi giai. Gi5 suf vdi mot so" nguyen difcJng n nko d6, ton tai mot cdch phan hoach thoa man de bai, ta c6

n n n

1=1

Suy ra tdng cac so' tijf 1 den 3n bang —^—^ ^ la so' chan. Do do

n = 0, l ( m o d 4 ) .

V d i n = 4 va n = 5 ta CO the phan chia nhiT ben dufdi

r5 , 7 ,12 r7 ,8 ,15

3,8,11

1,9,10

2,4,6

n = 4 : va n = 5 : < 5,9,14

3,10,13

1,11,12

Trirdng hdp n = 8 v^ n = 9, ta c6 the phan chia nhtf sau:

n = 8 : <

' 10,14,24

7,16,23

5,17,22

3,18,21

1,19,20

6,9,15

4,8,12

12,11,13

va n — 9:'

( 11,16,27

7,19,26

5,20, 25

3,21,24

1,22,23

8,9,17

6,12,18

4,10,14

12,13,15

Tit dSy ta c6 the xay diTng cdch phan chia cho triTdng hdp tdng quat.

Phan hoach nguyen vd phan IKKU li tap hap 269

Gia suf n = A; phan chia difdc, ta se chufng minh n = 4/c, 4A; + 1 cung phan chia du'cJc. ' "

V d i n = 4A;, ta c6 cac so' 1, 2, 3 , 12A; ta chia nhu" sau

' Qk- lSk + lA2k (v;,,

6fc - 3.6/c + 2.12/c - 1 ' l . ' j f t , . ,

- 6yt - 5, 6A; + 3,12fc - 2 1

l , 9 n , 9 n + 1

V d i n = 4A; + 1, ta c6 cac so 1, 2, 3 , . . . , 12A; + 2,12k + 3, ta chia nhu sau

' 6A; + 1,6/c+ 2,12/0 + 3

6 / c - l , 6 / c + 3,12A: + 2

- 6/c - 3,6/i; + 4,12/c + 1

l,9/c + 2,9A; + 3

Trong ca hai triTdng hdp, ta con diT la i cac so' chan 2, 4, 6 , . . . , 6/c (c6 the phan hoach dtfdc theo dieu gia suf).

Tiep theo, gia suf k phan hoach diTdc, ta xet each phan hoach cho tap { 1 , 2, 3 , . . . , 3A: + 1} nhu* sau:

a'i = 3ai - 1, h'i = 36 , c'j = 3ci + 1, vdi i = 1, n.

a'n+i = 3oi, h'n+i = 36i + 1, c'„+i = 3ci - 1, vdi z = Tji.

a'2n+i - 3ai + 1, fe'sn+i = 36i - 1, c'sn+z = 3Q , vdi i = Y j i .

Cuoi cung con la i cac so 1, 9A; + 2 va 9A; + 3. De thay phan hoach nay thoa man de bai.

Vay neu k thoa man thi 3k + 1, Ak, Ak + 1 cung thoa man. Ta c6 dieu phai chiJug minh. D

B a i 4. a) Hoi cd ton tai each phan hoach tap hap cdc so nguyen

duang thdnh cdc tap con cd 2 phan tii Au ^ 2 , A3,..., A^ sao cho vdi moi so nguyen duang n tong cdc phan tvc cua An Id 1391 + n ?


h) Hoi CO ton tai each phdn hoach tap hop cdc so nguyen duong thdnh cdc tap con c6 2 phan tvC Ai, A2, A3,... ,An sao cho vdi moi so nguyen ducfng n tdng cdc phdn tvt cua An Id 1391 + v? 1

Lori giai (a). Cau tra 15i la khong the (c6 the thay 1391 bdi mot so k bat ky) .

Vdfi moi n, tdng cdc phan tu" cua A i , A2, • • • , A n se la kn + n{n+ 1)

2 •

Hdn nCJa, gia tri nay phai nho nhat la -*-) g'j g Li

2n so nguyen dvTdng phan biet).

n (n + 1) ^ 2n(2n + 1) . , Suy ra kn H — - ^ vdi moi n, dieu nay khong

diing khi n —> + 0 0 .

(b) Cau tra Idi 1^ c6, ta xay dijfng nhif sau:

= { 1 , 1391},

An+i = j a ; , 1391 + {n + l f -x: m m | N * \ [ ^

• B a i 5. Chiing minh rang c6 the phdn hoach tap hap cdc so tu

nhien thdnh 100 tap khdc rdng sao cho vdi bat ky 3 so tie nhien a , 6, c thoa man a + 996 = c thl hai trong so chung thuoc cung mot tap hop.

Lofi giai. Vd i z = 1, 2, 3 , . . . , 99, goi A i la tap hdp tat ca cac so tif nhien x thoa man dieu kien x = 2{i- l ) (mod 198).

Tap hdp ^100 chtfa tat ca cac so le. Ta se chiJng minh rang each chia nay thoa man de bî . '

Neu b \h so chan thi a = c(modl98) hay a, a thuoc cung mot tap hdp.

Neu b le thi m6t trong hai so' a,c cung phai le, suy ra chiing cung thu6c Aioo-

Ta CO dieu phai chiJng minh. •

Phdn hoach nguyen vd phdn hoach tap hop 211

B a i 6. Chiing minh rang tap cdc so tU nhien c6 the duoc to hdi hai mau md hai dieu kien dudi day deu duac thoa man: ,

i) Vcti moi so nguyen top vd moi so tU nhien n thl cdc sop"^, vd p""*"^ khong duac to ciing mau. >

i i ) Khong ton tai mot cap so nhdn vo han cua cdc so tU nhien c6 cung mau.

Lcfi giai. Ta gia su* hai mau can to la A, B. Kh i do, xet 4 nhom mau AAB, ABA, BBA, BAB va xet mot day v6 han 5* cac mau nay bang each dat l ien tiep nhom mau thiJ nhat 1 Ian, nhom thu" hai 2 Ian. nhom thu" ba 3 Ian, nhom thu" Ui 4 Ian, quay ngufdc lai nhom thu" nhat 5 Ian, nhom thuf hai 6 Ian va cu" the, cu the la

S = AABABAABABBABBABBABABBABBABBAB... x l x2 x3 x4

Goi S{k) la mau thi? A; trong day, tinh tiif trai sang (chang han 5(2) = A. 5(5) - B) va C{k) la mau to cho so tiT nhien A:, tat nhien Sik), C{k) e {A, B] , k^ 1. Kh i do, ta se to mau vdi quy tac sau

Neu so tir nhien n = pl'p^^p^' •••pi' thi *

C(n) = 5(01 + a2 + 03 H h ak). 1.:,:'

Ta se chiJng minh quy tac nay thoa man de bai. '

Dieu kien thtf nhat la de thay do 3 mau l ien tiep trong S thi khong dong thdi bang nhau, con dieu kien thuf hai thi chiJng minh dtfdc bang phan chuTng. •

B a i 7. d mot ngoi lang no, moi ngifdi hoac luon noi that hoac luon noi do'i. Xet 5 ngUdi trong lang dang diJng thanh mot hang va mot ngufdi khach den tham lang nay hoi m6i ngtfdi trong ho rang CO bao nhieu ngufdi noi that trong hang. M 6 i ngiTdi se cho ngu'di khach mot con so tijf 0 de'n 5. Hoi c6 bao nhieu multiset cac cau tra Idi ma ngifdi khach nay c6 the nhan difdc ?

(Chu y rang multiset la tap hdp ma mot phan tuf c6 the xuat hien nhieu hdn 1 Ian).

272 Cdc p/uMng plidp gidi todn qua cue ky tin Olympic

Lcfi giai. TriTdc he't, ta c6 bo de ve so multi-sets: So cac multi-sets CO n phan tuT lay tuf mot tap hop c6 k phan tu* la C^+^. i -

Trd la i bai toan, ta se chpn cac bp c6 dang (a, 6, c, d, e) trong d6 0 ^ a ^ 6 ^ c ^ d ^ e ^ 5 s a o cho ton tai mot tmdng hdp trong thUc te thoa man. Ta xet 6 trifdng hdp sau:

(1) Neu a, h, c, d, e deu la cac so dtfdng thi ro rang luon ton tai triTdng hdp nhiT the neu tat ca 5 ngiTdi deu noi doi. So bp

/i nhtf the la C| = 126.

I Neu trpng cac so" a, h, c, d, e ma c6 mot so bang 0 thi phai CO mot ngUdi noi that.

(2) Neu a==6 = c = d = e = 0 t h i khong c6 trifdng hdp nao thoa.

(3) Neu a = 6 = c = = 0, e > 0 thi chi c6 bp (0, 0, 0, 0, 1) thoa v d i truTdng hdp 1 ngUcJi npi that, 4 ngu"5i noi doi .

(4) Neu a = 6 = c = 0, d, e > 0 thi ta xet cac trifdng hdp sau:

• Neu CO 1 ngifdi noi that thi d = 1, e 7 1, ta c6 4 bp nhif the.

• Neu CO 2 ngurdi npi that thi d = e = 2, c6 1 bp nhu" the.

De thay khong the c6 nhieu hdn 2 ngiTdi noi that. Do do, CO tat ca 5 bp thoa man de bai.

(5) Neu a = 6 = 0, c, e > 0 thi ta cung xet:

• Neu C O 1 ngufdi noi that thi c = 1, e 7 1 nen c6 tat ca C|+4_i = 10 bp.

• Neu C P 2 ngUdi npi that thi 2 trpng cac sp c, e = 2 va S P con la i khac 2, c6 tat ca 4 bp nhtfng c6 1 bp triing

L la i d tren la (0, 0, 2, 2, 1) nen chi xet 3 bp.

• Neu CO 3 ngUdi noi that thi c = d = e = 3 va c6 1 bp. Do do, CO tat ca 10 - h 3 - I - 1 = 14 bo thoa man.

• I (6) Neu a = 0, 6, c, d, e > 0 thi : , i

! • Neu CO 1 ngtfdi noi that thi 6 = 1 va c, d, e ^ 2, c6 C|+4_i = 20 bp.

Phan hoach nguyen va phdn hoach tap hap 273

• Neu CO 2 ngiTdi noi that thi 2 trong 4 so 6, c, d, e bang 2 va 2 SO c6n la i khdc 2.

o Neu hai so con la i Idn hdn 2 thi ta c6 the xet bp (0, 2, 2, d, e) vdi d, e > 2, cd tat ca C|+3_i = 6 bp.

o Neu trpng hai so con la i , cd 1 so bang 1 thi ta chi can xet 1 bp (0, 2, 2, 1, 1) v i cac bp con lai da duTdc dem d tren. Suy ra d day cd 6 - F l = 7 bp.

• Neu cd 3 ngu'di ndi that thi 3 so' trong 4 so" 6, c, d, e bang 3 va so' con la i khac 3.

o Neu so con la i Idn hdn 3 thi ta xet bp (0, 3, 3, 3, e), cd 2 bp thda man.

o Neu S P cdn la i be hdn 3 thi ta cd 2 bp (0, 3, 3, 3, 1) va (0, 3, 3, 3, 2) nhitag bp (0, 3, 3, 3, 1) da dxiac dem rpi nen chi cd 1 bp thda.

Suy ra d day cd 2 + 1 = 3 bp. \

• Neu cd 4 ngu'di ndi that thi 6 = c = d = e = 4 va cd 1 trtfdng hdp.

Do dd, cd tat ca 20 + 7 + 3 + 1 = 31 bp. j . '

Vay cd tat ca 126 + 0 + 1 + 5 - M 4 + 31 = 177 bp. "' ' •

7. Bai tap ap dung

7.1. Cac bai toan c6 hudng din giai ' ;

B a i 1. Chufng minh rang tpn tai cac phan hpach cua tap hdp so tu" nhien thanh hai tap hdp A, B sao cho vdi mpi n e N thi so' each Viet n = X -\- y v6\ y e A bang vdi so' each vie't n = a; -f- y vdi X, yeB.

B a i 2. Cd n diJa be muo'n chia m thanh keo thanh n phan sao cho moi thanh du'dc be khong qud 1 Ian. H o i v d i cac gia tri m, n the nao thi cd the lam diTdc dieu nay ?

B a i 3. „ B . . . . , ;

274 Cac phUcfng phdp gidi todn qua cac Icy thi Olympic

1. H o i CO the chia tap hop N thanh cac tap v6 han Ai sao cho: vdi mgi i, j thi ton tai so nguyen Xij ma Aj = {n + Xij\n G A, }?

2. Churng minh rang tap hdp cac so nguyen difdng khong the phan hoach thanh cac tap con Sa, S^, Sc v d i a, 5, c € R va

= { [ n a ; ] , n = 1, 2, 3. . . . } .

B a i 4. Phan hoach tap hdp M = { 1 , 2; 3 , . . . , 67} thanh 4 tap con. Chiing minh rang ton tai mot tap con trong so do chuTa ba so a, 6, c ma c = a + 6 (a, 6 khong nhat thie't phan biet).

B a i 5.

1. Cho cac phan hoach cua so nguyen dUdng n ma so cac so Icfn hdn hoac bang k bang so Idn thtf k. Chiang minh rang so phan hoach nhiT the bang so phan hoach cua n thanh cac thanh phan le phan biet.

2. Cho so nguyen diTdng n . Tinh so phan hoach cua n thanh n = oi + 02 + as + . . . + Cfc sao cho

fli ^ 02 ^ a3 ^ ... ^ Ofc ^ a i + 1.

Bai 6. Cho n ^ 3 va so' nguyen to p. Xet cic so' nguyen a i < 02 < aa < • • • < a „ . Chufng minh rang a\, 02, 03 a „ la cap

n

SO cong khi va chi khi ton tai mot phan hoach N = (J A , sao cho

ai+Ai = 0 2 + ^ 2 ^ ••• = a „ + A „ , trong do a+X = {a + x\x e X} .

Bai 7. K i hieu [n] = { 1 , 2, 3, . . . , n). Goi P, C la cac phan hoach khong tinh thur tu" cua tap hdp { 1 , 2, 3, . . . , n). Goi c(C, F, x) la so ham / tuf tap hdp [n] vao [x] vdi i

1. / ( a ) = f{h) neu a, 6 thupc cung ve mot phan cua C.

2. f{a) 7^ f{b) neu a, b thupc cung ve mot phan cua P.

Churng minh rang tong {-'^Y~\p - l ) ! c (C, P, x) khong phu thupc vao C, tong tinh tren tat ca cac phan hoach C vdi p k i hieu so phan cua phan hoach P.

Bai 8.

Phan hoach nguyen va plum hcarli tap hctp 275

a) Cho tap hdp M = { 1 , 2, 3, . . . , 40}. T im k nho nhat sao cho CO the phan hoach M thanh A:tap con Ax, A2, . . . , A^ ma khong ton tai a, b, c E A, nao do (vdi a, b, c khong nha't thie't phan biet) ma a + b = c.

b) Chufng minh rang c6 the phan hoach t$p hdp '

3 ' - l = {^•^•3 i ^ }

thanh A; tap con rd i nhau Ai, A2,..., Ak ma khong ton tai a, 6, c G Ai nao do (vdi a, 6, c khong nhat thie't phan biet) ma a + b = c.

Bai 9. Cho tap hdp 5 = { 1 , 2, 3, . . . , 3n} diTdc phan hoach thanh 3 tap con A, B, C c6 cac so phan tuf deu bang n. ChiJng minh rang ton tai a e A , 6 fc Z?, c e C sao cho

(a + - c)(fe + c - a ) ( c + a - 6) = 0. •; '

B a i 10. Cho so nguyen diTdng n > 1. Xet phan hoach tap hdp so nguyen dUdng thanh k tap con A i , A2, A 3 , . . . ,Ak. ChiJng minh rang ton tai 2 e { 1 , 2, 3, . . . , k] sao cho ton tai v6 han da thufc bat kha quy P{x) c6 bac n va c6 cac he so doi mot phan biet thupc tap hdp Ai.

B a i 11. Cho tap hdp A c N , A 7 0. Neu bat ky so tiT nhien nao du Idn deu c6 the bieu d i l n diTdc diTdi dang tong cua hai phan tuf cua A hoac bang 2 Ian mot phan tuf nao do cua A thi A du'dc gpi la tap day du. V d i moi so nguyen dtfdng x, gpi A{x) la tap hdp cac phan tuf nho hdn x trong A.

Chiang minh rang ton tai tap day du A va mot so^co dinh c sao cho vdi mpi x ^ 1 thi \A{x)\ C A / X .

Bai 12. T i m tat ca cdc so' nguyen diTdng n sao cho c6 the phan hoach tap hdp { 1 , 3, 5, 7, . . . , 2n - 1} thanh 12 tap con cd tong cac phan tuf deu bang nhau.

B a i 13. Ton tai hay khong mot phan hoach cua tap hdp cac so N * thanh A\, A^. . . . , A„ thoa man

1. Ai = +00. !


2. V d i m o i x, y,zA E At vdi cdc so x — y, z — t thupc cung mot X Z

tap khi va chi khi - = - . y t

7.2. Hifdrng d S n giai k

B a i 1 Xay difng cac tap hdp A, B theo quy tac sau

(1) leA. ' ;

(2) n va 2n — 1 thuoc cung mot tap hcJp. '

(3) n va 2n khdc tSp hcfp.

Ta chiJng minh difdc rangÂ, B thoa man bang quy nap.

777- k B a i 2. Cdc so thoa man \a m > n hoac n = 2m hoac — = n k + l

B a i 3. Xet A = {22"+i|n G N } va >1 = {22"|n e N } , M la tap hdp cac so' ma bieu dien nhi phan chi gom cac so thuoc A va N la tap hdp cdc so ma bieu dien nhi phan chi gom cac so' thuoc B.

Dat M = {mi < m2 < < • • •} va N = {ni < n2 < < • • •} va A; = {ui + m\m 6 M}. De thay cac tap hop xay difng theo each nay thoa man de bai.

B a i 4. Xet cac bd de sau:

(1) Do thi day du c6 17 dinh dtfcJc to cac canh bdi 3 mau thi ton tai mot tam giac c6 3 canh ciing mau.

(2) Do thi day du c6 68 dinh du-Oc to cdc canh bdi 4 mau thi ton tai mot tam gidc c6 3 canh ciing mau.

Trd lai bai todn, xet do thi 68 dinh thuoc { 0 , 1 , 2 , . . . , 67} va c6 canh ndi giiJa x, y difOc to mau ciia tap hop chila \x - y .

K h i d6, do thi nay diTcJc to bcii 4 mau va theo bd de tren thi ton tai mot tam gidc c6 3 canh ciing mau. Do do, ton iai x < y < z sao cho y — X, z — y , z — x ciing mau.

Dat a = y - X, b ^ z - y , c = 2 - a: thi a, b, c thuoc ciing mot tap va 0. + b = c . Ta c6 dieu phai chuTng minh.

Phan hogch nguyen va jj/um hogch tap hap 211

B a i 5. Suf dung bieu de Ferrers va chu y tinh ddi xufng qua dudng c h e o y = x. - '^ri 5/> ' J - ' - ' Bai 6. Trifdc het, ta chufng minh chieu thuan: • ;>i :

Gia su: (on) Id mot cap so cong cong sai d, tuTc la

= a i + ( i - l ) d , i = 1. 2, 3, n - 1.

Ta xet cdc tap hop sau: ' *

A„ = {0, 1,..., d - 1, n d , . . . , + d - 1, 2nd,

...,2nd + d-\,...]

An-i = {d, 2d-\, nd + d,...,nd + 2d-l,...]

Ai = {{n- \)d,..., (n - l ) d + of - 1, + (n - l)d,

...,nd+{n-\)d + d-\,...]

n Bang cdch xay diTng nay thi ta thay Ai nAj = 0 va \J A , = N .

i = i

Tiep theo, de chufng minh chieu dao, gia suf ton tai cdc phan hoach Ai, i = 1, 2, 3 , . . . , n cua N thoa man dieu kien de bai. Ta xet cac bd de sau:

(1) Goi mot cap {A, B) v6i A, 5 C N la (a. b) - phu neu nhir 1 1 = a, \B\ b va \A + B\ {0, 1, 2 , . . . . a6 - 1} vdi A +B = {a + b\ e A,b e B} .

Dat TO = m i n ( S \ { 0 } ) thi m | a va ton tai - phu

A', B' sao cho A = mA' + {0, 1, 2 , . . . , TO - 1} va 5 = mB' vdi mX — {mx \ e X} .

Ta chiJng minh dtfOc rang vdi moi h = im + j vdi i 0 , 0 ^ j ^ m - 1 thi h e A -i^ im. e A va h E B j = 0.

(2) Neu (A, B) la (a, b) - phu vdi a la so nguyen to thi A la mot ca'p so cong.

Ta chiirng minh bang quy nap theo b.


B a i 7. V d i n — I thi menh de hien nhien diing.

Co' dinh mot ham khac hang / thoa man (1), dong gop cua ham nay vao tdng d tren tren tat ca cac phan hoach P thoa man (2) la

{-ir\p-\)\. Ta se chufng minh rang dong gop nay bang 0 vdi moi ham khac

hang / nen khang dinh diing.

Xet mot ham khac hang / va mot phan hoach P_ thoa man (2) cua n n - 1], ta se t im so each de md rong phan hoach nay cho Ta CO the tao them phan mdi hoae them n vao baft cu" phan nao khong cMa k sao cho f{k) = f{n). Goi a la so cac gia tri k thoa man dieu nay, ta thay a phai du'dc chufa trong a phan khac nhau cua phan hoach P_ do dieu kien (2).

Do do, CO p-o, each de md rong P_ va giff lai so" phan nhu" cu, mot each khac la md rong no va tang so phan len. M d rong nay them vao tong da cho mot lu'dng la

{p - a)i-ir'{p - 1)! + {-lYp\ -a{-ir\p - 1)!. .

Do moi phan hoach P cua [n] thoa man (2) la mot md rong cua duy nhat phan hoach P_ cua [n - 1] thoa (2) nen tat ca cac dong gop cua cac phan hoach cua [n] thoa man (2) bang -a Ian tdng cac dong gop cua cac phan hoach cua [n - 1] thoa man (2). Hdn nffa, / khong phai hang so tren [n] va la hang so tren [n - 1] nen /(n) phai la mot gia tri mdi , do do a == 0 hay tich nhan diTdc luon la 0.

B a i 8. a) Tru'dc het, ta chu'ng minh A; 3 khong thoa man. Gia su" M = X U r U Z vdi X, r , Z la cac tap rd i nhau thoa man dieu kien bai toan va cd the la tap rong. Khong mat tinh tdng quat, ta gia suf \X\ \Y\ \Z\, suy ra \X\ 14.

Dat X = { x i , X2,...,Xm] v d i m = \X\, gia suf xx < X2 <

••• < Xm thi day so sau chiJa cdc so nguyen difdng phan biet thuoc M : X i , X2, ...,Xm, X2 - Xi, X3 - 1, . . . ,Xm - Xi.

Suy ra 2m - 1 ^ |M| 40 m ^ 20.

So cap ( x , y)eX xYVa

\X\\Y\^^-\X\{40-\X\) = ^.m-{40-m),

Phan hoach nguyin va phan ho(ich tdp hap 279

ma 14 ^ m ^ 20 nen " ' '

X\ m i n I - • 14 • (40 - 14), - • 20 • (40 - 20) j = 182.

Hdn niTa, tdng x + y nhan 79 giA t r i thuoc [2; 80] va do 182 > 2-79 nen ton tai ba cap so ( x i , t / i ) , (x2, ?/2), (a a, ys) ma xi-\-yi =

X2-\-y2 = + yz-

l\i day, chi ra su* mau thuan ve tinh cha't cua cac tap hdp X, Y, Z. Do do /c ^ 4. Ta chi can xay dUng 4 tap hdp thoa man yeu eau bai toan la xong. ' <•

b) Xet bieu dien tam phan m = (• • • 020100)3 va phan hoach tap S theo quy tac sau:

• Ne'u m = (• • • 020100)3 va oo = 1 thi m G AQ. %

• Neu m = (• • • 020100)3 va oo 7 0 thi : ton tai / ^ 1 sao cho a j_ i 7^ 0, o; = 0: chon / nho nhat thoa man tinh chat nay va xep m ^ Ai.

Theo tinh chat cua he tam phan thi ta chuTng minh diTdc neu m i , m2 e A,, thi m i + m2 ^ Ai.

Hdn nffa, m = ^ [ t - l ) G Ao va vdi 1 ^ m < ^ (3'= - l ) thi

chae chan trong bieu d i l n eua m eo it nhaft mot chu* sd 0 d v i tr i oj vdi ^ ^ A; - 1 hay m G A i , A 2 , . . . , Ak-i, dieu nay c6 nghia la moi m G 5 thi m thuoc mot trong A;tap hdp AQ, ^1, • • . , Ak-i-

Ta CO dieu phai chu'ng minh.

B a i 9. Gia suf ton tai each phan hoach S thanh A, B, C ma khong thoa man de bai. Gia suT 1 G A va A; la so nguyen diTdng sao cho 1, 2, 3 , . . . , k - l e Avk k ^ A. Gid suf A: G 5 thi suy ra 2 ^ C. Ne'u A: > 2 thi de thay 2 G A.

Ta c6 nhan xet: neu x G 5, x G C thi x - 1 G / l .

That vay, gia suf x G C, x - 1 ^ A thi x - 1 G S hoSe C. Ta xet cac bo ba:

• (1, X - 1, x) cd 1 G A , X G C ^ X - 1 G C. ' ^ '

280 Cdc phucmg phdp gidi todn qua cdc ky thi Olympic

• {x - k, k, x) CO k e B, X e C => X - k e B hoac C. \., , • {x - k, k - I, X - 1) CO k - I e A, X - I e C X - k e C. • {x-k-l,k,x-l) CO ke B,x-leC =^x-k-\eC.

Do x,x - 1 e C nen x ~ k,x - k - 1 e C. TifcJng W, ta cung c6 x - ik,x - ik - I e C. Ta chon i thoa man x - ik - 1 < k hoac X - ik < k thi dan den mau thuln. Nhan xet dufOc chufng minh. Tiep theo, dat C = {ci < C2 < C3 < • • • < c„} thi ta c6 ci - 1, C2 -I . C 3 - l,...,c„ - 1 G A

HcJn nffa, Q > 2, z T7n nen {1, ci - 1,..., Cn - 1} C A va |A| > n, mau thuan. Bai 10. Gia suT € A vdi moi i = 1, 2, 3 , . . . , /c. Chpn p la so nguyen to sao cho

p > m a x j o i , a 2 , . . . , 0 ^ } .

Suy ra ton tai i sao cho Ai chura v6 han so chia het cho p nhu'ng khong chia he't cho p'^.

Xet P{x) = O j X " + + 62x"~^ + ... + bo vdi 6, E A va

6, = P Q , ( C i , p ) = 1.

Theo tieu chuan Eisenstein ve da thiJc bat kha quy thi da thiJc P{x) nhiT tren ba't kha quy. Do ton tai v6 so' so' 6, nen ta c6 dieu phai chufng minh. Bai 11. Bat A = { 2"' + 2"2 + . . . + 2"'= I 0 ^ ai < 02 < • • • < Ofc.

ai - aj : 2,Vz 7 j •. Ta chiJng minh rang v6i so nguyen difdng N Idn hdn 1 bat ky

thi bieu dien difdc du'di dang tdng cua hai so thuoc A. X e t b i e u d i e n i V = 2">+2"2+-- -+2"" vdi 0 ^ oi < 02 < ... < an-

Phan hoach { 0 1 , 0 2 ^ O n } = U UV wdi U la tap hdp cac so chan, V la tap hdp cac so le.

Khi do iV = ^ 2" + ^ 2^ va ^ 2", ^ 2^ G A (neu mot trong u&U v&V ueU v^V

hai tap hdp U, V rong thi ta chia doi tap hdp con lai).

Phdn hoach nguyen va phan hoach tap hap 2 8 1

Tiep theo, vdi x ba't ky, ta x6t n G N* m^ 22"-2 < 2^", suy ra A{x) < A(22").

Hdn nffa, >1(2^") chlnh la so each chon 0 ^ o, < 02 < • • • < OA-cung tinh chSn le va bat dang thufc 2"^ + 2"^ + . . . + 2"* ^ 2^". Suy ra A(22") ^ 2"+i.

Do do, A{x) ^ .4(22") < 2"+i = 4 • 2"-^ ^ 4 ^ . Ta cd dieu phai chufng minh. Bai 12. Trirdc het, de thay dieu kien can la ( 1 + 3 + 5 + - • +271-1) =

: 12 hay n : 6. •n?

Dat n = 6k thi ta can phai cd — ^ 2n - 1 hay 3k'^ ^ 1 2 - 1 nen ^ 4.

Ta chiJng minh du'dc rang neu 6k thoa man thi 6(A; + 4) cung thoa man, tijf dd, chi can xet cac trUdng hdp nhd cua k de di de'n ket luan. 'i-^ Bai 13. Chu'ng minh rang phan hoach sau — {x \ = 2^{2k - 1)} thoa man de bai.

7.3. Bai tap tU luy^n Bai tap 1. Cho n Id so nguyen duang Idn han 1. Hai ngUdi chai A vd B chai mot tro chai bang viec xdy dUng cdc phdn hoach cua n dudi dang

n = oi + 0 2 H h Ofc vd oi ^ 0 2 ^ 0 3 ^ • • • ^ o/c ^ 1 A thdng neu nhu cd so le phdn trong phdn hoach (tvtc Id k Id so le) vd ngugc lai thi B thdng. Cdc so a^, 0 2 , o^ duac chon nhu sau: A bat ddu chon mot so a\ mot trong cdc so ti( 1 den n — 1. 5 chgn mot so 0 2 trong cdc so tie 1 den Oi sao cho Oi + 0 2 ^ f^- A tiep tuc chon mot so 0 3 givta cdc so' tC( 1 den 0 2 sao cho oi + 0 2 + 0 3 ^ ri vd cvC the den khi ndo xdy dUng hodn tat phdn hoach. Hay xdc dinh xem vdi mSi gid tri n> \ ai cd chien thudt de chien thdng. Bai tap 2. Hai doi ciing chai tro nhdt bong: Doi xanh nhdt bong xanh vd doi do nhdt bong do. Quy tdc nhdt bong nhu sau:

282 Cdc phuang phdp gidi toan qua cdc ky thi Olympic

• Cdu thu so 1 cua doi xanh se chi nhdt mot so bong chia he't cho 5.

• cdu thd so 2 cua doi xanh se chi nhdt mot so bong nho hon hodc bang 9.

• cdu thu so 3 cua doi xanh se ludn phdi nhdt mot so bong chdn.

• cdu thu so 4 cua dgi canh se nhdt 1 qua. i

• cdu thu so 1 cua doi do se ludn nhdt niQt so bong tuy y.

• Cdu thU so 2 cua doi do se ludn nhdt mot so bong tuy y.

• cdu thu so 3 cua doi do se hodc Id nhdt 5 qua.

Quy tdc chai nhu sau: Moi doi dugc chia cho mot so rat nhieu cdc gio bong, mSi gio c6 20 qua. Tgi moi vdng ddu, cdc cdu thu cua moi doi se thUc Men dSng tdc nhdt bong 1 idn hodc khong thUc Men, sao cho cd doi nhdt dung 20 qua bong. Dieu kien la bo cdc dgng tdc cua cdc cdu thu khong duoc lap Igi giCta cdc vdng ddu. KM ndo khong cdn cdch nhdt ndo nita thi dUng lai. Hoi trong hai doi, doi ndo se phdi ditng lai trUdc?

Bai tap 3. Mot ngUdi cd 210 vien soi, chia Idm mot so ddng soi. Sau do anh ta nhdt tit moi ddng ra mot vien vd tgo thdnh mot ddng soi mdi. Chitng minh rang ne'u anh ta lap lai mdi qud trinh nay thi se den luc anh ta cd cdc ddng soi vdi sd luang Id doi mot khdc nhau. Hay tdng qudt bai todn vdi cdc sd n thich hop.

Bai tap 4, Gid su: tap hap cdc sd nguyen duang dugc phdn hogch thdnh n cap sdcong cd cdng sai idn lugt Id di,d2,d3,... ,dn- Chvcng minh rang ton tgi i ^ j sao cho di — dj.

' Bai tap 5. Chvcng minh rang tap hap cdc sd nguyen duang cd the phdn hogch thdnh hai tap hap A, B sao cho A khong chiia cap sd cong ndo cd do ddi la 3 vd B khong chiia cap sd cgng ndo cd do ddi vd hgn.

Phdn hogch nguyen va nlum lioach tap hap 283

Bai tap 6. Xdc dinh sd nguyen duang k Idn nhdt thda man tinh chat sau: Tap hap cdc sd duang cd the dugc phdn hogch thdnh k tap con Ai, A2, •. •, Ak sao cho vdi moi sd nguyen duang n ^ 15 vd mgi

, 2 e { 1 , 2 , . . . , A;} thi ton tgi 2 phdn tv[ phdn Met cua Ai md tdng cua chiing ddng bang n. , • ;

Bai tap 7. Vdi moi phdn hogch nguyen cua sd nguyen duang n, ta ggi A{n) Id sd cdc sd 1 xudt Men trong phdn hogch vd B{n) la sd cdc sd khdc nhau xudt Men trong phdn hogch.

Chiing minh rang YIM^^) = YlB{n), tdng lay tren tat cd cdc phdn hogch. ,

Bai tap 8. Cho da thijCc f{x) e Z[x] sao cho cdc he sd cua f{x) deu nguyen duong vd deg / > 0. Vdi mgi sd nguyen duang n, ki Mew S{n) = { / ( I ) , / ( 2 ) , . . . , f{n)}. Chiing minh rang ton tgi sd nguyen duang n sao cho S{n) cd thi phdn hogch thdnh 10 tap con md tdng cdc phdn tv[ cua moi tap con ay bang nhau.

Bai tap 9. Tren bdn cd 10 qud bong vd 11 cdi hop ddnh sd tie 1 den 11-

1. Hdi cd bao nhieu cdch cho 10 qud bong vdo 11 cdi hop?

2. Cd bao nhieu cdch cho bong vdo hop sao cho vdi mgi \10 thi ta ludn cd it nhdt k qud bong trong k hop ddu tien?

3. Cd bao nhieu cdch cho bong vdo hop sao cho cd k cdi hop khong thda man dieu kien b vdi 1 ^ k ^ 10 ?

Bai tap 10. Tim sd phdn hogch cHa n thdnh cdc phdn {x, y, z) thda man tdng cua hai phdn bat ky khong nhd han phdn cdn Igi.

Bai tap 11. Cho (w„), Id hai day sd tu nhien thda man cdc dieu kien:

1. Ui = 1.

2. Vn = an - l-Un vdi a la sd tu nhien cho trudc.

3. Un+i la sd nguyen duang nhd nhdt khdc cdc sd ui, U2, • • Un-

Tinh U2on.


Bai tSp 12. Goi Sn Id cdc phdn hoach cua n ma moi thdnh phdn deu Idn Hon 1. Cdc phdn hoach co dang {xi, X2, • • •, Xk), {yi,y2, • • • ,yi) G

k I

Sn la ban be neu Uiî + '^) = U iVi + ôi S'{n) Id tap con 1=1 1=1

cua tap S{n) ma khong chiia bat cvc cap phdn hoach ban be nao. Chiing minh rang \S'{n)\ y/n.

Bai tap 13. Vdi mdi so nguyen duong n, ta goi day cdc so ai, a^, as,..., ak Id n-tot neu nhu ai + a2 + as H h afc = n, ai = = 1 vd \ai - a^+il ^ 1, z = 1, 2, 3 , . . . , fc - 1. Tinh gid tri nhd nhd't cua k theo n. ,

Tai lidu tham khao [1] Phan Thi Ha DiTOng, Bai gidng JMO 2010, nam 2010.

[2] Vu Van Tan, Tuyen tap cdc bai todn thi Olympic, nam 2006.

[3] Miklos Bona, Introduction to Enumerative combinatorics, 2007.

[4] h t t p : / / m a t h o v e r f l o w . n e t /

[5] h t t p : / / i n a t h l i n k s . r o /

[6] h t t p : / / e n . w i k i p e d i a . o r g /

[7] h t t p : / / m a t h s c o p e . o r g

Ldl GIAI OE THI C H O N H S G T O A N Q U O C G I A NXM 2013

Bai 1. Gidi he phuang trinh: m i l l , F.' \

sm X + 1

sin^ X + \ +

/sin^ y + + J cos x + —^

20y_ x + y

x + y

{x, y e

Lcri giai. D i l u kien xac dinh: x ~ , y ~ (k, m e Z) wk

xy>0. Nhan theo ve hai phiTdng trinh cua he, ta thu diTdc:

V {x + yf

trong &6 A=\x + — ^ + ./cos^y +

/ I — I : — \

X sin^ y + + Wcos2 x + - \

Su' dung baft ding thiJc Cauchy-Schwarz va AM-GM, ta c6 7

y) X

sin^ X + 1 \

V sin^ X J cos X +

1 \ COS^ X J

sm X cos X

( sin 2x

+ 1

sm X cos X

+ 1

Hoan toan tu'Ong tu, ta cung c6

+

' • 2 1 \sm^ y + cos y +

cos" y J

2|sin2x| 2|sin2x

1 \^

285

286 Cac phucfng phdp gidi todn qua cdc ky> thi Olympic

D o d 6 , theo ba t d ^ n g thuTc A M - G M :

y ( « i n ^ ^ + - ^ ) ( c o s 2 ^ + ^ ) ( s i n 2 y 4 - (cos^ y + ^

^ 4 . / 5 \

= 10 ^ 20 xy

{x + y) 2' .

D a n g tMc x a y ra k h i va c h i k h i | s i n 2 x | = 1 va x = y, tvCc x = y =

! [ + /c e Z . T h i i l a i , ta thay r a n g k h i x = ? / = - + — , A; e Z t h i

4 2 1 x 1

^ 2 x + j / x + y 2 s in X = cos^ X = s m y

do ca h a i ve cua m 6 i phiTcJng t r i n h t rong he da cho d e u bang VTO.

V a y X = y = ^ + Y ' ^ ^ ^ ^ ' '^^^ n g h i e m cua he phtfdng

t r i n h da cho. 1^

NhSn xet. Day la bai toan de nhat va cung la bai t o d n . . . dd nhat cua ky th i . D6 v i y tirdng dung bat dang thiJc la qua ro rang, con dd la v i viec trpn Ian giffa cac bieu thiJc lUdng giac va bieu thiJc dai so da lam cho bai toan trd nen xau x i va thieu tiT nhien. Ngoai cac danh gia da neu trong hai 15i gidi tren con c6 the dung nhieu each khac, c6 the dat them an phu de de bien ddi hdn.

Y tirdng dung bat d i n g thUc trong gidi phiTcfng trinh Id khong m d i va gan day nhat da diTdc suf dung trong ky V M O 2009: Gidi he phuctng trinh:

1 2 1

V x ( l - 2x) + ^y{l - 2x) = -

Sir Idp la i trong cdc de thi V M O cua chung ta c6 the noi da trd thanh he thong. C6 the ddn ciJf mot so v i du trong ITnh viTc he phiTdng trinh de minh chuTng cho luan diem nay:

(1) (VMO, 1996) Gidi he phumg trinh:

1 A - 2

I ^ V x + yj

Ldi gidi de thi chon HSG todn quo'c gia nam 2013 287


1 +

1 -

/ lib :'.elb i:yr\-^: 12 \^

3x + y/


j x^ + Srcy^ = - 4 9

I x2 - 8xy + y2 ^ By - 17x

(4) (VMO, 2010) Gidi he phumg trinh: , j

( x^-y^ = 240

\x^-2y^ = 3 {x^ - 4y2) - 4 (x - By) *

(5) (VMO, 1994) Gidi he phumg trinh sau tren tap cdc so thuc:

x^ + 3x-3 + ln{x^ - x + l ) ^ y

y^ + Sy-3 + ln(y2 _ + i ) = ^

{z^ + 3z-3 + ln{z^ - z + l) = x

(6) (VMO, 2006) Gidi he phumg trinh sau tren tap cdc so thUc:

' _ 2a; + 6 • Iog3(6 - y) = x

v / y 2 _ 2 y + 6 - l o g 3 ( 6 - 2 ) = y

^ v ' ^ 2 - 2 z + 6 - l o g 3 ( 6 - x ) - z

..ir Idp la i nay cho thay nhiJng sdng tao, nhffng y ttfdng m d i gan day da thieu v^ng rat nhieu, thay vao do la nhffng bai toan "cho du dpi h inh" . B e thodt kh6i tinh trang nay, ra't can phdi c6 siT dau tiT manh me cho cong |tdc xay diTng nguon de de xuat, ma gidi phdp khd thi nhat la manh dan siuf dung sir d6ng g6p cua cac ban trd la ciTu V M O , ciTu I M O nhu" kinh nghiem ciia cac nirdc c6 phong trao Olympic Toan manh nhir Nga, M y , Romania . . .

Cuoi cung, cung can noi them Id bdi todn nay tuy de nhuîg cung c6 lam nhieu ban ma't diem d khau trinh bay. Dieu kien xac dinh d dau

288 Cdc phUcfng phdp gidi todn qua cdc ky thi Olympic

bai todn va thuf la i triTdc khi ket luan la nhu-ng phan khong the thieu. Nhieu thanh vien trong dpi tuyen thi I M O 2012 da quen trinh bay phan thuT la i day du va mat di 1 diem d bai 4 (phu'dng trinh ham), day la mot dieu dang tiec va la bai hpc quy gia cho cac hpc sinh chuyen Toan, du gik\c bai nhiTng the hien 15i giai cung phdi that can than.

B a i 2. Cho day sothuc ( a „ ) xdc dinh bdi a i = 1 vd a„+i = 3 —

vdi moi n ^ 1. ChiCng mink rdng day (an) c6 gidi han hUu hgn. Hay

tlm gidi han do. ,,

Lcf i g i a i . Tn fdc he't, bang quy nap, ta se chiJng m i n h :

' a„ > 1, V n > 1. (*)

3 Tha t vay , v d i n = 2, ta c6 U2 = - nen (*) dung . G ia suf (*) d i ing v d i

afc + 2 n = A; > 1, tiJc la > 1, ta can chuTng m i n h Uk+i = 3 > 1,

hay:

2""+^ >ak + 2.

X e t h a m so u{x) = 2^+^ - x - 2 t r en ( 1 , + oo ) . Ta c6

w ' (x ) = 2 ^ + M n 2 - l > 0 , V x > l .

D O do u{x) la h a m dong bie'n, suy ra u{ak) > u{l) > 0. Tijf do suy

ra (*) d i i n g v d i m o i n > 1.

BSy g i d , do a„ ^ 1 v d i m o i n ^ 1 n6n :

a„+ i = 3 - ^ ^ < 3 , V n ^ l .

T i e p theo, ta se chuTng m i n h ( a „ ) la day tang. X e t h a m so f{x) =

3 - v d i 1 < X < 3. Ta CO 2x

, I n 4 + x l n 2 - 1 ^

/ (^) = 2- ^ '

3

nen f(x) dong b i e n t r en ( 1 , 3) . Ta cung c6 02 = - > Oi nen day

tang. D o d d , day ( a „ ) da cho tang va b i chan t r en b d i 3 nen c6

Ldi gidi de thi chon HSG todn quoc gia nam 2013 289

g i d i ban hffu han . Gii sur g i d i han do la L e ( 1 , 3) . T rong cong

thiJc xac d inh cua day, chuyen qua g i d i han , t a c d L = 3 - ^ " ' " ^ 2^ •

Ta se chiJug m i n h phi fdng t r i n h :

x = 3 - ^ - (1)

CO n g h i e m duy nhat t r en ( 1 , 3) . T h a t vay , x e t h a m so g{x) = 3 -X + 2

- a; v d i X e ( 1 , 3) , ta cd 2x

_ I n 4 + a ; l n 2 - 1 ^ l n 4 + x l n 2 - 1 - 2^

Ta xe t t i ep h a m so h{x) = l n 4 + a ; l n2 - 1 - 2^, a; € ( 1 , 3) t h i :

/i ' (x) = l n 2 ( l - 2 ^ ) < 0 , V a : € ( l , 3 ) , " '

nen day la h a m ngh ich bie'n. Suy ra h{x) < h{\) = l n 8 - 3 < 0, hay l n 4 + a ; l n 2 - l - 2 ^ < 0 v d i m o i x G ( 1 , 3) . D o dd , h a m so g{x) nghich bie'n t r en ( 1 , 3) va phiTdng t r i nh g{x) = 0 cd khong qua m o t ngh i em. H d n nffa, ta cd g{2) = 0 nen a: = 2 la n g h i e m duy nha't c i ia phiTcfng t r i n h (1) . V i vay day so da cho cd g i d i han la 2. •

Nh3n xet . Ngoai each danh gia chSn mien gia tr i cua day thupc (1 , 3) nhu tren, ta cung c6 nhieu c6 nhieu danh gia khac chat hdn giup qua trinh quy nap thuan tien hdn.

Bai n^y cung kha cd bdn v^ quen thupc. Bai todn khdo sdt sif h6i tu ciia day so dang a„_|_i = / (a „ ) da trd thanh giao khoa, dac biet la kh i / 1 ham tang. Van de chiJng minh day bi chSn cung khong c6 kho khan vi chi can chufng minh difdc a„ > 1 thi hien nhien ta c6 a„ < 3. Cudi cdng, viec giai phu'dng trinh x = f{x) gay doi chiit kho khan do'i vd i mot so ban, dac biet la cdc ban hpc Idp 11 (va Idp 10 - nSm nay c6 mot so ban Idp 10 tham gia ky thi).

Bo sung them cho y kien ve sir lap lai trong cac de thi V M O , chiing toi tiep tuc du'a ra mot so bai loan c6 hinh thiJc va each tiep can tu"dng tu" bai V M O nam nay da dUdc suT dung trong cdc ky V M O trifdc day:

(1) (VMO, 1998) Choa^l la sothuc. Ddt:

xi = a, Xn+i = 1 + In xt

, V n = l , 2, . . .

Chiing minh rang day {a;„} c6 gidi han hHu han vd tim gidi han do.

290 Cdc phUCn^ pliap i>uii loan qua cac ky tin Olympic

(2) (VMO, 2008) Cho day sothuc (x„) duac xdc dinh bdi x i = 0, X2 = 2

Xn+2 = 2 - " " + - , Vn = 1, 2, 3, . . .

Chiing minh rang day so tren c6 gidi han hitu han khi n dan den v6 cung vd tim gidi hgn do.

d day chiing toi chi neu hai M i todn c6 dang ci each giî rat gidng vdi bai nam nay, vi cic bai day so cac nam 2000, 2001, 2005 cung c6 dang day so - /(a;„) nhifng deu c6 nhiJng net thii vi rieng do khai thac diTqJc nhiing y hay.

B a i 3. Cho tarn gidc khong can ABC. Ky hieu (/) la dudng tron tarn I noi tiep tarn gidc ABC vd D, E, F Ian luat la cdc tiep diem cua dudng tron (/) vdi cdc canh BC, CA, AB. Dudng thdng di qua E vd vuong gdc vdi BI cat {!) tgi K {K ^ E), dudng thing di qua F vd vuong gdc vdi CI cat (/) tgi L {L ^ F). Goi J Id trung diem cuaKL.

(a) Chiing minh rang D, I, J thdng hang.

(b) Gia sii cdc dinh B vd C co dinh, dinh A thay ddi sao cho ti so AB

= k {k khong ddi). Goi M, N tuong ling la cdc giao diem IE, IF vdi (/) {M E, N F). MN cat IB, IC idn luat tgi P, Q. Chiing minh rang dudng trung true cua PQ ludn di qua mot diem cd dinh.

Lcfi giai. (a) Do D, E Ian li/cJt la cac tiep diem cua (/) vdi cac canh BC, CA nen ta c6 DE ± CI. Theo gia thiet thi FL ± CI nen suy ra DE \\ Uokn tohn ti/dng tiT, ta c6 DF \\

Tijf cdc cung bang nhau trong dufdng tron, ta c6 DK = DL = EF hay D nam tren difdng trung triTc cua KL. Hdn nffa, / cung nam tren di/dng trung tn^c cua KL nen suy ra D, I, J thang hang.

Ldi gidi di thi chon MSG todn qudc gia ndm 2013 291

(b) Goi T la trung diem BC, G la giao diem cua AI vdi BC, U \li d iem doi xiJng vdi T qua G. Ta se chtfng minh rang du-dng trung trtfc cua PQ luon di qua U. Goi X, Y Ian liTdt la giao diem cua EF vdi BI va CI. Ta cd

XIC = 180° - BIC = 180° -

= AEI - EFI = AEF = XEC.

DoJ d^C, 7, E, X dong vien, suy ra BXC = 90°. Tudng tu", ta suy ra BYC = 90°.

V i BXC = BYC = 90° nen bon diem B, C, X, Y n^m tren dufdng trdn tam T, difdng kinh BC. Suy ra T n^m tren du'dng trung true cua XY.

De thay rang cac cSp diem M, E wa N, F doi xvCng nhau qua / nen EF \\ hay XY \\ Tuf do, ta cd AIEX = A / M F , suy ra X v^ P ddi xuTng vdi nhau qua / . TufcJng txi, ta cung cd Y va Q ddi xufng nhau qua I. Do dd hai duTdng trung trufc cua XY va PQ ddi xiJng v d i nhau qua AI. TiJf day suy ra du'dng trung trufc cua PQ di qua U.

292 Cdc phucfng phdp gidi todn qua cdc ky thi Olympic C Fi A R

Vi B, C CO dinh nen T co dinh. Ti so — = — = khong ddi n e n G co dinh. Do dd U co dinh. Vi vay dUdng trung trirc cua PQ luon di qua d i e m co dinh U. •

Nh§n xet. Vi$c d\i dodn diem co dinh d tren BC la tiTcfng doi ddn giSn vi chi can lay A' doi xiJng vdi A qua BC, ta se diTdc hai dUdng trung trifc doi xiJng nhau qua BC. Do d6 chiing se c^t nhau tai mot diem tren BC va la diem co dinh can tim (de thay r ing neu diem A th6a man dieu

AB kien —— = k thi diem A' doi xuTng vdi A qua BC cung se th6a man dieu AC A'B kien = k). Cdc ket qui cd bin cung tUdng doi de nhin ra vi mo hinh cua bai n^y da tufng xuat hien trong bai 3, de VMO 2009.

Tuy vay, cii mdi v^ chinh la cai khd cua bai toan nay n i m 6 cho: Viec xdc dinh M , N 1 ph6p doi xiJng tam / , nhiTng de gidi bî todn ta can chuyen qua ph6p doi xiJng true AI. Dieu nay doi h6i hoc sinh phdi n^m vffng tinh chat cua phep bien hinh mdi co the lap luan nhanh chong diTdc, neu khong hoc sinh co the dung mot so bien doi goc, tinh toan dai so cung c6 the giSi quyet diTdc bai todn nhitog kha vat vd. Mot kho khan khdc cua bdi nay chinh la kho ve dUOc hinh chinh xac do cac vi tri diem khd gan nhau, khong ro rang dan den tinh trUc quan bi mat di, nhieu khi dan den ngp nhan. Do do, bai todn nay doi h6i nhieu cl kha nang phan dodn ciia thi sinh.

AB GiS thiet —— khong ddi chi 1 de suy ra chan duTdng phan gidc cd AC dinh nhiftig v6 tinh 1dm mot so hoc sinh nghi den diJdng tron Apollonius. Viec chi ra diem cd dinh difa theo ti sd chia doan BC la c6 the lam diTdc nhiTng cung tiTdng ddi khong ddn gian.

Bai 4. Cho trade mot so so tu nhien duac viet tren mot dudng thing. Ta thyCc hien cdc budc dien so len dudng thdng nhu sau: tai mdi budc, xac dinh td't cd cdc cap so ke nhau hien co tren dudng thdng theo thii tu tic trdi qua phdi, sau do dien vdo giita mdi cap mot so bang bang tdng cua hai so' thuoc cap do. Hoi sau 2013 budc, so 2013 xudt hien bao nhieu idn tren dudng thdng trong cdc trUdng hap sau:

(a) Cdc sd cho trudc la 1 va 1000

Ldi gidi de thi chon HSC loun qudc gia nam 2013 293

(b) Cdc sd cho trudc Id 1, 2, . . . , 1000 va duac xep theo thii tu tdng ddn tic trdi qua phdi?

Lcfi giai. (a) Ta se chu'ng minh rang co dung hai sd 2013 trong day nhan difdc sau 2013 Ian thiTc hien. That vay, chii y rang ta chi quan tarn den sd Ian xuat hien ciaa 2013 nen khi co hai sd diJng ke nhau

tdng cac sd deu Idn hdn 2013 hoac cac sd da xuat hien tiir iuidc d6 rdi (ta da hoan toan xac dinh sd do co bang 2013 hay khong) thi ta khong can liet ke nffa.

Day ban d iu 1 : 1, 1000.

Sau birdc 1:

Sau birdc 2:

Sau birdc 3:

1, 1001, 1000.

1, 1002, 1001, 2001, 1000.

1, 1003, 1002, 2003, 1001, 3002, 2001, 3001, 1000. Tiep theo, ta chi giff lai 4 sd dau tien. Sau bufdc 4:

1, 1004, 1003, 2005, . . . Sau birdc 5:

Sau birdfc 6:

Sau birdc 7:

Sau birdc 8:

1, 1005, 1004, 2007, . . .

1, 1006, 1005, 2009, . . .

1, 1007, 1006, 2011, . . .

1, 1008, 1007, 2013, . . . Sau birdc thu" 8, sd 2013 xuat hien Ian dau tien. Sau dd, truf tdng cua cap dau tien con be hdn 2013 thi cac cap con lai deu idn hdn 2013. Hdn nffa ngay tijf dau, sau mdi Ian thyc hien thi sd thff hai trong day tang dung mot ddn vi nen sau 1013 bffdc thi sd thff hai

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