vmo 2015 loigiai&binhluan
DESCRIPTION
vmo2015TRANSCRIPT
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Trn Nam Dng (ch bin),Nguyn Tt Thu, V Quc B Cn, L Phc L
LI GII V BNH LUN
THI VMO 2015
Thnh ph H Ch Minh, ngy 16 thng 01 nm 2015
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Cc tc gi xin chn thnh cm n cc bn:
1. Trn Quc Lut (Gio vin THPT Chuyn H Tnh, TP H Tnh).
2. Nguyn Vn Linh (Sinh vin H Ngoi Thng H Ni).
3. Hong Kin (Sinh vin H KHTN, HQG H Ni).
4. Nguyn Huy Tng (Sinh vin H KHTN, HQG H Ni).
5. Trn Anh Ho (Hc sinh THPT Chuyn Thoi Ngc Hu, An Giang)
6. Trn Phan Quc Bo (Hc sinh THPT Chuyn L Qu n, KhnhHa)
7. Trn Quc Anh (Hc sinh THPT Chuyn HSP H Ni)
cng nhiu gio vin, hc sinh v cc bn am m Ton khc ng gpcc ni dung, gp xy dng qu bu cho ti liu!
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1. thi
Ngy thi th nht. (08/01/2015)
Bi 1. (5,0 im) Cho a l s thc khng m v (un) l dy s xc nhbi
u1 = 3,un+1 =1
2un +
n2
4n2 + a
u2n + 3 vi mi n > 1.
a) Vi a = 0, chng minh rng dy s c gii hn hu hn v tm giihn .
b) Vi mi a [0; 1], chng minh rng dy s c gii hn hu hn.Bi 2. (5,0 im) Cho a,b, c l cc s thc khng m. Chng minh rng
3(a2 + b2 + c2) > (a+ b+ c)(ab+
bc+
ca)+
+(a b)2 + (b c)2 + (c a)2 > (a+ b+ c)2.
Bi 3. (5,0 im) Cho s nguyn dng k. Tm s cc s t nhin n khngvt qu 10k tha mn ng thi cc iu kin sau:
i. n chia ht cho 3;
ii. cc ch s trong biu din thp phn ca n thuc tp hp {2, 0, 1, 5} .
Bi 4. (5,0 im) Cho ng trn (O) v hai im B,C c nh trn (O),BC khng l ng knh. im A thay i trn (O) sao cho tam gic ABCnhn. Gi E, F ln lt l chn ng cao k t B,C ca tam gic ABC.Cho (I) l ng trn thay i i qua E, F v c tm l I.
a) Gi s (I) tip xc vi BC ti im D. Chng minh rng DBDC
=
cotBcotC
.
b) Gi s (I) ct cnh BC ti hai im M,N. Gi H l trc tm tam gicABC v P,Q l cc giao im ca (I) vi ng trn ngoi tip tamgic HBC. ng trn (K) i qua P,Q v tip xc vi (O) ti im T(T cng pha A i vi PQ). Chng minh rng ng phn gic trongca gc MTN lun i qua mt im c nh.
Ngy thi th hai. (09/01/2015)
Bi 5. (7,0 im) Cho fn(x) l dy a thc xc nh bi
f0(x) = 2, f1(x) = 3x, fn(x) = 3xfn1(x) + (1 x 2x2)fn2(x) vi mi n > 2.
Tm tt c cc s nguyn dng n fn(x) chia ht cho x3 x2 + x.
Bi 6. (7,0 im) Vi a,n l cc s nguyn dng, xt phng trnh a2x +6ay+ 36z = n, trong x,y, z l cc s t nhin.
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a) Tm tt c cc gi tr ca a vi mi n > 250, phng trnh cholun c nghim t nhin (x,y, z).
b) Bit rng a > 1 v nguyn t cng nhau vi 6. Tm gi tr ln nhtca n theo a phng trnh cho khng c nghim (x,y, z).
Bi 7. (6,0 im) C m hc sinh n v n hc sinh nam (m,n > 2) thamgia mt lin hoan song ca. Ti lin hoan song ca, mi bui biu din mtchng trnh vn ngh. Mi chng trnh vn ngh bao gm mt s biht song ca nam n m trong , mi i nam n ch ht vi nhaukhng qu mt bi v mi hc sinh u c ht t nht mt bi. Haichng trnh c coi l khc nhau nu c mt cp nam n ht vinhau chng trnh ny nhng khng ht vi nhau chng trnh kia.Lin hoan song ca ch kt thc khi tt c cc chng trnh khc nhauc th c u c biu din, mi chng trnh c biu din ng mtln.
a) Mt chng trnh c gi l l thuc vo hc sinh X nu nh hy ttc cc bi song ca m X tham gia th c t nht mt hc sinh khckhng c ht bi no trong chng trnh . Chng minh rngtrong tt c cc chng trnh l thuc vo X th s chng trnh cs l bi ht bng s chng trnh c s chn bi ht.
b) Chng minh rng ban t chc lin hoan c th sp xp cc bui biudin sao cho s cc bi ht ti hai bui biu din lin tip bt kkhng cng tnh chn l.
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2. Nhn xt chung
K thi chn hc sinh gii quc gia mn Ton nm hc 2014-2015 (VMO2015) din ra trong 2 ngy 08 v 09/01/2015. V tng quan, ta c ththy rng thi nm nay hay, ph hp vi vic tuyn chn hc sinh gii.
Cc vn t ra trong kh cn bn, quen thuc nhng cng c nhngkh khn nht nh tng bi. V d bi t hp khai thc ch quenthuc v cc s chia ht cho 3 c cc ch s thuc 1 tp hp (dng cnbc 3 ca n v hoc truy hi), nhng a ch s 0 vo gy cht rcri. Bi 6 s hc th khai thc nh l Sylvester v biu din dng t hptuyn tnh ax+by. Bi bt ng thc rt nh nhng (so vi bt ng thckhng nm ngoi), c th gii bng kin thc THCS nhng cng gy khcho khng t th sinh.Nm nay c mt im c bit l ch c mt bi hnh hc, li l bi khkh nn c th s khin mt s bn gii hnh cha c c hi bc l ht strng ca mnh. Thay vo mt bi hnh l mt bi t hp di, khng qukh v bn cht nhng i hi kh nng c hiu ca th sinh. Nhiu thsinh chia s rng: "Em khng hiu bi t hp h hi g?". Qu tht, khnng c hiu, xy dng v chuyn i m hnh l im yu c hu cahc sinh Vit Nam.i vo chi tit, ta c th im qua tng bi nh sau:Ngy 1 kh c bn, gm cc vn t nhiu u c gii thiu trongchng trnh Ton chuyn mt cch i tr.
Bi 1 (dy s) c cu a qu d v quen thuc, cu b li kh khnk thut nht nh, cn s dng nh l kp, n iu hoc b nh x co. Nhiu th sinh b ln su vo cu b ca bi ton dn nthiu thi gian gii quyt cc cu cn li.
Bi 2 (bt ng thc) th khng qu d cng khng qu kh, n lmt bt ng thc i xng, ng bc, dng tng i ph bin vicc hc sinh. i bin kh cn xong l c th nhn ngay ra btng thc Schur bc 4 v AM-GM. Bi ny c rt nhiu cch giinn s rt tic cho hc sinh no b v sau, v y l v chnh cabi ton. y l mt bi ton kh hp l tng xng vi v tr ca ntrong thi.
Bi 3 (t hp) khai thc ch quen thuc ( xut hin trong cc thi Romania 2003, Ph thng nng khiu 2009, Lm ng 2014).C hi rc ri ch s 0 nhng li c "gii" bng iu kin n < 10k
(ch khng phi c m ch s). Phng php cn n v gii quytgn nhng cng cn trnh by cht ch. Phng php truy hi sgy kh mt cht v c n 3 dy. bi ny, c l th sinh lm trnvn khng nhiu, nhng gim kho chm s kh mt.
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Bi 4 (hnh hc phng) l bi hnh duy nht, c hai cu u mc trung bnh kh, nhng s hc sinh lm trn vn bi ny s khngnhiu. Ch v phng tch, trc ng phng vn ng vai trch o trong sut cc nm gn y.
ngy thi th hai, thi so vi ngy u tin "gy sc", kh c v kthut ln t duy. Nguyn nhn l khng c bi hnh v bi t hp phtbiu qu di. Hai bi 5, 6 tuy quen thuc nhng li l phn m cc thsinh t . V tng th ngy 2 hay hn.
Bi 5 (dy s, a thc) tng i c bn. Dng phng trnh ctrng hoc quy np d dng tm c fn = (2x1)n+(x+1)n. Kiu biton chia ht ny kh ging vi nhng bi chia ht trong s nguyn.Cch lm truyn thng l khai thc tnh tun hon ca s d. Tuynhin, thc t nhiu th sinh khng c phng hng g.
Bi 6 (s hc) s khng kh khn lm nu quen vi nh lSylvester. C hai bi ton mu trc l IMO 1983 v Vietnam TST2000. Cng nh cc nm, bi s hc t khi xut hin, nhng nu cth n s l mt bi kh. Tuy nhin, nu cha bit nh hng sdng nh l ny th y qu l mt th thch thc s.
Bi 7 (t hp) kh di, quan h gia cc khi nim kh ri v ddn n hiu nhm. Hc sinh cn chuyn v mt m hnh ton hcno (bng, graph hoc hm s) thy r hn vn . Thc ra vbn cht th n ch l mt bi ton m c th gii bng song nh vquy np.
Nh vy, nhn chung th nm nay hay. Ngy 1 ra tht c bn v quen cho i tr. Ngy 2 gy kh v phn loi. Qua kho st mt s i tuyn,th sinh nh gi ny va sc v t nhiu cng gii quyt c mt sni dung trong bi. D on nm nay im t gii khuyn khch s vo khong 13-15, cn im lt vo vng 2 thi chn i tuyn IMO l24. Nm nay chc s t gii nht.
Ti liu ny chng ti tng hp, bin tp li t nhiu ngun, c bit lmathscope.org gii thiu cho hc sinh, thy c Chuyn Ton tham kho.Do hon thnh trong thi gian gp rt v khi lng tnh ton ln nnc th cn nhiu sai st. Mi gp , thc mc xin gi v cc a ch mail:[email protected], [email protected], [email protected],[email protected]. Xin chn thnh cm n!
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3. Li gii chi tit v bnh lun
Bi ton 1. Cho a l s thc khng m v dy s (un) c xc nh bi
u1 = 3, un+1 =1
2un +
n2
4n2 + a
u2n + 3 vi mi n > 1.
a) Vi a = 0, chng minh rng dy s (un) c gii hn hu hn v tm giihn .
b) Vi mi a [0; 1], chng minh rng dy s (un) c gii hn hu hn.
Li gii.
a) Vi a = 0, ta c dy s un+1 = 12un +14
u2n + 3.
Xt hm s f(x) = 12x+ 1
4
x2 + 3 vi x > 0. Ta c
f (x) =1
2+
1
4
xx2 + 3
> 0, x > 0.
Suy ra hm s f (x) ng bin trn (0;+).T cng thc truy hi ta suy ra vi mi n th
un > 0 , un+1 = f (un) v u2 =3
2+
3
2< u1
nn bng quy np ta suy ra c (un) l dy gim v b chn di bi 0nn n c gii hn hu hn.
t limun = x th x > 0 v l nghim ca phng trnh
x =1
2x+
1
4
x2 + 3
x2 + 3 = 2x
{x > 0x2 + 3 = 4x2
x = 1.
Vy gii hn cn tm l limun = 1.
b) Vi a [0; 1] ta c n24n2+1
6 n24n2+a
6 14. Do
1
2un +
n2
4n2 + 1
u2n + 3 6 un+1 6
1
2un +
1
4
u2n + 3.
Xt hai dy (xn) v (yn) c xc nh bi x1 = 3xn+1 = 12xn +
1
4
x2n + 3
v
y1 = 3
yn+1 =1
2yn +
n2
4n2 + 1
y2n + 3
Theo kt qu cu a) ta c lim xn = 1.
Ta s chng minh limyn = 1 v t ta suy ra c limun = 1 v liu cn chng minh. Di y ta s nu hai cch x l gii hn ny.
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Cch 1. Ta c
yn+1 1 =1
2(yn 1) +
n2
4n2 + 1
(yn 1) (yn + 1)y2n + 3+ 2
+2n2
4n2 + 1
1
2
=1
2(yn 1) +
n2
4n2 + 1
(yn 1) (yn + 1)y2n + 3+ 2
1
2(4n2 + 1)
N2
4N2 + 1.
Suy ra
1
2yN+1 +
(N+ 1)2
4(N+ 1)2 + 1
y2N+1 + 3 >
1
2yN +
N2
4N2 + 1
y2N + 3
hay yN+2 > yN+1.
Do , bng quy np ta chng minh c dy (yn) tng t s hng th Ntr i. Do , dy (yn) hi t. t limyn = y ta tm c y = 1.
Nu (yn) l dy gim k t s hng th n0 tr i, kt hp vi dy (yn) bchn di bi 0 ta suy ra dy (yn) hi t. t limyn = y ta tm c y = 1nn trng hp ny loi.
Vy limyn = 1 v bi ton c chng minh.
Cch 2. Ta s chng minh bng quy np rng yn > 1 2n vi mi n > 2. ()
Tht vy, vi n = 2, d dng thy rng () ng.
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Gi s () ng vi n > 2 th ta c yn > 1 2n > 0. Suy ra
1
2yn +
n2
4n2 + 1
y2n + 3 >
1
2
(1
2
n
)+
n2
4n2 + 1
(1
2
n
)2+ 3.
Ta cn chng minh
1
2
(1
2
n
)+
n2
4n2 + 1
(1
2
n
)2+ 3 > 1 2
n+ 1
n2
4n2 + 1
(1
2
n
)2+ 3 >
(1
2
n+ 1
)
1
2
(1
2
n
) 2n
4n2 + 1
n2 n+ 1 > n
2 n+ 2
2n(n+ 1)
4n2(n+ 1)n2 n+ 1 > (4n2 + 1)(n2 n+ 2)
16n4(n+ 1)2(n2 n+ 1) > (4n2 + 1)2(n2 n+ 2)2
Ta thy bt ng thc trn ng vi n = 2, 3, ta xt n > 4. Ch rng
16n2(n+ 1)2 > (4n2 + 1)2 4n(n+ 1) > 4n2 + 1, ng v
n2(n2 n+ 1) > (n2 n+ 2)2 n(n 2)2 > 4 cng ng vi mi n > 4.T ta c xn > un > 1 2n ng vi mi n, m lim xn = lim
(1 2
n
)= 1
nn ta c limun = 1.
Nhn xt. th nht ca bi ton rt c bn, thuc dng mu mc v lni dung quen thuc ca phn l thuyt gii hn dy s m hu ht cchc sinh c gii thiu. Ta ch cn chng minh dy s gim bng quynp l coi nh bi ton c gii quyt xong.
Tuy nhin, th hai kh mi l v c th coi y l cu kh nht trongngy thi u tin. Vic nh gi dy (un) kp gia hai dy (xn) v (yn) lsuy ngh rt t nhin. T kt qu cu a, ta thy rng lim xn = 1 nn tasuy ngh n vic chng minh limyn = 1 c c gii hn. lm ciu ny, ta i chng minh dy (yn) l dy gim v b chn di.
Do 2 (yn+1 yn) = 2n2
4n2+1
y2n + 3 yn nn c (yn) gim th ta cn c
y2n > 12n4
12n4+8n2+1.
Tuy nhin, khng sun s nh vy, vic chng minh tnh cht ny qukh do phi thc hin quy np vi biu thc phc tp. Cn nu chngminh dy ny b chn di bi 1 th cng quy np khng thnh cng bih s bin thin theo n gy ra kh nhiu rc ri.
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iu ny gi cho ta mt nh gi nh nhng hn l yn > 1 kn vi kdng no . thc hin c quy np, ta cn c nh gi
1 k2
2+
n2
4n2 + 1
3+ (1
k
n)2
> 1 kn+ 1
n2
4n2 + 1
3+ (1
k
n)2
> n2 2kn+ n+ k
2n(n+ 1).
Xt vi k = 1, ta thy bt ng thc trn chnh l
56n7 52n6 + 48n5 37n4 + 18n3 11n2 + 2n 1 > 0, ng do n > 1.
Trong li gii theo cch 2, ta s dng nh gi dng trn vi k = 2.
Di y l mt s bi ton tng t:
Bi 1.( chn i tuyn HSP 2010) Cho dy s (xn) tha mn: x1 = x2 = 1,xn+2 = x2n+1 12xn,n > 1Chng minh rng dy s ny c gii hn hu hn v tm gii hn .
Gi . Ta nh gi |xn| < 1n ,n > 5.Bi 2. (VMO 2012) Cho dy s (xn) tha mn: x1 = 3,xn = n+ 2
3n(xn1 + 2) ,n > 2
Chng minh rng dy s ny c gii hn hu hn v tm gii hn .
Gi . Chng minh dy gim trc tip hoc nh gi xn > 1+ 3n vi n > 2.Bi 3. (Hng Yn, 2011) Cho dy s (xn) xc nh bi
x1 = a > 0,
xn+1 = xn +x2nn2
,n > 1
Tm tt c cc gi tr a sao cho dy s c gii hn hu hn.
Gi . Ta chng minh rng 1xn+n2
6 1an(n+1) .
Ni chung, cch nh gi cc dy s thng qua mt hm bin thin theon khng phi qu mi m v kh ph bin trong cc bi dy s c h sthay i theo n hoc l tng ca nhiu s hng ca dy.
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Ngoi ra, ta c th tip cn bi ny theo kiu dng chng minh ccdy nghim ca phng trnh thng qua im bt ng. Di y l mtcch nh th:
t fn(x) = x2 +n2
4n2+a
x2 + 3 vi n > 1 v x > 0.
R rng hm s c im bt ng duy nht l
an =
3
(2+ a2n2
)2 1
Nu x > an th fn(x) > an v ngc li, nu 0 < x < an th fn(x) < an. Ngoira, ta thy rng an xc nh nh trn tng ngt v c gii hn 1.
Xt dy s (un), nu tt c cc s hng u ln hn 1 th u2n+ 3 < 4u2n nn
suy ra
un+1 uN+k. Suy ra dy (un) tng ngt k t 1 v n b chn trn bi 1nn c gii hn. Chng minh kt thc.
Ta thy rng cch lp lun chia trng hp trn cng kh th v, n cngtng t cch 1 ca li gii c gii thiu. Tuy nhin, im mnh cali gii ny l n khng ph thuc vo gi tr a [0; 1] nn c th thyrng bi ton vn ng vi mi a khng m. Ngoi ra, n cho ta mt cchnh gi nh nhng hn khi kho st tnh hi t ca dy nh tng: dcha bit tnh bin thin ca dy c th nh th no, ta c th xt mi
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tnh hung: nu ri vo trng hp thun li th tt; nhng nu khngthun li th bng cc lp lun thch hp, ta vn chng minh c; suyra n lun ng trong mi trng hp.
Lp lun ny tng c dng gii quyt bi ton kh th v sau:
Bi 4. (VMO bng A, 2005) Xt dy s thc (xn) xc nh bi cng thc{x1 = a,
xn+1 = 3x3n 7x
2n + 5xn
,n = 1, 2, 3, . . .
Xc nh tt c cc gi tr a sao cho dy s trn c gii hn hu hn vtrong tng trng hp, hy xc nh gii hn .
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Bi ton 2. Cho cc s thc a, b, c > 0. Chng minh rng
3(a2+b2+c2) > (a+b+c)(ab+bc+ca)+(ab)2+(bc)2+(ca)2 > (a+b+c)2.
Li gii. V tri ca bt ng thc kh n gin. Dng pht biu ca nvi tng cc bnh phng gi cho ta ngh n ng nht thc Lagrange mt hng ng thc quen thuc c dng chng minh bt ng thcCauchy-Schwarz:( n
i=1
a2i
)( ni=1
b2i
)
( ni=1
aibi
)2=
16i 0.
C th hn, ta c ng thc sau:
3(a2 + b2 + c2) (a+ b+ c)2 = (a b)2 + (b c)2 + (c a)2.
Do , bt ng thc v tri c th c vit di dng:
(a+ b+ c)2 > (a+ b+ c)(ab+
bc+
ca).
n y th c l bn no cng s ngh n vic s dng bt ng thcquen thuc x2 + y2 + z2 > xy + yz + zx (p dng cho x =
a, y =
b v
z =c) hon tt php chng minh.
y, ta s dnh s quan tm nhiu hn cho bt ng thc v phi.Nhn xt ban u cho thy y l mt bt ng thc tng i cht vdu bng xy ra ti hai trng hp a = b = c v a = b, c = 0 (cng cchon v tng ng). Do , ta cn phi rt cn trng trong cc nh gica mnh.
Ngoi ra, ta cng thy rng ch kh ca bi ton chnh l cc cn thc.Nu ta c th ph c du cn a bt ng thc v dng n gin hnth chc chn bi ton cng s tr nn sng sa hn. n y, c hai tng chnh nh sau:
1. t n ph kh cn: y l mt hng i kh t nhin v cc cnthc y cng n gin, cc biu thc di du cn ch c dngbc mt. Do , ch cn mt ln t n ph x =
a, y =
b, z =
c l
ta c th kh c ht cc cn thc v a v xt mt bt ng thcthun nht bc 4 i vi x, y, z. Bc ca bt ng thc mi cngkhng qu cao nn y l hng i hon ton kh thi.
2. S dng nh gi kh cn: y l tng thng thy khi x lcc bi ton c cn. Vn c t ra y l ta phi la chnnh gi cht sao cho cc iu kin du bng phi c mbo.
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Cc hng tip cn c trnh by di y hu ht u s dng hai tng trn lm t tng ch o.
Cch 1. Khai trin trc tip.
y c l l hng i t nhin nht cho bi ton ny. Ta ch vic tx =
a, y =
b, z =
c ri nhn tung ht ra. Khi , bt ng thc cn
chng minh c th c vit li di dng:x4 + xyz
x+
xy(x2 + y2) > 4
x2y2. (1)
n y, nu bn no c tm hiu s ngh ngay n bt ng thc Schurbc 4:
x2(x y)(x z) + y2(y z)(y x) + z2(z x)(z y) > 0.Dng khai trin ca n chnh l:
x4 + xyz
x >
xy(x2 + y2). (2)
S tng ng gia hai bt ng thc (1) v (2) gi cho ta ngh n vicdng (2) nh gi cho (1). Ngoi ra, (2) cng c du bng ti x = y = zv x = y, z = 0 (cng cc hon v) tng ng vi trng hp ng thc ca(1). Do , y s l mt nh gi kh n v ta c th yn tm v anton ca n. Tht vy, sau khi nh gi, ta ch cn xt bt ng thc:
2
xy(x2 + y2) > 4
x2y2
xy(x2 + y2) > 2
x2y2
v n ch l mt h qu trc tip ca bt ng thc AM-GM:xy(x2 + y2) >
(xy 2xy) = 2
x2y2.
Li bnh. t n ph l mt trong nhng k nng c bn cn c trong btng thc. Nhiu bi ton c hnh thc cng knh phc tp, tuy nhinsau nhng bc t n ph n gin, ta c th a bi ton tr v dngmi m nhiu tng (m trong cng c th l gc ca bi ton)s c phi by ra.
C nhiu kiu t n ph, trong c ba kiu sau rt thng dng: t nph lm n gin hnh thc bi ton, t n ph thun nht hahoc i xng ha, v t n ph lng gic da vo du hiu t iukin gi thit.
Cch 2. Phng php SOS.
y l hng i t nhin th hai sau phng php khai trin. Trc ht,ta cng s t n ph x, y, z nh hng 1 trn kh cn tin cho vicquan st. Ta a bi ton v chng minh:
(x2 + y2 + z2)(xy+ yz+ zx) +
(x2 y2)2 > (x2 + y2 + z2)2.
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Trong bt ng thc trn, c hai s hng cng cha nhn t x2 + y2 + z2.Mt cch t nhin, ta ngh n vic ghp hai s hng vi nhau. Lcny, bt ng thc c vit li thnh:
(x2 y2)2 > (x2 + y2 + z2)(x2 + y2 + z2 xy yz zx).
S xut hin ca tng bnh phng(x2 y2)2 bn v tri v phn tch
qu quen thuc: x2
xy =
1
2
(x y)2
gi cho ta ngh ngay n vic dng phng php phn tch bnh phngSOS x l bi ton. C th, ta vit c bt ng thc cn chng minhdi dng:
Sx(y z)2 + Sy(z x)
2 + Sz(x y)2 > 0,
trong Sx = f(x, y, z) = 2(y + z)2 (x2 + y2 + z2) = y2 + z2 + 4yz x2, cnSy = f(y, z, x), Sz = f(z, x, y) c nh ngha tng t.
n y, ta ch vic s dng cc tiu chun ca phng php l c.Gi s x > y > z, khi ta c
Sy = z2 + x2 + 4zx y2 > 0, Sz = x2 + y2 + 4xy z2 > 0
vSx + Sy = 2z
2 + 4zx+ 4yz > 0.Do x z > y z > 0 nn (z x)2 > (y z)2. T suy ra
Sx(y z)2 + Sy(z x)
2 + Sz(x y)2 > (Sx + Sy)(y z)2 > 0.
Li bnh. Mt iu cn ch l khi s dng phng php SOS, cc bncn phi chng minh li cc tiu chun ca n. Nhiu bn cu th chghi gn l Sx + Sy > 0, Sy + Sz > 0 ri suy ra iu phi chng minh. Nhth l cha c.
Ngoi cch s dng cc tiu chun SOS nh trn, ta cng c cch bini m khng phi s dng tiu chun no da trn ng nht thc ngin:
(x y)2(x z)(y z) = 0. (3)
C th thy im mu cht gy kh khn trong vic x l tng:(x y)2(x2 + y2 + 4xy z2) > 0
chnh l phn s m mi s hng, chng hn nh z2 trong s hng(x y)2(x2 + y2 + 4xy z2). Nu ta em cng vi tng
(x y)2(x z)(y z)
vi mt s lng thch hp vo s lm tng s lng z2 ln s hng ny
15
-
v rt c th s thu c mt i lng khng m. C th, ta hy vng sc s k sao cho:
x2 + y2 + 4xy z2 + k(x z)(y z) > 0 (x+ y)2 + (2+ k)xy kz(x+ y) + (k 1)z2 > 0.
Quan st mt cht, cho th thy ngay nu chn k = 2 th ta s vit cbiu thc (x+ y)2 kz(x+ y) + (k 1)z2 di dng bnh phng. T , tathu c mt li gii ngn gn th v sau: Bt ng thc cn chng minhtng ng vi
(x y)2(x2 + y2 + 4xy z2) + 2
(x y)2(x z)(y z) > 0,
hay (x y)2
[(x+ y z)2 + 4xy
]> 0.
ng nht thc (3) gip chng ta x l c bi ton theo mt li SOSrt th v a n mt bt ng thc hin nhin. y cng l mt kinhnghim ca chng ti tch ly c khi tm hiu v phng php SOS.Tt nhin, ng nht (3) ch hiu qu cc bt ng thc i xng bc4. Vi cc bt ng thc bc cao, chng ta cn mt ng nht thc tngqut hn tng cng tnh hiu qu. Chng ta c mt kt qu th vsau (bn c c th t chng minh): Cho f(x, y, z) l mt a thc i xngvi hai bin x, y. Khi , ta c th phn tch:[
(x y)2(x z)(y z) f(x, y, z)] = (x y)2(y z)2(z x)2 g(x, y, z)trong g(x, y, z) l mt a thc i xng vi ba bin x, y, z.
Nh vo ng nht thc trn m chng ti x l thnh cng rt nhiubt ng thc bng phng php SOS rt n gin ch khng cn phidng tiu chun phc tp no.
Cch 3. nh gi kh cn.
Mt hng i khc thay cho t n ph l tm cch nh gi ph cnthc. C th, ta s tm cc nh gi thch hp cho
ab,bc,ca vi
chiu > ph du cn. Thng th vi cc dng cn tch nh th ny,cch ph cn thng dng l s dng bt ng thc AM-GM. Tuy nhin, bi ton ny, n li cho nh gi vi chiu ngc li:
ab 6 a+ b
2,bc 6 b+ c
2,ca 6 c+ a
2
khng phi chiu ta cn. C cch no iu chnh khng nh?
16
-
Mt tng th v y l s dng nghch o. Nh bit, vi bt ngthc dng th nghch o ca n s o chiu. Do , ta c th ngh nvic vit
ab thnh ab
abri nh gi:
ab > 2ab
a+ b. (4)
Nh vy l s ph c cn thc vi chiu ta mun. Tuy nhin, mt iucn lu y l s 0 khng c nghch o. Th nn nu mt trong cccn thc c mt s bng 0 th ta khng th dng cch ny c. Do ,cn phi xt trng hp loi tr tnh hung ngoi mun ny.
Nu trong a, b, c c mt s bng 0, chng hn c = 0, th bt ng thc cnchng minh s tr thnh:
(a+ b)ab+ (a b)2 + a2 + b2 > (a+ b)2.
Cng vic y l kh n gin v ta c (a+ b)ab > 2ab.
Tip theo, ta xt trng hp a, b, c > 0. Lc ny, ta c th s dng c(4). Bi ton c a v chng minh
2(
a)( ab
a+ b
)+
(a b)2 >(
a)2
.
Sau khi thu gn, n c dng:
2(
a)( ab
a+ b
)> 4
ab
a2.
V a + b + c c th tch ra cc i lng a + b, b + c, c + a lin quan nmu ca cc s hng ca tng
aba+b
nn ta c th x l rt gn v tritheo cch sau:
VT = 2 ab(a+ b+ c)
a+ b= 2
ab(1+
c
a+ b
)= 2
ab+ 2abc 1
a+ b.
Khi , bt ng thc c th vit li thnh:
2abc( 1a+ b
+1
b+ c+
1
c+ a
)> 2(ab+ bc+ ca) a2 b2 c2.
n y th tng t nhin l s dng bt ng thc Cauchy-Schwarzdng cng mu lm gim s lng cc phn thc:
1
a+ b+
1
b+ c+
1
c+ a> 9
2(a+ b+ c)
17
-
v a bi ton v xt mt bt ng thc mi:
9abc
a+ b+ c> 2(ab+ bc+ ca) a2 b2 c2.
Tuy nhin, y chnh l bt ng thc Schur bc ba.
Li bnh. y, chng ti mun ch vi cc bn v cch tch cc tchab, bc, ca, abc c s dng trong li gii trn.
Nh ta bit, nhng bt ng thc m trong cc trng hp du bngca chng c trng hp khng ti tm th thng kh nh gi hn ccbt ng thc bnh thng. Nguyn nhn l cc b hon v. Mt btng thc i xng (hoc hon v) nu c du bng ti b (A, B, C) thcng s t c du bng ti cc hon v ca n l (B, C, A) v (C, A, B).Do , nh gi thnh cng th ta phi tm c mt nh gi sao chon m bo c c ba trng hp. R rng rt kh!
i vi cc bi ton c du bng ti bin th cch tch trn cho ta mtk thut x l c bit hiu qu. Tht vy, gi s ta cn chng minh btng thc f(a, b, c) > 0 vi du bng l a = kb (k 6= 0), c = 0 (v cc honv) chng hn ( y ch xin ly v d mt trng hp c th phn tch,cn nhiu trng hp khc cng c th x l tng t). Khi , nu vitc bt ng thc trn di dng:
ab g(a, b, c) + bc g(b, c, a) + ca g(c, a, b) > 0
th ta ch cn quan tm nh gi biu thc i din g(a, b, c) theo dubng a = kb, c = 0 l m khng cn ch nhiu n cc hon v cab ny. Nu nh gi thnh cng th sau khi nhn thm ab vo hai v, tas thu c mt nh gi cho s hng ab g(a, b, c) vi du bng xy rati ab = 0 v a = kb, c = 0. Hin nhin nh gi ny s m bo c cba trng hp hon v ca a = kb, c = 0.
T y, ta thy rng cc iu kin s cng thun li hn nu ta tch rac s hng c dng abc h(a, b, c). Lc ny, h(a, b, c) c th c nhgi kh l v t, bi l ti trng hp bin th tch abc bng 0 mtri, th nn khi nhn vo th kiu g cng m bo c du bng bin.
Cch 4. S dng hm li.
Chc hn bn c yu Ton u bit n tnh cht th v sau ca hmli: Nu hm s f(x) lin tc v li trn on [a, b] th gi tr ln nht ca ns t c mt trong hai im x = a hoc x = b, cn i vi hm lm th sl gi tr nh nht.
Th nhng, li khng c nhiu bn ngh n vic s dng tnh cht nyvo gii ton. Mt trong nhng nguyn nhn c l l tnh cht ca hm
18
-
li. Nh trn cp, hm li s t cc i ti bin v hm lm st cc tiu ti bin. Nhng bi ton ny th li khng c bin r rng,cc bin c bin di nhng li khng c bin trn.
Mt iu na cng cn phi ni n l hu ht cc bn hc sinh u chc tm nhn v m m cha c n ci nhn vi m. C ngha l bicho bt ng thc bao nhiu bin th cc bn ch nhn bng ng bynhiu bin ch khng ngh n tm nhn khc i. l mt tm nhn sailm. Trn thc t, c c li gii thnh cng th ta nn bt u bngnhng th nh nht nht, ch n mi kha cnh.
Bt ng thc cho c dng i xng vi ba bin a, b, c, th th n cngl bt ng thc i xng vi hai bin bt k no trong ba bin trn.Chn hn cc bn vn con nh chng ta hc t cp 2 rt nhiu rngcc bi ton i xng hai n c th c x l hiu qu bng php t nph tng-tch S = x+ y, P = xy nh vo quan h ca chng: S2 > 4P.
C th hn, nu ta c nh c th bt ng thc s c dng i xng vi av b. Khi , nh php t S = a + b, P = ab, ta c th chuyn bt ngthc v dng:
g(S, P) > 0.Khi , nu ta c nh S na th y s ch cn l mt bt ng thc vimt bin l P v lc ny P c chn min vi bin trn ln bindi l
[0, S
2
4
].
n y, nu ta c th suy xt c tnh n iu hoc tnh li lm cagP(S, P) th cng c th a ra c kt lun v tnh cht cc cc tr can ri t i n li gii.
Li gii chi tit theo hng ny nh sau: t f(a, b, c) = VT VP. Khi ,ta phi chng minh f(a, b, c) > 0. C nh c v S = a+ b. t P = ab th tac 0 6 P 6 S2
4. Ta c bin i:
a+b =
a+ b+ 2
ab =
S+ 2
P
v (ab)2+(b c)2+(ca)2(a+b+ c)2 = h(P), trong h(P) l mt biuthc bc nht ca P. Do :
f(a, b, c) = (S+ c)
(P +
S+ 2
P
)+ h(P) = g(P).
Nu c hai s a, b u bng 0 th bt ng thc hin nhin ng nn tach cn xt trng hp S > 0 l (l lun ny l m bo khong(0, S
2
4
)tn ti, m bo cho vic xt o hm ca g(P)). D thy hm s
g(P) lin tc trn[0, S
2
4
]. Ngoi ra, vi mi P (0, S2
4
), ta tnh c:
g (P) = (S+ c)
(1
2P+
1P
2S+ 2
P
)+ k
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-
trong k l h s cao nht ca h(P). R rng g (P) l hm gim ngt trn(0, S
2
4
)nn g(P) l hm lm trn
[0, S
2
4
]. T suy ra
g(P) > min{g(0), g
(S2
4
)}, P
[0,S2
4
].
Nh th, chng minh bt ng thc cho, ta ch cn chng minhg(0) > 0 v g
(S2
4
)> 0. Mt khc, ta li d thy:
g(0) = f(S, 0, c), g
(S2
2
)= f
(S
2,S
2, c
).
Do , t nhng l lun trn, c th thy rng ta ch cn xt bt ngthc ti hai trng hp: c mt s bng 0 hoc c hai s bng nhau, l. hng 3, ta chng minh c bt ng thc ng ti trnghp th nht. Nh vy, ta ch cn phi kim tra trng hp th hai nal c.
Gi s a = b. Khi , bt ng thc tr thnh
(2a+ c)(a+ 2
ac)+ 2(a c)2 > (2a+ c)2.
Sau khi khai trin v rt gn, ta phi chng minh
4aac+ 2c
ac+ c2 > 7ac.
V ln c dng tng, cn v b c dng tch gi cho ta ngh ngay n btng thc AM-GM nh gi:
4aac+ 2c
ac+ c2 > 7 7
(aac)4(cac)2c2 = 7ac.
Li bnh. Bng cch s dng hm li v bt ng thc Karamata1, ta sc thm cch nhn nhn tng quan hn cho nhiu vn , nm bt cbn cht tt hn.
1Bt ng thc Karamata c xy dng da trn khi nim b tri v tnh cht tiptuyn ca hm li:
Cho hai b s khng tng A = (a1, a2, . . . , an) v B = (b1, b2, . . . , bn), trong ai, bi ucng thuc vo mt min I. Ta ni rng A tri hn B, k hiu A B, nu cc iu kin sauc tha mn ng thi:
a1 > b1a1 + a2 > b1 + b2. . . . . . . . .a1 + a2 + + an1 > b1 + b2 + + bn1a1 + a2 + + an1 + an = b1 + b2 + + bn1 + bn
Lc ny, nu f(x) l mt hm kh vi bc hai v li trn I th:
f(a1) + f(a2) + + f(an) > f(b1) + f(b2) + + f(bn).
20
-
Nhn xt. Qua tm hiu v nghin cu ti liu, chng ti suy on rngbi ton ny c xut x t bi ton sau y: Cho a, b, c l cc s thckhng m tha mn iu kin a+ b+ c = 1. Chng minh rng
a+ (b c)2 +b+ (c a)2 +
c+ (a b)2 >
3. (8)
Tt nhin, y ch l nhng suy on c tnh ch quan, nhng nu suyxt k, cc bn s thy bi VMO 2015 chnh l mt mu cht quan trngtrong chng minh bt ng thc (8). Tht vy, bnh phng hai v ca (8),ta thy n tng ng vi
2[
a+ (b c)2][b+ (a c)2
]+
(a b)2 > 2.
n y, s dng bt ng thc Cauchy-Schwarz, ta c[a+ (b c)2
][b+ (a c)2
]>[
ab+ (a c)(b c)]
=
ab+
(a c)(b c)
=
ab+1
2
(a b)2.
Do , chng minh (8), ta ch cn chng minh cab+
(a b)2 > 1,
hay (a
)(ab
)+
(a b)2 >(
a
)2.
y chnh l bt ng thc v phi trong bi s 2 ca VMO nm nay.Cn v tri c l tc gi t thm ra vi mc ch g im cho cc thsinh tham d k thi.
Ni ring v bt ng thc (8), n cng c mt xut x rt th v... t hnhhc. Chnh xc hn l t s tng t ha mt bt ng thc v ngtrung tuyn ca tam gic. Ta bit rng, trong mt tam gic vi di bacnh l a, b, c th:
ma +mb +mc 6 2p.Xt cc tam gic c na chu vi p = 1. t a = y + z, b = z + x v c = x + yvi x, y, z > 0. Khi , ta c x+ y+ z = 1 v:
m2a =2b2 + 2c2 a2
4=
2(z+ x)2 + 2(x+ y)2 (y+ z)2
4
= x(x+ y+ z) +(y z)2
4= x+
(y z)2
4.
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-
T php bin i ny, ta thu c bt ng thcx+
(y z)2
4+
y+
(z x)2
4+
z+
(x y)2
46 2.
C th thy bt ng thc (8) chnh l mt s tng t ha bng cchthay i h s ca cc bnh phng di du cn.
Nh cp, phn cui ny, chng ti xin c xut mt tng qutcho bt ng thc v phi trong VMO nm nay nh sau (trng hpbi VMO chnh l ng vi n = 3):
Cho s t nhin n > 2. Xt n s thc khng m x1, x2, . . . , xn. Chng minhrng
2
n 1
( ni=1
xi
)( 16i 4
16i
-
Bi ton 3. Cho s nguyn dng k. Tm s cc s t nhin n khng vtqu 10k tha mn ng thi cc iu kin sau:
i) n chia ht cho 3.
ii) cc ch s trong biu din thp phn ca n thuc tp hp {2, 0, 1, 5} .
Li gii. V 10k khng chia ht cho 3 nn ta ch cn xt cc s t 0 chon 999 . . . 9
k ch s
, tc l cc s c khng qu k ch s.
t S = {2, 0, 1, 5}, b sung cc ch s 0 vo trc nu cn thit, ta a vxt cc s c dng a1a2 . . .ak vi ai S. Ta cn m cc s nh vy v chiaht cho 3. Ch l a1a2 . . .ak chia ht cho 3 khi v ch khi a1+a2+a3+. . .+akchia ht cho 3, ta a bi ton v vic m s cc b (a1,a2,a3, . . . ,ak) Sksao cho tng chia ht cho 3.
n y ta c cc hng gii quyt nh sau.
Cch 1. Dng cng thc truy hi.
Vi i = 0, 1, 2, ta t
A(n, i) = {(a1,a2,a3, . . . ,an) Sn |a1 + a2 + a3 + . . .+ an i(mod3) }
v tan = A(n, 0),bn = A(n, 1), cn = A(n, 2).
D dng thy rnga1 = 1,b1 = 1, c1 = 2.
Xt phn t (a1,a2, . . . ,an,an+1) ca A(n+ 1, 0).
Nu an+1 = 0 th (a1,a2, . . . ,an) A(n, 0). Nu an+1 = 2 hoc 5 th (a1,a2, . . . ,an) A(n, 1). Nu an+1 = 1 th (a1,a2, . . . ,an) A(n, 2).
T y ta suy ra an+1 = an + 2bn + cn (1). Hon ton tng t, ta cng c
bn+1 = an + bn + 2cn (2)
vcn+1 = 2an + bn + cn (3)
T y ta tnh c a2 = 5,b2 = 6, c2 = 5,a3 = 22,b3 = 21, c3 = 21. Cng ccng thc (1), (2), (3) li, v theo v, ta c an+1+bn+1+cn+1 = 4(an+bn+cn).T y suy ra
an + bn + cn = 4n
(Ch rng iu ny cng c th suy ra d dng bng quy tc m).
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Tr (1) cho (2), tr (2) cho (3), tr (3) cho (1), ta c
an+1 bn+1 = bn cn,bn+1 cn+1 = cn an, cn+1 an+1 = an bn.
Do , an+3 bn+3 = bn+2 cn+2 = cn+1 an+1 = an bn. Tng t th
bn+3 cn+3 = bn cn, cn+3 an+3 = cn an vi mi n.
S dng cc gi tr ban u ai,bi, ci vi i = 1, 2, 3 v tnh cht trn, tasuy ra rng
Nu k chia ht cho 3 th bk = ck = ak 1. Nu k chia 3 d 1 th ak = bk = ck 1. Nu k chia 3 d 2 th ak = ck = bk 1.
T y, kt hp vi ng thc ak + bk + ck = 4k , ta suy ra kt qu
aK = 4k13 nu k khng chia ht cho 3; aK = 4k+23 nu k chia ht cho 3.
Cch 2. Dng a thc v s phc
Xt a thcP(x) = (x2 + 1+ x+ x5)
k
Ta cP(x) = (x2 + 1+ x+ x5)
k=
(a1,a2,...,aK)Sk
xa1+a2+...+ak
Ta thy tng cc h s ca P(x) bng s cc b (a1,a2, . . . ,ak) Sk v bng4k. Hn na s cc b (a1,a2, . . . ,ak) Sk sao cho a1 + a2 + . . . + ak bngtng cc h s ca cc s m chia ht cho 3 trong khai trin ca P(x).
t P(x) = a0 + a1x + a2x2 + a3x3 + . . . + a5kx5k. Ta cn tnh T =a3m l
tng cc h s chia ht cho 3 trong khai trin. Gi l nghim ca phngtrnh x2 + x+ 1 = 0 th ta c 3 = 1. T d dng suy ra 1+ k + 2k = 0 vimi k khng chia ht cho 3 v 1+ k + 2k = 0 vi k chia ht cho 3. ()Ta c
P(1) = a0 + a1 + a2 + a3 + . . .+ a5k,
P() = a0 + a1+ a22 + a3
3 + . . .+ a5k5k,
P(2) = a0 + a12 + a2
4 + a36 + . . .+ a5k
10k
p dng tnh cht (), ta suy ra P(1) + P() + P(2) = 3T . Suy ra
T =P(1) + P() + P(2)
3=
4k + 2k + 4k
3
Cui cng, li p dng tnh cht (*) ta suy ra T = 4k13
nu k khng chiaht cho 3 v T = 4
k+23
nu k chia ht cho 3.
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Nhn xt. y l mt bi ton kh quen thuc. Dng ny gp mts thi ca Rumani (2007), PTNK (2009) hay Lm ng nm va ri.Chng hn thi PTNK 2009:
Cho s nguyn dng n. C bao nhiu s chia ht cho 3, c n ch s vcc ch s u thuc {3, 4, 5, 6}?
Ci mi v cng l ci kh ca bi ton ny l ch s 0 v xt tt c ccs nh hn 10k (ch khng phi l s c k ch s). Tuy nhin, nh thytrong li gii trn, kt hp hai ci kh v mi li ta c mt ci c (vd!). Theo phng php cng thc truy hi, c th s c mt s bn gpkh khi gii h phng trnh truy hi (nu c 2 dy th d hn, 3 dy khri).
Nu khng x l c ci mi bng cch l lun nh trn v phi m scc s c n ch s lp t {2, 0, 1, 5} v chia ht cho 3 th ta s a n cchgii sau:
Gi An,Bn,Cn l s cc s c n ch s lp t {2, 0, 1, 5} v chia 3 d 0, 1, 2tng ng. Khi ta c A1 = 1, (s 0 ta coi l s c 1 ch s), B1 = 1,C1 = 2v
An+1 = 2(Bn + Bn1 + . . .+ B1) + (Cn + Cn1 + . . .+ C1) (1),
Bn+1 = (An +An1 + . . .+A1) + 2(Cn + Cn1 + . . .+ C1) (2),
Cn+1 = 2(An +An1 + . . .+A1) + (Bn + Bn1 + . . .+ B1) (3)
(hc sinh rt d b nhm ch ny)
Suy ra A2 = 4,B2 = 5,C2 = 3.
Trong (1) thay n bng n+ 1, ta c
An+2 = 2(Bn+1 + Bn + . . .+ B1) + (Cn+1 + Cn + . . .+ C1) (4).
Ly (4) tr (1) ta c
An+2 An+1 = 2Bn+1 + Cn+1,
tc lAn+2 = An+1 + 2Bn+1 + Cn+1.
Tng t,
Bn+2 = An+1 + Bn+1 + 2Cn+1,Cn+2 = 2An+1 + Bn+1 + Cn+1.
Do vy, ta c A2 = 4,B2 = 5,C2 = 3 v
An+1 = An + 2Bn + Cn,Bn+1 = An + Bn + 2Cn,Cn+1 = 2An + Bn + Cn
vi mi n > 2.
25
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Tip theo ta gii gn ging nh li gii 1. Ch p s ca bi ton lA1 +A2 + . . .+Ak v theo (1), (2), (3) th tng ny bng
4CK+1 + BK+1 2AK+19
.
Trong tnh hung yu cu chia ht cho s khc, khng phi 3 th ligii s th v hn nhiu v trong tnh hung , vai tr ca hm sinh thhin r hn nhiu. Ta xem xt th mt s bi ton tng t sau:
Bi 1. ( tp hun i tuyn IMO 2014) C bao nhiu s t nhin c 9ch s, trong khng cha ch s 0 v chia ht cho 11?
Bi 2. (VMO 2008) C bao nhiu s t nhin c 2008 ch s, chia ht cho9 v trong biu din ca n c khng qu 2 ch s 9?
Bi 3. C bao nhiu s t nhin chia ht cho 7, c n ch s v mi ch sc ly t tp hp {1, 3, 4, 6, 7, 9}?
Bi 4. C bao nhiu s t nhin b hn 5000 c tng cc ch s chia htcho 4?
26
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Bi ton 4. Cho ng trn (O) v hai im B,C c nh trn (O), BC khngl ng knh. Mt im A thay i trn (O) sao cho tam gic ABC nhn.Gi E, F ln lt l chn ng cao k t B,C ca tam gic ABC. Cho (I) lng trn thay i i qua E, F v c tm l I.
a) Gi s (I) tip xc vi BC ti im D. Chng minh rng DBDC
=
cotBcotC
.
b) Gi s (I) ct cnh BC ti hai im M,N. Gi H l trc tm tam gicABC v P,Q l cc giao im ca (I) vi ng trn ngoi tip tam gicHBC. ng trn (K) i qua P,Q v tip xc vi (O) ti im T (T cngpha A i vi PQ). Chng minh rng ng phn gic trong ca gcMTN lun i qua mt im c nh.
Li gii.
a) Gi s im D nm trong cnh BC, trng hp im D nm ngoi chngminh tng t. Ta c hai cch x l nh sau:
Cch 1. Gi R,S ln lt l giao im ca (I) vi BC (cc giao im ny cth tng ng trng E, F trong trng hp tam gic ABC cn). Ta c
AR AF = AS AE ARAS
=AE
AF=AB
AC RS BC.
Do (I) tip xc vi BC ti D nn
BD2
CD2=BF BRCE CS =
BF
CE BRCS
=BF
CE ABAC
=BF
CE BECF
=cotB
cotC.
Vy ta c DBDC
=
cotBcotC
.
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Cch 2. Gi X, Y ln lt l giao im ca (I) vi BE,CF. Ta c BD2 =BX BE,CD2 = CY CF (phng tch vi ng trn (I)).Hai tam gic BXF,CYE c XBF = YCE,BXF = CYE nn ng dng. Suyra
BX
CY=BF
CE=
cosB
cosC.
Do ,BD2
CD2=BX
CY BECF
=cosB
cosC sinCsinB
=cotB
cotC
hayBD
CD=
cotB
cotC.
b) Trng hp tam gic ABC cn ti A, bi ton hin nhin ng.Xt trng hp tam gic khng cn A, khng mt tnh tng qut, gi sAB < AC. Gi G l giao im ca EF v ng thng BC.
Xt cc ng trn (BHC), (I) v ng trn ng knh BC. Ta thy:
Trc ng phng ca (BHC), (I) l PQ. Trc ng phng ca (I) v ng trn ng knh BC l EF. Trc ng phng ca (BHC) v ng trn ng knh BC l BC.
Do , PQ,EF,BC ng quy ti tm ng phng ca 3 ng trn ny. Tac GT 2 = GP GQ = GM GN nn ng trn (TMN) cng tip xc vi (O)ti T .
Do , ta c GTM = GNT (cng chn cung TM ca ng trn (K)).
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Theo tnh cht gc ngoi ca tam gic th GNT = NTC + NCT . Hnna, do GT tip xc vi (O) nn GTB = GCT . Tr tng ng tng v 2ng thc, ta c BTM = CTN.T y d thy phn gic ca gc MTN v BTC l trng nhau hay phngic ca MTN i qua trung im J ca cung BC khng cha A, y lim c nh.
Ta c pcm.
Nhn xt. V cu a ca bi ton, c th ni y l mt phn tng ihay, n khng qu n gin nhng cng d hc sinh c th t tinv thoi mi bt tay vo th sc. Vic s dng tnh cht c bn caphng tch ca ng trn s dn n s xut hin mt cch t nhinca cc im R,S,X, Y nh trn.
Ngoi ra, vi hc sinh no khng quen vi vic k ng ph, ta cn cth lp lun bng t s lng gic trc tip, khng cn v thm bt c gnh sau:
K hiu E l gc DEF = BDF, bng bin i gc, ta c
B = AEF = 180(E+CED) = 180(E+180(C+F)) = C+FE.
Suy ra B+ E = C+ F. Do BFD = CED,CFD = BED.Xt t l din tch cc tam gic
BD
CD=SFBD
SFCD=FB FD sinBFDFC FD sinCFD = cotB
sinBFD
sinCFD
vBD
CD=SEBD
SECD=EB ED sinBEDEC ED sinCED =
1
cotC sinBEDsinCED
.
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Suy ra (BD
CD
)2=
cotB
cotC BDCD
=
cotB
cotC.
Li gii ny pht hin ra mt s cp gc bng nhau na v bin icng kh nh nhng.
Khai thc thm m hnh ny, ta cng pht hin ra thm kh nhiu iuth v:
1. Bn im R,S,X, Y l cc nh ca mt hnh ch nht.
2. ng thng ni giao im ca DE,CF v DF,CE song song vi BC.
Ta cng thy rng, ngay t u, hc sinh cng s thc mc v cch dngca ng trn (I), tc l ng trn i qua 2 im cho trc v tip xcvi 1 ng thng. Nhiu bn v hnh bng cm tnh, dng ng trn (I)trc v chnh iu ny khin h khng i su vo bn cht ca biton v trn c s , gii quyt c cu b.
chnh l giao im G ca EF,BC trong li gii ca bi ton nu. Tathy rng GE GF = GD2 hay GD chnh l trung bnh nhn ca hai onGE,GF, hon ton dng c. Ch rng, c 2 im D nh vy, mt imnm pha trong on BC v mt nm pha ngoi on BC, hai im nycng vi B,C lp thnh mt hng im iu ha.
n y, ta pht hin ra rng cn phi c cu nhn xt nh u tin hn ch vic ph thuc hnh v, do khng nu r l tip xc vi cnhhay vi ng thng BC.
Ta cng rng nu gi L l chn ng cao gc A ca tam gic ABCth cotB
cotC= LB
LCnn bi a tng ng vi LB
LC= DB
2
DC2.
Mt cch t nhin, ta t ra cu hi tng qut: iu kin no i vi bim A ,B ,C nm trn cc cnh BC,CA,AB ca tam gic ABC theo tht, AA ,BB ,CC ng quy v ng trn i qua B ,C tip xc vi BC tiA1 th
A BA C
=A1B
2
A1C2?
Ngoi ra, nu xt cc ng trn tng t theo gc B,C th c ba ngtrn v c s c c im g? y l cc khai thc rt th v cho biton ny.
V cu b ca bi ton, cng tng t nh nhng bi ton v yu t cnh trong cc k thi VMO, TST gn y, bi 4 ln ny i hi hc sinhkh nng phn on, suy lun tt v nht l kh nng "n gin ho" biton, tm ra yu t quyt nh trong mt lot gi thit c v phc tp v"t tc dng".
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Bi ton ny c l c pht trin t b sau: Cho tam gic ABC vM,N l cc im thuc BC, mt im T nm trn ng trn (O). Khi ,nu ng trn (TMN) tip xc vi (O) ti T th BTC,MTN c chungng phn gic trong.
B ny c th chng minh bng cng gc nh li gii trn, hoc sdng php v t tm T binM,N thnh X, Y thuc (O) th XY BC v ta ckt qu trn.
Do vy, vi nhng hc sinh nhn ra tnh cht GT tip xc vi c (TMN)v (O) th ch cn mnh dn b qua nhng "yu t tha" trong l s cngay li gii.
Ta c th d on im c nh thng qua vic xt hai im A i xngqua trung trc BC, t suy ra im c nh cch u B v C, s d an kt lun hn.
C kh nhiu ng trn gp mt trong bi ton, nhng vi nhng hcsinh c kinh nghim th vic s dng tnh cht trc ng phng ccc ng ng quy l iu d hiu. D bi c pht biu tng idi v phc tp, nhng khng phi l kh x l nhng chuyn ny.
Trn thc t, ta c th gii m khng cn s c mt ca E, F. Vn vi xutpht t trc ng phng, xt 3 ng trn (BHC), (O), (TPQ) th s ctip tuyn ti T i qua giao im ca BC,PQ.
S kin EF cng i qua G c th s dn n nhiu tnh cht mi hoc nukt hp vi cc tnh cht c cng s a ra nhiu bi ton khc khaithc. Di y l mt bi ton m rng do bn Nguyn Vn Linh xut:
Cho tam gic ABC ni tip ng trn (O). Mt ng trn bt k i quaB,C v mt ng trn (J) khc thay i ct BC tiM,N v ct (I) ti P,Q.ng trn i qua P,Q v tip xc vi (O) ti T nm cng pha vi A so viBC. Khi , phn gic gc MTN lun i qua trung im cung BC khngcha A.
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Bi ton 5. Cho dy a thc fn(x) c xc nh bi
f0(x) = 2, f1(x) = 3x, fn(x) = 3xfn1(x) + (1 x 2x2)fn2(x),n > 2.
Tm tt c cc gi tr n a thc fn(x) chia ht cho a thc x3 x2 + x.
Li gii. T cng thc truy hi cho, ta bin i nh sau
fn(x) = 3xfn1(x) +(1 x x2
)fn1(x)
fn(x) (x+ 1) fn1(x) = (2x 1) fn1(x) (2x 1) (x+ 1) fn2(x) fn(x) (x+ 1) fn1(x) = (2x 1) [fn1(x) (x+ 1) fn2 (x)]
Suy ra
fn(x) (x+ 1) fn1(x) = (2x 1)n1 [f1(x) (x+ 1) f0(x)] = (2x 1)
n1 (x 2)
fn(x) (2x 1)n = (x+ 1)[fn1(x) (2x 1)
n1]
T y ta c
fn(x) (2x 1)n = (x+ 1)n[f0(x) (2x 1)
0] = (x+ 1)n
hayfn(x) = (2x 1)
n + (x+ 1)n.
t Q(x) = x3 x2 + x = x(x2 x+ 1).
V fn (x) chia ht cho a thc g(x) = x3x2+x nn fn (0) = 0 hay 1+(1)n= 0
nn n l.
V fn(2) = (5n + 1) (do n l) chia ht cho (2)2 (2) + 1 = 7. Do 125
1(mod7) nn ta xt cc trng hp sau
n = 3k, k l, ta c 5n + 1 = 53k + 1 (1)k + 1 = 0 (mod7) n = 3k+ 1, k chn, ta c 5n + 1 = 5 53k + 1 6 (mod7) n = 3k+ 2, k l, ta c 5n + 1 = 25 53k + 1 24 3 (mod7).
T , ta suy ra iu kin cn ca n l n = 3k vi k l.
Khi fn (x) = (x+ 1)3k+ (2x 1)3k
...(x+ 1)3 + (2x 1)3 vi mi k.
Nhn thy (x+ 1)3+ (2x 1)3 = 9x3 9x2+ 9x chia ht cho a thc g(x) nnta c n = 3k vi k l s t nhin l tha mn. t k = 2m+ 1 vi m nguynth n = 6m+ 3.
Vy tt c cc s n cn tm c dng 6m+ 3 vi m l s nguyn dng.
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Nhn xt. Ta c th tm ra c cng thc tng qut ca dy a thc cho bng cch coi x l hng s v xy dng s hng ca dy s tng ng,tc l xt dy s {
u0 = 2,u1 = 3x,
un = 3xun1 +(1 x 2x2
)un2
vi x l tham s thc no .
Xt phng trnh c trng t2 3xt+ 2x2 + x 1 = 0 t = x+ 1, t = 2x 1.Suy ra un = (x+ 1)
n+ (2x 1)n. Da vo u0,u1 ta tm c = = 1.
Do , fn (x) = (x+ 1)n+ (2x 1)n.
Mt cch khc lm l d on cng thc thng qua vic tnh ton trcvi gi tr:
f0 = 2; f1 = 3x; f2 = 5x2 2x+ 2; f3 = 9x3 9x2 + 9x; f4 = 17x4 37x3 + 30x2 4x+ 2.
Ta thy h s bc cao nht ca cc a thc c dng 2n+ 1 v h s cui l0 v 2 lun phin nn d on a thc cn tm c dng (2x 1)n+(x 1)n.
Kim tra trc tip, ta thy a thc cn tm l (2x 1)n + (x+ 1)n v ny ch cn quy np l xong. Ngoi ra, ta c th gii quyt cn li l tmn c s chia ht bng cch dng s phc nh sau:
Do fn(x) chia ht cho x3 x2 + x nn fn(0) = fn () = fn (2) = 0 vi l sphc tha 2 + 1 = 0.
Ta cng c 3 = 1 v / R. Ta c fn(0) = 0 nn n l s l. t n = 2k + 1th
fn() = (2 1)2k+1
+ (+ 1)2k+1
= (2 1)(42 4+ 1
)k+ (+ 1)
(2 + 2+ 1
)k= (2 1) (3)k + (+ 1) (3)k = (3)k
((2 1) + (+ 1)()k
)Do fn () = 0 nn ta cn c (+ 1)()
k+ (2 1) = 0. Ta xt cc trng hpsau:
Vi k = 3t th (+1)()k+21 = (+1)()3t+21 = (+1)+21 =3 6= 0, khng tha.
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Vi k = 3t + 1 th ( + 1)()k + 2 1 = ( + 1)()3t+1 + 2 1 =(2 ) + 2 1 = 0, tha mn.
Vi k = 3t + 2 th ( + 1)()k + 2 1 = ( + 1)()3t+2 + 2 1 =(2 1) + 2 1 6= 0, khng tha.
T suy ra n = 2(3t + 1) + 1 = 6t + 3 tha mn. Bc th li thc hintng t.
Ta thy bi ton ny thuc dng tng i chun v tnh chia ht ca athc kt hp vi a thc xc nh bi h thc truy hi. Bi ton t vtr u tin ca ngy 2 cng vi hai bi khc kh hn hn khin cho biton ny dng nh l la chn duy nht cho cc th sinh.
Dng a thc xc nh bi truy hi ny cng khng phi qu xa l, tnht l vi a thc Chebyshev ni ting; tuy nhin, nhiu th sinh t ralng tng v khng gii quyt c trn vn bi ton ny l iu rt ngtic.
Di y l mt bi tng t:
Cho dy a thc fn(x) xc nh bi cng thc{f0(x) = 2, f1(x) = 2x+ 2,
fn+2(x) = (2x+ 2)fn+1(x) (x2 + 2x 3)fn(x),n > 1
Tm tt c cc gi tr n sao cho fn(x) chia ht cho x2 + 2x+ 5.
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Bi ton 6. Vi a,n nguyn dng, xt phng trnh a2x + 6ay + 36z = ntrong , x,y, z l cc s t nhin.
a) Tm tt c cc gi tr ca a vi mi n > 250, phng trnh cholun c nghim (x,y, z).
b) Bit rng a > 1 v nguyn t cng nhau vi 6. Tm gi tr ln nht can theo a phng trnh cho khng c nghim (x,y, z).
Li gii. Ta s s dng b quen thuc sau:
Cho cc s nguyn dng a,b nguyn t cng nhau. Khi , s nguynln nht khng biu din c di dng ax + by vi x,y t nhin lN0 = ab a b. Ni cch khc, N0 = ab a b + 1 l s nguyn dngnh nht sao cho phng trnh ax + by = m c nghim t nhin vi mim > N. (nh l Sylvester).
a) Gi s a l mt gi tr tha mn iu kin bi. Ta s ni ngn gn sn l biu din c nu tn ti x,y, z t nhin sao cho
a2x+ 6ay+ 36z = n.
R rng (a, 6) = 1 v nu ngc li, gi s (a, 6) = d > 1 th a2x + 6ay + 36zchia ht cho d. Ta chn n > 250 v n khng chia ht cho d th dn nmu thun.
D thy rng iu kin cn mi n > 250 biu din c l (a, 6) = 1. rng
a2x+ 6ay+ 36z = a2x+ 6(ay+ 6z) = a(ax+ 6y) + 36z
p dng b trn, ta thy cc s
6a2 a2 6 = 5a2 6 v 36a+ 36 a = 35a+ 36
tng ng l cc s ln nht khng biu din c trong tng cch biudin. Suy ra
5a2 5 < 250, 35a+ 36 < 250
Suy ra a < 7. Kt hp vi iu kin (a, 6) = 1 suy ra ch c a = 1,a = 5 thamn.
Ta s chng minh a = 1,a = 5 cng chnh l cc s cn tm. Tht vy,
Vi a = 1, ta c x+ 6y+ 36z = n, phng trnh lun c nghim l (n; 0; 0).
Vi a = 5, ta chng minh mi s n > 250 u biu din c di dng25x+ 30y+ 36z vi x,y, z t nhin. Ta vit
25x+ 30y+ 36z = 25x+ 6(5y+ 6z 20) + 120.
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p dng b , mi s t nhin khng nh hn 20 u c th biu dinc di dng 5y + 6z nn suy ra mi s t nhin u c th biu dindi dng 5y+ 6z 20 vi y, z t nhin.
Tng t, mi s t nhin khng nh hn 120 u biu din c didng 25x+6u. Vim > 250 thm120 > 120 > 120 nn tn ti x v u t nhinsao cho m 120 = 25x + 6u. Li chn y, z t nhin sao cho u = 5y + 6z 20th c m 120 = 25x+ 6(5y+ 6z 20). Suy ra m = 25x+ 30y+ 36z.
Vy tt c cc gi tr a cn tm l a = 1,a = 5.
b) Ta chng minh bi ton tng qut sau: Cho cc s nguyn dng a vb nguyn t cng nhau. Khi
N = a2b+ ab2 a2 b2 ab+ 1
l s nguyn dng nh nht phng trnh a2x + aby + b2z = m cnghim t nhin vi mi m > N.
Tht vy, nu m > N, ta vit phng trnh a2x+ aby+ b2z = m di dng
a (ax+ by (ab a b+ 1) + b2z = m a2b+ a2 + ab a
Do m a2b+ a2 + ab a > b2a b2 a+ 1 nn theo b cu a, tn tiu, z t nhin sao cho
m a2b+ a2 + ab a = au+ b2z.
Li p dng b , tn ti x,y t nhin sao cho u+abab+ 1 = ax+by.Khi
ma2b+a2+aba = a (ax+ by (ab a b+ 1))+b2z m = a2x+aby+b2zCui cng, ta chng minh s a2b+ab2a2b2ab khng biu din cdi dng a2x+ aby+ b2z vi x,y, z t nhin.
Tht vy, gi s ngc li, nu tn ti x,y, z t nhin sao cho
a2b+ ab2 a2 b2 ab = a2x+ aby+ b2z
th a2(b 1 x) + b2(a 1 z) = ab(y + 1). Suy ra hoc b 1 x > 0 hoca 1 z > 0. Khng mt tnh tng qut, gi s b 1 x > 0. Ta vit ngthc trn di dng a2(b 1 x) = b(ay+ bz ab+ b+ a).
Do (a2,b) = 1 nn t y suy ra b 1 x chia ht cho b. iu ny muthun v 0 < b 1 x < b. Do , a2b + ab2 a2 b2 ab khng biu dinc di dng a2x+ aby+ b2z vi x,y, z t nhin.
Bi ton c chng minh.
Cu b ca bi ton ny chnh l trng hp c bit ca bi ton vapht biu vi b = 6 nn s n ln nht khng biu din c di dnga2x+ 6ay+ 36z vi x,y, z t nhin l s 5a2 + 30a 36.
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Nhn xt. Ch nh l Bezout, thut ton Euclid l mt ch quantrng trong chng trnh s hc. Nu nm vng ch ny th cng snm c nh l Sylvester v lm quen vi bi ton Frobenius v ccng xu. y l ni dung gp trong kh nhiu thi, v d nh thiIMO 1983, VN TST 2000. C th l:
Bi 1.(IMO 1983) Cho a,b, c l cc s nguyn dng i mt nguyn tcng nhau. Chng minh rng 2abcabbcca l s nguyn ln nht khngbiu din c di dng abx+ bcy+ caz vi x,y, z l cc s t nhin.
Bi 2. (VN TST 2000) Cho a,b, c i mt nguyn t cng nhau. S nguyndng n c gi l s bng bnh nu khng biu din c di dngabx + bcy + caz trong x,y, z l cc s nguyn dng. Hi c tt c baonhiu s bng bnh?
Bi 3. (Titu Andreescu, 104 Number Theory Problems) Cho a v n lcc s nguyn dng. Chng minh rng s nghim nguyn khng m(x,y, z) ca phng trnh ax+ by+ z = ab l
1
2[(a+ 1)(b+ 1) + (a,b) + 1] .
nh l Sylvester thc ra c pht biu y nh sau (xem thm tiliu S hc qua cc nh l v bi ton):
Cho a,b l cc s nguyn dng nguyn t cng nhau. Chng minh rngN0 = ab a b l s nguyn ln nht khng biu din c di dngax + by vi x,y l cc s nguyn khng m. Hn na, vi mi p,q nguynvi p + q = N0, c ng mt trong hai s p,q biu din c di dngax+by vi x,y l cc s nguyn khng m (m ta gi tt l biu din c).
nh l ny c chng minh thng qua b sau:
B . Cho a,b l cc s nguyn dng nguyn t cng nhau v b > 1.Chng minh rng vi mi s nguyn N, tn ti duy nht cp s nguynx,y tha mn iu kin N = ax+ by v 0 6 x < b.Chng minh b .
Do (a,b) = 1 nn theo nh l Bezout, tn ti x,y sao cho N = ax + by.By gi chia cho b, ta c x = bq + x0 vi 0 6 x0 < b v ta c N =a(bq+ x0) + by = ax0 + b(y+ qa). Tnh tn ti c chng minh. Tnh duynht l hin nhin v nu c 2 biu din nh th th ta c ax+by = ax +by ,suy ra a(x x ) = b(y y). Suy ra x x chia ht cho b. M 0 6 x, x < bnn iu ny v l.
Quay tr li nh l, nu n > N0. p dng b , tn ti x,y nguyn saocho n = ax+by vi 0 6 x < b. Suy ra by > ababax = aba(b1) = b.Suy ra y > 0. Vy ta tm c x,y t nhin sao cho n = ax+ by.
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Nu N0 biu din c th ta c ab a b = ax+ by vi x,y > 0. Ta c haibiu din ab a b = a(b 1) + b(1) v ab a b = ax+ by. T tnh duynht trn, ta suy ra y = 1, mu thun.
Vy N0 khng biu din c. Do N0 khng biu din c nn nu p+q =N0 th ch c nhiu nht 1 trong 2 s p,q biu din c. Cui cng, nup khng biu din c th chn x,y sao cho p = ax + by vi 0 6 x < b thta phi c y < 0. Lc
q = N0 p = ab a b ax by = a(b 1 x) + b(y 1)
biu din c do b 1 x > 0 v y 1 > 0.Ni thm v bi ton Frobenius v nhng ng xu. y l bi ton xcnh s tin ln nht khng th tr c khi ch s dng cc ng xu cmnh gi c nh no . V d vi cc ng xu mnh gi 3 v 5 n v ths tin ln nht khng tr c l 7 n v. S ln nht vi mi b s nhth ta gi l s Frobenius.
Mt cch ton hc, bi ton c pht biu nh sau: Cho cc s nguyndng a1,a2,a3, ...,an c c chung ln nht l 1. Tm s nguyn ln nhtkhng biu din c di dng k1a1+k2a2+ ...+knan vi k1,k2, ...,kn l ccs nguyn khng m. S nguyn ln nht ny c gi l s Frobenius vthng c k hiu l g(a1,a2, ...,an).
Vi n = 2, bi ton ny c li gii trn vn v chnh l nh l Sylvester nu. Xa hn na, ngi ta chng minh c mt s kt qu linquan nh:
1. nh l Schur khng nh rng vi iu kin cc s nguyn dnga1,a2,a3, ...,an c c chung ln nht l 1 th s Frobenius tn ti.
2. g(a,b, c) >3abc a b c
3. Vi n bt k, tm ra c cng thc hm g cho cc cp s cng, cps nhn, ngoi ra l mt s nh gi chn trn, chn di.
Nh vy vi n > 2 v cc bt k th tnh ti nay cha tm c cng thctng minh cho hm g(a1,a2, ...,an) v c v hng i ny khng kh thi.Trong khi , vi mt s lng ng xu c nh, tn ti thut ton tnhs Frobenius trong thi gian a thc (tnh theo logarith ca gi tr tin xuc trong d liu vo). Hin cha c thut ton thi gian a thc theo sng xu, v bi ton tng qut khi gi tr ng xu l ln ty l bi tonNP-kh.
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Bi ton 7. C m hc sinh n v n hc sinh nam (m,n > 2) tham gia mtlin hoan song ca. Ti lin hoan song ca, mi bui biu din mt chngtrnh vn ngh. Mi chng trnh vn ngh bao gm mt s bi ht song canam n m trong , mi i nam n ch ht vi nhau khng qu mt biv mi hc sinh u c ht t nht mt bi. Hai chng trnh c coi lkhc nhau nu c mt cp nam n ht vi nhau chng trnh ny nhngkhng ht vi nhau chng trnh kia. Lin hoan song ca ch kt thc khitt c cc chng trnh khc nhau c th c u c biu din, mi chngtrnh c biu din ng mt ln.
a) Mt chng trnh c gi l l thuc vo hc sinh X nu nh hy ttc cc bi song ca m X tham gia th c t nht mt hc sinh khckhng c ht bi no trong chng trnh . Chng minh rng trongtt c cc chng trnh l thuc vo X th s chng trnh c s l biht bng s chng trnh c s chn bi ht.
b) Chng minh rng ban t chc lin hoan c th sp xp cc bui biudin sao cho s cc bi ht ti hai bui biu din lin tip bt k khngcng tnh chn l.
Li gii.
a) Ta nh s cc hc sinh n theo th t t 1 n m v cc hc sinh namt 1 n n. ng vi mi chng trnh vn ngh, ta biu din vic ghp cpca cc cp nam n song ca thnh mt bng m n gm m hng n ctnh sau: Bng s c nh s 1 hoc 0, trong nm hng i ct jc in s:
S 1 nu hc sinh n th i v hc sinh nam th j c ht vi nhau. S 0 nu hc sinh n th i v hc sinh nam th j khng ht vinhau.
0 0 1 0 11 1 0 0 00 0 1 0 01 0 1 1 11 0 1 1 1
Mt bng gi l tt nu trn mi hng v mi ct u phi c t nht mts 1. R rng theo bi th tt c cc bng biu din cho chng trnhu l tt v hc sinh no cng c biu din.
Xt mt hc sinh X no , gi s l n; trng hp hc sinh namchng minh tng t. Chng trnh no l thuc hc sinh X nu nhtrn bng tng ng ca n, tn ti t nht 1 ct c ng mt s 1 nm
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trn hng ca X, ta gi bng ny l l thuc X v ct nh th l ct lthuc X.
Ta cn chng minh rng, trong cc bng l thuc X, s bng c s cc s1 chn bng s bng c s cc s 1 l.
Tht vy,
Xt trng hp trong bng c k ct l thuc X th r rng k < n v nukhng, ngc li, k = n th ton b cc trn hng X u l 1, cn tt ccc cn li ca bng u l 0. Do m > 2 nn tn ti mt dng ton l s0, mu thun. Vi k < n, ta b k ct ra khi bng th trn bng s mti ng k s 1. Mi trong n k cn li ca hng X s c in s 0hoc 1 ty v cc ct cn li u cn t nht mt s 1 na khng thuchng X. Do , nu ta b lun hng X i th bng cn li vn l tt.
Suy ra s bng l thuc X trong trng hp ny s l 2nk nhn vi slng bng tt c kch thc (m 1) (n k) cn li. Trong mi bng ,ta chn mt bt k ca hng X v thay i s t 0 1, 1 0 th s dnn thay i tnh chn l ca s cc s 1 trn bng.
Do , r rng tn ti mt song nh i t tp hp cc bng l thuc X cs cc s 1 chn n tp hp cc bng l thuc X c s cc s 1 l.
Do , s lng hai loi bng ny l bng nhau. ng vi mi k = 1,n 1v cc cch chn k ct ph thuc X th s lng bng c s 1 l v chnu bng nhau, v th nn tng s bng c s cc s 1 l bng vi bng cs cc s 1 chn. Ta c pcm.
b) Tip theo, ta t f(m,n) v g(m,n) ln lt l s cc bng tt m n cchn v l cc s 1. Xt mt hc sinh n ty , t l X. Ta xt cc trnghp sau:
Nu tn ti mt ct no l thuc X th theo cu a, s bng c scc 1 chn bng s bng c s cc s 1 l, t gi tr ny l h(m,n).
Nu khng tn ti ct no l thuc X th b hng tng ng ca Xi, ta cn li mt bng tt c m 1 hng v n ct.
Mt khc, s trng hp m hng X c s l v c s chn in s 1 lnlt l
L =
a1( mod 2)Can v
C =
a0( mod 2),a>0Can (do hng X khng th ton l s 0)
Ta bit rng (1+ x)n = C0n + C1nx + C
2nx
2 + ... + Cnnxn nn vi x = 1, ta c
ngay L = C+ 1.
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Ch l tnh chn l ca s cc s 1 thuc dng X s quyt nh n tnhchn l ca bng cn li nn ta c cng thc truy hi sau{
f(m,n) = h(m,n) + L g(m 1,n) + C f(m 1,n)g(m,n) = h(m,n) + L f(m 1,n) + C g(m 1,n)
Do f(m,n) g(m,n) = (L C) (g(m 1,n) f(m 1,n))
= g(m 1,n) f(m 1,n)
Lp li qu trnh ny n khi s hng v s ct nh nht c th, tc lm = n = 2, ta c
f(m,n) g(m,n) = (1)m+n4 (f(2, 2) g(2, 2)) .
m trc tip, ta thy c f(2, 2) = 3,g(2, 2) = 4 nn suy ra f(m,n)g(m,n) =(1)m+n3. T ta thy rng s lng ca hai loi bng khng vt qu1 v c th sp xp cc bng theo th t chn, l an xen c iukin bi.
Ta cng c pcm.
Nhn xt. C th ni y l mt bi t hp di nht trong lch s cc kthi VMO. Cch y ng 10 nm, trong VMO 2005 bng A c mt bilin quan n nh l Turan m trn tp ch THTT tng nhn xt: "yl mt bi ton ri rc, khng phi kh song hu ht hc sinh u b tay!Ti sao vy? C l vn l tm l. Trc ht khi c ra hc sinh thychong ngp bi cc khi nim a ra trong bi ton v cc mi quan hlogic gia chng, nu thong qua th cha th hiu c ngay." Trong biton ny, nhn xt trn dng nh vn ph hp.
V n cng vi bi 5 trong VMO 2009 l cc bi ton dng m bnghai cch, m hnh ha thnh bng c th ni l kh nht trong nhiunm tr li y.
Bi ton ny t cui thi, c di v s im nh th khin nhiuth sinh b cuc ngay t bc c . K thc, bi ton cng khngn ni qu kh hiu v c l tc gi rt c gng trong vic din t bi mt cch r rng nht c th. Tuy nhin, trong bi c gi thit mii nam n ch ht vi nhau khng qu mt bi v mi hc sinh u cht t nht mt bi khin nhiu th sinh hoang mang v khng hiu lthng tin ny m t cho mt chng trnh hay l c k lin hoan ny. Num t cho chng trnh th c v hi ri rm v lung tung, mi hc sinhc th ln din nhiu ln trong mt chng trnh v cc tit mc c thc lp i lp li trong nhiu chng trnh khc nhau. Nhng k thc,n li chnh l mun ca bi.
tin hnh x l bi ton, ta cn phi tm cch m hnh ha n thnhdng thch hp, trong , mi rng buc u c th biu din c.
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y, vic lp thnh bng nh phn cho mt chng trnh c th nh trnc l l la chn sng sa, n gin nht. Cch ny cng quen thuc vihu ht cc hc sinh v bi ton v t mu bng, in s trn bng cngtng i quen thuc (t nht l trong kim tra ca Vin Ton trongthng 12 va qua cng c mt bi nh th).
Bn cnh la chn , ta cng c th dng cc cch tip cn khc nh:
Dng bipartite graph hay cn gi l th lng phn, th 2 pheri thao tc trn cc nh v cnh.
Dng cch chia thnh cc b (nam, n, chng trnh) ri m bng2 cch v tnh tng, hoc cch ni khc l xt hm sf(a,b, c) : (A B C) {0, 1} vi A,B,C ln lt l tp hp n, nam v cc chngtrnh.
Suy cho cng, cc cch ny cng l song nh, chuyn i cch tip cnnhng bn cht vn th: bng tt chnh l mt ma trn cnh k ca th v n cng chnh l tng hp cc gi tr m hm trn nhn c vicng mt gi tr c C.Mt nhn xt c bn nhng mang tnh quyt nh trong c 2 a v b cabi ton l: Trong cc tp hp con ca mt tp hp c n > 1 phn t, stp hp con c l phn t bng s tp hp con c chn phn t v cngbng 2n1. ny c th gii quyt d dng bng song nh hoc m trctip v dng nh thc Newton.
So snh vi cc bi ton trong nhng nm gn y, c th coi y l biton m phn a h tr rt tt cho phn b vi vic chia mt "trng hpln" thnh hai trng hp nh. Phn a va c tc dng gi , va nggp mt phn vo lp lun ca phn b. Ci kh ca bi ton ny c l ch s dng tng truy hi quy np, v i n cng.
Chng ta s t ra cu hi sau y mt cch t nhin: C tt c baonhiu chng trnh trong lin hoan vn ngh?. y l mt bi ton khngd v khng cho ra kt qu dng tng minh nhng c th c th sinhno m trong bi thi ca mnh, c gng m s lng ny.
Ta c th gii quyt bng nguyn l b tr nh sau (ta vn xt bng c mhng v n ct): Ta gi mt bng m mi hng u c t nht mt s 1 lbng tt theo hng, gi tp hp cc bng ny l P. Gi A l tp hp ccbng tt theo hng nhng li c mt ct no khng c s 1 no, Ai ltp hp cc bng tt theo hng nhng ct th i li khng c s 1 no. Tathy
A = A1 A2 A3 ... An.R rng s bng tt cn tm chnh l |P| |A|. Trc ht, ta tnh s bngtt theo hng.
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Ch rng mi c hai cch in l 0 hoc 1 nn hng c n s c 2n
cch, tuy nhin, loi tr trng hp tt c cc u l 0 ra th c 2n 1cch. Cc hng khc cng tng t th nn c tng cng (2n 1)m bngtt theo hng hay |P| = (2n 1)m. Do cc ct bnh ng vi nhau nn tac th m i din mt trng hp no suy ra cc trng hp cnli. Gi s c k ct no trong cc ct 1, 2, 3, ...,n l khng cha s 1 novi 1 6 k < n. Trn mi hng s cn li n k in vo cc s 0 hoc 1sao cho c t nht mt s 1 (do k thuc cc ct kia u c in s 0),s cch in cho mi hng l 2nk 1 v cho c bng l (2nk 1)m.
T , theo nguyn l b tr, ta c
|A| = |A1 A2 A3 ... An| =nk=1
Ckn(1)k(2nk 1
)m.
Do , ta c s bng tt cn tm l
(2n 1)m
nk=1
Ckn(1)k(2nk 1
)m=
nk=0
Ckn(1)k(2nk 1
)mVi tnh bnh ng ca m,n, ta cng suy ra c ng thc sau
nk=0
Ckn(1)k(2nk 1
)m=
mk=0
Ckm(1)k(2mk 1
)n.
Trn bng nh phn m n, ta cn c th t ra nhiu cu hi tng tnh:
Bi 1.(Columbia Olympiad) C bao nhiu bng nh phn m n sao chotrong mi hnh vung con 2 2 ca bng th s cc s 0 bng s cc s 1?p s l: 2m + 2n 2.
Bi 2.( kim tra ca Vin Ton 2014) C bao nhiu bng nh phnmnm mi hng v mi ct u c s chn s 1? p s l: 2(m1)(n1).
Bi 3.(AIME 2007) C bao nhiu bng nh phn 6 4 sao cho c ng 12s 1 v mi hng c ng 2 s 1 v mi ct c ng 4 s 1? p s l: 1860.
Cui cng, nhn xt chung v bi ton ny, ta thy pht biu ca n cnhi khin cng dn n bi ton cng knh, nng n. C l xut phtt mt kt qu nghin cu c no , tc gi chuyn i t ngn ngTon hc sang mt tnh hung thc t trong i sng. Nhng do phi mbo nhiu quan h ni ti trong vn gc nn cui cng, tnh hung li phn tc dng tr nn rc ri, thiu t nhin. C l nu pht biuthnh dng bng hoc dng th nh phn tch trn th c nhiuhc sinh s nm bt tt tng hn v s mnh dn tip cn hn, nhiukhi kim tra nng lc, kh nng t duy ca hc sinh th cng ch cn mc nh th m thi.
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thiNhn xt chungLi gii chi tit v bnh lun