Trn Nam Dng (ch bin)

V Quc B Cn Trn Quang Hng L Phc L

Hong Kin Nguyn Huy Tng

LI GII V BNH LUN

CHN I TUYN QUC GIAD THI IMO 2014

Thnh ph H Ch Minh, ngy 05 thng 06 nm 2014

Vit Nam Team Selection Test 2014 2 thi chnh thc

Ngy thi th nht - 25/03/2014

Bi 1. Tm tt c cc hm s f : Z Z tha mnf2m + f (m) + f (m) f (n)

= n f (m) +m, m, n Z.

Bi 2. Trong mt phng ta vung gc Ox y, xt cc im nguyn c ta thuc tphp sau

T =(x ; y) : 20 x , y 20, (x ; y) 6= (0; 0) .

T mu cc im thuc T sao cho vi mi im c ta (x , y) T th c ng mt tronghai im (x ; y) v (x ;y) c t mu. Vi mi cch t nh th, gi N l s cc b(x1; y1), (x2; y2) m c hai im ny cng c t mu v

x1 2x2, y1 2y2 (mod 41).Tm tt c cc gi tr c th c ca N.

Bi 3. Cho tam gic ABC ni tip trong ng trn (O). Trn cung BC khng cha A ca(O) ly im D. Gi s CD ct AB E v BD ct AC F . Gi (K) ng trn nm trongtam gic EBD, tip xc vi EB,ED v tip xc vi ng trn (O). Gi (L) l tm ngtrn nm trong tam gic FCD, tip xc vi FC , FD v tip xc vi ng trn (O).

a) Gi M l tip im ca (K) vi BE v N l tip im ca (L) vi CF . Chng minhrng ng trn ng knh MN lun i qua mt im c nh khi D di chuyn.

b) ng thng qua M v song song vi CE ct AC P, ng thng qua N v songsong vi BF ct AB Q. Chng minh rng ng trn ngoi tip tam gic AMP,ANQcng tip xc vi mt ng trn c nh khi D di chuyn.

Ngy thi th hai - 26/03/2014

Bi 4. Cho tam gic ABC nhn, khng cn c ng cao AD v P thuc AD. Cc ngthng PB,PC ln lt ct CA,AB ti E, F.

a) Gi s t gic AEDF ni tip. Chng minh rngPA

PD= (tanB+ tanC) cot

A

2.

b) Gi CP ct ng thng qua B vung gc AB ti M . BP ct ng thng qua C vunggc AC ti N. K l hnh chiu ca A ln MN. Chng minh rng BKC + MANkhng i khi P di chuyn trn AD.

Bi 5. Tm tt c a thc P(x ),Q(x ) c h s nguyn v tha mn iu kin: Vi dy s(xn) xc nh bi

x0 = 2014, x2n+1 = P(x2n), x2n = Q(x2n1) vi n 1.

th mi s nguyn dng m l c ca mt s hng khc 0 no ca dy (xn).

Vit Nam Team Selection Test 2014 3Bi 6. Cho m,n, p l cc s t nhin khng ng thi bng 0. Khng gian ta c chiathnh cc mt phng song song cch u nhau. Mt cch in vo mi khi lp phngn v mt trong cc s t 1 n 60 c gi l cch in in Bin nu tha mn: trongmi hnh hp ch nht vi cc mt trn h mt cho v tp hp kch thc ba cnh (shnh lp phng trn trn cnh) xut pht t mt nh l 2m + 1,2n + 1,2p + 1, khi lpphng n v c tm trng vi tm ca hnh hp ch nht c in s bng trung bnhcng ca cc s in tm ca 8 hnh lp phng cc gc ca hnh hp . Hi c tt cbao nhiu cch in in Bin?

Nhng cch in l ging nhau nu trong mi cch, ta chn mt hnh lp phng lm tmth cc s c in vo cc khi lp phng n v c cng v tr tng i vi tm tronghai cch in l bng nhau.

Vit Nam Team Selection Test 2014 4Nhn xt chung

Xt v cu trc, thi chn i tuyn thi IMO nm nay gm 6 bi ton, trong c 2 bihnh hc phng, 2 bi i s v 2 bi t hp. C th nh sau:

Bi 1: i s (phng trnh hm trn tp s nguyn). Bi 2: T hp (bi ton m). Bi 3: Hnh hc phng (ng v im c nh). Bi 4: Hnh hc phng (hnh hc tnh ton). Bi 5: i s (phng trnh hm a thc trn tp s nguyn). Bi 6: T hp (bi ton m)

Ngoi hai bi hnh ra th tnh cht s hc xut hin c bn bi cn li nhng ch ngcc vai tr lm nn xy dng cc vn ch khng thc s l cc bi s hc. Trong mingy, cc bi ton c sp xp mt cch tng i theo kh tng dn. Bi 1 v 4 c thcoi l hai bi d. Tip theo l bi 2, bi 5 mc trung bnh v kh nht l bi 3 v bi6. Theo kin ca chng ti, lt vo i tuyn, cc th sinh chc chn s phi gii quytc bi 1,4 v t nht lm c hon chnh 1 bi trong cc bi 2,5 (cng l cc bi quytnh), cn li kh c th sinh no gii quyt c trn vn bi 3 v 6.

Hai bi hnh hc nm nay, xt v tnh phn loi th n, c mt bi d v mt bi kh, trong cu 3b s l thch thc ng k. Bi 4 kh d khi xt li cc bi cng v tr k thi ccnm trc. Tuy nhin, nu t vo chung vi cc bi 5 v 6 trong mt ngy thi th li khhp l v s to mt th thch nh cho th sinh trc khi bc vo hai th thch ln hn. Vbn bi cn li, kin thc c s dng ni chung rt quen thuc, i hi t duy thun tyv trnh by cn thn ch khng dng cc nh l kh. Bi 1 l phng trnh hm trn tps t nhin c tnh vo i s, nu nm vng cch x l phng trnh hm th s vtqua bi ny khng my kh khn. Bi 5 v a thc, cng l i s nhng c lin quan ncc tnh cht s hc, gii tch. Bi 2 v 6 u l cc bi t hp nhng c t trn nn shc, xu hng thng thy trong cc VMO, TST nhng nm gn y.

Ta c th thy mt im kh trng hp ba bi 2,5 v 6 l mi bi u c khai thc yut chiu v v vy m bi ton c th hnh dung d hn khi a v trng hp 1 chiu. Babi ny cng c nhiu im kh, c v cch tip cn, lp lun cht ch trong trnh by ligii ln p s cui cng ca bi ton. So vi thi cc nm trc th c th thy TST2014 c phn nh hn, khng c bi qu kh v hai bi d ca k thi cng d hn. Tuy vy, thi vn m bo tnh phn loi cao, cc bi hnh i hi nm cc k thut nht nh x l cn cc bi khc th li i hi t duy l chnh, hn ch s dng cc nh l, cng cmnh l mt c im ni bt v cng l u im ca thi ny.

thi, li gii v bnh lun di y c tham kho t bnh lun ca GS. Nguyn TinDng trn page SputnikEducation v s ng gp ca cc bn: traum (L Hng Qu),chemthan (Nguyn Ngc Trung), LTL (Nguyn Vn Linh), blackholes (Hunh Vic Trung),haojack123 (Trn Anh Ho) cng mt s thnh vin khc ca din n mathscope.org.

Vit Nam Team Selection Test 2014 5Li gii chi tit v bnh lun

Bi 1.

Tm tt c cc hm f : Z Z tha mnf2m + f (m) + f (m) f (n)

= n f (m) +m, m, n Z. ()

Li gii. t a = f (0). R rng f 0 khng phi l nghim ca phng trnh, do tnti m0 Z sao cho f (m0) 6= 0. T , bng cch thay m = m0 vo ng thc () ta ddng suy ra f n nh. Mt khc, thay n = 0 vo (), ta c

f2m + (a+ 1) f (m) = m, m Z. ()

Kt qu ny chng t f l ton nh. Do , b Z sao cho f (b) = 1. Thay m = n = bvo (1), ta c f (2b) = 0. Mt khc, thay m = n = 0 vo (), ta cng c f (a2 + a) = 0.T y, kt hp vi tnh n nh ca f , ta suy ra

b =a2 + a

2.

By gi, thay n = b vo (), ta c

f (2m) =a2 + a

2 f (m) +m, m Z. ( )

Tip tc, thay m = 0 vo (), ta cfa f (n) + a = an, n Z.

n y, bng cch thay m = an vo () ri so snh vi ng thc trn, ta suy ra(a+ 1) f (an) + 2an = a f (n) + a, n Z.

Cn nu thay n = b, ta ca(a2 + a)

2= f (0) = a, t suy ra a {0, 1, 2}. Ta xt cc

trng hp sau:

Nu a = 1 th d dng suy ra f (n) = 1 2n, n Z. Tuy nhin, khi th li ta thyhm ny khng tha mn yu cu bi.

Nu a = 0 th t (), ta suy ra f (2m) = m, m Z. So snh kt qu ny vi (),ta c f (m) = 0, m Z, v l v f phi l mt hm khc hng.

Nu a = 2, ta cf (2n) + 4n = 2 f (n) + 2, n Z.

Vit Nam Team Selection Test 2014 6Mt khc, theo ( ), ta li c f (2n) = f (n) n. Do , ta c

f (n) + 3n = 2 f (n) + 2, n Z.Thay n bi n vo ng thc trn, ta cng c f (n) 3n = 2 f (n) + 2, n Z.Kt hp hai iu ny li, ta suy ra f (n) = n 2, n Z. Th li, ta thy hm nytha mn yu cu bi.

Vy hm s cn tm l f (n) = n 2, n Z.Nhn xt.

bi ton trn, tnh n nh ton nh c nhn thy kh d dng v n cn ng vimi s thc ch khng ch trn tp s nguyn. y cng l iu tin quyt cn c victhay th cc gi tr c th thc hin d dng.

Ngoi cch gii trn, ta cng c th d on ra biu thc ca hm s v bng cch chncc s c bit, ta c th x l nhanh chng c nhu cu ny. Vic pht hin ra hm sf (n) = n2 ni chung khng kh v nu hm s cn tm l a thc th d thy bc ca nl 1 v t thay vo ng nht h s. Bng cch thay cc gi tr thch hp, ta tnh c

f (4) = 2, f (3) = 1, f (2) = 0, f (1) = 1, f (0) = 2.Thay m = 3 vo cng thc cho, ta c ngay

f7+ f (n)

= n + 3 vi mi n.

T y c th quy np c kt qu trn. Mt cch khc cng kh nhanh nh sau:

t a = f (0) v u, v,w l cc s nguyn tha f (u) = 0, f (v) = 1, f (w ) = 1. Ta thchin thay cc gi tr thch vo biu thc ca hm s cho nh sau: Nu m = w,n = uwth f

2w + 1+ f (u w ) = u.

Nu m = u,n = 0 th f (2u) = u. Suy ra

2w + 1+ f (u w ) = 2u f (u w ) = 2(u w ) 1.Tip tc thay m = x Z,n = u w th f 2(x + (u w ) f (x )) = x + (u w ) f (x ).Cui cng, thay m = x + (u w ) f (x ),n = v th

x + (u w ) f (x ) = u hay f (x ) = u xu w .

y l biu thc tuyn tnh theo x , ta d dng chn c cc gi tr thch hp.

Ngoi ra, ta cn c th tip cn theo cch sau: Ta chng minh c rng f (2m f (m)) = mv do f song nh nn f (m)+ f 1(m) = 2m. y l bi xut hin trong k thi APMO 1989:Xc nh tt c cc hm s f : R R tha mn

f n iu tng. f (x ) + g (x ) = 2x vi g l hm ngc ca f .

Vit Nam Team Selection Test 2014 7Bi 2.

Trong mt phng ta vung gc Ox y, xt cc im nguyn c ta thuc tphp sau

T =(x ; y) : 20 x , y 20, (x ; y) 6= (0; 0) .

T mu cc im thuc T sao cho vi mi im c ta (x , y) T th c ng mttrong hai im (x ; y) v (x ;y) c t mu. Vi mi cch t nh th, gi N l scc b (x1; y1), (x2; y2) m c hai im ny cng c t mu v

x1 2x2, y1 2y2 (mod 41).Tm tt c cc gi tr c th c ca N.

Li gii. Trc ht, ta chuyn bi ton t 2 chiu thnh 1 chiu.

Ta c 210 1(mod 41) nn 40 s nguyn khc 0 c gi tr tuyt i khng vt qu 20c th chia thnh 2 dy, mi dy c di 20 sao cho nu s hng u chia 41 d x th shng sau chia 41 d 2x .

T mu cc s ca dy ny sao cho trong 2 s i nhau th c ng 1 s c t mu. Taquan tm n s lng cc cp s lin tip cng c t mu trong dy.

D thy 220 1(mod 41) nn nu thm s hng u ca mi dy vo cui th dy migm 21 s vn tha mn tnh cht trn nn ta c th chuyn thnh vng trn v pht biuli bi ton nh sau: Cho a gic u 20 nh ni tip trong mt ng trn sao cho trong2 im i xng qua tm th c ng 1 nh c t mu. Tnh s cc cp nh lin tipc t mu c th c.

Vi n l s chn, gi Sn l tp hp s cc cp k nhau cng c t mu c th c ca agic c n nh. Ta s chng minh bng quy np rng

S4n = {2k + 1|0 k n 1} v S4n2 = {2k|0 k n 1}.iu ny c th chng minh bng quy np nh sau:

Vi n = 2 th d thy nhn xt ng.

Gi s nhn xt ng n n 2. Xt a gic c 4n + 2 nh. a gic ny c th to thnhbng cch thm nh A vo gia hai nh th 2n, 2n + 1 v thm nh B vo gia hai nhth 4n, 1. Ta xt cc trng hp:

Nu nh 2n, 2n+1 u c t v A khng c t th tng ng: nh 4n, 1 khngc t v B c t. S cp k nhau cng c t gim i 1.

Nu nh 2n, 2n + 1 u c t v A cng c t th tng ng: nh 4n, 1 khngc t v B khng c t. S cp k nhau tng ln 1.

Nu trong hai nh 2n, 2n + 1 c 1 nh c t v A cng c t th tng ng:trong hai nh 4n, 1c 1 nh c t v B khng c t. S cp k nhau tng ln 1.

Nu trong hai nh 2n, 2n + 1, c 1 nh c t v A khng c t th tng ng:trong hai nh 4n, 1 c 1 nh c t v B c t. S cp k nhau tng ln 1.

Vit Nam Team Selection Test 2014 8Do , S4n+2 = {x 1|x S4n} hay S4n+2 = {2k|0 k n}.Tng t, ta cng c S4n+4 = {x 1|x S4n+2}. Tt nhin khng xy ra trng hp S4n+4c cha s 1 v c trng hp 0 cp s 4n+2, cc nh phi c t xen k v trnghp gim i s b khng xy ra, suy ra S4n+4 = {2k + 1|0 k n}.R rng trong cch chng minh trn, ta cng ch ra c cch xy dng cc trng hp c th t mu tha mn c tt c cc gi tr trong tp hp tng ng. Nhn xt cchng minh.

T suy ra S20 = {1, 3, 5, 7, 9} .Do , kt qu cho bi ton ph s l 2S20 vi nh ngha 2S = S+ S = {a+ b|a, b S}.Quay tr li bi ton ban u, chuyn t 1 thnh phn x trn thnh 2 thnh phn(x , y), ta c th thc hin nh sau:

ng vi mi vng trn cha cc s thuc dy A, ta ly mt vng trn mi cng gm cc sthuc dy A t ln sao cho mi s thuc ng trn c khp vi ng mt s thucng trn mi. Vit cc cp s khp nhau thnh mt dy, dy chnh l dy cc ta im m lin sau ca (x1, y1) l (2x1, 2y1) theo mod 41.

D thy c tt c 20 cch ghp nh th (c nh vng trn c v xoay vng trn mi).Tng t vi vic ghp cc dy A B,B A,B B nn c tng cng l 80 cch ghp tothnh 80 dy. Tuy nhin, ta cng xt thm 4 dy c bit, tng ng vi cc im nm trntrc tung v trc honh.

C th l xt thm dy C gm 20 s 0 v xt 4 cch ghp: A C ,C A,B C ,C B. Do, tng cng c 84 dy cc ta .Theo chng minh trn th mi dy, s cc cp c th c l S20 vy nn p s ca biton l 84S20, cng chnh l cc s chn t 1 84 = 84 n 9 84 = 756. Bi ton c giiquyt hon ton.

Nhn xt.

gii c bi 2, ta phi cm nhn c bi ton. Mt cch t nhin khi ly x , y thngh n vic tm 2x , 2y (thay v ngc li l xt cc im cng c t mu). T , khibit cc s c th xoay thnh mt vng trn th s nhn thy c min gi tr kia chnh ls cnh v vic a thnh mt m hnh nh nu s d trnh by nht, mi th u sngsa hn nhiu. S c ch khi ta pht biu bi ton tng qut sau lm vic vi cc tham snh tm ra quy lut. Bi ny bn cht khng kh, nhng rt d sai nht l khi kt lunvi vng thng qua mt s nhn xt nh.

Theo kin chng ti, vic chm bi ny kh l kh khn trong vic nh gi cc bi giitheo hng 7 hay 0+ . im mu cht y l vic tch ring honh , tung , mhnh ha di dng a gic u v ch n ng d thc 210 = 1024 1(mod 41).Bi ton ny c phong cch ca bi 4 VMO 2012, bi 3 VMO 2013, u l cc bi kh mim, l dng t hp m c kt hp s hc, tuy pht biu hi c phn gng p trong vict vo trc ta nhng li gip hc sinh d tng tng hn.

Vic gii quyt bi ton trong trng hp 1 chiu mang tnh quyt nh x l bi tonvi nhiu chiu hn. Chng hn ta pht biu li bi ton trn nh sau:

Vit Nam Team Selection Test 2014 9Trong khng gian Ox yz, xt cc im nguyn c ta thuc tp hp

T =(x ; y; z) : 20 x , y, z 20, (x ; y; z) 6= (0;0;0) .

T mu cc im thuc T sao cho vi mi im c ta (x ; y; z) T th c ng mttrong hai im (x ; y; z) v (x ;y;z) c t mu. Vi mi cch t nh th, gi N ls cc b (x1; y1; z1), (x2; y2; z2) m c hai im ny cng c t mu v cc ta cachng tha mn iu kin x1 2x2, y1 2y2, z1 2z1(mod 41). Tm tt c cc gi tr cth c ca N.

Lp lun tng t nh trn nhng y, ta li t thm mt vng trn na trn hai vngtrn c. trn ta c ord41(2) = 20 nn nu hp thay 20 v 41 bi cc s khc. Ta thy ccs 41,20 quyt nh di chu k ca cc dy trn v trong trng hp s lng cc schia ht cho chu k ny, ngha l ta c th chia u dy s cho thnh cc vng trn thbi ton c gii quyt tng t. Nh th, ta i n bi ton tng qut:

Vi k,n, p, r l s nguyn dng m ordp(r ) (2k), xt tp hp T cc b s nguyn khng

ng thi bng 0 l (x1, x2, , xn) sao cho |x i | k, i = 1,n. T mu cc s thuc T saocho c ng mt trong hai b (x1, x2, , xn) v (x1,x2, ,xn) c t mu. Vimi cch t nh th, gi N l s cc b (x1, x2, , xn) v (x 1, x 2, , x n) tha mn:

C hai b ny cng c t mu. x i = r x i(mod p). vi i = 1,2,3, ,n.

Khi , tt c cc gi tr c th c ca N l RSordp(r ) vi R =1

k

2k2

ordp(r )+ 1

n 1

v

s S2k c nh ngha nh trn.

Tt nhin, c nhng trng hp dy khng lp li th phi tnh ton theo cch khc v rrng ta khng th xy dng c thnh mt m hnh khp kn nh a gic u trn. Biton vn cn c th pht trin theo nhiu hng rt th v!

Bi 3.

Cho tam gic ABC ni tip trong ng trn (O). Trn cung BC khng cha A ca(O) ly im D. Gi s CD ct AB E v BD ct AC F . Gi (K) ng trn nmtrong tam gic EBD, tip xc vi EB,ED v tip xc vi ng trn (O). Gi (L) ltm ng trn nm trong tam gic FCD, tip xc vi FC , FD v tip xc vi ngtrn (O).

a) Gi M l tip im ca (K) vi BE v N l tip im ca (L) vi CF . Chngminh rng ng trn ng knh MN lun i qua mt im c nh khi D dichuyn trn (O).

b) ng thng qua M v song song vi CE ct AC P, ng thng qua N vsong song vi BF ct AB Q. Chng minh rng ng trn ngoi tip tam gicAMP,ANQ cng tip xc vi mt ng trn c nh khi D di chuyn.

Vit Nam Team Selection Test 2014 10Li gii. a) Trc ht, ta chng minh b sau:

B (nh l Sawayama v Thbault m rng). Cho tam gic ABC ni tip ng trn(O). D l mt im thuc tia i tia BA. ng trn (K) tip xc DA,DC ln lt ti M ,Nv tip xc ngoi (O). Chng minh rng MN i qua tm bng tip ng vi nh A ca tamgic ABC .

Chng minh. Gi s ng phn gic ngoi ti nh C ca tam gic ABC ct ng trn(O) ti E khc C . ng thng CE ct MN ti J . ng trn (K) tip xc vi ng trn(O) ti F . D dng thy rng E,N, F thng hng. Tht vy, v E l trung im cung ABcha C ca (O) nn OE AB. Ngoi ra, ta cng c MN AB nn MN OE. Mt khc,K ,O, F thng hng nn E,N, F cng thng hng.

T y, ta c tam gic EAF v ENA ng dng, suy ra EA2 = EB2 = EN EF .

O

BC

A

D

J

N

K

M

F

E

Ta li c

FMN =1

2F KN =

1

2FOE = FAE = FC J

suy ra t gic CFMJ ni tip. Do ,

EFJ = 180NFJ = 180(NFM+MFJ) = 180(JMC+MCJ) = MJC .T , ta c 4EFJ 4EJN, suy ra EJ2 = EN EF = EA2 = EB2.T y, bng bin i gc, d dng suy ra AJ l phn gic ca BAC nn J l tm bngtip ng vi nh A ca tam gic ABC . Ta c iu phi chng minh.

Vit Nam Team Selection Test 2014 11Tr li bi ton,

Gi s ng trn (K) tip xc ED ti G v ng trn (L) tip xc FD ti H . Theo b trn th MG,NH cng i qua tm bng tip J ng vi nh A ca tam gic ABC .

Ta s chng minh rng MJN = 90. Tht vy,

KEF+LFE =1

2DEB+

1

2DFC+DEF+DFE = 90BAC+180BDC = 90

do EK LF . Ta cng c MG EK v LF NH nn MG vung gc NH ti J . Vyng trn ng knh MN lun i qua im J c nh. Ta c pcm.

O

B C

D

A

E

F

J

MK N

L

G

H

P

Q

Y

Z

R

b) D thy tam gic EMG cn v MP CE nn EMG = EGM = GMP. Do , MJchnh l phn gic ngoi tam gic AMP m AJ l phn gic gc BAC nn suy ra J cng ltm bng tip gc A ca tam gic AMP.

Gi (R) l ng trn tip xc vi CA,AB ti Y,Z v tip xc ngoi vi ng trn (O).Theo b ban u th khi cho D hoc E chy ra xa v tn th cc ng trn (K), (L)trng vi ng trn (R) v cng c on ni hai tip im i qua J . T y, d dng suy raJ l trung im YZ. Do J l tm bng tip ca AMP nn ng trn (R) tip xc AP,AMti Y,Z cng tip xc ng trn ngoi tip tam gic AMP.

Vit Nam Team Selection Test 2014 12Chng minh tng t, ta cng c (R) cng tip xc vi ng trn ngoi tip tam gicANQ. Vy ng trn ngoi tip tam gic AMP,ANQ lun tip xc ng trn (R) c nh.Ta c pcm.

Nhn xt. Vi bi ton ny, nu ai quen thuc vi nh l Sawayama v Thbault vcc m rng ca n th gii quyt khng qu kh khn. Nhng ni chung th vi hu htcc th sinh, mt mt phi xc nh c m hnh nu ra trong bi chnh l dng mrng; mt khc, cc bn phi chng minh li thnh cng nh l ny trong thi gian tngi ngn, bi ton ny xng ng v tr s 3. y l ln u tin trong thi chn ituyn ca Vit Nam c mt bi hnh v tr ny (mt lut bt thnh vn l bi 3 chnh lbi kh nht thi).

Trc ht, ta nhc li nh l Sawayama v Thbault dng gc: Cho tam gic ABC ni tipng trn (O) . D l mt im thuc on BC . ng trn (K) tip xc DA,DC ln ltti M ,N v tip xc trong vi (O) . Chng minh MNi qua tm ni tip tam gic ABC .

Trc ht, ta th tp trung vo khai thc a. R rng c th thay ng trn ngoi tipthnh ng trn bt k qua B,C , ta thu c mt bi ton rt th v. Ch rng cua) ni ln MG,NH i qua im Jc nh. Ta c cc bi ton sau:

Bi 3.1. Cho tam gic ABC v mt ng trn (O) bt k c nh i qua B,C . D l imdi chuyn trn (O) sao cho A,D khc pha vi B,C . Gi s CD ct AB E v BD ct AC F. Gi (K) l ng trn tip xc vi EB,ED ln lt ti M ,Nv tip xc trong vi (O) .Gi (L) l ng trn tip xc vi FC , FD ln lt ti P,Q v tip xc trong vi (O) . Chngminh rng giao im MN,PQ lun nm trn mt ng trn c nh khi D di chuyn.

Bi 3.2. Cho tam gic ABC ni tip ng trn (O) . Gi P l im di chuyn trn cungBC cha A ca (O) . PB,PC ct CA,AB ln lt ti E, F. ng trn (K) tip xc vi onEA,EB v tip xc trong vi (O) . ng trn (L) tip xc vi on FB, FC v tip xc vicung BC khng cha A ca (O) . Chng minh rng ng trn ng knh MN lun i quamt im c nh khi P di chuyn.

Bi 3.3. Cho tam gic ABC v (O) c nh i qua B,C . Gi D l im di chuyn trn cungBC ca (O) sao cho D,A cng pha vi BC . DB,DC ct CA,AB ln lt ti E, F. ngtrn (K) tip xc vi on EA,EB ti M ,N v tip xc trong vi (O) . ng trn (L) tipxc vi on FB, FC ti P,Q v tip xc (O) ti mt im khng cng pha A so vi BC .Chng minh rng giao im ca MN v PQ lun thuc mt ng trn c nh.

Li gii chi tit ca cc bi ton trn c th tham kho trong [2].

th hai ca bi ton l mt xy dng th v khi i hi phi xy dng thm ng trnMixtilinear bng tip gc A ca tam gic ABC . Tnh cht s dng trong li gii trn khquen thuc v do chng minh c nh l Sawayama v Thbault nn khng cn thchin li chng minh tnh cht ny. Mt bi ton c lin quan xut hin cch y kh lutrong chn i tuyn nm 1999 ca Vit Nam:

Bi 5 VN TST 1999. Cho tam gic A1A2A3 ni tip trong ng trn (O). Mt ng trn(K1) nm trong gc A2A1A3 ca tam gic A1A2A3, tip xc vi cc cnh A1A2,A1A3 vtip xc trong vi ng trn (O) ln lt ti cc im M1,N1,P1. Cc im M2,N2,P2 vM3,N3,P3 xc nh mt cch tng t. Chng minh rng cc on thngM1N1,M2N2,M3N3ct nhau ti trung im mi on.

Vit Nam Team Selection Test 2014 13Bi 4.

Cho tam gic ABC nhn, khng cn c ng cao AD v P thuc AD. Cc ngthng PB,PC ln lt ct CA,AB ti E, F.

a) Gi s t gic AEDF ni tip. Chng minh rngPA

PD= (tanB+ tanC) cot

A

2.

b) Gi CP ct ng thng qua B vung gc AB ti M . BP ct ng thng quaC vung gc AC ti N. K l hnh chiu ca A ln MN. Chng minh rngBKC +MAN khng i khi P di chuyn trn AD.

Li gii. a) Gi Q l giao im ca EF v BC th hng im (Q,D,B,C) iu ha. T ysuy ra chm (AQ,AD,AB,AC) v (DE,DF,DA,DQ) cng iu ha.

Hn na, do DA DQ nn DA l phn gic trong ca EDF . Do t gic AEDF ni tipng trn nn d thy AE = AF .

A

B CD

PF

E

GQ

Gi s ng trn ngoi tip t gic AEDF ct BC ti G khc D. Ta c DG l phn gicngoi ti nh D nn GE = GF do 4AGE = 4AGF nn AG l phn gic BAC . Theonh l Menelaus cho ct tuyn BPE ca tam gic ADC v ct tuyn CPF ca tam gicABD th

EA

EC=

PA

PD BDBC

,FA

FB=

PA

PD CDBC

T y suy raFA

FB+

EA

EC=

PA

PD

BDBC

+CD

BC

=

PA

PD.

Do , ta c

PA

PD=

FA

GF.GF

FB+

EA

GE.GE

EC= cot

A

2tanB+ cot

A

2. tanC = (tanB+ tanC) cot

A

2.

Ta c pcm.

Vit Nam Team Selection Test 2014 14b) Gi L l giao im ca BM v CN th r rng im L c nh. T gic ACKN cACN = AKN = 90 nn ni tip. Tng t, ABKM cng ni tip. Do , ta cMAN +BKC = MAK +NAK +BKC = MBK +NCK +BKC = BLC

khng i. Ta c pcm.A

BCD

P

N

M

K

L

Nhn xt. y l bi ton mc d v th sinh c th x l mt cch nh nhng vi cckin thc quen thuc v hnh phng. Bi ton mang mu sc i s kh nhiu v ni chungi hi th sinh tnh ton, bin i (tng t nh bi 3 trong IMO 2013) hn l tip cntheo hng hnh hc thun ty. Hai cu a) v b) ca bi ton khng lin quan n nhau.Ta s tch ring thnh hai bi ton v phn tch tng bi ton mt. Cu a) pht biu iukin di dng mt biu thc lng gic nh vy cha p, ta hon ton c th c mt hthc lng thun ty hnh hc ca cu a). Ta xt bi ton sau:

Bi ton 4.1. Gi thit tng t nh bi ton. Chng minh rng nu t gic AEFD ni tip

thAD

PD=

AB AC + AD2DB.DC

.

y chnh l dng tng qut ca cu a ca bi ton gc v c th x l d dng bng nhl Van-Obel: Cho tam gic ABC c D thuc BC v P thuc on AD. Tia BP ct AC ti N,

tia CP ct AB ti M . Khi , ta cAD

AP=

AM

AB+

AN

AC. Bn cnh , li gii nu trn gi

cho ta nh n bi ton trong IMO Shortlist 1994 vi ni dung nh sau:

Vit Nam Team Selection Test 2014 15Bi ton 4.2. Cho tam gic ABC c ng phn gic AD. Gi E, F l hnh chiu ca D lncc cnh CA,CB. Gi s BE giao CF ti P. Gi H l hnh chiu ca P ln BC . Chng minhrng HP l phn gic EHF.Cui cng, ta cc ni dung ny, ta c th pht biu bi ton tng qut nh sau:

Bi ton 4.3. Cho tam gic ABC vi P l im bt k trong tam gic ABC . Cc ngthng PA,PB,PC ct AB,BC ,CA ln lt ti D,E, F. Gi H , K l hnh chiu ca F,E lnAD v L,N l hnh chiu ca F,E ln BC . Gi s t gic AEDF ni tip. Chng minh rngPA

PD=

FH

FL DADB

+EK

EN DADC

.

Ni v phn b, ta thy y ch l mt bi ton ng gic quen thuc. Ta c th thm mt s vo bi ton ny vn phong ph hn:

Bi ton 4.4. Cho tam gic ABC ng cao AD. im P di chuyn trn AD. Gi s PB,PCln lt ct cc ng thng qua C vung gc vi CA v qua B vung gc vi AB ti N,M .Gi K l hnh chiu ca A ln MN.

a) Chng minh rng MAN +BKC khng i khi P di chuyn.

b) Chng minh rng MAC = NAB.

c) Chng minh rng KA l phn gic BKC .

Di y l hai bi m rng theo con ng ng gic cho bi ton trn:

Bi ton 4.5. Cho tam gic ABC ni tip (O) . Cc im E, F c nh thuc (O) sao choEF BC . im P di chuyn trn AE. Gi s PB,PC ln lt ct FC , FB ti N,M . v trungtrc BC ct MN ti K . Chng minh rng MAN +BKC khng i khi P di chuyn.Bi ton 4.6. Cho tam gic ABC ni tip ng trn (O) . E, F c nh thuc (O) sao choEF BC . Cc im P,Q ln lt thuc AE,AF. Gi s PB,PC ln lt ct QB,QC ti N,M .Chng minh rng MAB = NAC .Li gii chi tit ca cc bi ton trn c th tham kho trong [3].

Bi 5.

Tm tt c a thc P(x ),Q(x ) c h s nguyn v tha mn iu kin: Vi dy s(xn) xc nh bi

x0 = 2014, x2n+1 = P(x2n), x2n = Q(x2n1) vi n 1.

th mi s nguyn dng m l c ca mt s hng khc 0 no ca dy (xn).

Li gii. Ta s chng minh rng nu cc a thc P(x ),Q(x ) tha mn bi th chngphi u c bc l 1. Tht vy,

Xt trng hp mt trong hai a thc P(x ),Q(x ) l a thc hng. (1) Nu P(x ) a,x Z th (xn) c dng:

x0 = 2014, x1 = a, x2 = Q(a), x3 = a,

Vit Nam Team Selection Test 2014 16v d thy mi s hng ca dy ch nhn 1 trong 3 gi tr (2014,a,Q(a)).

(2) Nu Q(x ) a,x Z th (xn) c dngx0 = 2014, x1 = P(2014), x2 = a, x3 = P(a),

v d thy mi s hng ca dy ch nhn mt trong ba gi tr (2014,P(2014),a,P(a)).

C hai iu ny u khng tha mn iu kin bi do mi s nguyn dng m phi lc ca mt s hng khc 0 no ca dy s xn.

Tip theo, nu mt trong hai a thc P(x ),Q(x ) c bc ln hn 1, khng mt tnh tngqut, ta gi s l Q(x ). R rng khi Q(P(x )) cng c bc ln hn 1.

Ta thy nu R(x ) l a thc c bc ln hn 1 th vi mi k > 0 ln ty , tn ti x c gi trtuyt i ln sau cho |R(x )| > k |x | . iu ny l d thy do khi x + th ta c giihn

limx+

|R(x )||x | = +.

Ta s chng minh rng tn ti N ln sao cho |Q(P(x ))| > |P(x )|+ |x | vi mi x > N.Ta ch cn xt 2 trng hp: Nu P(x ) l bc 1 th d thy tn ti k ln sao chok |P(x )| > |x | vi mi x Z.Suy ra, tn ti N ln sao cho

|Q(P(x ))| > (k + 1) |P(x )| = |P(x )|+ k |P(x )| > |P(x )|+ |x | vi mix > N.Nhn xt c chng minh.

Theo gi thit th trong dy s cho, phi tn ti s hng |x i | ln ty v r rng, ta cngphi c j > 0 sao cho

x2 j > N + 1 v x2 j c gi tr tuyt i ln nht trong 2 j s hngu tin ca dy (xn).

Tht vy, ta thy rng tn ti v s s hng x2 j tha mn iu kinx2 j = max

0k2 j{|xk |},

gi T l tp hp cc ch s tha mn. Nu nh trong cc s hng nh th, khng c s hngno tha mn

x2 j > N + 1 th vi mi t T , ta c |x i | |x t | N + 1 vi mi i t T .Tuy nhin, do |T | v hn nn iu gi s trn l v l v nhn xt c chng minh. Vix2 j l s hng tha mn iu kin trn, chn m =

x2 j+2 x2 j th ta thym =

Q(P(x2 j)) x2 j Q(P(x2 j)) x2 j > P(x2 j) > x2 j vx2n+2 = Q(x2n+1) > x2n+1 .Do , trong 2 j + 1 s hng u tin ca dy, khng c s hng no chia ht cho m.

Mt khc, x2k+2x2k = Q(P(x2k))Q(P(x2k2)) ...P(x2k) P(x2k2) ...x2k x2k2 =m v tng t th x2k+3 x2k+1 = P(x2k+2) P(x2k) ...x2 j x2 j2 = m. T y suy ravi k j th x2k+2 x2k v x2k+3 x2k+1 u chia ht cho m, tuy nhin x2 j+1 v x2 j+2 ukhng chia ht cho m nn xk khng chia ht cho m vi k > 2 j + 2. Do , trong dy cho, khng c s hng no chia ht cho m.

Vit Nam Team Selection Test 2014 17iu mu thun ny cho ta thy nhn xt ban u l ng v degP(x ) = degQ(x ) = 1.t P(x ) = ax + b vi |a| > 1 v Q(x ) = cx + d vi a, b, c,d Z v ab 6= 0 th ta c

x2n+1 = ax2n + bx2n+2 = cx2n+1 + d

vi mi n 0.

Suy ra x2n+2 = cax2n + bc + d v x2n+3 = cax2n+1 + ad + b vi mi n 0.C hai dy ny u c cng thc truy hi dng yn+1 = kyn + h vi k = ac,h Z. Gi sk 6= 1 th cng thc tng qut ca dy ny l yn = kn y0 + h kn1k1 vi mi n. R rng nuk = 1 th dy s tng ng khng tha mn, ta xt k 6= 1.

Nu h = 0 th ta c yn = kn y0, r rng khng tha mn iu kin.

Nu h 6= 0 th do gcd

k,

kn 1k 1

= 1 vi mi n nn gi s gi s t l s m ln

nht m k t |h th cc s nguyn dng c dng k s vi s > t u khng l c cabt c s hng no ca dy, khng tha mn. T y, suy ra k = 1 hay ac = 1.

Cui cng, ta ch cn xt 2 trng hp:

Nu P(x ) = x+a v Q(x ) = x+b vi a, b Z th bng quy np, ta chng minh cx2k = 2014+ k(a+ b) v x2k+1 = 2014+ a+ k(a+ b). D thy rng nu a+ b = 0th dy ny khng tha mn. Nu nh a+ b 6= 0 th gi S, T,R ln lt l tp hp ccc nguyn ca a+ b, 2014 v 2014+ a. Ta xt cc trng hp sau:

+ Vi m / S th gi s a+ b chia m d t vi t 6= 0; do , khi k chy qua mt hthng d y modulo m th tn ti mt s hng ca dy chia ht cho m vs lng cc s hng nh th l v hn. R rng s cc s hng bng 0 ca dyl hu hn (khng qu 2) nn tn ti s hng khc 0 ca dy chia ht cho m.

+ Vi m S m m / T,R th dy s tng ng khng tha mn.+ Vi m S v m T hoc m R th tng ng, mi s hng c ch s chn

hoc mi s hng c ch s l ca dy u chia ht cho m v d thy, tnti s hng khc 0 ca dy chia ht cho m. Do , cc s a, b phi tha mn(S\T) (S\R) = hay mi c ca a+ b phi l c ca 2014 hoc l c ca2014+ a.

Nu P(x ) = x + a v Q(x ) = x + b th cng c lp lun tng t v cc s hngca dy tng ng khi l x2k = 2014k(ab) v x2k+1 = 2014+a+k(ab).iu kin ca a, b l a b 6= 0 v mi c ca a b phi l c ca 2014 hoc lc ca a 2014.

Vy tt c cc a thc P(x ),Q(x ) cn tm l

P(x ) = x + a,Q(x ) = x + b trong a, b Z tha mn a + b 6= 0 v mi c caa+ b phi l c ca 2014 hoc l c ca 2014+ a.

P(x ) = x + a,Q(x ) = x + b trong a, b Z tha mn a b 6= 0 v mi cca a b phi l c ca 2014 hoc l c ca 2014+ a.

Vit Nam Team Selection Test 2014 18

Nhn xt.

y l bi ton lin quan n cc tnh cht s hc ca a thc. Trong cc bi ton ny,c tnh cht c bn sau y kh hu dng: Nu P(x ) l a thc c h s nguyn th(a b)| (P(a) P(b)) vi a 6= b. Hc sinh cng phi vn dng tt tnh n iu ca athc khi bin ln v mt cht kho lo trong vic x l cc k thut nhng vn nh.Bi ton yu cu s cn thn, chi tit; suy lun mt cch logic, khoa hc, c th t hp l.C th ni y l mt bi ton hay, mi l v c nhiu by trong .

Thot nhn vo bi ton ny, ta c th thy khng d hnh dung ra m hnh v cch tipcn v c ti 2 a thc khng lin quan n nhau xut hin. Mt suy ngh t nhin l nngii th vi trng hp P,Q trng nhau.

Khi ta c bi ton ch vi mt a thc v cch tip cn cng d nhn ra hn, thm chcc hc sinh c th x l tt hn nu c bit v bi ton sau:

(Arthur Angel, Problem Solving Strategies) Cho a thc P(x ) vi h s nguyn tha mniu kin P(n) > n vi mi s nguyn dng n. Xt

x1 = 1, x2 = P(x1), x3 = P(x2), Bit rng vi mi s nguyn dng N, tn ti mt s hng ca dy s chia ht cho N. Chngminh rng P(n) = n + 1.

Ta c c xn+2 xn+1 chia ht cho xn+1 xn vi mi n, ta suy ra cc phn t ca dy sxn+1, xn+2, xn+3... khi chia cho xn+1 xn c cng s d.Cng vi tnh tng ca a thc, ta s thy nu xn+1 qu ln so vi xn th a thc tngng s khng th tha mn iu kin bi ton. Cch tip cn ny chc hn s mang li mtci nhn va bao qut, va n gin hn cho bi s 5.

Li gii trn c gng lm sng t tt c cc vn t ra trong bi ton nn tng idi. Ta thy rng t tng chnh l loi trng hp bc 0 ca cc a thc a v xt 2trng hp quan trng:

t nht mt trong hai a thc c bc ln hn 1. Bc ca c hai a thc u bng 1.

Tuy nhin, ta cng cn hiu nguyn nhn ti sao cn phi chia ra nh th, tc l ti sao lplun trng hp (1) li khng p dng c cho trng hp (2)? Cu tr li nh sau:cch lp lun (1) i hi cn c c hai bt ng thc sau

|Q(P(x )) x | > |x | v |Q(P(x )) x | > |P(x )|u phi ng (*). Khi , dng tnh cht n iu ca a thc ch ra mt s nguyndng m khng l c ca s hng khc 0 no ca dy v dn n mu thun.

T vic xc nh r bn cht ca vn y, ta c th x l bi ton theo mt cch ngngn hn (a v trng hp h s ca c hai a thc u c gi tr tuyt i l (1) nh sau:

Vit Nam Team Selection Test 2014 19p dng lp lun tng t trng hp (1), ta c th gii quyt trng hp (2) nh sau:

Nu h s cao nht ca mt trong hai a thc P(x ),Q(x ) c gi tr tuyt i ln hn hocbng 3 th tnh cht (*) vn ng.

Nu h s cao nht ca c hai a thc P(x ),Q(x ) u khng vt qu 2 th ta xt hiuxn+4 xn v lp lun tng t.Cc bn th t tng trn tm ra mt cch lp lun ngn gn hn cho bi ton trn!V mt hng m rng, chng ta th xem xt bi ton trong trng hp thay 2 a thc bi3 a thc nh sau:

Tm tt c a thc P(x ),Q(x ),R(x ) c h s nguyn v tha mn iu kin: Vi dy s(xn) xc nh bi x0 = 2014, x3n+1 = P(x3n), x3n+2 = Q(x3n+1), x3n+3 = R(x3n+2),n 0,th mi s nguyn dng m l c ca mt s hng khc 0 no ca dy (xn).

Cc bn th xem xt c im chung c bn cng nh ci kh ca bi ton mi ny hocbi ton tng qut khi thay 2,3 bi k 4 a thc so vi bi ton gc l g nh!

Bi 6.

Cho m,n, p l cc s t nhin khng ng thi bng 0. Khng gian ta cchia thnh cc mt phng song song cch u nhau. Mt cch in vo mi khi lpphng n v mt trong cc s t 1 n 60 c gi l cch in in Bin nutha mn: trong mi hnh hp ch nht vi cc mt trn h mt cho v tp hpkch thc ba cnh (s hnh lp phng trn trn cnh) xut pht t mt nh l2m + 1,2n + 1,2p + 1, khi lp phng n v c tm trng vi tm ca hnh hpch nht c in s bng trung bnh cng ca cc s in tm ca 8 hnh lpphng cc gc ca hnh hp . Hi c tt c bao nhiu cch in in Bin?

Nhng cch in l ging nhau nu trong mi cch, ta chn mt hnh lp phnglm tm th cc s c in vo cc khi lp phng n v c cng v tr tng ivi tm trong hai cch in l bng nhau

Li gii. Khng mt tnh tng qut, gi s cc mt phng song song trong h mt cch unhau mt khong l 1 n v. Ta chn mt hnh lp phng no ri ly tm O ca nlm tm ca khng gian vung gc, cc trc Ox ,Oy,Oz song song vi cc ng thngca h cho v chiu dng cc trc ta chn ty . Khi , ta quy c ta camt hnh lp phng trong h mt cho l khong cch t hnh chiu ca n ln cc trcOx ,Oy,Oz n O (chn du ty vo chiu m hay chiu dng ca cc trc ta ).

Xt mt cch in c tnh cht in Bin no v ta gi hai hnh lp phng phi cin cng mt s trong cch in ny l c quan h vi nhau. m s cch in inBin, ta m s ln nht cc hnh lp phng n v c th c chn sao cho cc hnh imt khng c quan h vi nhau, t S l s lng hnh ny.

Trc ht, ta s chng minh rng vi mi hnh hp ch nht c kch thc cc cnh l2m + 1,2n + 1,2p + 1 th 8 s in gc v s in tm l bng nhau.

Tht vy, Do cc s c dng in l nguyn dng nn phi tn ti mt s nh nht,gi s l a v hnh lp phng A no c in s a ny. Khi , hnh hp ch nht c

Vit Nam Team Selection Test 2014 20kch thc 2m + 1,2n + 1,2p + 1 v cc hnh c A l tm s c 8 s in 8 nh ukhng nh hn a m li c trung bnh cng bng a. Suy ra 8 s in cc nh v s in tm ca hnh hp ch nht ny l ging nhau. Lp lun tng t, ta c th m rng ratrong khng gian v c mt tp hp T cc hnh lp phng in s a.

Vi cc hnh lp phng khng thuc T , gi s b > a l s nh nht c in cho cc hnhlp phng cn li v tng t trn, ta c mt tp hp T gm cc hnh lp phng ins b. R rng c nh th, ta thy nhn xt trn c chng minh v y cng chnh l iukin cn mt cch in l c tnh cht in Bin.

Nh th, hnh nm ta (x , y, z) s c quan h vi hnh nm ta

(x + (1)rm, y + (1)sn, z + (1)t p).Cc s r, s , t {0;1} c chn mt cch c lp vi nhau (l 8 nh ca hnh hp chnht c tm l (x , y, z) v cc cnh l 2m + 1,2n + 1,2p + 1). Hn na, y ta xthnh hp ch nht c cnh 2m+1,2n+1,2p +1 tng ng song song vi cc mt phngOyz,Ozx ,Ox y v ta hon ton c th hon i th t ny.

T y suy ra, hnh lp phng ta (x , y, z) s c in s ging vi hnh lp phng ta

x + a1m + a2n + a3p, y + b1m + b2n + b3p, z + c1m + c2n + c3p

. Tuy nhin,

vi hai hnh c quan h vi nhau, khi xt s thay i cc thnh phn x , y, z tng ngth cc i lng m,n, p s hoc tng, hoc gim ch khng gi nguyn v mi ln nhth th tnh chn l ca a1+ a2+ a3, b1+ b2+ b3, c1+ c2+ c3 ng thi thay i. Ban u,xt v tr (x , y, z) th ba tng ny cng tnh chn l (cng bng 0) nn tnh chn l caa1 + a2 + a3, b1 + b2 + b3, c1 + c2 + c3 l lun ging nhau.

Mt h qu ca nhn xt trn l: hnh lp phng (x , y, z) s c quan h vi hnh lpphng ta (x + x1, y + y1, z + z1) vi (x1, y1, z1) l mt hon v ca b cc s(2km, 2kn, 2kp) v k Z,d m,n, p. ()Hn na, theo nh l Bezout th trong cc t hp tuyn tnh ca m,n, p, ta c d =gcd(m,n, p) > 0 chnh l t hp tuyn tnh c gi tr tuyt i nh nht. T y suyra cc hnh lp phng n v trong mt hnh lp phng ln hn c kch thc d d ds i mt khng c quan h vi nhau (v hai hnh lp phng n v bt k trong hnh nys c chnh lch ta nh hn d). T suy ra S l bi ca d3.

Khi , ta thy vi mi hnh lp phng d d d ty , t tn l A, th c 6 hnh lpphng c chung mt vi n nh hnh v di y:

Vit Nam Team Selection Test 2014 21t m1 =

m

d,n1 =

n

d, p1 =

p

d, ta xt cc trng hp sau:

Nu c ba s m1,n1, p1 u l, t m1 = 2m2 + 1,n1 = 2n2 + 1, p1 = 2p2 + 1 th rrng m = 2dm2 + d,n = 2dn2 + d, p = 2dp2 + d.

Do , theo (), hnh lp phng ta (m,n, p) s c quan h vi hnh cc ta (m d,n d, p d) (cc hnh k nh vi A). Trong m hnh trn, nu ta chnhnh A v 3 hnh k mt vi n, mi chiu mt hnh th 4 hnh ny i mt khngc quan h vi nhau v khng th chn s lng no ln hn 4 c. T , ta cS = 4d3.

Nu trong ba s m1,n1, p1 c 2 s l v 1 s chn, ta c th gi s m1,n1 l v p1chn. Ta li tm1 = 2m2+1,n1 = 2n2+1, p1 = 2p2 th r rngm = 2dm2+d,n =2dn2 + d, p = 2dp2.

Do , theo (), hnh lp phng ta (m,n, p) s c quan h vi hnh cc ta (m d,n d, p) (cc hnh k cnh vi A). Ta li hon i v tr ca m,n, p, tcl thay i chiu ca (x , y, z) th suy ra thm (m,n d, p d) v (m d,n, p d)c quan h vi A. Trong m hnh trn, nu ta chn hnh A v 1 hnh k mt vi n thhai hnh ny khng c quan h vi nhau v khng th chn s lng no ln hn 2c. T , ta c S = 2d3.

Nu trong ba s m1,n1, p1 c 1 s l v 2 s chn, ta c th gi s m1 l v n1, p1chn. Ta li t m1 = 2m2 + 1,n1 = 2n2, p1 = 2p2 th r rng m = 2dm2 + d,n =2dn2, p = 2dp2.

Do , theo (), hnh lp phng ta (m,n, p) s c quan h vi hnh cc ta (md,n, p) (cc hnh k mt vi A). Ta li hon i v tr ca m,n, p, tc l thayi chiu ca (x , y, z) th suy ra thm (m,n d, p) v (m,n, p d) c quan h viA. Trong m hnh trn, ta ch chn c hnh A ch khng chn c thm hnh nokhng c quan h vi n. T , ta c S = d3.

Cui cng, ta thy mi c lp vi nhau s c 60 cch in s (t 1 n 60) cho nnn s cch in in Bin tng cng l 60S vi S c xc nh nh sau: Trong ba s

m

gcd(m,n, p),

n

gcd(m,n, p),

p

gcd(m,n, p), nu c

3 s l th S = 4d3. 2 s l th S = 2d3. 1 s l th S = d3.

Nhn xt.

Con s 60 v thut ng in Bin trong bi ton c l nhc chng ta v nm 2014 l knim 60 nm chin thng in Bin Ph (07/05/1954 07/05/2014). Tuy nhin, cc bn

Vit Nam Team Selection Test 2014 22th sinh chc cng nhanh chng nhn ra rng s 60 khng c nhiu nh hng trong vici tm li gii.

Bi ton ny kh l v kh. C th ni khi x l cc bi ton m t hp th cng vic utin chnh l on p s v nu on thnh cng th nhiu khi mi th s d dng hn.im bt ng nht y li chnh l p s khi m trn thc t, c nhiu p s chahon ton chnh xc c a ra. iu ny c l xut pht t vic x l hai iu kin khichuyn v dng s hc cui cng khng cn thn. Nu thc s bnh tnh, v c thi gian, tac th xt th vi v d nh vi m,n, p thy c cc kt qu tng i khc nhau chkhng cng mt dng.

Tuy nhin, trong phng thi, cng vi vic bi 5 lm mt khng t thi gian, y thc s lmt th thch khng d.

y l mt bi ton hay, c nhiu t cho cc bn hc sinh th hin kh nng. Bi ton ihi nhng t duy n gin, nhng tinh t phn suy lun t hp, v s cn thn, khoa hctrong phn x l iu kin s hc cui cng. Nu cha c mt hnh dung tt v khng gian3 chiu, ta hon ton c th chn cch tip cn t vic, a bi ton v 2 ( xt mt phngv hnh ch nht 2 cnh l) hay thm ch 1 chiu ( ng thng v on thng cha lim). C th nh di y:

Cho k,a l cc s nguyn dng. Trn trc s nguyn, ngi ta nh s cc im nguynbng cc s t 1 n k sao cho: Mi on thng c hai u mt nguyn v c cha ng2a + 1 im nguyn th trung im ca n c nh s bng trung bnh cng s nh hai u mt. Hi c tt c bao nhiu cch nh s khc nhau?

Bng lp lun theo nguyn l cc hn tng t nh li gii bi ton gc, ta thy rng:

Vi cc on thng c cha ng 2a + 1 im nguyn th s c nh cho 2 umt bng vi s nh cho trung im.

Cc im cch nhau t hn a n v di th khng c rng buc vi nhau. Nh tha im c ta t 0,1,2, ...,a 1 s i mt nh s c lp vi nhau v kt quca bi ton s l ka.

Tip tc pht trin ln, ta xt bi ton sau:

Cho k,a, b l cc s nguyn dng. Trn trc s nguyn, ngi ta nh s cc im nguynbng cc s t 1 n k sao cho: Mi on thng c hai u mt nguyn v c cha ng2a+1 hoc 2b+1 im nguyn th trung im ca n c nh s bng trung bnh cngs nh hai u mt. Hi c tt c bao nhiu cch nh s khc nhau?

Tip tc lp lun nh bi trn, ta thy cc v tr 0,a, 2a, b, 2b c nh cng mt s hayni cch khc, cc v tr cch nhau a hoc b n v th c nh cng mt s. T suy ratt c cc im v tr xa+ yb vi x , y Z v x , y khng ng thi bng 0 s c nhcng mt s vi nhau.

Theo nh l Bzout th trong cc s c dng xa+ yb th c = gcd(a, b) > 0 c gi tr tuyti nh nht. V th nn cc s cch nhau mt khong c n v th c in cng mt sv cc s cch nhau mt khong nh hn c n v th khng c rng buc g vi nhau. Saubc ny, r rng ta ch c s cch t mu l kgcd(a,b).

Vit Nam Team Selection Test 2014 23Bng cch tng t, vic thay 2 s a, b thnh nhiu s vn c gii quyt tng t v biton lun c mt kt qu theo dng ly tha trn. Tuy nhin, nu thay bi bi ton 2 chiu,ta li c n 2 kt qu: Cho k,a, b l cc s nguyn dng. Trong mt phng ta , ngita nh s cc im nguyn bng cc s t 1 n k sao cho: Mi hnh ch nht c cc cnhsong song vi 2 trc ta v trn mi cnh c 2a+ 1,2b+ 1 im nguyn th trung bnhcng 4 s in nh bng s in tm. Hi c tt c bao nhiu cch nh s khc nhau?

Kt qu ca bi ton l: t c = gcd(a, b), nu trong hai sa

c,b

cc:

2 s l th kt qu l k2c2 . 1 s l th kt qu l k c2 .

trn, ta lp lun da trn nh l Bzout v c chung ln nht l mt nh l cbn trong s hc. Ta cng c th lp lun da theo nh l Sylvester v biu din cc scng cho kt qu tng t. iu ny tng ng vi kt qu trong trng hp 3 chiu cabi ton gc. Nh th ta thy rng s lng s tng ln khng nh hng n mc khca bi ton m chnh s chiu mi l vn , nu tip tc xt vi s chiu tng ln l n,liu c n p s hay khng? Cu tr li xin dnh cho cc bn.

Vit Nam Team Selection Test 2014 24Ti liu tham kho

1. Topic thi v bnh lun VN TST 2014 ca din n mathscope.org

http://forum.mathscope.org/showthread.php?t=46988

2. Trn Quang Hng, Xung quanh bi hnh hc trong k thi chn i tuyn VN ngy 1

http://analgeomatica.blogspot.com/2014/03/xung-quanh-mot-bai-toan-hinh-hoc-trong.html

3. Trn Quang Hng, Xung quanh bi hnh hc trong k thi chn i tuyn VN ngy 2

http://analgeomatica.blogspot.com/2014/03/tom-tat.html

4. Nguyn Th Hng, Lng nh Nguyt, Lng Th Thanh Mai, o Th QunhNga, nh l Sawayama v Thbault

http://analgeomatica.blogspot.com/2014/02/inh-ly-sawayama-va-thebault.html

5. Functional equation APMO 1989

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=450331

6. Bnh lun ca GS Nguyn Tin Dng trn trang Sputnik Education

https://www.facebook.com/sputnikedu/posts/720502957971422

7. Bi ton IMO Shortlist 1994, G4

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=352892

8. Mu Latex ca thy Chu Ngc Hng

https://www.writelatex.com/read/htndbgqrjqzp

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