Tạp trí toán học và tuổi trẻ số 283 tháng 1 năm 2001

ln i?*fr uat,,? rr ' rA* frJc # VA T}AS TAO * HCItrcAt{ Hoc vl*T FIAM fif,,*xt/ 8t Y @"; i*st d r &U;;*4 ;Th;a,r*,,,t*i,,1{*,#,{*;, "r llha bac hoc, vat li nguoi My B I I phorangco!in (B Franklin 1706 1790) da sap x6p 64 s6 tr"r 12 d6n 75 trong nhrrng phAn giao nhau cua cac hlnh vanh khdn dd tao thanh mO cung cac con s6 vd s6 otdm la 348 CQIS B sO theo m6i bdn kinh duong tron lon nhAt ta d6u dugc 34B, nhung ngoiri B tdng kei tr6n con nhi6u b6 8 sO khac cUng co tdng la 348 Cric hinh vanh khdn, m[.]

*fr uat,,? rr ' rA* frJc

# VA T}AS TAO * HCItrcAt{ Hoc vl*T FIAM

fif,,*xt/

8t

Y @";

i*st .: d

r.&U;;*4

l\lha bac hoc, vat li nguoi My B

I I phorangco!in (B Franklin 1706-1790) da

sap x6p 64 s6 tr"r 12 d6n 75 trong nhrrng phAn

giao nhau cua cac hlnh vanh khdn dd tao

thanh mO cung cac con s6 vd s6 otdm la 348.

CQIS B sO theo m6i bdn kinh duong tron lon

nhAt ta d6u dugc 34B,.nhung ngoiri B tdng kei

tr6n con nhi6u b6 8 sO khac-cUng co tdng la

348 Cric hinh vanh khdn, m6i hlnh m.6t rnau,

s6 goi ! cho cdc ban cac bd s6 nhLt th6

Ddnhchobgndgc Hay tim cac b6 B s6 co

ffiffiffi phdm se danh cho ban

nao tlm dLlorc nhi6u nh6t B b6 sO nhrth6

Gitii Cdp biii: CHUYEN CHTNG

MII\IH DTNH Li PITAGO

NHIJNC \IONC TRON IV AO

RAt tiiic ring kh6ng c6 ban nao chtrng minh din} li pitago bdng cA 3 c6ch Vi trang bdo nAy c6 han n6n m5i cdch chi dua ra m6t ldigi6i

Cdch 1 Tfnh di6n tich cdc hl nh va bi6n d6i dai sO fr t ;

')b o*ol = sABCD=sr + & s, = I & + zt)aoi

.-**b2=CCdch 2 PhAn chia hai hinh vu6ng nh6 thirnh ciic phAn ritigh6p laithinh hinh vu6ng lon (h.2)

hinh vu6ng nh6 vA vdo hinh vu6ng lon dd cirng duoc m6J

HOt NGHt cOruc rAc vtEr.r

roAN Hec rudr rRE 2ooo

S.lltg 22.122000 rai FIOi rnrdng 8t 'l'ril IIurg Dao, lli NOi

drr diin_ril Hr)i nghi coug rlc r,'i0r-r nirn 2000 cri7r t4p it,i Tofu,

Iroc vr)'fu0i tre Vli tlu cr1 GS D6 Long V0l, Chri ticli IIOi'l.oal

Itgc Vigt Niuri P(iS Vu Dultg 'l'huv I)lrti

(]i6rn doc NXII Giiro duc, cric ri1, r,ien

IIOi ddng bien tAp cr)ng hou 100 cOng tric viOn cic tinh, rhiurh tir'lhr-m Tllien -llue

trd ra v?i chc chn b0 troug toh sotui IrGS

Hohlg (lhfing vil hien tAp vi0n Trin Chi IIie{u 1ii 1'p Itd Chi Minli cunr cldn rhtun

tlu hqi

rrgtr; Hi,i rrslri diurlr'!ili rrhrins tharrlr trt'lr vlr rrllul! r itrt clrrnilarn .lrr,,r

cua nlirn 2000 r'i d0 xudr nhr-niu biCn phip

nharn tiip tUc dOi rnrii ti0i duus vir'hinir

thirc cr'ra iirp clii l.rour niuri 200 l'r,i nhunp rurrD l-it{p r.Jr.it Nlrieu i.r)rr-g liie vit rr .lll plriit

biiu rrlrtnr.u y kitit di' xuit vl lti r r.rrrs tlrrrclri s0 cr1 brr<ic chuytin lrrrnlt clrulin bi

currg dit nuric btrric vilr rrin kirrlt rtr tri

thfc Chc cOng tlc vien curls hoain rrglrenlr su ra ddi cua Torin Tu6"i tlrit Nhtur dip niy, (iS NeuyOn (lanh 'trtin

vh P(i,S Vu l)ur:rng 1-huy dir rrao llhng

danh du vi giai thuung clio circ ban doi

doqrt giZri Xuit sric vi gi?ri Nhdt cuQc tiri

giAi -'l'odn vA Var Li rCl) lap chi

nf,rn 2000.

vK'r'

Mathernatics and Youth

Ndm thd 38 sazas(t-20a1)

Tda sogn : Ngol87, phd Gidng Vo, HA NQi

DT : 04.51 42648-A4.51 42650FAX: @a).4.5142648

Hoi iting bian tSp :

rucuvEru cAruH ToAN,

HoANG cHUNG, ttcO o4r

rucuvEru vrEr uAr, orNH

oUANG HAo, rucuvEru

xuAtt HUY, PHAN HUY

ruAr, vU THANH xntEr, te HAt KHOI, ttcuyEru vAru

MAu,

"

uoAt'tc r-E M|NH, NGUYEN KHAC MINH, TRAN

vAT.I NHUNG I.ICUYEru

9ANG PHAT, PHAN THANH

ouANG, rn H6ruc ouAttc,

oAt'tc Huttc tui(ruc,

rnAru rnArus rRAr,

r-E eA xuAr*+r rR'iNH,

Dai didn phia Nam :

NguyZn Khdnh NguyAn - Vb tinh chAt cta m6t tri

gi6c n6i ti6p clSc bi6t

@ fieng Anh qua cfc bii torin - English through Math

Problems -Ng6 ViAt Trung

@ Ntrin ra th6 gicii - Around the \Vorld

DA thi Olympic to6n cria Li6n bang Dric

@ Cn.rdn bi thi vio Dai hqc - University Entrance

PreparationDd.ng Thanh H&i - MOt s6 tap tich phAn ddc bi6t

LTN-Tr?in Phuong - Db thi m6n To6n vho DH Ngoaithuong He N6i, kh6i D, ndm 2000

@ Nguye'n Khd.c Minh - Ldi giAi c6c biri to6n thi chon

- hgc sinh gi6i qu6c gia m6n to6n - l6p 72 THPT nim

Giii c6c bdi cria sd 279

@ firn hidu sfiu thOm todn hoc scr cdp - AdvancedElementary Mathematics

Triin Nam Dfr,ng - Song 6nh vir c6c bdi to6n giAi

tfch t6 hop

@ foan hoc vi tlli sdng - Mathematics and LifeXud.n Trung - Td lich tr6n m6t ng6n tay

@ Cau lac bQ - Math Club

Sai ldm ri tIAu ? Where's the Nlistakes ?

Bia 1: Tru sd ViQn Todn hgc ViQt Nam Bia 2 : ToSn hoc mu6n miu - Nh0ng vong trbn ki'Ao

H6i nghi C6ng tdc vi6n THTT

Bia 3 : Giei tri ts6n hoc - Math Recreation

Qdnh eho efr.e

IRIIilG HtlG GO SO

Trong c\c tA gi6c nQi ti6p c6 mQt tti gi6c d{c

biQt Ta hly x6t bLi to6n sau :

Bdi torin 1.: Tr) mQt didm M ngotri du0ng trbn

y( cic ti6p tuydn MB MD vd c6t tuydn MAC ddn

dn0ng trdn Chung rninh rlng AB.CD = AD.BC

Tri gi6c nQi tidp ABCD c6 tinh chft d[c biet le

tich cdc cqnh ddi bdng nhau"

Ddo lai :

BAi tor{n 2: Gi?a sri tri gi6c A,BCD nQi tidp trong

drrong trdn vtr AB.CD = BC.AD, khi d6 cfc tidp

truydn tai B vd C hbflc cdt nhau tren dudng thing

AC, ho{c song song.

Chtng minh:

- Ndu BD lh,

dutrng kinh ctta

dudng trOn thi

cic ti€p tuydn tpi

B vd D se song

song vdi nhau vb

song song vdi

- Gi[ srl BD khOng lh dutng kinh vh gi6 srl cfc

ti6p tuydn tai I vl D ldn lwt cft tia CA t?i M vd

VQn dung kdt qul 2 bdi tap ren ta c6 thd giti

mQt sd btri to6n thri vi sau dhy.

Bii todn 3 Tri didm A ngotri dutng trdn v0 c6c:

tidp tuydn AM, AN vd cdt :rry€n AKL ddn dumg

ffOn Ve il$ng thtng d b{t ki song song vdi AM.Gi6 sri d cfu c6c dudng thing KM vd LM tai P va

B tuong rmg Chr:ng minh ring iluOng thflng MN

di qua trung didm do1n PQ.

Chftng minh

Se ktrOng trnh hutrng ddn kdt qu6 btri toAn khi

cho dulng thtng d di qua N (h.3) Lric d6 :

P Hitth 3LMNPa LMKN-U=Y NM

Bii tor{n 4 Cho MBC nQi tidp trong dudng

trdn tam o, c6c tidp tuy6n tai B va c cdt nhau tpi

N V0 diy AM song song vdi BC MN c{t duOng

trdn ldn nta t4i P Chring minh rlng dudng thEng

PA di qua trung didm c:0;a BC.

Chftng minh

Cdch 1 Gitr str D ld trung didm ctra BC vd AD

cit Crfng rOn mm O Qi K Ta s0 chttng minh

cic t16p tuydn t4i B vd C c6t nhau tron duOng

thhng MK, hic d6 K trirng vdi P Xet tA gi6c

Theo kdt quA btri

tofun 2 ta c6 di6u phti

T{tcL li do dd xly ra c6c cltng thric trong ldi

glbi cdc ban cd thd tg gi6i thich don gitrn !.

- Dd kdt thric cho mOt tinh chdt dgp cira mot

tu gi6c nQi tidp d[c biQt, xin mdi cdc bAn hay gi6i

bdi to6n sau ddy :

BAi torin 5 Cho hai dudng trdn (O1) vd (Oz)

cit nhau tai A vtr B TI) mOt didm C irran duong

thlng AB vtr nlm ngoii hai dudttg trdn v0 hai

ti6p tuydn CM, CN vdi dudng trdn (O1) V0 c6c

dncrng thtrng AM, AN cit dudng trdn (O, ldn nria

tai P vh Q tr-rong tmg Chimg minh ring duong

th&ng MN di qua trung didm crta PQ.

IHAM GrA CUqC GHor DnU THIEil lllEl{ Ki

Ngudi trong Anh ?

Problem Let ABCD be a quadrilateral and

EFGH a square inscribed in ABCD such that EB

=FC=GD=IlAasin

the following Picture.

Prove that ABCD is also

a square,

Solution Rotate the

figure counterclockwise

through lr/2 about the

center of the square so

that ll falls on G, G falls

onF,Afalls on A'andD

falls on D' Since

IFGC + IGFD' = ZFGC + ZHGD = nlZ,

the line FD' is perpendicular to GC Since FD'

= GD = FC, D' is on either GC at on the side of

the line GC opposite to F, and hence A' is oneither GD or on the same side of GD as F Hence

IEHA = IHGA' does not exceed IHGD Tlrus,

applying the argument all around the square we

get Z EHA< IHGD < ZGFC < IFEB < IEHA

with the result that all these angles are equal Hence

IHGD + ZDHG = ZEHA n aPTlQr = nl2

so ICDA = xl2 Likewise, IDAB = IABC =IBCD = ICDA = xl2 The result now follows

easily.

Trimdi.

quadrilateral = tA gi6csquare = hinh vuong

inscribe = nOi tidp (dOng tu)

picture = hinh ve, [nhrotate = eua] (dQng tt)counterclockwise = theo chi6u quay ngrryc

kim ddng hd (trang tit)

center = t6fiI, trung tam

fall = roi, tr[ng (dOng tt)

perpendicular = vuOng g6c (tinh tt)

opposite = nguqc phia, ddi hp (tinh tu)

exceed = vuqt, ltrn hon (dOng tu)

apply = 6P dung (dOng tu)

argument = li le , lufln o1

result = kdt qulangle = g6c

likewise = ciing nhu vfly, tuong tu Gidi tu)

NGO VIBT TRUNG

vtr y sao cho viOn d6 c6 thd ddn vi tri 4 sau mOt

sd hnu han lin chuy.dn Y

Bei 2 Tr0n mQt dopn thing dO dei 1 dcrn vi

c6 mOt sd hnu han do4n nh6 r0i nhau duoc t0

mhu sao cho khOng c6 hai tlidm cdch nhau 0,1

dOu duo c t0 meu Chring minh ring tdng dO

dtri cfia c6c donn dugc t6 m)u khOng thd vucrt qu6 0,5

Bei 3 X6t ngti gi6c mi mdi dudng ch6o song

song vdi mOt canh cta n6 Chrlng minh ring ti

sd cria d0 dei dudng ch6o vdi c4nh ttrong l1ng

song song vdi n6 li nhu nhau ddi vdi c6 5

c4nh Hxy x6c dinh ei|tri crha ti sd nhy

Bei 4 Chfng minh ring m6i sd nguy6n k(k > 1) c6 mQt bOi sd nh6 hon ka vI bQi sd niy

vidt trong hQ th&p phAn b&i nhi6u nhdr li 4 chtr

sd kh6c nhau.

VONG 2

BAi 5 C6 thd phri kin mOt hinh vuOng cpnh

dli 5 dcrn vi b&i 3 hinh vuOng cqnh dtri 4 don

vi duoc khOng ?

Bni 6 C6c 0 cria mOt btrng

hinh vuOng kichthufc n x n

dnqc d6nh sd trl t d€n n2,

ching han nhu hinh v0 bOn

vli n = 5 Bpn c5 thd chgn n

0 sao cho khOng c6 2 0 ntro

cing cQt honc cing hing rdi

cQng cdc sd trong cdc 0 dx chon Gi6 tri cfla

tdng ndy c6 thd bing bao nhiOu ?

Bii 7 C6 4 dulng thing tr0n mOt mat phingsao cho cri 3 ducrng thing trong chring x6c

dinh dugc mOt tam gi6c

MQt trong c6c dudng thing ntry song song

vdi mot trung tuydn cira mQt tam gi6c dugc

hinh thenh bOi 3 dtrdng thing kh6c Chfmg

minh rdng m6i dudng thing kh6c (vOi dulng

thdng vria n6i trOn) cung c6 tinh chdt nly.

Bii 8 Tim tAp hqrp tdt c6, cdc sd nguyOn

duong n sao cho n.2n-1 ld sd chinh phrxrng

Trong c6c ki thi tuydn sinh vtro c6c trudmg dai

hoc vhcao d8ng thudng gip cfuc bhi to6n vd phdp

tinh tich phan Ndu bidt khai th6c vtr m0 rong c6c

bli to6n dO ta tnu dugc c6c kdt qui kh6 dpp vd c6c

tinh chdt qia tich phAn vh vQn dung c6c tinh chdt

nly girip ta gitri duqc mQt s6luqng l&t c6c bhi t$p

vd ph6p tich ph6n cudi cdp TIIPT Sau d0y ltr mQt

sd tinh chdt cira tich ph0n vtr c6c vfln dung cira n6.

(GV khoa cobdn Hqc vi€n Phbng kh1ng Kh6ng qudn, HdTdy)

bi gidi.ra c6 / =j!ro* ='yo* *i Xo*.

DAt : x = -l ) dx= -dt, dofl.r) l} hemle ="(-0

j.(ri*la, = Jflcosx)dx00

Tinh chdt 2: Nduflx) ren hlc ve ta ham ch6n Apdwrgtinhchdt 3=I=h#.**

fien [-a, a] thi a a

uvrr L 6' -' -'-

, =i nrr* = zi xga* - ,, j o*o- a* *T

- "** , a*chring minh tuSmg tg nhf tinh chdt 1, chti y Jo

coszr + sinu'r I sinrx + cosux

(D0 thi tuydn sinh Hgc viQn COng nghQ Buu

chinh Vi0n thOng - 1999)

Ldi gidi VQn dqng tinh chdt 4 v1i a = 2 > 0,

f(x) = f ld ham ch5n, lien ruc tren R ta c6 :

,, Tfnh chAt S.,rNdu flx) hen tpc tren [0, l]thifa4ri*;a, = J [,r(sinx)dx

lni gidi Dfltx = rc - t + dx = -dt

,!0

3 7 = Jx,4siw)dx =- i f" - t) frsil\x - t)ldt0z

-r_ n, I,f d(2cosx) n I 2cov1n

= ' = i\- / Jo(r""*)r y=- +'i o" '8 z l o

SONG ANH (Tidp trang 22)

Pturcng phdp thtz hai cnng dnh sr rhong qua viQc tiqFlla:^tP hqp G -= {(rr, 12, "', rn)lri ltr c6c xdu

fnh f(i, k), tuy nhi€n vdi mQt c6ch nirin f<rrac : vOi r' nhi Plan d0 dai k) vd t0'ng cdn tinh bing

phdntfrnyn2, ,r,thuQc {1,2, ,n}taddmxemc6 L d(rr 12, ,rn)vtrd(ry12, ,r,)l)sdcdcx6u

bao nhieu b0 (Ar, A2, ., Ap) th6a man di6u kiQn 41 u e,,rr ,r) e G

Az w A*= {nr, n2, , n) (3), tD d6 tinh duqc I(1, t) r, kh6c (0, 0, ., 0) Vtr suy nghi xa hon mQt chrit theo

vh So hu0ng nly_, ta c6 lli gili thri ba cdn ngdn gqn vd bdt

Ta cho c6c phdn t& n1, n2, , z, ,,d6ng ki" c6 mf,t '*o}ff :fiil'r?* ou (At.A2 As) sao cho i e 41 u

trong c6c tap hsp A, theo quy tdc : n6u, chfrng h4n n, n

dang ki c6 mat trong A,, 42 vd kh0ng c6 mar rong c6c A.t w v Au thi sr = I.!o(i) (bp dpc hay ts kidm ha

6p cOn I3i thi phi6u dang ki ghi lI (1, 1,0, ,0[cdn '-z - "' - "r *

n6u n, chi c6 rirat trong,{o thi-gtri pliidu ia (0, ,'0, t) tai sao) ,=1

triX'- $frlfl H J*Xf# lf'*%il'.t'*rJtr:t J lltJ I :o ", ct.(Z')k b0 gr, l?, ,Aotvoi A, c {t,2, ,

Azi '.wA).Ylt"iptri'ciu Oang friiai1,i;-:.:;,4 n) Trong sdntry, c6 (2n-t7* b0 vdd A1w A2 \) A*

lflp dnqc b0 (Ar, Az, A) D6 thdy ring vdi hai bQ khOng chrla i (vi hic n)y vdi m6i A, chi cdn 2n-1 c6ch phidu

{Ang ki khdc nhau, ra_c6.hai bO tap hgrp (A,, Ar, chgn) Nhu th€ c6 2k - 2(n-1)k b0 (Ar, A2, , A1) vOt i

,A1)kh6cnhau v)nnut]r6sdb0 (Ar, Ar, ,eol:clr6a e ,+1u Az vAo Nhuvfly,so(i) =2i"-r)*.12r- t; vi

Ttr (3) btrng sd bQ- phidu dlng- ki.hqrp IQ Vi phi6u s,. o.2rn:r* ok - D.

aane

-ti :fi? "y'p = l' "'' i gdm 't sd 0 hoac I vl-e^Y "-

no-rang .ain giai cudi ctng rdt dn tr:qng v6 ss don c6 it.nhdt I sd 1 ncn^n, c6 2k-l cach ghi phi6u ddng ki gia, J,iJ,i3."fiiluig c6 duqc s[ doir gitrn d6, chung ra

yi_rhq vfly c6 tdtce(zk-l)i bQ phi6u ddng ki hq'p lQ 6aprr6i mtry m0 vf suy nghirh nnie"u, phtri'c6 nhimg

khdc nhau t<i nang ccr bAn, nh0ng t<1nh nghipm thu nhfln duqlcudi cr)ng, chri f rxng c6 c'In c6ch chen i phdn tri rD [i"#,rr8:3 Stl,.dji,lSt,J,1,t3,1Htli f$;f:6c quan

n phdn trl nen ta c6 T(i, k) = CL(2k - l), va tt ddy suy ""*i;;;g;;;"

sau s€ rrinh b1y tidp c6ch gitu mQt s0

ra so = n(2k - l).2k(n-t) uu toan [d trq,p co ap oung pnooig pfiap son"c anh.'

oE rfu mou TofiN vfio Dfl Hgc lrcofil THUonc Hf, HqE

a psAtq oAmr cno rAr cA cAc rni smn

(chuv0n ban ' vtr chua Phfln ban).

Cdo I 1 Khlo s6[ su bidn thiOn vd ve d6 firi

hdm sd:

!=x3-6x2+9x-l

2 Tri mQt didm bft ki trOn cludng th&ng x = 2 ta

c6 thd kE dugc bao nhiOu tidp tuydn tdi dd thi

1 Giei phunng trinh:

sin8,x + cossr = 2(sin10r + cos10x) +l*ra

2 Cho MBC Gi6 sir G lL giao di6m cdc

dndng trung tuydn ctra tam gi6c Ki }lrigu ZGAB

= u,2GBC = p, IGCA = y Ctu:mg minh ring :

costg., + cotgB + cotgy =U-#4

trong d6 a, b, c lA d0 dAi ba c4nh vtr S ltr diQn tich

cira tam gi6c.

KH6ID.I{AM2tltlo

Tinh tich phtn :

HUONG oAnr crAr

cAu IV cho parabol !2 = 8x vd di6m I(2' 4)

nim tren parabol Xdt g6c vuong thay d6i quay

ouanh di6m 1 vtr hai cAnh cfra g6c vuOng c[t

parabol tai hai d\€m M vd N (kh6c vdi di6m f.

'Ct,org minh ring dulng thf,ng MN luOn lu0n di

qua mQt didm cd dinh

2 Cho parabol ! = x2 + 1 v) duOng thtng y =

mx+2

Chtmg minh ring khi m thay^ddi, Oudng ttrEng

luOn luOn cdt parabol t4i hai ili6m phdn biQt Hay

xfc dinh m sao cho phin dign tich hinh phlnggidi hpn bdi du0ng th8ng vI parabol lI nh6 nhft

B PHAN DANH cHo ruNG LoAI P6t

TUqNG THi SINH

CAu Va (cho th{ sinh thi theo chaong trinh

CAu I 1) Ban dqc hJ gili

2) GiL srt M(2; ru) lh didm thuQc ttudng thing

x = 2 Dudng thtng di qua M ld y = a(x-Z) + m.

Ddng thtng ntry ld tidp tuydn cira dd thi o HQ

bi6n tren R 3 (3) c6 1 nghiQm duy nhft v6i mgi

m = ti M kE clugc duy nhft mQt tidp tuydn tdi

Tt d6 suy ra : phuong trinh c6 nghiQm duy

nhdtx= 1.

Cnu III.

1) Ta vidt phrrng trinh v6 dAng :

sins(l - Zsnzx)- cos8r(2cos2x - t) =]cosX

e cosax(sin8r - cos8x - i, = o

Vi : sinsx < 1 < I *

"or8, nOn phrrng trtnh

4 hrong dtrng vdi :

cos2r= 0 ex =;.q(keO.

2) Gqi M li trung didm cria BC Tt cOng thric

tinh d0 dli dudng tnurg tuydn ta c6 :

Tuong tUddi v6'i cotgB, cotgy Trld6 ta c6 dpcm.

C8u IV 1) Ta c6 I(2,4) e (P) : !2 = 8x

coi u (* , 4, - G, z) cung rhuoc (p)

Gi[ sr] A (xo, yJ ltr di6m c0 dinh e dudng

ldn[ye tutN Khi d6 ta c6 he sau dring Vm, n

= Di6m cd dinh le A (10, -4)

^2) Holnh dQ giao didm cria 2 dulng ll nghiQm cfia phrrng trinh :

P+1=mx+2af-mx-l=0(*)

Phtrong trinh (*) c6 P = -1 < 0 nen luon c6 2

nghiQm.ro 0 ,, = 2 dudng nly luOn cft nhau

tAi2 di6mA, B tuung rlng (xem hinh ve).

Gqi S ltr diQn tich hinh phfrng gidi hnn b&i

Lfl GIAI CAC BAI TOAN THI CHQN HQC SINH CT6T QUOC GIA

(VyTrung hqcphd thing)

NOn tD (7) suy ra [/'(.r)l < 1 V.r e tO; f] no

d6, tt (6) ta duqc g'(x) < 0 V.r e [0; trgl Suy

ra g@) lI hlm nghich bidn r0n tO;{cl Kdt

hsp vdi (5) suy rt a = b, haY

limx21ra1 Do d6 (,rr) lh day hQi tU khi n -+ @

(dpcm)

Nhsn x6t: - KhOng it thi sinh dL, vi6t : "D6 thdy

Bei 1 Cho sd thvrc c > 2 Ddy sd (xn), n = 0, l,

2, , daqc xhv d.wg theo cdch sau ! xo = \[s,

x,n*t = G-W (n = o, 1,2, ) idu cac

bidu thtlc dudi cdn ld khAgg dm.

Chrmg minh ring d6y (xn) duo c x6c dinh vdi

mqi gi6 tri n vI tdn tai gidi h4n htru h4n limx,

- Nhi0u thi sinh dt chrng minh a = b x:udtphAt fe

gih, thiltfra) = a vdflb) = b xrn luu f : gia thi6t d6

nrtrrg duong vdi vigc thua nhan (xz) le dAy hoi tU

khi a-+o, hay, n6i mQt c6ch khdc, thr)a nh$n a = b.

Bai 2 Tran mdt phdng cho trufuc cho hai

iludng trdn (O1, rl vd (Oz, r) TrAn dudng

trbn (O1, rl ldy m\t didn Ml vd trAn dadng

ffdn (O; 12) ldy m1t di€m M2 sao cho dudng

thdng A1M1 cdt dudng thdng O2M2 tqi mQt

didm Q Cho Ml chuydn dQng tAn dudng trbn

(Ot, ri, M2 chiydn dQng tAn dudng trdn (Ov

r2i ctiig thZo chidu kim ddng hd vd vhi vQn tdcgdc nhu sau :

1) Tim qu! tfch tung didm tloqn thdng MtMz

2) Chtm7 minh rdng giao didm thrt hai cilatlwdng trdn nSoQi tidp tam gidc M1QM2 vhidudng trdn ngoqi tidp tam gidc OIQO2ld mQt

didm cd dinh

Ldi gif,i (ban dqc $ ve hinh) :

1) Gqi O ll trung dr€m Op2 Hidn nhiOn O

lI didm cd tlinh !-dy cdc- didm Ml1, M'2 saro

cho : Ollu\ = OVvI't, [fi?2 = OA2 Y, Mv

M2fiiong ring chuydn dQng trOn (Oy r1), (O2,

,2f trreol,:ng chidu vtr vdi crlng vfn tdc gpc

i6n U' 1, M'2 s€ quay quanh O theo cing chi6u

vI vd'i ctng vfn tdc g6c (8)

Ta c6 : M lt trung didm MrM, e

1 ol",* dq o olu = ltolu', * o

)(orM' t+ o2M 7)

Suy rarl dugc xdc dinh ( 1).

Git str xp (k 2 1) da duvc x6c dinh Khi d6,

hi6n nhiOn cd 0 < x1r<{ < c.

Vi thd : 0 < c + x1r1 2c +'{*-+ x1r ^{k ,

+c-.f,*rrrO.

Suy ra.rp*1 duo c x6c dinh (2)

Tri (1) vtr (2), theo nguyOn li quy nap, suy ra

Suy raflx) nghfch bidn trOn t0; {cl D6n tdi

(x2) vd (xz**) ltr cic dly dcrn diQu Kdt hsp

vdri (3) suy ra (x2) vn @z**) ltr cfc day hQi tt;

'e > Mh,trune Untrung di€m di€m cflaM'['cl0la M' ,M', (9)

Tt (8) vI-(9) ta dusc: Quf tich ctn M \d

2) Ggi P li giao didm thri hai cria dtdng trbn

ngoni tidp LMTQMz vi dudng trbn ngoq.i ridp

LO1)O2 De dtrng chring minh duo.c :

PO,fr MO1M1- LPOyMy Suy ra

d= -4 Do d6,

P thuoc duong trdn Apoloniut dlrng tron doan

OlO2cddinh, theo ti sd khong Odi 1 (10)

DE thdy : (POy POz) = c{, = const Suy ra, P

thuOc cung chria g6c dinh hud'ng khOng ddi cr

drmg trOn do4n O1O2 cd dinh (11)

Tt) (10) vI (11) suy ra P Ii didm cO dinh

(dpcm)

Ghi chri q_Uiq d f, ltr g6c dfnh hu6ng

NhAn x6t: Do kh0ng sfr firng g6c dinh hu6'ng

nOn nhidu thi sinh c6 ldi gitri ho{c khOng dny dri,

thidu chinh x6c ho{c rudm rh, phrlc t!p

Bni 3 Cho dathftc

P(x) = x3 - 9P + 24x- 27.

Chtm:g minh rdng vhi mdi sd nguyAn duong n

lu1n tdn tqi m\t sd nguy€n duong an sao iho

P(an) chia hdt cho 3n.

Ldi giii Ta c6 P(1) = 81 Do d6, khi chon

an = | thi P(an): 3n (n = I, 2, 3, 4) (I2)

= P(a) a 3n-t67o2 - 6ab + 8b)(modln+ly

Y\ a2 - 6a + 8= 3 (mod 9) (do a : I (mod9)

Bd dA 2 Yn> 5 lan:1 (mod 9) sao cho :

P(a") :0 (mod 3u)

Chftng minh : YOr n = 5 ta c6 a5 = l0 th6a

man cAc yOu cdu cria bd dC Gie su, Ud oa z

dfng vdi n = k (k > 5), nghia ll 1ap : 1

(mod 9) sao cho P(ap) = 3k.v, v e N*

X€t: apal = ak* 3k-2b, b e Z

Theo bd dd 1, ta c6 :

P(a**r) = P(ap) + 3kbh = 3k(v + bh) (moA3k+11

Vi (ft, 3) = 1 nOn c6 thd chqn b e N* sao cho

bh = -v(mod 3) V6'i b d6 ta c6 : ap*1: (mod9) vh, P(a1r*1) = Q 1ms6 3/.+1;.

Theo n-guyOn li quy nap, bd dd 2 duo c chung

minh B$ dd,2vd,(12) cho th{y :

Vn e N lan e N sao cho P(an):3" (dpcm)

NhAn x6t Chi c6 mQr rhi sinh gihi duqc bAi trOn.

Bei 4 Cho trahc gdc a vhi 0 < a < n Tim

P 3@) =x3sina - .rsin3cr + sin2cr,

= sino(,r + 2cosa)(.r2 - 2xcosa + l)

Tn d6 suy ra : vfii flx) = *2 - 2xcoso + I thi

Hom nta, Yn 2 3 ta c6 :

Pn+{x) = Tn]-l sincr - "rsin(n+1)cr + sinno

= .r'+lsincr + r[sin(n-l)a - 2sinncrcosa] +

sinacr

= xlx'zsina - xsinnc- + sin{n-l)crl +

+ (x' - 2xcoso + l)sinna

= xPn(x) + (; - 2xcoscl + l)sinncr (L4)

Tt (13) vI (14) suy rafl.r) = x2 - 2-rcosa + 1

Itr tam thric cin tim

Nhin x6t M[c dn bili to6n tren khong phli

le bei todn kh6 nhung ciing c6 tdi 107 trongtdng sd 216 thi sinh du *ri U| Cidm O.

Bei 5 Cho tri di€n ABCD cd bdn kinh dudng

tr\n ngoei tidp cdc mdt ddu bdng nhau Chtugminh rdng cdc cqnh ddi di€n cnit* diQn ABCD

bdng nhau

Ldi gifri Ki hiQu cic milt BCD, CDA, DAB

vd ABC tuong ring ld m{t 1,2,3,4 Vdi X e

{A, B, C, Dl, i e {1,2, 3, 4} takt hiQu X; ln sfi

do g6c ph6ng rai dinh X & mdt i

a

4443

Ia,*Zs,* Ic,* LDi= +n

i=2 i=l i=l i=l

Bei 6 Tlm tdt cd cdc hdm sA'fG) th1a man

didu kiQn : xzf7) + f( l-x) = 2x - f , vhi mpi sd

fa frai ngniem cria phuong trinh.r2 - x - 1 = 0.

Theo Gnh li Viet tac6'.

Dt = n - Aai D2= B4vd Dl = C+'

D6n tdi : D2 +.D3= 84 * C4= n - A4= Db

trii vdi tinh chdt cria g6c tam dien Tt d6 suy

raA4<f ve vi th€, MBC ltr tam gi6c nhsn Do

d6, tt (17) ta c6 : D1 = A4, Dz = B+ vh

Dt= Cq SuY ra ' D1+ D2* D3= v'

Kdt hatp didu ntry vdi (16) ve (15) ta dugc :

tdng tdt ie, cec g6c phing tai m6i dintr crla trl

dien Uing r Ddn diy, bing clch trlr cdc p[t

nto, cde vb, DAB 6tra tu oien ten m[t phf,ng

(ABq, d6 dlng chrtng minh duo.c ; AB = CD;

BC = DA vd, AC = DB (dpcm)

Nh{n x6t Tdt ch chc thi sinh ddu gitri bhi to6n

theo c6ch : Xudt ph6t tt gie thidt, chimg minh tam

O cira m{t cdu ngopi ti6p trl difu ABCD cdch ddu

cac mAt phang chua c6c mat cira tri diQn d6' V) tdi

dlv, tdt ia OCu hoflc khong x6t ddn kha nrng O th

mm mfi ciu blng tidp cua t0 diQn hoflc c6 xdt ddn

nhrmg khOng loai trD dugc khtr ntrng ntry.

Lin luqt thay x = a, x = b vio ding thric crha

bii ra, vdri luu y t61 (20), ta du-o c :

trpng d6 o, bll.hai nghiQm cria phuong rinh

x'2 lx - 1 = 0 Ngugc l4i, vdi luu f tdi dinhngtria cria a, b vd (20), dC kidm tra thdy flx)

dio c x6c dinh bOi (O th6a mtrn didu kiQn dd

bli.

Vfly tdt cL cLc hnm sdflx) duqc x6c dinh b&i

(4 ln tdt cL cilc him sd cdn tim

Nh$n x6t Chi c6 1 tli sinh gitri quydt trQn vgn

bli todn ften Rdt nhidu ttri sinh cho ldi giAi khong

diy dri, thi6u chinh x6c do dl dua ra c6c kdt luQn sai vd nghipm cua mOt hQ phr:ong trinh b0c nhdt hai

fln c6 ilinh thrrc blng O.

TIIONG BAO

Do thAnh pn5 tfiay odi oi6n s6 nhd,

-ban docqiJt thu vO fOa soan Todn hqc & tudi tr6 va

iodn fudlthd xin ghitheo dia chi :

" Ngd 1s7 PhO Gidng V6, He Nfli

Dia didm, s6 OiCn thoai vd Fax khOng thay ddi'

TI{TT

a+b=lvdab=-l (20)

l1

T'C RR III IIRY

CAC LCIP THCS

BAi T1/283 Chring minh ring vdi m6i sd

nguyOn duong n thi sd (3 + n3)(3.n3 + 1) ho{c

chia hdt cho 49 hopc kh6ng chia hdt cho 7.

PHAMNGQCQUANG

(GV trudng THPT Lam Son, Thanh Hda)

B iT2l283 Tinh ginfi bidu thrlc

\fx4-yp44 + t$1+-a11a-g * ^[rt+-i$-y)

th6a mln didu kiQn : a + 2b + 3c = 1 Chring

minh ring it rf:rdt m0t trong hai phunng trinh

sau c6 nghiQm thgc :

+i - +1Za+ 1)r+ 4a2 + l9Labc + 1 = 0

4* - 4Qb+ 1).r+ 4b2 + 96abc+ 1 = 0

rnANvANnrfovn

(GV trudng THCS T1t Cudng, TiAn lnng, Hdi Phbng)

Bdi'141283 Cho tu giec (l6i) ABCD vh, M lh,

trung didm ctn AB Xdt didm P thuQc doAn

thilng AC sao cho'hai du0ng thilng MP vd BC

cdt nhau, ggi giao didm d6ld L Ggi Q lI didm

thuoc dean thhng BD sao cho * QD = O'PC

Chtmg minh ring dudmg thLngTQ luOn di qua

mOt di6m cd dinh khi P chay trOn dopn AC

NGUYENPHUdC

(GV rrudng THCS Kim Inng, Hud)

BAi T5/283 Cho tam gi6c ABC c6 diQn tich

bing 1 (dcrn vi) Gpi R vtr r tunng rmg ltr b6n

kinh dudng trdn nggai tidp vl nQi tidp tam gi6c

A.BC Chring minh ring

a = (2Q2000 a 1 120001200t

vI b=(2Q2o0r a 112tt01120o0

oAoquaNcortm

(Phbng GD-DTVdn Giang, HungYAn)

B i T71283 Cho k (k > 2) sd thuc dunng a1,

a2, , ao.Chtmgminh ring

(GV trudng THPT chuy€n Hodng VdnThu, Hda Binh)

Bei T8/283 X€t day sd thUc ay a2, a3,

th6a m6n c6c di0u kiQn : 0 1 on < I vh

(ViAn COng nghp rhdng tin)

Bei T9l283 Ggi A, B, C ll ba g6c cfia tam

gi6c ABC Chring minh ring

1 r *.o.{) (, *4) ( r *.o,{r)

, {3 ':{l

'( t *7.)

NGUYENTH6siNH

(GVtrudng THPT chuyZn Hd Giang)

Bei T10/283 Chring minh rIng ndu hai hinhn6n crlng nQi tidp mQt hinh ciu vtr c6 dign tichxung quanh blng nhau thi thd tich cta chring

cnng blng nhau Di6u ngugc l4i c6 dring

khOng ?

HOANGCONGTHANH

(GV trudng TIIPT Nguy4n Chi Thanh, ThiaThian - Hu€)

cAc oB v4r r,i Bni LY283 MQt c6i xich dugc gi{r ffOn mflt

bln nim ngang kh6ng ma '4 s6t ma I OO dli cria

xich duqc thf, thong bOn cAnh btrn Xich c6

chidu dli ll l, c6 khdi luqng ltr m duo.c ph6n bd

*