[vnmath.com]-combinatorics via problems&theorems
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T hp qua cc nh l v bi tonTrn Nam Dng
Trng i hc KHTN Tp HCM
1. Cc quy tc mTa ni tp hp A c n phn t nu tn ti song nh f: A --> {1, 2, ..., n}. K hiu | A | = n.
Quy tc cng: Nu cng vic A c hai phng n thc hin (loi tr ln nhau), phng n 1 c n1 cch thc hin, phng n 2 c n2 cch thc hin th cng vic A c n1 + n2 cch thc hin. Trn ngn ng tp hp: Nu A ( B = ( th |A ( B| = | A | + | B |.
Quy tc nhn: Nu cng vic A c th chia thnh 2 cng on tip ni nhau, cng on 1 c n1 cch thc hin, cng a 2 c n2 cch thc hin th cng vic A c n1n2 cch thc hin. Trn ngn ng tp hp: |A B| = | A |.| B |.Quy tc phn b: , trong l phn b ca A trong X.1. a) C bao nhiu s c 3 ch s?
b) C bao nhiu s c 3 ch s khc nhau?
c) C bao nhiu s chn c 3 ch s khc nhau?2. a) C bao nhiu s c 3 ch s chia ht cho 3
b) C bao nhiu s c 3 ch s, chia ht cho 3 nhng khng cha ch s 3?
c) C bao nhiu s c 3 ch s khc nhau v chia ht cho 3?3. Trong mt trng hc mi mt hc sinh nam quen vi 32 hc sinh n v mi mt hc sinh n quen vi 29 hc sinh nam. Hi trong trng hc nam nhiu hn n hay n nhiu hn nam, v nhiu hn bao nhiu ln. 4. Xt bng ch nht m n . Hi c tt c bao nhiu hnh ch nht c cnh song song vi cnh ca bng?5. Cho p l s nguyn t v a l s nguyn dng. Mt ng trn c chia thnh p qut bng nhau.
a) C bao nhiu cch t p qut bng a mu, nu ta c nh ng trn (khng xoay).
b) Nu ta cho php xoay ng trn, v hai cch t c coi l nh nhau nu c th thu c t nhau qua mt php quay th c tt c bao nhiu cch t?Xt cc tp hp A, B, C ... thuc X. Ta nh ngha hm c trng ca A, B, C,... l cc nh x t X vo {0, 1} c xc nh nh sau:
Hm c trng hon ton xc nh tp hp, v ta c cc tnh cht c bn sau:
1) A = B (A(x) = (B(x) vi mi x thuc X (v khi ta vit (A = (B)
2) A ( B (A(x) (B(x) vi mi x thuc X (v khi ta vit (A (B)
3) (A(B = (A.(B
4)
6. a) Chng minh rng (A(B = (A + (B - (A.(B.b) Chng minh rng (A(B = (A + (B - 2(A.(B.c) p dng chng minh A((B(C) = (A(B)(C vi mi A, B, C.
Hm c trng lin quan trc tip n php m thng qua cng thc quan trng (v hin nhin) sau:
(1)7. p dng cc tnh cht ca hm c trng v cng thc (1), hy chng minh
a) Quy tc cng
b) Quy tc nhn
c) (Cng thc bao hm v loi tr cho n = 3) |A ( B ( C| = |A| + |B| + |C| - (|A(B| + |B(C| + |C(A|) + |A(B(C|.
d) (Quy tc m theo phn t) Cho F l mt h cc tp con ca X. Vi mi k = 0, 1,...,|X| gi nk l s tp con thuc F c k phn t, vi mi x thuc X, gi c(x) l s cc tp con thuc F cha x. Khi ta c
e) (p dng quy tc m theo phn t) C 20 th sinh tham gia cuc thi Vietnam Idol. BGK s chn ra 5 gng mt xut sc nht, cn khn gi cng s chn ra 5 gng mt c u thch nht. Nu cc danh sch c chn mt cch ngu nhin th trung bnh s c bao nhiu th sinh c gp mt trong c hai danh sch?
Hng dn: Gi F l tp tt c cc cp (A, B) vi A,B ( [20], |A| = |B| = 5. Bn cht ca bi ton l tnh gi tr ca
.2. Cc i tng t hp c bnXt tp hp X gm n phn t. T tp hp c bn ny, ta c th xy dng cc i tng t hp phong ph.
Tp cc tp con ca tp X: Tp cc tp con ca X c k hiu l P(X). D thy |P(X)| = 2n. Cc tp con ca mt tp hp l mt i tng xut hin kh nhiu trong cc bi ton m. Chnh hp: Chnh hp chp k ca mt tp hp l mt b k phn t phn bit c sp th t ca tp hp y. V d nu X = {1, 2, 3} v k = 2 th ta c cc chnh hp l (1, 2), (1,3), (2, 1), (2, 3),(3, 1),(3, 2). S cc chnh hp chp k ca n phn t c k hiu l .
Hon v: Hon v ca n phn t l chnh hp chp n ca n phn t , ni cch khc, l mt cch sp th t cc phn t . Hon v ca X cn c th nh ngha nh mt song nh t X vo X. S cc hon v ca n phn t c k hiu l Pn.T hp: T hp chp k ca mt tp hp l mt b k phn t phn bit khng sp th t ca tp hp y. Ni cch khc, l mt tp con k phn t. V d nu X = {1, 2, 3} v k = 2 th ta c cc t hp l {1, 2}, {1,3}, {2, 3}. S cc t chp k ca n phn t c k hiu l .
Chnh hp lp: Chnh hp lp chp k ca mt tp hp l mt b k phn t khng nht thit phn bit c sp th t ca tp hp y. V d nu X = {1, 2, 3} v k = 2 th ta c cc chnh hp lp l (1, 1), (1, 2), (1,3), (2, 1), (2, 2), (2, 3),(3, 1),(3, 2), (3, 3). S cc chnh hp lp chp k ca n phn t c k hiu l .
T hp lp: T hp lp chp k ca mt tp hp l mt b k phn t khng nht thit phn bit khng sp th t ca tp hp y. V d nu X = {1, 2, 3} v k = 2 th ta c cc t hp lp l {1, 1}, {1, 2}, {1,3}, {2, 2}, {2, 3}, {3, 3}. S cc t hp lp chp k ca n phn t c k hiu l .
tin li, ta thng ly X = {1, 2, ...,n} v ta k hiu tp ny l [n].
1. a) Dng quy tc nhn, hy chng minh rng .b) Chng minh rng .2. Dng nh ngha t hp ca hy chng minh cc ng thc sau:
a)
Hng dn: Chia cc tp con k phn t ca [n+1] thnh 2 loi: cha n+1 v khng cha n+1.
b)
Hng dn: Hy tr li cu hi: C bao nhiu tp con k phn t ca [n]. V tng cng [n] c bao nhiu tp con k c ( v chnh n?
c) (Cng thc nh thc Newton) .
Hng dn: C th chng minh bng quy np da vo a) hoc chng minh trc tip bng cch xt (x+y)n = (x+y)(x+y)...(x+y). to ra mt n thc xn-kyk, ta phi ly x t n-k du ngoc v y t k du ngoc cn li. C bao nhiu cch ly nh vy?
d) (Quy tc lc gic)
3. Hon v lp v nh l a thc
a) Bng ch ci c k k t 1, 2, ..., k. Ch ci th i c ri phin bn. Bit r1 + r2 + ...+ rk = n. Hi c bao nhiu t khc nhau c di n?
b) Chng minh rng
, trong
. 4. Cho tp hp X c n phn t. C bao nhiu cch chn cc cp c th t (A, B) cc tp con ca X sao cho:
a) A ( B = (;
b) A ( B = X;
c) A ( B;
d)* A v B khng cha nhau. 5. Phng trnh x1 + x2 + x3 = 100 c bao nhiu nghim nguyn khng m?6. B bi c 52 qun, trong c 13 gi tr: 2, 3, 4,..., 10, J, Q, K, A vi 4 cht: c, r, chun (tp), bch. C, r mu , chun, bch mu en. Chn ra 5 qun t b bi. Ta bit rng c cch chn nh vy. Hi trong cc cch chn , c bao nhiu cch chn trong :
a) Khng c qun bi c gi tr ging nhau;
b) C 3 qun bi gi tr ging nhau v hai qun bi khc ging nhau.
c) C 5 qun cng cht;
d) C 2 mu;
e) C 4 cht.
7*. Trong n gic li k tt c cc ng cho. Bit rng khng c ba ng cho no ng quy ti mt im. Hi a gic li c chia ra thnh bao nhiu phn? Cc ng cho ct nhau ti bao nhiu im?
Hng dn: Giao im ca hai ng cho xc nh mt cch duy nht bi 4 nh ca a gic. Mi lin h gia s phn ca a gic c chia ra v s giao im nh th no?
3. Phng php song nh
Nu tn ti song nh f: A --> B th | A | = | B |. Nguyn l n gin ny rt c ch trong cc bi ton m. Chng ta s thng xuyn gp tnh hung sau: m s phn t ca tp hp A, ta xy dng mt tp hp B c cu trc quen thuc (v c th m d dng) v thit lp mt song nh t A vo B, t | A | = | B |. 1. Xt cc tp hp A = {(x1, x2,..., xn) ( Nn | x1 + x2 +... + xn = k}
v B = { (y1, y2, ..., yn) ( N*n | 1 y1 < y2 < ...< yn-1 k + n - 1}. Xt tng ng
f(x1, x2, ...,xn) = (x1+1,x1+x2+2,...,x1+x2+...+xn-1+n-1)
a) Chng minh rng f l mt nh x t A vo B;
b) Chng minh rng f l song nh;
c) Kim tra li rng
d) T suy ra
Kt qu bi ton trn c gi l bi ton chia ko Euler:
nh l: S nghim nguyn khng m ca phng trnh x1 + x2 + ...+ xn = k bng Nhiu bi ton m c th m hnh ha a v bi ton ny. Ch khi s dng, cn chng minh li nh mt b . 2. a) (S ng i ngn nht trn li nguyn) Chng minh rng s ng i ngn nht trn li nguyn t im A(0; 0) n im B(m, n) bng
b) Cho m n, tm s ng i ngn nht t im A(0, 0) n im B(m, n) v i qua cc im c honh khng nh hn tung .
Hng dn: Hy chng minh rng s ng i ngn nht t im A(0, 0) n im B(m, n) v i qua t nht mt im c honh nh hn tung bng s ng i ngn nht t im (-1, 1) n B. c) (Bi ton v s Catalan) C 2n ngi xp hng mua v. Gi v l 50.000, c n ngi c tin 50.000 v n ngi ch c tin 100.000, trong quy ban u khng c tin l. Mi ngi vo mua v theo mt th t ngu nhin. Tnh xc sut tt c mi ngi u c th mua v m khng phi ch ly tin tr li. Nu trong quy c sn k t tin 5.000 th sao? 3. a) C n ngi xp thnh mt hng dc. C bao nhiu cch chn ra k ngi, sao cho khng c hai ngi k nhau c chn?b) C n ngi xp thnh mt vng trn. C bao nhiu cch chn ra k ngi, sao cho khng c hai ngi k nhau c chn?
4. (VMO 2012) Cho mt nhm gm 5 c gi, k hiu l G1, G2, G3, G4, G5 v 12 chng trai. C 17 chic gh c xp thnh mt hng ngang. Ngi ta xp nhm ngi cho ngi vo cc chic gh sao cho cc iu kin sau c ng thi tha mn:1/ Mi gh c ng mt ngi ngi;2/ Th t ngi ca cc c gi, xt t tri qua phi, l G1, G2, G3, G4, G5; 3/ Gia G1 v G2 c t nht 3 chng trai;4/ Gia G4 v G5 c t nht 1 chng trai v nhiu nht 4 chng trai.
Hi c tt c bao nhiu cch xp nh vy?
(Hai cch xp c coi l khc nhau nu tn ti mt chic gh m ngi ngi chic gh trong hai cch xp l khc nhau).5*. (MOP 2006) Cho cc s nguyn dng n v d vi d | n. Gi S l tp hp cc b n s 0 x1 x2 ... xn n sao cho d | x1 + x2 + ... +xn. Chng minh rng ng mt na s phn t ca S c tnh cht xn = n.6. (Ngh An 2009) Cho n l s nguyn dng ln hn hay bng 2. K hiu A = {1, 2, , n}. Tp con B ca tp A c gi l 1 tp "tt" nu B khc rng v trung bnh cng ca cc phn t ca B l 1 s nguyn. Gi Tn l s cc tp tt ca tp A. Chng minh rng Tn n l 1 s chn.
Hng dn: C n tp con tt c 1 phn t. Chia cc tp con tt cn li thnh 2 loi, loi 1 l cc tp tt c cha trung bnh cng, loi 2 l cc tp tt khng cha trung bnh cng. Hy chng minh 2 loi ny c s phn t bng nhau. 7. (M, 1996) Gi an l s cc xu nh phn di n khng cha chui con 010, bn l s cc xu nh phn di n khng cha chui con 0011 hoc 1100. Chng minh rng bn+1 = 2an vi mi n nguyn dng. 4. Cng thc bao hm v loi tr
Khi ta cn tm s cc phn t ca mt tp hp X tha mn mt trong cc tnh cht P1, P2, ..., Pk ta c th t Ai = {x ( X| x tha mn tnh cht Pi} v tnh . tnh s phn t ca hp ny, ta cn n cng thc bao hm v loi tr.
1. a) Cho A, B l hai tp hp bt k, chng minh rng |A ( B| = | A | +| B | - |A ( B|.
b) Cho A, B, C l ba tp hp bt k, chng minh rng |A ( B ( C| = |A| + |B| + |C| - (|A(B| + |B(C| + |C(A|) + |A(B(C|. c) (Cng thc bao hm v loi tr) Cho A1, A2, ..., An l cc tp hp bt k, khi ta c
Hng dn: Chng minh bng quy np hoc bng cch s dng hm c trng.
2. Trong 100 s nguyn dng u tin c bao nhiu s
a) Hoc chia ht cho 2, hoc chia ht cho 3, hoc chia ht cho 5?
b) Chia ht cho ng 2 trong 3 s 2, 3, 5?3. Mt lp hc c 20 hc sinh. C gio mun t chc 4 chuyn du kho cho hc sinh sao cho
a) Mt hc sinh tham d t nht mt chuyn du kho;
b) Hai chuyn du kho bt k c t nht mt thnh vin chung.
Hi c bao nhiu cch t chc cc chuyn du kho nh vy?Hng dn: Hy cho cc hc sinh ng k tham gia cc chuyn i. Mi hc sinh c bao nhiu cch ng k?
4. (Bi ton v v hnh phc) V xe but c dng a1a2a3a4a5a6 trong a1, a2, a3, a4, a5, a6 l cc ch s thuc E = {0, 1, 2, , 9}. V a1a2a3a4a5a6 c gi l v hnh phc nu nh a1 + a2 + a3 = a4 + a5 + a6. Hy tm s v hnh phc trong cc v t 000000 n 999999 theo s sau:
a) Chng minh rng s nghim ca phng trnh
a1 + a2 + a3 = a4 + a5 + a6
(a1, a2, a3, a4, a5, a6) ( E6(1)
bng s nghim ca phng trnh
a1 + a2 + a3 + a4 + a5 + a6 = 27 (a1, a2, a3, a4, a5, a6) ( E6 (2)
b) Chng minh rng s nghim ca phng trnh (2) bng s nghim ca phng trnh
a1 + a2 + a3 + a4 + a5 + a6 = 27 (a1, a2, a3, a4, a5, a6) ( N6tr i s phn t ca , trong
Mi = { (a1, a2, a3, a4, a5, a6) ( N6, a1 + a2 + a3 + a4 + a5 + a6 = 27, ai ( 10}
c) Dng cng thc bao hm v loi tr v bi ton chia ko Euler, hy tm s v hnh phc trong cc v t 000000 n 999999.5. a) Cho E = {1, 2, ...,n}. C bao nhiu song nh f: E --> E tha mn iu kin f(i) ( i vi mi i ( E.
b) Cho bng vung T gm n n , B l mt bng con ca T (mt tp con cc ca bng). Gi rk(B) l s cch t k qun xe i mt khng n nhau ln B. Chng minh rng s cch t n qun xe i mt khng n nhau ln T \ B (tc l khng c t vo cc ca B) c th tnh theo cng thc
trong ta quy c r0(B) = 1.
6. Chng minh rng s cc ton nh t mt tp hp c m phn t vo mt tp hp c n phn t c th tnh c tnh theo cng thc
.7. a) Trn mt phng cho n hnh. Gi l din tch phn giao ca cc hnh vi ch s i1,..., ik, cn S l din tch phn mt phng c ph bi cc hnh trn; Mk l tng tt c cc s . Chng minh rng
i) S = M1 - M2 + M3 - ...+ (-1)n+1Mn;
ii) S M1 - M2 + M3 - ... + (-1)m+1Mm vi m chn v
iii) S M1 - M2 + M3 - ... + (-1)m+1Mm vi m l.
b) Trong hnh ch nht din tch 1 c 5 hnh c din tch 1/2 mi hnh. Chng minh rng tm c
i) hai hnh c din tch phn chung khng nh hn 3/20;
ii) hai hnh c din tch phn chung khng nh hn 1/5;
iii) ba hnh c din tch phn chung khng nh hn 1/20.
5. Xy dng cng thc truy hi
Mt trong cc k thut quan trng gii quyt cc bi ton m l chia bi ton thnh cc bi ton nh hn, gii cc bi ton nh ri kt hp li. K thut nh th c gi l chia tr. V k thut ny c th s dng thit lp cc h thc truy hi: gii bi ton m vi tham s n, ta chia bi ton thnh nhng bi ton nh hn vi nh hng l cc bi ton nh ny lin quan n bi ton ban u vi tham s nh hn.
Trong mt s trng hp, ta c th t thm cc bi ton ph to ra cc dy s truy hi ln nhau.
Ch , cc h thc truy hi thng s c dng phng trnh sai phn tuyn tnh h s hng m ta bit cch gii.1. a) Tm s cc xu nh phn di n khng cha hai bit 1 k nhau;b) C bao nhiu cch lt ng i kch thc 2 n bng cc vin gch kch thc 1 2 (vin gch c th xoay).
c) Cng cu hi nh trn vi ng i kch thc 3 2n.
2. (PTNK 2009) Cho s nguyn dng n. C bao nhiu s chia ht cho 3, c n ch s v cc ch s u thuc {3, 4, 5, 6}?3. C bao nhiu tp con khc rng ca {1, 2, 3, ..., 2012} c tng cc phn t chia ht cho 3? 4. Tm s cc dy s (x1, x2, , x2012) tha mn iu kin: xi ( {1, 2, 3}, x1 = x2012 = 1, xi+1 xi vi mi i = 1, 2, , 2011.5. C 2n ngi xp thnh 2 hng dc. Hi c bao nhiu cch chn ra mt s ngi (t nht 1) t 2n ngi ny, sao cho khng c hai ngi no ng k nhau c chn. Hai ngi ng k nhau l hai ngi c s th t lin tip trong mt hng dc hoc c cng s th t hai hng.
6. Tm s tt c cc b n s (x1, x2, , xn) sao cho
(i) xi = ( 1 vi i = 1, 2, , n.
(ii) 0 ( x1 + x2 + + xr < 4 vi r = 1, 2, , n-1;
(iii) x1 + x2 + + xn = 4.
7. C bao nhiu s nguyn n, 0 n < 1011 c tng cc ch s chia ht cho 11?6. a thc v ng dng trong bi ton m
Mt tnh cht rt n gin ca a thc l li c nhng ng dng rt hiu qu a mt s bi ton m s nghim ca phng trnh tuyn tnh v bi ton tm h s ca xn trong khai trin ca mt a thc.
1. (Bi ton m u) Chng minh rng s nghim ca phng trnh
a1 + a2 + a3 + a4 + a5 + a6 = 27
vi ai l cc s nguyn 0 ai 9 bng h s ca x27 trong khai trin ca a thc
(1+x+x2+...+x9)62. a) S dng ng thc ng vi mi x c | x | < 1, hy chng minh rng
b) p dng kt qu bi 1 v cu a), hy tm ra mt li gii khc cho bi ton 4.4.
Mt s bi ton m s quy v vic tnh tng cc h s ca cc ly tha l bi ca mt s t nhin n. Trong nhng trng hp nh th, cn bc n ca n v s rt hu dng.S phc (( c gi l cn bc n ca n v nu (n = 1. Ta c tnh cht n gin nhng hu ch sau:Nu ( ( 1 l cn bc n ca n v th 1 + ( + (2 + ...+ (n-1 = 0.
3. a) Cho a thc . Tng c th tnh theo cng thc sau.
.
trong l cn nguyn thy bc n ca n v.
b) Chng minh rng s cc s c n ch s lp t cc ch s {3, 4, 5, 6} v c tng chia ht cho 3 bng tng cc h s a3k trong khai trin ca (x3 + x4+ x5 + x6)n. T a ra mt li gii khc cho bi ton 5.2.4*. (IMO 1995) Cho p l mt s nguyn t l. Tm s cc tp con A ca tp hp {1, 2,, 2p}, bit rng
(i) A cha ng phn t;
(ii) tng cc phn t ca A chia ht cho p.
Hng dn: Xt a thc Q(x) = (x-()(x-(2)...(x-(2p) vi ( l nghim ty ca phng trnh xp-1 +xp-2 + ...+ x + 1 = 0. 5. Vi mi tp hp A = {x1, x2, ..., xn} cc s thc, ta gi A(2) l b cc tng xi + xj vi i 1 xt S = {1, 2, 3, , n}. T cc s ca S bng 2 mu, u s mu v v s mu xanh. Hy tm s cc b (x, y, z) thuc S3 sao cho
a) x, y, z c t cng mu;
b) x + y + z chia ht cho n.
Trong bi tp 4.5, ta thy rng s cch t n qun xe i mt khng n nhau ln T \ B s hon ton c xc nh nu ta tnh c rk(B). Bi tp di y cho chng ta phng php tnh rk(B) thng qua khi nim v tnh cht ca a thc xe.Vi rk(B) c nh ngha trong bi 4.5, ta t . 7. (Tnh cht c bn ca a thc xe)a) Cho B v C l hai bng con "khng n nhau" ca T (bng vung n n) (tc l mt con xe nm mt bt k ca B v mt con xe nm mt bt k ca C khng n nhau), khi ta c
rB(C(x) = rB(x).rC(x).
b) Cho B l mt bng con ca T, x l mt thuc B, C = B \ {x} v D l bng thu c t B bng cch xa i dng cha x v ct cha x. Khi ta c
rB(x) = rC(x) + xrD(x).
c) p dng tm s cch t 8 qun xe i mt khng n nhau ln bn c 8 8, trong khng c t xe ln hai ng cho.
7. Quy np trong cc bi ton t hp
Quy np l mt phng php suy lun quan trng trong gii ton. mc 5, chng ta s dng tng quy np trong vic xy dng cc cng thc truy hi gii bi ton m. Di y, chng ta s lm quen vi mt s ng dng ca quy np trong vic chng minh cc nh l, tnh cht, d on cc chin thut trong tr chi.
1. Cho S = {(x, y) ( Z2 | 0 ( x ( m, 0 ( y ( n, x + y > 0}. Chng minh rng ph tt c cc im ca S bng cc ng thng khng i qua gc ta , ta cn t nht m + n ng thng. 2. a) (nh l Mantel Turan) Chng minh rng th n bc n khng cha tam gic c khng qu nh.
b) Trong mt giai bong a co 20 i tham gia, thi u vong tron mt lt (kt thuc giai mi i a vi mi i con lai ung mt trn). Tim s k ln nht sao cho sau mi k vong u (mi i u k trn) lun tim c 3 i i mt cha a vi nhau.
3. a) Chng minh rng bng vung 2n 2n khuyt mt bt k lun c th ph kn c bng cc qun trimino hnh ch L.b)* Chng minh rng nu n ( 5 l s nguyn dng khng chia ht cho 3 th bng vung n n khuyt mt bt k c th ph kn c bng cc qun trimino hnh ch L.
Hng dn: Hy quy np nhy cch, t n --> n+6.
4. Mi mt con ng Sikinia u l mt chiu. Mi mt cp thnh ph c ni bi ng mt con ng trc tip. Chng minh rng tn ti mt thnh ph m t mi thnh ph khc ta c th n thnh ph bng con ng trc tip, hoc i qua nhiu nht mt thnh ph khc.Ghi ch: Trn ngn ng th, c th pht biu bi ton nh sau - Chng minh rng trong mt th c hng y , tn ti mt nh m khong cch t 1 nh bt k khc n n 2. Pht biu mt cch khc na: C n i bng chuyn thi u vng trn mt lt. Khi tn ti mt i bng A sao cho nu A thng B th tn ti C sao cho C thng A v thua B.
Cc bi ton tr chi chnh l dng ton s dng n quy np ton hc nhiu nht. Ch l quy np ton hc y bao gm hai phn: d on cng thc v chng minh cng thc v trong rt nhiu trng hp, vic d on cng thc ng vai tr then cht.
5. Hai ngi A v B cng chi mt tr chi. Ban u trn bn c 100 vin ko. Hai ngi thay phin nhau bc ko, mi ln c bc k vin vi k ( {1, 2, 6} . Hi ai l ngi c chin thut thng, ngi i trc hay ngi i sau?
6. a) Trn bng c s 2010. Hai ngi A v B cng lun phin thc hin tr chi sau: Mi ln thc hin, cho php xo i s N ang c trn bng v thay bng N-1 hoc [N/2]. Ai thu c s 0 trc l thng cuc. Hi ai l ngi c chin thut thng, ngi i trc hay ngi i sau.
b) Cng cu hi vi lut chi thay i nh sau: Mi ln thc hin, cho php xo i s N ang c trn bng v thay bng N-1 hoc [(N+1)/2].
7. An v Bnh chi tr on s. An ngh ra mt s no nm trong tp hp X = {1, 2, , 144}. Bnh c th chn ra mt tp con bt k A ca X v hi S ca bn ngh c nm trong A hay khng?. An s tr li C hoc Khng theo ng s tht. Nu An tr li c th Bnh phi tr cho An 2.000 ng, nu An tr li Khng th Bnh phi tr cho An 1.000 ng. Hi Bnh phi tt t nht bao nhiu tin chc chn tm ra c s m An ngh?
8. Phn chng trong cc bi ton t hp
Phng php phn chng c 4 s c bn nh sau
1)
(chng minh bng mu thun)
2)
(chng minh bng mnh phn o)
3)
(s 1 p dng cho mnh A --> B)
4)
(chng minh bng mu thun vi gi thit)
Php phn chng c bit hiu qu trong cc bi ton t hp, v cc bi ton ny thng c cu hnh lng, pht sinh nhiu trng hp, r nhnh. Phn chng gip chng ta gim bt ti a cc r nhnh.
1. Chng minh rng t 8 s nguyn dng tuy y khng ln hn 20, lun chon c 3 s x, y, z la dai 3 canh cua mt tam giac.
2. (IMO 1983) Cac im trn chu vi tam giac u ABC c t bng mt trong hai mau xanh va o. Chng minh rng tn tai mt tam giac vung co cac inh c t cung mau.
Trong vic chng minh mt s tnh cht bng phng php phn chng, ta c th c thm mt s thng tin b sung quan trng nu s dng phn v d nh nht. tng l chng minh mt tnh cht A cho mt cu hnh P, ta xt mt c trng f(P) ca P l mt hm c gi tr nguyn dng. By gi gi s tn ti mt cu hnh P khng c tnh cht A, khi s tn ti mt cu hnh P0 khng c tnh cht A vi f(P0) nh nht. Ta s tm cch suy ra iu mu thun. Lc ny, ngoi vic chng ta c cu hnh P0 khng c tnh cht A, ta cn c mi cu hnh P vi f(P) < f(P0) u c tnh cht A. 3. Cho ng gic li ABCDE trn mt phng to c to cc nh u nguyn.
a) Chng minh rng tn ti t nht 1 im nm trong hoc nm trn cnh ca ng gic (khc vi A, B, C, D, E) c to nguyn.
b) Chng minh rng tn ti t nht 1 im nm trong ng gic c to nguyn.
c) Cc ng cho ca ng gic li ct nhau to ra mt ng gic li nh A1B1C1D1E1 bn trong. Chng minh rng tn ti t nht 1 im nm trong hoc trn bin ng gic li A1B1C1D1E1.
4. Trn mt phng nh du mt s im. Bit rng 4 im bt k trong chng l nh ca mt t gic li. Chng minh rng tt c cc im c nh du l nh ca mt a gic li. Phng php phn chng thng c kt hp hiu qu vi cc nguyn l chng minh khc nh nguyn l bt bin, nguyn l Dirichlet, chng minh bng t mu, m bng hai cch. 5. Xt hnh vung 7 ( 7 . Chng minh rng ta c th xo i mt phn cn li khng th ph kn bng 15 qun trimino kch thc 1 ( 3 v 1 qun trimino hnh ch L. Hng dn: T mu!
6. Hnh trn c bi 5 ng knh thnh thnh 10 bng nhau. Ban u trong mi c 1 vin bi. Mi ln thc hin, cho php chn 2 vin bi bt k v di chuyn chng sang bn cnh, 1 vin theo chiu kim ng h v 1 vin ngc chiu kim ng h. Hi sau mt s hu hn ln thc hin, ta c th chuyn tt c cc vin bi v cng 1 c khng? Hng dn: T mu v bt bin!
7. Ta vit vo cc ca bng 10 ( 10 cc ch s 0, 1, 2, 3, ..., 9 sao cho mi ch s xut hin 10 ln.
a) Tn ti hay khng mt cch vit m trong mi hng v mi ct xut hin khng qu 4 ch s khc nhau?
b) Chng minh rng tn ti mt dng hoc mt ct trong c t nht 4 ch s khc nhau. Hng dn: Phn chng kt hp vi m bng hai cch. Gi an l s dng cha n v bn l s ct cha n th an.bn 10. 9. Nguyn l DirichletNguyn l chung v th (hay cn c gi l nguyn l Dirichlet) khng nh mt s kin hin nhin rng n+1 con th khng th c xp vo n chung sao cho mi con th u ring mt chung. Mt cch tng qut hn, nguyn l chung v th khng nh rng:
Nu mt tp hp gm nhiu hn kn i tng c chia thnh n nhm, th c mt nhm no c nhiu hn k i tng.
Chn l ny rt d kim tra: nu nhm no cng c nhiu nht k i tng th tng cng ch c nhiu nht kn i tng c chia ra.
y l mt trong nhng nguyn l khng xy dng (non-constructive) lu i nht: n ch ni n s tn ti ca mt chung trong c nhiu hn k vt m khng ni g n cch tm ra chung ny. Ngy nay chng ta c nhng tng qut ha rt mnh ca nguyn l ny (cc nh l kiu Ramsey, phng php xc sut).
Mc d nguyn l chung v th c pht biu rt n gin, n c hng lot cc ng dng khng tm thng. Ci kh ca vic ng dng nguyn l ny l xc nh c xem th l g v chung l g.1. Trong mt gii bng chuyn c 8 i tham gia, thi u vng trn 1 lt. Chng minh rng tm c 4 i A, B, C, D sao cho A thng B, C, D, B thng C, D v C thng D.
2. a) Chng minh rng trong 6 ngi bt k lun c 3 ngi i mt quen nhau hoc 3 ngi i mt khng quen nhau. Chng minh iu ny ni chung khng ng vi 5 ngi.b) Chng minh rng trong 9 ngi bt k lun tm c 3 ngi i mt quen nhau hoc 4 ngi i mt khng quen nhau. Chng minh iu ny ni chung khng ng vi 8 ngi.
Vi mi cp s nguyn dng (m, n), ta chng minh c tn ti s R(m,n) nh nht sao cho trong R(m,n) ngi bt k, lun tm c m ngi i mt quen nhau hoc n ngi i mt khng quen nhau. S ny gi l s Ramsey. Theo hai bi tp trn th R(3, 3) = 6, R(3, 4) = 9.
c) Chng minh rng R(3, 5) = 14, R(4, 4) = 18, R(3, 6) = 18.
d) (Erdos) Chng minh rng R(r, s) R(r-1,s) + R(r, s-1), t suy ra .3. (IMO 1972) Chng minh rng t 10 s c hai ch s, ta lun c th chn c hai tp con khc rng ri nhau c tng cc phn t bng nhau. Khi ng dng Nguyn l Dirichlet trong cc bi ton c yu t hnh hc, ta thng xy dng chung bng cch chia hnh ra thnh cc thnh phn nh hn.
4. a) Chng minh rng trong 6 im bt k nm trong hnh ch nht 3 4, lun tm c 2 im c khong cch khng ln hn b) Cho 9 im nm trong hnh vung cnh 1. Chng minh rng tn ti mt tam gic c nh ti cc im cho c din tch khng vt qu 1/8.
c) Vi i = 1, 2, ..., 7, ta c ai, bi l cc s khng m sao cho ai + bi 2. Chng minh rng tn ti hai ch s i ( j sao cho |ai - aj | + |bi - bj| 1.d) (VMO 2011) Cho ng gic li ABCDE c di mi cnh v di cc ng cho AC, AD khng vt qu . Ly 2011 im phn bit ty nm trong ng gic . Chng minh rng tn ti mt hnh trn n v c tm nm trn cnh ca ng gic cho cha t nht 403 im trong s cc im ly.
e) Chng minh rng trong bi ton trn, kt qu vn ng nu thay 403 bng 503.Nguyn l Dirichlet cn c mt dng khc: Nu A, B l cc tp con ca X v |A| + |B| > |X| th A v B c phn t chung.
5. a) Chng minh rng t n s nguyn bt k lun tm c mt s hoc mt s s c tng chia ht cho n.
b) Chng minh rng t 9 s nguyn bt k lun tm c 5 s c tng chia ht cho 5.
c) Chng minh rng trong 70 s nguyn dng khng ln hn 200, lun tm c hai s c hiu bng 4, 5 hoc 9. d) Chn ra 69 s nguyn dng t tp hp E = {1, 2, , 100}. Chng minh rng tn ti 4 s a < b < c < d trong 4 s c chn sao cho a + b + c = d. Kt lun bi ton cn ng khng nu ta thay 69 bng 68?
6*. Trn bn c quc t c 8 qun xe, i mt khng n nhau. Chng minh rng trong cc khong cch i mt gia cc qun xe, c hai khong cch bng nhau. Khong cch gia hai qun xe bng khong cch gia tm cc vung m qun cc qun xe ng.
7. (nh l Erdos-Szekeres) Cho A = (a1, a2,, an) l dy gm n s thc phn bit. Nu n ( rs + 1 th hoc A c dy con tng di s+1 hoc A c dy con gim di r+1 (hay c hai).
10. Nguyn l cc hnMt tp hp hu hn cc s thc lun c phn t ln nht v phn t nh nht. Mt tp con bt k ca N lun c phn t nh nht. Nguyn l n gin ny trong nhiu trng hp rt c ch cho vic chng minh. Hy xt trng hp bin! l khu quyt ca nguyn l ny.1. Cho n im xanh v n im trn mt phng, trong khng c 3 im no thng hng. Chng minh rng ta c th ni 2n im ny bng n on thng c u mt khc mu sao cho chng i mt khng giao nhau.
2. Co 3 trng hoc, mi trng co n hoc sinh. Mi mt hoc sinh quen vi it nht n+1 hoc sinh t hai trng khac. Chng minh rng ngi ta co th chon ra t mi trng mt ban sao cho ba hoc sinh c chon i mt quen nhau.
Nguyn ly cc han co th c ng dung chng minh mt qua trinh la dng (trong bai toan lin quan n bin i trang thai), trong bai toan v thi, hay trong cac tinh hung t hp a dang khac. Cac i tng thng c em ra xet cc han thng la: oan thng ngn nht, tam giac co din tich ln nht, goc ln nht, inh co bc ln nht, chu trinh co dai ngn nht
3. (inh ly Sylvester) Cho tp hp S gm hu han cac im trn mt phng thoa man tinh cht sau: Mt ng thng i qua 2 im thuc S u i qua it nht mt im th ba thuc S. Khi o tt ca cac im cua S nm trn mt ng thng.
4. (Trn u ton hc Nga 2010) Mt quc gia c 210 thnh ph. Ban u gia cc thnh ph cha c ng. Ngi ta mun xy dng mt s con ng mt chiu ni gia cc thnh ph sao cho: Nu c ng i t A n B v t B n C th khng c ng i t A n C. Hi c th xy dng c nhiu nht bao nhiu ng?
5. Trong quc hi M, mi mt ngh s c khng qu 3 k th. Chng minh rng c th chia quc hi thnh 2 vin sao cho trong mi vin, mi mt ngh s c khng qu mt k th.6. 2n+1 ngi tham gia tr chi bn sng sn trn mt cnh ng. Bit rng khong cch gia h i mt khc nhau v mi ngi bn vo ngi gn mnh nht. Chng minh rnga) C t nht mt ngi khng b bn;
b) Khng ai b bn qu 5 pht;
c) ng i ca cc vin n khng ct nhau;
d) Tp hp cc on thng to bi ng i cc vin n khng cha mt min ng.
7. Trong mt phng cho 100 im, trong o khng co ba im nao thng hang. Bit rng ba im bt ky trong chung tao thanh mt tam giac co din tich khng ln hn 1. Chng minh rng ta co th phu tt ca cac im a cho bng mt tam giac co din tich 4.
11. Nguyn l m bng hai cchK thut m bng hai cch l mt k thut kh thng dng trong ton hc da trn nguyn l c bn: mt i lng lun c nhiu cch tnh khc nhau, ty theo cch ta nhn nhn i tng. V iu quan trng l tt c cc cch tnh u cho ra mt kt qu nh nhau. Nh vo iu ny m ta c th thit lp ra cc mi lin h, cc phng trnh, cc bt ng thc, chng minh cc hng ng thc.1. Ti mt hi ngh c 100 i biu. Trong s c 15 ngi Php, mi ngi quen vi t nht 70 i biu v 85 ngi c, mi ngi quen vi khng qu 10 i biu. H c phn vo 21 phng. Chng minh rng c mt phng no khng cha mt cp no quen nhau.
Bn cht t hp ca chnh l s cch chn ra k phn t (khng sp th t) t mt tp hp gm n phn t, hay ni cch khc s tp con k phn t ca mt tp hp gm n phn t. Hiu r bn cht ny, chng ta c th chng minh hng lot cc cng thc cha bng cch gii cng mt bi ton bng hai cch khc nhau.2. Chng minh cc ng thc sau bng phng php t hp
a)
b)
c)
d)*
e)
f)* Trong t hp l s cc cp to thnh t n phn t, l s cnh ca th y bc n. Trong nhiu bi ton, s dng ngha t hp ny cng vi cch m bng hai cch gip chng ta tm ra cha kho cho li gii.3. (Bulgarian MO 2006) Mt quc gia c 16 thnh ph v c 36 tuyn bay ni gia chng. Chng minh rng ta c th t chc mt chuyn bay vng quanh gia 4 thnh ph.4. a) C 8 ci hp, mi ci hp c 6 vin bi thuc mt trong n mu. Bit rng khng c hai vin bi cng mu trong 1 hp v khng c hai mu xut hin trong qu 1 hp. Tm gi tr nh nht ca n.
b) Trong quc hi c n thnh vin. Ngi ta t chc 8 cuc hp (tip ni nhau), mi cuc hp c 6 ngi tham d. Bit rng 2 thnh vin bt k khng hp chung vi nhau qu 1 ln, tm GTNN ca n.c) Trong Duma quc gia c 1600 i biu, lp thnh 16000 y ban, mi y ban c 80 i biu. Chng minh rng c t nht hai y ban c khng di 4 thnh vin chung.L thuyt v phng php m bng hai cch c tp hp trong bi ton di y
5. (Ma trn lin thuc v ng dng)
a) Cho A = (aij) l ma trn r c vi Ri, i = 1, 2, ..., r l tng cc dng; Cj, j = 1, 2,..., c l tng cc ct. Khi ta c .
b) Cho F l h cc tp con ca X. Vi x thuc x, ta gi d(x) l s phn t ca F cha x. Khi ta c: .
c) Cho A = (aij) l ma trn (0, 1) r c vi Cj l tng cc ct. Gi s rng vi hai dng bt k, c ng t ct cha 1 trong c hai dng , khi ta c:
d) Cho A = (aij) l ma trn r c vi Ri, i = 1, 2, ..., r l tng cc dng; Cj, j = 1, 2,..., c l tng cc ct. Chng minh rng nu Ri > 0 vi mi i = 1, 2, ..., r th . Tng t nu Cj > 0 vi mi j = 1, 2, ..., c th .e) Cho A = (aij) l ma trn (0, 1) r c vi Ri, i = 1, 2, ..., r l tng cc dng; Cj, j = 1, 2,..., c l tng cc ct, Ri > 0 vi mi i = 1, 2, ..., r; Cj > 0 vi mi j = 1, 2, ..., c. Chng minh rng nu Cj Ri khi aij = 1 th r c.6. (IMC 2002) 200 hc sinh tham d mt cuc thi gii ton. H gii 6 bi ton. Bit rng mi mt bi ton c t nht 120 hc sinh gii c. Chng minh rng tn ti hai hc sinh m hp li gii ht c 6 bi ton. 7*. Cho X l tp hp vi | X | = n v A1, A2, , Am l cc tp con ca X sao cho
i) | Ai | = 3 vi mi i = 1, 2, , m.
ii) | Ai ( Aj | 1 vi mi i j.
Chng minh rng tn ti A thuc X, A cha t nht phn t sao cho A khng cha Ai vi mi i = 1, 2, , m.
12. Bt bin v n binNhiu bi ton tr chi hay bin i trng thi c gii quyt mt cch kh hiu qu nh khi nim bt bin, n bin.
Cho tp hp ( (tp hp cc trng thi) v tp hp T (tp hp cc php bin i) cc nh x t ( ( (. Hm s f: ( ( R c gi l bt bin i vi cp ((, T) nu ta c f(t(()) = f(() vi mi ( thuc ( v vi mi t thuc T.Nguyn l bt bin: Nu f l mt bt bin ca ((, T) v f(() ( f(() th ( khng th thu c t ( thng quan cc php bin i T.1. Xt mt bng vung 4 x 4 . Ti mi ca bng vung c cha du + hoc du -. Mi mt ln thc hin, cho php i du ca tt c cc trn cng mt hng hoc cng mt ct. a) Gi s bng vung ban u c 1 du + v 15 du -. Hi c th a bng ban u v bng c ton du cng c khng?b) Gi s bng vung c ton du "+". Hi c th a bng ban u v bng c 2 du - c khng?
c) Gi s bng vung c ton du "+". Vi nhng gi tr no ca k th c th a bng ban u v bng c k du -?
2. Trn bng c cc s 1/96, 2/96, 3/96, , 96/96. Mi mt ln thc hin, cho php xo i hai s a, b bt k trn bng v thay bng a + b 2ab. Hi sau 95 ln thc hin php xo, s cn li trn bng l s no? Bt bin f i vi cp ((, T) c gi l bt bin ton nng nu:
Trng thi (f c th a v t trng thi (s bng cc php bin i T khi v ch khi f((f) = f((s).
Bt bin ton nng s gip chng ta gii quyt trn vn bi ton chuyn c. Tuy nhin, vic xy dng mt bt bin nh vy khng n gin. Trong nhiu trng hp, s d dng hn khi chng ta xt n mt h bt bin ton nng.
nh ngha. H cc bt bin (f1, f2, , fk) i vi cp ((, T) c gi l h bt bin ton nng nu: Trng thi (f c th a v t trng thi (s bng cc php bin i T khi v ch khi fi((f) = fi((s) vi mi i = 1, 2, , k.
Mt khi nim quan trng khc c nhiu ng dng trong vic nghin cu cc bt bin, l khi nim qu o. Trn (, ta a ra quan h chuyn c nh sau: Ta ni trng thi (s c th chuyn c v trng thi (f bng cc php bin i T nu tn ti mt dy cc php bin i t1, t2, , tm thuc T sao cho
(f = tk(tk-1((t1((s)))
Khi ta vit (s (T (f.
Trong nhiu trng hp, quan h (T c tnh phn x ((s c th a v (s bng cch khng lm g c), bc cu (thc hin php hp cc php bin i) v i xng (nu cc php bin i T l kh nghch). Trong trng hp (T l mt quan h tng ng v ta s vit (s ~T (f thay v (s (T (f. Vi quan h tng ng ny, ( s c chia thnh cc lp tng ng, c i din l (1, (2, , (p, k hiu l (i = { ( ( (| ( ~T (i }. Ta gi (i l cc qu o sinh bi (i. D thy hai qu o bt k hoc trng nhau, hoc khng giao nhau.
3. Xt tip bi ton 1.a) Tp trng thi c bao nhiu phn t? C bao nhiu qu o?
b) Hy tm mt h bt bin ton nng.
4. Hnh trn c chia thnh n . Trn mi c mt vin si. Mi mt bc i cho php chn hai vin si v chuyn sang bn cnh, mt vin chuyn theo chiu kim ng h, mt vin chuyn ngc chiu kim ng h. Ta mun bng nhng bc i nh vy chuyn tt c cc vin si v cng mt .
a) Chng minh rng nu n l th iu ny lun thc hin c.
b) Chng minh rng nu n = 4k+2 th iu ny khng thc hin c.
c)* Nu n = 4k th c thc hin c khng?Ngc li vi bt bin, n bin l i lng lun tng (hoc khng gim, hoc lun gim, hoc khng tng) trong qu trnh bin i. n bin thng c dng chng minh mt qu trnh phi dng, hoc nh gi kt qu cui cng ca mt qu trnh.
5. (PTNK 2010) Xt s t nhin n ( 2. Bt u cc s 1, 2, , 2n-1, 2n ta thc hin cc php bin i nh sau: Chn 2 s a, b sao cho a b ( 2, xa hai s ny v thay bng hai s a 1, b + 1; vi b s thu c, ta li thc hin php bin i tng t, v c nh vy.
a) Chng minh rng sau mt s ln thc hin cc php bin i nh trn, ta phi t n trng thi dng, tc l khng th thc hin c mt php bin i no na.
b) Gi k l s php bin i cn thc hin t n trng thi dng. Hy tm gi tr ln nht v gi tr nh nht ca k.
6. a) Trn bng c n s 1. Mi ln thc hin cho php xa i hai s x, y bt k v thay bng . Sau n-1 ln thc hin, trn bng cn li s a. Chng minh rng
b)* Khi n = 10, hy tm gi tr nh nht ca a.
c) ( Nng 2012) Cho cc s 1, 2, 3,, n. Chng ta thc hin vic xa hai s bt k trn bng v thay bng s mi bng 2 ln tng ca hai s . C tip tc qu trnh nh vy cho n khi trn bng ch cn li mt s. S cui cng trn bng chnh l s ko m hc sinh nhn c. Chng minh rng s ko lun ln hn
7.* (IMO 1986) Ti mi nh ca mt ng gic li ta vit mt s nguyn sao cho tng ca tt c 5 s l dng. Nu 3 nh lin tip c ghi cc s x, y, z tng ng v y < 0 th ta c php thc hin bin i sau: cc s x, y, z c thay bng cc s x+y,y, z+y tng ng. Cc php bin i c lp li khi m c t nht mt trong 5 s l m. Xc nh xem qu trnh ny c nht thit dng li sau mt s hu hn cc bc hay khng? 13. Chng minh bng t mu
Nhiu bi ton v ph hnh c th gii quyt kh hiu qu bng cch t mu. S dng bao nhiu mu v t nh th no s ph thuc vo cc iu kin c th ca bi ton. Chng minh t mu thng kt hp tt vi cc phng php khc nh phn chng, m bng hai cch.
1. a) Chng minh rng bn c quc t 8 8 b i a1 v h8 khng th ph kn bng 31 qun -mi-n.
b) C bao nhiu cch chn 2 ca bn c quc t phn cn li c th ph kn bng 31 qun -mi-n?c)* Cng cu hi trn vi iu kin b sung: Hai cch chn c coi l ging nhau nu chng c th thu c t nhau qua mt php quay.2. a) Chng minh rng bn c 10 10 khng th ph kn bng 25 qun tetramino hnh ch T.b) Chng minh rng bn c 8 8 khng th ph kn bng 15 qun tetramino hnh ch T v 1 qun tetramino hnh vung.
c) Chng minh rng bn c 10 10 khng th ph kn bng 25 qun tetramino hnh ch I.3. Xt hnh vung 7 ( 7 . Chng minh rng ta c th xo i mt phn cn li khng th ph kn bng 15 qun trimino kch thc 1 ( 3 v 1 qun trimino hnh ch L.
4. nh ca mt ng gic li l cc im nguyn v cc cnh ca ng gic l cc s nguyn. Chng minh rng chu vi ng gic l s chn. Trong mt s bi ton, thay v t mu, ta c th dng cch in cc lu tha ca cn bc n ca n v vo cc cn bc ri tnh tng cc s c in bng hai cch suy ra cc kt lun. Tnh cht cn bn ta dng y l: Nu ( l cn nguyn thy bc n ca n v th 1 + ( + ... + (n-1 = 0.
5. a) Chng minh rng hnh ch nht a b ph knh c bng cc hnh ch nht 1 n khi v ch khi hoc n | a, hoc n | b.
b) Hnh vung 8 8 c ph bng 21 hnh ch nht 1 3. Hi trng cn li c th l no?6. Hnh vung 6 6 c ph bi cc qun -mi-n 2 1. Chng minh rng lun c th tm c 1 ng thng ct ngang hoc ct dc hnh vung m khng ct qua bt c mt -mi-n no.7*. (Vietnam TST 2010) Gi mi hnh ch nht kch thc 12 (hoc 21) l mt hnh ch nht n. Gi mi hnh ch nht kch thc 23 (hoc 32) b khuyt hai vung 11 hai gc i nhau l mt hnh ch nht khuyt.
Bit rng ngi ta c th ghp kht mt s hnh ch nht n v mt s hnh ch nht khuyt vi nhau to thnh mt hnh ch nht kch thc 20082010. Hi ghp c nh vy, cn phi dng t nht bao nhiu hnh ch nht n ? 14. Bi tp tng hp 11. (VMO 2001, bng B) Cho bng vung kch thc 2000 2001. Hy tm s nguyn dng k ln nht sao cho ta c th t mu k vung con ca bng tha mn iu kin: hai vung con no c t mu cng khng c nh chung. 2. (VMO 2002, bng B) Cho S l tp hp tt c cc s nguyn trong on [1, 2002]. Gi T l tp hp tt c cc tp con khc rng ca S. Vi mi tp hp X thuc T, k hiu m(X) l trung bnh cng tt c cc s thuc X. t , hy tm gi tr ca m. 3. (Vng Tu 2009) Trn bn c vua kch thc 8x8 c chia thnh 64 vung n v, ngi ta b i mt vung n v no v tr hng th m v ct th n. Gi S(m;n) l s hnh ch nht c to bi mt hay nhiu vung n v ca bn c sao cho khng c no trng vi v tr ca b xa b ban u. Tm gi tr nh nht v gi tr ln nht ca S(m;n).4. (Hng Yn 2012, vng chung kho) Trong mt phng cho n im A1, A2, ....., An (n4) sao cho khng c 3 im no thng hng v khng c 4 im no cng nm trn 1 ng trn. Gi at (1 t n) l s cc ng trn ngoi tip tam gic AiAjAk (1 i < j < k n) cha im At. t N = a1 + a2 + + an. Chng minh rng cc im A1, A2, ...., An l cc nh ca mt a gic li khi v ch khi .
5. Hai ngi A v B cng chi mt tr chi. Ban u trn bn c 100 vin ko. Hai ngi thay phin nhau bc ko, mi ln c bc k vin vi k ( {1, 2, 6} . Hi ai l ngi c chin thut thng, ngi i trc hay ngi i sau?
6. Co n i bong thi u vong tron 1 lt. Hay lp lich thi u gm n-1 vong u sao cho trong mi vong, mi i chi thi u nhiu nht 1 trn.
7. Cho 2n+1 may tinh. Hai may tinh bt ky c ni vi nhau bi mt si dy. Chng minh rng co th t cac may tinh va cac si dy bng 2n+1 mau sao cho:
i) Cac may tinh c t mau khac nhau
ii) Cac si dy xut phat t cung mt may tinh c t mau khac nhau
iii) Hai may tinh va si dy ni chung c t mau khac nhau.
8. (Vit Nam TST 2009) C 6n+4 nh ton hc tham d 1 hi ngh, trong c 2n+1 bui tho lun. Mi bui tho lun u c 1 bn trn cho 4 ngi ngi v n bn trn cho 6 ngi ngi. Bit rng 2 ngi bt k khng ngi cnh nhau hoc i din nhau qu 1 ln.a. Hi c th thc hin c khng vi n=1?b. Hi c th thc hin c khng vi n>1?9. Co ba lp hoc A, B, C, mi lp co 30 hoc sinh. Bit rng mt hoc sinh bt ky u quen vi it nht 31 hoc sinh khac lp. Chng minh rng tn tai ba hoc sinh a, b, c ln lt thuc lp A, B, C sao cho ho i mt quen nhau.10. Cho n la s nguyn dng. Ta noi s nguyn dng k thoa man iu kin Cn nu tn tai 2k s nguyn dng phn bit a1, b1, ..., ak, bk sao cho a1 + b1, ..., ak+bk cung phn bit va nho hn n.
(a) Chng minh rng nu k thoa man iu kin Cn thi k ( (2n-3)/5.
(b) Chng minh rng 5 thoa man iu kin C14.
(c) Gia s rng (2n-3)/5 nguyn. Chng minh rng (2n-3)/5 thoa man iu kin Cn.
11. Cho n l mt s t nhin 2. Xt tp hp W = {(x1, x2, , xn) | xi (( Z). Vi mi x = (x1, x2, , xn) ( W ta t
Ux = { y = (y1, y2, , yn) ( W | |x1 y1| + |x2 y2| + + |xn yn| 1} Chng minh rng tn ti mt tp con W0 ca W sao cho vi mi x thuc W ta c
| Ux ( W0 | = 1.
12. Trong mt nhm n ngi c 3 ngi i mt quen nhau v mi mt ngi ny quen nhiu hn 1 na s ngi trong nhm. Tm s t nht c th s b ba ngi i mt quen nhau.13. a) (AIME 1996) Hai ca hnh vung 7 ( 7 c t bng mu vng. Cc cn li c t bng mu . Hai cch t c coi l tng ng nhau nu chng c th thu c t nhau bng mt php quay trn mt phng ca hnh vung. m s cc cch t mu khng tng ng.
b) (VMO 2010) Cho bng 3 ( 3 v n l mt s nguyn dng cho trc. Tm s cc cch t mu khng nh nhau khi t mi bi 1 trong n mu.
Hai cch t mu gi la nh nhau nu 1 cch nhn c t cch kia bi 1 php quay quanh tm.
15. Bi tp tng hp 2
1. (i hc S phm 2010) Cho tp hp S = {1; 2; 3; ...; n}. Tm s cch chia tp S thnh ba tp con khc rng sao cho mi tp con khng cha hai s nguyn lin tip.
2. (ng Nai 2010) Tp hp cc s nguyn dng c t bi hai mu en v trng. Gi thit rng, tng ca hai s khc mu lun b t mu en v c v hn s b t mu trng.
Chng minh rng tng v tch ca hai s b t mu trng cng s b t mu trng.3. (H Tnh 2012) Trong k thi chn i tuyn hc sinh gii ton ca mt tnh c 20 em tham gia. Mi hc sinh phi thi 2 vng, mi vng c gi l mt bi thi. im ca mi bi thi c cho l mt s t nhin t 1 n 10. Phng thc chn i tuyn l so snh kt qu im ca tng bi thi tng ng (vng 1, vng 2 ) gia cc th sinh. Th sinh A gi l so snh c vi th sinh B nu im mi bi thi ca A khng nh hn im mi bi thi tng ng ca B. Bit rng khng c hai th sinh no c cng cp im s tng ng. Chng minh rng c th chn c ba th sinh A, B, C sao cho A so snh c vi B v B so snh c vi C.4. (Nga 2007, Bc Ninh 2009) Trong bng hnh vung gm 10 x 10 vung (10 hng, 10 ct), ngi ta vit vo cc vung cc s t nhin t 1 n 100 theo cch nh sau: hng th nht, t tri sang phi, vit cc s t 1 n 10; hng th hai, t tri sang phi, vit cc s t 11 n 20; c nh vy cho n ht hng th 10. Sau ct bng hnh vung thnh nhng hnh ch nht c 1 x 2 hoc 2 x 1. Tnh tch s ca hai s trong mi hnh ch nht ri cng 50 tch li. Cn phi ct hnh vung nh th no tng tm c nh nht ?
5. (Nht Bn 2007) Ta c 15 tm th c nh s 1, 2, , 15. C bao nhiu cch chn ra mt s (t nht 1) tm th sao cho tt c cc s vit trn cc tm th ny u ln hn hoc bng s tm th c chn.
6. Trn mt ng thng nm ngang, cho 2005 im c nh du trng hoc en. Vi mi im, xc nh tng tt c cc im trng bn phi v im en bn tri ca n. Bit rng, trong 2005 tng trn c ng mt s xut hin s l ln. Hy tm tt c cc gi tr c th c ca s ny.7. Sau khi khai trng c ng 10 ngy, mt nhn vin th vin cho bit
(1) Mi ngy c ng tm ngi n c sch;
(2) Khng c ngi no n th vin mt ngy qu mt ln ;
(3) Trong hai ngy bt k ca 10 ngy th c t nht l 15 ngi khc nhau cng n th vin.
Cn c ng thi c ba iu kin m nhn vin th vin cung cp hy cho bit s ngi ti thiu n th vin trong 10 ngy ni trn l bao nhiu?8. ( Nng 2012) C bao nhiu cch chia 13 vt i mt khc nhau cho 6 ngi, sao cho trong s c ng 2 ngi khng nhn c vt no?
9. Cho 51 s nguyn dng c hai ch s. Chng minh rng c th chn ra 6 s trong chng sao cho khng c 2 s no trong cc s c chn c cc ch s khc nhau bt k hng no. 10. Cho n im trn mt phng (n 3). Gi d l khong cch dng nh nht trong cc khong cch gia cc im ny. Chng minh rng s cc khong cch bng d khng vt qu 3n - 6. 11*. (Hi Phng 2010) Cho tp A = {1; 2; 3; ...; 2009}. Chng minh rng, c th t mu mi phn t ca tp A bng mt trong hai mu en trng sao cho mi cp s cng cng sai khc 0 gm 18 phn t ca A u c t bi c hai mu.12*. (PTNK 2012) Cho s nguyn dng n v tp hp X = {1, 2, , 4n}. Hai tp con A, B ca X c gi l khng ging nhau, nu | A ( B | 2n+1 ( y A ( B = (A \ B) ( (B \A) l hiu i xng ca A v B).
Xt tp M = {A1, A2, , Am} gm m tp con i mt khng ging nhau ca X.
a) Chng minh rng m 2n.b) Chng minh rng .13. (VMO 2012) Cho s nguyn dng n. C n hc sinh nam v n hc sinh n xp thnh mt hng ngang, theo th t ty . Mi hc sinh (trong s 2n hc sinh va nu) c cho mt s ko bng ng s cch chn ra hai hc sinh khc gii vi X v ng hai pha ca X. Chng minh rng tng s ko m tt c 2n hc sinh nhn c khng vt qu [RIGHT][I][B]Ngun: MathScope.ORG[/B][/I][/RIGHT]16. Ti liu tham kho[1]. Arthur Engel, Problem-Solving Strategies, Springer 1998. [2]. Alfutova N.B, Ustinov A.B, i s v l thuyt s dnh cho cc trng chuyn ton, NXB MCCME 2003. (Ting Nga)[3]. Nguyn Vn Mu, Trn Nam Dng, V nh Ha, ng Huy Run, ng Hng Thng, Chuyn chn lc t hp v ton ri rc, NXBGD 2009.
[4]. Trn Nam Dng (ch bin), Li gii v bnh lun thi cc tnh v cc trng i hc nm hc 2009-2010, Ebook. [Ti liu c hon thnh ngy 10/11/2012]_1413563212.unknown
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