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Tạp chí toán học và tuổi trẻ số 297 tháng 3 năm 2002

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Trang 2

t ' " Tlorrg thtrc lien thLrorrg g,iip trtrollg h(Ip co lnot

tam 96, tAm ton mreng bia td giay hlnh vttottg

md phrii x6c dinh mot hinh da gii:rc cldu ben trongc6 diOn tich khd l6n nhurg khOng cd dung cu do ve kT thuat B t: tt

srrtr (h.l r : ('hoA tlirng D, B trirng C de c6 HK satt dochoABtrurrg voi I I /I

AP r cri P nirn trcn i IK d.i co AL roi gAp thco ,4C cle co AF I I /

I

J

'I'ztng phim clilrh cho lt)i girii ddy dLr gihi thich 16 rlng l{itrlt I

Gitii ddp Dril; DUdN(; DI t:u,r quAN NII lR.tiN ll,\x c(l

1) Ta chia ba:ur ci-,- 1x 2001o thr\nh cilc hinh chlr nhAt Ar, Az, , Ar.r.1k(ch thrr6c 4 x 3 O Cl6 nhidLr

ciich cli quAu rnii trr o diu (D) dcn 6 cuoi (C) lien tiep c1-ra tat cii ciic O cua bin cil, drrcli ciay clii rrr 2ciich o hinh 2 r'ir hinh 3 trong rlo roi rn6i ctudng di qda ciic hinh chlr nhAt o gilta chi ghi-hai cli0m

diiu boi cirng rnot so.

M6i bLr6c di (-+) cfia quAn mti tulin theo cdc quy tac sau :

a)D +'l'holc'l'-+ D (h.4) ;b) B-'> LvIL -+ Bhoac L+ L

Neu trong circh cli c6 it nhat I llin chuy€in tit h)ng trOn cita so dd xudng hirng drrdi thi so t) I- nhidtr

hon so o 11, cdn neu chi di theo h)rng tr:en thi khong bao giir cli qua ci'rc o I-D vi\ B'[' ; dcitr triri r'(vi

dieu ki0n ('r').

Clhc biin sau cih chi ra chch di cira clLrAn md til o D, duoc nhin tlng ph;irn :

l) Plrunt NgocTLi,l2[1, trudng 1'lII'}T DAn toc nQitrf, Dak T0, Kon'I'urt-t

2) N guvin Mittlr DLi'c, I li\, trudng THm l'hanh Chuong I, Ngh0 ;\n

3) N,qrrr,/rr Tlti Chtrrtg, clOi 7 xa Thanh Ngg., Thanh ChLtong, Ngh0 An

Trang 3

Todn hoc vd Tudi trd

Nam tht39

So'297 (3-2002) Tda soqn : 1878, ph6'Gidng V6, He Noi

@

Dinh cho Trung hoc co s& - For Lower

Secondary Schools

LA Duy Ninh - Ph5.t tridn tt m6t bii

to6n hinh hocTim hiiu siu th6m toSn hoc so cdP - Advanced ElementarY Mathematics

Trd.n Thi Nga - Vdi phuong ph6P gi6i

phrrong trinh him

Vil Einh Hba, Nguydn Trqng Tudn,Nguydn Anh Qud.n - Ding" flliong

pf,ap phan hoalh tap hqp dd giii baito6n sd hoc

Ha Huy Khodi - Giai thti8ng L6 Vdn

Tieng Anh qua c5c bii to6n'English @

through Math Problems

NgO ViQt Trung - Bdi sd SrToin hoc vh ddi s6ng - Math and Life

Phan Thanh Quang - Bai to6n ki

nh6ng ddi mauD6' thi tuydn sinh 16p 10 h9 THPT

Nhin ra the gidri - Around the World

Dd thi chon d6i tuydn to6n Rumani

(4 - 2001)

Ban doc tim tdi - Reader's Contributions

Trinh Xudn Tinh, D6 Ud,n Drrc - Vi0c

rr6c lrrqng gi6 trf m6t bidu thtlc gita c6c

HQi diing biai tQP : .

*o*e* 6ANH T6AN HoANG cHuNc, NGO DA; iU, r-g rnXc BAo NGITYEN nuv.oonN ryor-rrQNvIEr HAr DrNH ouANG HAo, r{cuyEN xuAr.r H1-rv FueN rruv KHAL vU THANH KHIET, tE HAI KI,{0I'

Ndr]vtN'vlrrr-r"rIu uonNc LE MrM{, Ncr.ryEN xriAc un'ls, rnAN vAN NHIING ltcqm{ oAxcpn{r pHeN ru$\rH eUANG TA HdNc eUANG.DAN6 HLING rsANc, W DWNG THUY, TR:1N THANH

rner lE BA KHANH rniNn, NGO vIEr rRUNc

Trudry ban biertr4ip ; NGUYEN Vtrf ffAf Biitt rdp : V0 KIM THLTY Hd qUeNC Vfmt

Tri sIT : VIJ ANH TIIIJ,TTiNh bdY : NGUYEN TI{I OANH

Dqiii4n pttia Nam:rniv cUlutEU,ZSt NguyinVdnCil,Tp.HdChlMinh.DT:08.8323044

Forum

Nguydn Anh Dfing - Vd bei to5.rr

p[.rot g trinh tidp tuYdn

Ng6 ViQt Trung - Giai thuAng Khoa hoc

ViQn To6n hoc nlm 2001 CAU lac bQ - Math Club

Sai ldm 6 dAu ? - Where's the Mistakes ?

Bia 2 : To6n hoc mu6n mhu - Multifarious

Mathematics

cdp gidy tld tao da gi6c ddu Bia 3 : Giii tri to6n hoc - Math Recreatir-rn Bia 4 : Chi nh5nh Nhir xudt b6n Gi6o dr;c tai thdnh ph6 Dn NEng

Chiu tnich nhi\m xudt biin : Gi6m d0c^NXBoGi6o Duc :

Trang 4

Trong bli brio nly chfing t6i mudn gi6i thi€u

vdi cdc ban mOt sd kOt qui khai th6c tir m6r bai

todn girip ban doc hidu duo c 16 qud trinh tao lAp

bii todn mdi ti bdi toi{n ban ddu, tim duo c

ngu6n g6c cira bii to6n vi budc ddu llm quen

v6i tap duot s6ng tao to6n hoc Tru6c hdt ta hdy

x6t m6t kdt qui hdt srlc quen thu6c

Bii to6n l Clto tam gidc ABC vd diiim I

cdt cdc canh BC, CA, AB ldn lrcn tui A', B', C'

Chung minh rang

c6c tam gi6c IBC,

ICA, IAB vd ABC

I) Van gUng c6c h€ thrlc vd ti so di€n tich vi

cdcbdt ding thric COsi, Bunhiac6pxki d0 chrrng

minh duo.c ldn lugt cdck€tqui sau :

Qhit

LE DUY NINH

Til (1) vd BDT (10) ta c6 BDT (12)thi n=2:

(Nguy€n @ trudng DHSP Hd Noi 2)

+-* IAIBIC2 _2- (5)

IA' IB' IC'

cing ldn hon21' kh6ng cDng nh6 hon 2 (6)

(7)IAIBIC2'

Dau ding thrlc xiy ra rrong m5i BDT ffen khi

vh chi khi I h trong tAm LABC Xin huong d6n

\ u! IB" IC'

Trang 5

Mdi c6c ban chrlng minh tidp cr{c BDT cdn lai.

II) Ta c6 the m& r6ng crdc bhi to6n n€u tr€n

sang kh6ng gian bing phuong ph6p tuong tu .

Bdi torin trong kh6ng gian duoc chring minh nhd

iip dung he thfc ti sd thfj tich giSng nhu vi6c 6p

clqlg hQ thrlc ti so diOn tich trong hinh hoc

phang Crich llm nly g5p phdn rdn luy€n kr

nang chuydn h5a cdc tinh chat hinh hoc giffa

mar phang va khong gian, ti' d6 thdy duo c mdi

Ii€n h0 khang khit gifta hai loai bhi tap hinh

hoc nhy.

Bii to6n 2 Clrc fi di€n ABCD vd didm I ndm

DI cat cdc ndt d6'i di€n ltitt lLrot tui A', B', C',

D' Chmg mirfi rdng

IA' IB' IC' _ r_ I _=l ID' (t)

AA' BB' CC' DD'

r !: ^:a: IA' l\ _!1pco IB _VACD

LOI glal - .-:; AA' ha = -.- - -VABCD BB'

Vosr.,,

- Vtaco +Vtaco +Vteao +VAac -,

VtscoTuong trJ nhu 0 phdn I, chfng ta c6 cdc k6t

quA sau v6i n lI sd nguyOn duong :

IA IB IC ID

-+-+

AA' BB' CC' DD' AA' BB' CC' DO'

IA' IB' IC' ID'

IA

+ tB +!9*!L rz (4t

IA' IB' IC' ID'

IA' IB' IC' ID' 4

IAIBICID3

IA IB IC ID IB' IC' ID'

3, khOng cirng nh6 hon 3 (6)

Xin mdi c6c ban tu chrlng minh ciic bdt ding

thfc (1) - (15) Rdt mong cdc ban tim duo.c cdc

kOt qui tuong tu gifra hinh hoc ph&ng vh hinh

hoc kh6ng gian tr) c6c bhi todn khi{c

LTS: Zda soan nhQn duoc ntdt s6' kAlt qudtttong ttt, cirtg rtdi dung voi bdi b(ro ndy cila batt

Nguydn Hrtu Hiiu, K38D, klrctaTodn, DH 1/inh.

Mottg nhdn daoc' cdc bdi btio kluic cila ban.

iili \

7 '"\l ;7

Trang 6

ffi*a pHAp GIAI puUoNc rninH HAM

(Tidp theo)

phuong phdp gidi phuong trinh hdm (PTH), bdi

ndy gidi thi(u ti€p cdc phuong phdp khdc

3 Phuong phdp x6c dinh nghiGm duy nhdt

C[ng gi6ng nhu c6ch giii phuong trinh th6ng

nghiQm cita PTH vd chrlmg minh rang ngoii c6c

nghiOm d6 ra PTH khOng c6 nghi€m n)ro khdc.

ThOng thuong ta hay thir cdc hlm s6 dac biOt :

hlm f,ang Ueh clOng nhat, hirm tuyar tinh, dd

xem chitng co phai li nghi€m cira PTFI hay

khorrg.

Y{ du 4 Tim tdt cd cdc hdm s6' f .' R -+ R thda

mdnf(f)+f(y))=y+(f(x))2 (1)

vdi moi s6'thuc x, y.

(Dd thi toi{n qu6c rd,1992)Gidi.D€, thdy/(x) =.r li nghigm cira IrfH (1)

Ta sE chrlmg minh d6 lh nghiOm duy nhdt cfra

(1) Dat a = f(O).Thay ldn luot x = 0 rdi y = 0

viro (1) ta duo c

.f(*2 + a)=lf(x)12 v6i moix e R (3)

Thay y = 0 vio (2) duac f(a) = az (4)

i.ry'1*\2 + f(a)l = a + W$)f)z = o * (r * o')'

Thay crlc kdt qui bidn ddi niy vio (5) dugc

12 + o + o4 =, * (* n o')' suy ra a = o.

VaY/(0) = Q.

Thay a = 0 vlo (2) dugc /(/(y)) = y hay ld, z =

lb) thi y = f(z) v6i moi y, z e R (6)

TRAN THI NGA

TU (7) suy ra ndu x = /2 > 0 thi/(x) > 0.

Nduflx) = 0 thi theo (1), (7) c5:

x = )( + lf@))z = f1i + f@)) = f(*2)= [f(x)]2 = 0Y4y x = 0 <3 f(x) = 0, vh ndu x > 0 thi

Vdi.r > 0 bat ki c1at.r = I 2, tir (6), ( 1), (7) c6

l(.r+.r,) =.7\r1 +l(;)) =: + U(r)lr =./(y) + f(t2) =

l(x) +l(-v) r,6i moi -r, e R r').v > 0 (9)Dar y = -x thi 0 =fl0) - f(, + (-x)) = f(x) +

J?x) => lGx) = -f(x) vdi x bdt ki

GiA str/(x) = x + tt v6i bat ki x e R Xdt cdc

trudng hqp sau :

Ndu u > 0 thi theo (6), (8), (9) c6

x = flfl,x)) - flx+u) = f(x) + f(u) = x + u + f(u)

d6n ddn m0u thu6n v\ u> 0,f(u) > 0 theo (8)oNdu u<0 thi-r>0n6n x=.|(f(x))=

Ta c6 thd dua vio dn phu u = h(f(xD dd,

g(u) = u, titcld tim didm bdt d6ng cira him g(u).khi xac dinh duoc a ta sE tim duo c nghiom ctra

PTH d5 cho

Vidtt5.ChoS= {xl x €R,x> -l} Timtd't

cd cdc hdm sd f :S -+ S thda rntin 2 di6u kiett :

(1) f(x + f(y) + .rf(t')) = 1' + f(x) + vf(-r) voimoi.r,) sS,'

Q) IJi- ld luint rittrg, t'o'i -l <x<0 vii 0<x'

(D€ thi todn qudc ft'1991)Gidi.Thay ) = x vio di6u kien (1) ta c6

Trang 7

t crir rHuoNG tE uflru rruEm ufim ztltll

GiAi thuong LO V[n Thi0m cira H6i To6n hoc

Vi€t Nam dJoc giinh dd tang cac thdy cO gi6o

c5 thhnh t(ch xuat sic trong giAng day mOn to6n

o bAc phd thOng vi cdc hoc sintr phd thOng dat

thinh tich xudt sic trong hoc tAp mOn To6n

GiAi thuong L0 Vln ThiOm nim 2001 duoc trao

cho ciic thdy gi6o vd hoc sinh sau dAy:

I) Thdy gi6o

Thu, Hda Binh

- Da 11 nim day To6n tai tinh midn nfi Hoi

Binh, 1i0n tuc lI giiio viOn day gi6i cdp tinh

- Tt i996 ddn nay, phu trrich giing day tdt ci

Binh C6c hoc sinh cta D6iiuydn dd dat32 gihi

qudc gia, trong d6 c6 2 giiti nhat, 10 giii nhi

2.Vfi Qu6'c Luong, sinh nf,m 1953, td tru&ng

td Todn trudng TT{CS Chu Vrn An, Hi NOi.

- D6 c6 thdm ni0n giAng day 29 nim.liOn tuc

chu nhi€,m cr{c 16p chuydn, chon cfia trudng

- Lien ruc tham gia bdi dudng hoc sinh gioi vh

DOi tuydn quAn Ba Dinh, trong d6 c6 nhi6u em

dat giii cao 6 c\c cudc thi Thinh phd, Qu6c gia,

HA HUY KHOAI

nhidu em v6 sau du-o c giii trong ci{c ki thiOlympic qudc td.

- Li ddng tic giit ctra 2 cudn sdch bdi dudng

hoc sinh gi6i (NXB Girlo duc) vi c6 nhidu bdidang trOn tap chi Toiln hoc & Tu6i tre.

II) Hoc sinh:

l.Vfi Ngqc Minh, hgc sinh lop 11 (nam hoc

-Tin, trudng DHSP HA NOi.

- Gi6i nhat ki thi HS gi6i thinh phd ldp 7,8,9.

- Giai nhdt ki thi hoc sinh gi6i Kh6i chuy0n

To6n-Tin DHSP He Noi ldp 10, ll,12.

- Giii ba Ki thi HS gi6i todn quoc nim 2001.

- Huy chuctng vdng CuOc thi Olympic Todnqudc td2001

2000-2001), Khdi Phd thOng chuyOn Toiin-Tin,(hiQn lh hoc sinh Khdi Ct nhAn khoa hoc tdinang), trudng DHKHTN-DHQGHN

- Giei nhi ki thi HS gi6i todn qu6c 1999-2000.

- Giai nhi ki thi HS gi6i todn qu6c 2000-2001.

qudc td2001

l-r5 trao siii thuons L0 Vin Thi€m nam 2001 dI

duoc td Jhrlc ngly'"3 -3-2OOZ tai Hoi trudng LeVan Thi6m 19 L0 Th6nh TOng Hh NOi.

G- thi/(r) = u vdi m6i z c6 dang (11) Thay

x = n vdo(10) ta c6 fluz + 2u) = uz + 2u.

r Ndu -I < u< 0 thi cfrng c6 -l < u2 +2u <O

mi theo didu ki0n (2) thi chi c6 1 didm bAt dOng

trong (-1, 0) non u2+ 2u - v D u = 0 hoic

u = -l didu ndy khOng xiy ra.

Ndu z > 0 thi u2 + 2u > 0 Lap luan nhu tr€n,

didu niy cfing kh6ng xhy ru.

.Y4y u = 0lh didm bdt dong duy nhdt thu6c 5

cita him l nghia lI x + (l+.rXx) = 0

c6c phuong trinh hlm sau ddy :

Bni 1 Giii cr{c PTH:

/ r\

a)f(x) - zfl ^ : I =xvdi.r*0, -1, +1.

\-r+I/

b) 2t(2x) = f(x) + r, trong d5/lipn tuc trOn R.

Bni 2 Tim tdt cil cdc hlm sd/: R -+ R th6a

Trang 8

DuNG PHgoNG wap pnfrn HgficH TfP trgP

)\r/

DE GIAI BAI TOAN SO HOC

.r1t^

LTS BAi tudn s6' 6 trong cu\c thi Todn qu1ic

te'\MO) ldn turt 36) ndm 1995 h mot bdi todn

kh6 ddi h6i tinh s6'cdc tdp con cfia m6t tQp hqp.

Ldi gidi bdi ndy cila ban Ngd Ddc Tudit (Huy

chuong vdng cila IMO ldn tht 36) dd duo c ddng

trong THTT sd 224 (2/1996) Trong bdi bdo

"Ding cdi do dd dem cdi thdt" THTT s6'250

(4119"98) tdc gid Ddng Hingit,a,,g cld sir dturg

sd phftc dd gidi bdi todn tdng qudt hon sau ddy:

Bii to6n 1 Cho p 1I m6t sd nguyOn td 16 Tim

sd ci{c tap con A cfia tAp hqp X = {I,2, , n} c6

tinh chdt sau :

i) A chrla dring p phdn ttr

ii) Tdng tdt ch cr4c phdn tir cira A chia hdt

cho p

Bdi todn s6'6 cia IMO ldn tht 36 ld trudng

hop ri€ng khi n = 2p.

Tda soan dd nhdn duoc thu cila cdc ban :

Todn, trudng THPT NK Trdn Phrt, Hdi Phdng,

HC bac cila IMO ldn thft 42 ndm 2001) v0 ldi

gidi bdi todn n€u tr€n dua ten sil phdn hoqch

tdp hW.i' tudng ndy dd daoc rinh bdy t6m tdt

trong hudng ddn gidi cfia naoc ra di thi ld Ba

sft dung phuong phdp ndy dd gidi nhidr.t bdi

todn khdc

Ta goi su phAn chia tAp hqp A thdnh ci{c tAp

hgp con kh6c r5ng A1, A2, , A*(k > 1) cira A 1I

2) Cdc t&p con dOi m6t rdi nhau, nghia lI

4 oAi = A vdi moi i *7, i, j e {1,2, , k}

Dd hp mOt phdn hoach cira tAp hqp A ta xdc

dinh m6t quan h0 T gltra cdc phdn tir ctra A c6

ba tinh chdt sau vd goi li quan h0 tuong duong :

1) Tinh phin xa : iTxvii moi x thu0c A

2) Tinh ddi xung : n6u xIy thi yTx vdi moi x,

y thuOc A

6

3) Tinh bac cdu : n€u iTy vit yTz th\ iTz vdimoi x, y, z lhulc A

Ngudi ta dd chrrng minh rang : MOt tap hqp A

duoc phdn hoach thhnh ci{c tap hqp con Ay A2, , Ar khi vI chi khi tdn tai quan h€ tuong duong

T grira c6c phdn tit cfia A th6a mdn : ify e x, ycing thu6c mOt tAp con A; ndo d6 (1 < i < ft)

Ta sE vAn dung phuong phdp phdn hoach tAp

hqp dd giii ciic bhi to6n li€n quan ddn bdi sd

hoc s6 6 trong ki thi IMO ldn thf 36.

V6i m6i qp hqp s6 hfru han A ki hicu lAl li sd

c6c phdn tir'bira''tap f,qp A'vI S(A) ld tdng cric phdn ttl cira tAp hqp A

Bii to6n 2 Cho p ld mil sd nguy€n fi'ld vd

sd nguy€n duong t < p Tim s6' cdc tdp con Dcfia tdp Y = {1,2, , p} c6 tinh chdt sau

D6 thdy quan h6 D q*D^ gitta cdc tAp gdm r

phdn tir trong J(ld quan h0 tuong ducrng vit,7(duo c phAn hoach thinh c6c lop rdi nhau, m6ildp g6m p mp (do m = 0.l, , p-11 vi mdi tAp

c6 dring r phdn tit

ct

Vi v4y s6 ci{c lop cira .7/ h 2

,p

Tinh tdng S(D*) = f(D) + mt lmodp), ta thay

mt kh6ng chia hdt cho sd nguy6n tdp n6n S(D,n)

vh S(D) c6 s6 du kh6c nhau khi chia cho p md

m6i ldp c6 dfng p t4p (m = 0, 1, , p-1) nen

trong mdi lop phii tdn tai dring m6t tAp c6 so du

bang r khi chia cho p

Vdi m6i r mi 0 < r < p - lgqi s, lh sd cdc tdp

D ctra ./(md S(D) = r (mod p) ta ket luAn duocSo=51 = =Sp-l =

C,,

p

Trang 9

Ldi giii biri todn 1 Gie sfl n = kp + r vdi

Trong tap € cdc mp con A c5 dring p phdn tit

cta tdp Xta xdc dinh mOt quan hC 7 nhu sau :

ATA'ndt th6a mdn 2 didu ki0n

1) nOu Aw A' c C;v6i chi sd i lon nhdt (1 < I

< t) thi A a Ci*1= A' (\ C'i+r

2) t6n tai dnh xa f : An B; + A' a Bi (i l6n

nhdt nhu ren) th6a mdn : vdi moi x e A n B; thi

f(x) = x + m (mod p), trong d6 m li hlng sd,

0<m<p-l;l<f(x)<p.

Dc dhng chrlng minh duo c quan hO I ld quan

h€ tuong duong trong mp 6) vd €; duoc phan

hoach thlnh lop c6c tAp con c6 ding p phdn tft.

Trong phAn hoach niy c5 ft lop g6m dring 1 tAp

con li {ar\ , {nr}, , {r* } xet lop c6 it nhdt

hai tap hqp ln A vit A'vdi chi sd i lon nhdt nlo

d5, mI Aw A' c C;,lic d6 A n Ci*t = E = A'

A Ci*r vi A n Bi = D, A' n B, = D'.Lic d6

A = D w E ; A' = D' w E v6i D vi E rdi nhau, D'

vd E rdi nhau GiA sil lDl = lD,l = t = p_ lEl .

Lric d6 x6t quan hO 7 han chd trong cdc ldp D

g6m dring r phdn trl trong tdp ,7{lcdc mp con D,

ta 6p dung duoc kdt qui bhi todn2:

M5i l6p cdc mp D cta ,7( ch(ra d:frrg p tap

hqp, suy ra m6i l6p cdc tAp A cira d cfing chrla

dring p qp hgp, do d6 sd cdc l6p nhu thd cira d

/-P t.

le "' -^ (doc5klop {8,}).

p

Ndu S(E) = p - r (mod p) vi S(D) = r

(mod p) thi S(A) = S(D) + S(E) : 0 (mod p)

nghia lI trong m6i lop ctra tilp A c6 dring mOt

4p h-d-p c6 tdng c6c phdn tir chia hdt cho p

T6m lai, s0 c6c tap con c6 dring p phdn trl vi

c6 t<ing cr{c phdn ttl chia hdt cho p bang

k+ci-t ,rrons d6k=lol ,

rr[No Ar{H ouA eAc BAI I0AI{

BN Sd 51

Problem Given are two skew lines in the

space and two segments of fixed lengths on the

lines Prove that the volume of the tetrahedron

determined by the endpoints of the two

segments is unchanged by sliding the segments

along the lines

Solution Recall that a base of the tetrahedron

is one of its triangular faces and the

H'J"IJ"JIJ t#";

tetrahedron 1s

proportional to the area

of a base and the

corresponding height,

we can solve the

problem by showing thateach of these ingredients

remains unchanged

Obviously, it is enough to show this claim whenone of the segments is held fixed and the other

is permitted to move .

Let AB be the fixed segment and let CD move

along a line x Then the area of the base ACDremains constant for all positions of CD Thedistance from B to the base ACD is the distance

from B to the plane spanned by A and the line x,

which is constant, too

recalltriangularheight

distance vertex

= lQch, ch6o nhau (tinh tt)

= khong thay d6i (tinh tU)

= trugt (dong tit)

= dqc theo (ph6 tit)

= nh6 lai, nh6c lai (dong tU)

= d?ng tam gi6c (t(nh tit)

= dQ cao

= khoing c6ch

= dinh

proportional = ti lQ thuan (tinh tir)

area = dign tich

solve = gi6i, gi6i quyOt (dQng til)

ingredient = thinh Phdn

obviously = mqt cdch 16 rhng (ph5 til)

hold = n6m, gi[, dfing (dOng tit)

permit = cho ph6p (dong tU)houe = chuidn dong, di chuydn (dOng tit)

constant = bit bidn (tinh tti)

position = vi tr(

ipan = c6ng ra, trii ra (d6ng tt)

Trang 10

V{D- VA*N NHS

Ta thtt tim mdi li6n h0 gifa c6c mlu sau khi

xiy ra m6t sd cu6c gap gifla ci{c con ki nh6ng.Gii sir sau a cu6c gap kidu u, ta c6

Suy ra sau a + b + c ldn gap nhu thd c6

N-X-N,,- a+2b -c-(X, -a-b+2c)

=3(b-c)+N.-\

Vdi N - Xn = 2tac6:

N-X-2=bgisdciia3

Ndu chring c6 miu tr6ng ciL thi N = X = 0 =

N - X - 2 = -Z kh6ng phii bOi sd ctra 3.

Ndu chring c6 miu xdm ch thi N = 0 + N - X

- 2 = -45 - 2 = -47 kh6ng phii b6i sd cria 3.

NCu chring c6 mdu nAu ci thi X = 0 =

N - X - 2 = 45 - 2 = 43 khOng phii boi sd

cira 3.

VAy vdi 13 con xr{m, 15 con nAu viL 17 con

tr6ng thi kh6ng thd xAy ra trudng hop 45 con

cing mILr sau nhffng ldn gAp nhau duoc

C{c ban hay di tU kdt luAn ddn gii thidt dd xdtxem vdi nhfng miu nho cira 45 con ki nhOng rhi

rnOt miu ? Til bei todn nhy ban c6 nghi ra mOr

Ki'nhOng lh lohi bb si4t ludng cu ddo hang

sdng 6 cdc bdi cdt, bd bidn d m6t hdn dno tur

DO Duong c6 m6r loli ki nh6ng dAc biet, rdr

d6c dr{o Chfng c6 ba miu : x6m (X), n0u (hat

d6) (N), trSng nhat (T) Di6u thri vi le khi

hai con ctng miu gdp nhau thi chring gifr

thi chfng ddi thhnh miu thrl ba Vi du 2 con

mhu x6m vd tring g[p nhau thi chfng c[ng ddi

thdnh mlLu nAu.

Tai m6t thdi didm ngudi ra ddm thAy c6 45

con ki nhOng vdi cdc mlu : 13X, 15N, 17T.

Biri to6n dit ra th : Li0u sau m6r sd ldn gap

nhau gita chfing, 45 con ki nhOng v6i cdc mdu

nhu rren c6 thd bidn thinh ctng mot miu duoc

kh6ng ? Ndu duo c, thi mhu gi ? Ndu kh6ng, thi

tai sao ?

Ldi giii :

Lric ddu tiOn : Xo = 13, No = 15, To = 17.

C6 ba trudng hqp don giAn nhdt khi c6c con ki

nh6ng gip nhau dd chring d<ii mhu.

Kidu a : 1 xi{m gip 1 nAu thdnh trSng c6,

De thay rlng thrl tu ciic kidu g[p 1I khOng

quan trong : oB = Bcr nghia lh kdt qui ctra hai

kidu gap cx r6i B vI B rdi cr li nhu nhau.

C6 thd kidm tra dd thay rang kdt qui ctra cr{c

kidu gap nhau khdc cflng vAy, nghia l] oy = ycr

vd By = yB.

,.:*a."f;

Trang 11

vtON THI : roAN vdxc r

(Ddrfi clto tli sinlt thi vdo clu.ty€nTodn,

Ly, H6a, Sinh)

(Thdi gian ldmbdi : 150 phtu)

CAu I Tim c6c gi6 trr nguyOn ,r, y th6a mdn

dang thric :

(Y+2)x2+l=Y2CAu II 1) Giai phuong trinh :

CAu III Cho nira vbng trdn dudng kinh AB =

2a Tran doan AB ldy didm M Trong nira m6t

phing bd AB chfa nta vdng trdn, take 2 tia Mx

vd My sao cho fff*=fu, = 30" Tia Mx cit

nira vdng trdn & E, tia My c6t ntta vdng trbn & F

Ke EE' , FF' vu6ng g6c xudng AB

1) Cho AM = !, ti* diOn tfch hinh thang

2 vrr6ng EE'F'F theo s

2) Khi di€m M di r16ng tr6n AB, chfi'ng minh

dudng th6ng EF luOn tidp xric vdi mOt vdng trdn

c0 drnh.

thuc khdc

CAu V Vdi x, y, z ld nhfrng sd thuc duong,

hdy tim gi6 tri 16n nhdt cfia bidu thrlc

(Thdi gian ldm bdi : 150 phit)

CAu VI 1) Cho f(x) = axZ + bx + c c6 tinh

chdt : /(r) nhAn gi5 tri nguyOn kfti x li sd

nguy0n H6i cr{c hC s0 a, b vd, c c6 nhat thidt

phii ln cdc sd nguy€n hay khOng ? Tai sao ?2) Tim ciic sd nguy6n kliOng 8m;r, y th6a mdn

=9

t"

loro *bya =17Hdy tfnh gi6 tri c:iia cdc bidu thrlc :

A= ax5 + bys

B-a;oor*by'0"

CAu IX Cho doan thing AB c6 trung didm ld

O Goi db d2ldc6c dudng th&ng wOng g6c vdi

AB tttong fng tai Avit B M6t g6c wOng dinh O

c6 m6t canh cit dr b M, cdn canh kia cit d2 b

;V Ke OH vu'lng g6c xu6ng MN Vdng trdnngoai tidp tam gi6c MHB cat r/, & didm thrl hai

E khdc M MB cat NA 61, dudng thhng HI cit

EB b K Chrmg minh rang K nam tr6n mOt vbng

trdn cd dinh khi g6c vuOng quay xung quanh

dinh O.

duoc son mOt mit b[ng miu d6 vd mat kia bang

vdng trdn sao cho tdt ch cdc d6ng tidn d6u c6

mit xanh ngira l€n phia tr€n Cho phep m6i ldnddi mat ddng thdi 5 ddng ti6n liOn tidp canh

nhau H6i vdi ci{ch ldm nhu thd, sau m6t sd hfru

han ldn ta c6 thd lim cho tdt ch c6,c ddng tidnddu c6 m[t d6 ngr]a lOn phia tr0n duo c hay

khOng ? Tai sao ?

Cnu IV Gi[ su x, y, z ld cr{c sd

khong thoa m6n h0 ding thrlc :

Trang 12

ni ttt at\t n|t rurit ntt Butttttt

Bei 2 Cho tAp hqp c6c s6 nguY0n Z

a) Cho f, I : Z > Z lit c6c song 6nh Chtmg

minh rlng hdm sd h: Z > Z, duoc x6c dinh boi

h(x) flx)S@) v6i moi x e Z, kh6ng thd la

tohn 6nh.

b) Cho f : Z -+ Z ld todn 6nh Chtmg minh

rang tdn tai cdc tohn 6nh g, h : Z -+ Z thbamdn

f(x) = S@)h(x) v6i moi x e Z

Bni 3 C6c canh ctra m6t tam gi6c c6 dQ ddi lh

a, b, c Chung minh"bdt d8ng thrlc :

+ (a+b-c)(-a+b+c) < J"urdi * Ja * Jil

Bni 4 C6 ba trudng hoc, m6i trudng c6 200

hoc sinh M6i hoc sinh (trong ba tru&rg ndy) c6

ft nhdt mOt ngudi ban trong m6i truhng (ndu a li

ban cta b th\ b lI ban cia a) Bidt rang tdn t4i

mot mp hqp E g6m 300 hoc sinh trong c6c

trudng n6i tr6n sao cho vdi m6i trudng S vI v6i

hai em hoc sinh x * y b{t ki trong E md x' y

kh6ng thuOc trudng S thi s6 ban cita x vI sd ban

ctra y trong trudng S li hai sd kh6c nhau.

duo c m6t hoc sinh sao cho chring li ban

cfia nhau.

Ngudi gioi thi\u: v0 oilur gde

l0

tdc giri Trdn Tudn DiQp dd gioi thi€u ldi gidi

Bdi todn 1 duoi dAy tong d€ thi tuy€it sinh vdo

trudng EH Vinh ndm 2001 cilng voi sr.t phdn

Tda sogn dd nhdn duoc tha cfia cdc banTrinh

Hd Tafl vd ban Dd Vdn Dftc (@ todn trildng

dd chuttg'minh k€'t-qud crta cdc bdi todn tdng

qudt hon dudi ddy, d6ng thdi cfing n€u ldi bdn

iham cfiaTS NguydnVi€t Hdi

BAi 3 (cira ban TrinhXudnTinh)

Chung minh rang :

2 r, =a2 + b2 + 12 + rout ro, e)

BAi 4 (ctra ban DdVan O*c1

Chfng minh rang vdi2p = 3, vI s6 thuc t,

Trang 13

vfi vffic rloc nUoruc cd rHI mor mrdu umtla

'Gltla

cflccrroItmot a^ rnlir cilc

b) '4p N€u, = 2- thi a2 + b2 + c2 + tabc >

\,i\t)

16 32 ^ ( zo\2( - 8 ) +f = ip +fin\.[.-t.J [rc+6 -;* )

a) Ndu 2-=,

=!,x6t c le canh ng6n nhdt'4p3p

thic< 2o 3 =rr=L.4 3p3 =zvit+r-2>or nOn tU (7a) suy ra 4f> 8p',titc lh c5 (6a)

Giistr c <b < avd a+ b + c= 3 thi0<c < 1.

Nduo <k<? 22 thiftc< I t.6=vdphii

(5) duong TiI d6 3a2 + 3b2 + 3c2 + kabc >

Ldi birn th€m C6 ttrd sir dung cdch bidn ddi

cta ban D5 Van Drlc dd u6c luong gi6 tri bidu

Trang 14

T'C RR III TR?

cAc l6p rHCs

BdiTll297 Tim ba sd nguy€n a sao cho bidu

thrlc sau c6 cirng mot gi6 tri: a3 - l\az + 47a

l-a I-b a+b

trong d6 a, b ld hai sd thuc duong th6a mdn didu

ki6na+b<l

TRr.xJNc Ncoc DAc

(W THft li Quy Dbn, Qu1, Nhon, Binh Dinh)

B iT3l297 Chung minh bat ding thrlc

trong d6 a1, a2, a7, ct4, c5 li 5 sd duong th6a

man oreu Kren

01 +a2+a3*rj+o!>t

TRI,XJNG CAO DIING

(sY top DiQn *i;fli,f, Bdch Khoa

BiiT4l297 Duong trbn mm O bdn kinh R vi

duong trbn mm O'bdn kfnh R'(ft > R) tidp xric

ngoii tai didm A G6c vu6ng xAy citt hai dudng

trdn 6 c6c didm B vd C (khdc A) Gqi H ll hinh

chidu cira A trOn duong thing 8C H6y xrdc dinh

vi tri cdc didm B, C dd do diLi AII i6n nhat vi

tfnh gi6 tri d6 theo R, R'.

NGUY:ENDI,IC TA].I

(ffi Tp.H6 Chi Minh)

kinh EF Ldy hai didm N, P trOn duong thing

EF sao cho ON = OP Tir didm M ndo d6 nam

dudng thd,ng MN cit dudng trbn tai A vd C,

duong thhng MP cat dudng trdn tai B vi D sao

cho B vd O nam khr{c phfa ddi vdi AC Ggi K ld

giao didm crta OB vd AC, Q ld giao didm cira

nguyOn td cirng nhau, chring minh rdng c6 dring

BniT71297 Tim gir{ tri l6n nhdt cira bidu thrlc

ol + al + + rl, Qt > l), trong d6 ci4c sd thuc

e1, a2, , a,, thu6c [0; 2] vn th6a mdn

o t * o' 7::_' i _1_': _.n_* _

IRINH XUAN TINH

(G/ THPT Phil Xty€n B, HdTay)

B iT8l297 Tim moi gi6 tri cfia tham sd a dd

phuong tr\nh ai + 2cosx = 2 c6 ding2 nghiOm

BiiTgl2gT Duong trbn tdm 1b6n kinh r tidp

xtic vdi ba canh BC = a, CA = b, AB = c cia

tam gi6c ABC ldn lugt 6 c6c didm M, N, P Goi

S li dicn tich MBC vh ho, h6, fu lh d0 dli duongcao ctra LABC tuong rlmg vdi cric dinh A, B, C.

(GV KhoaTodn, DH Vinh, Ngh€ Att)

(OA, OB, OC vuOng g6c ttng d6i mQt) Goi cr,

B, y ldn lugt li g6c hqp bOi c6c mat OBC, OCA,

OAB vdi mdt ABC Tim giri tri l6n nhdt cira

bidu thrlc

Itgcr.tgB tgp.tgy tgy.tgcr ) tgzo.tgzB.tgzy

BUI LINH PHUONG

(S/ K338 khoaTodn DHSPThdi Nguy€n)

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