thtt so 279 thang 09 nam 2000

7/17/2019 THTT So 279 Thang 09 Nam 2000

http://slidepdf.com/reader/full/thtt-so-279-thang-09-nam-2000 1/28

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)

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;i

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l/il

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Ban-

hAy

lAm

m6t bdng

giAy

c6 cac nilp

gAp. (cac

n6t ch6m

tr6n

h)nh 1)

tao thirnh

10 tam

gi6c

d6u

vd

t6 3 mAu.

.GAp.

bdng

giay

ldn lugt

O

theo

cdc doan

thdng BC,

DE

FA

(h)nh

2) rOi

ddn

hai tam

gidc

kh6ng

c5

miu

vdo nhau

dd

duo.

c

luc

gidc

ddu

ABCDEF'

c5

mdt

phdi

toan

mdu

do, cdn

mit

trii

todn

Mqt

phAi

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1

Hinh 1

AB

Hinh 2

mdu xanh

(hinh

3 : cdc

didm

cirng

chLt

gAn

trirng

nhau,

cdc

n6t

drit

chi duong

khuat).

Ddy ciic dinh B, D,

F

xu6ng

dudi

vAo

phia

trong

(m0i

t6n d hinh a)

dd

ch0ng

gAn

trUng nhau,

rdi tit

dinh

O mo

AB

Hinh 3

A

Hinh 4

ffi3:rtl3;

;ft"TBJ:i"Y"r3i,:i'i#

nsoai,

ta rai

duoc ruc

gidc

deu

nhLrns c6

1 mdt da

ddi

mau. Luy6n nhanh

tay, ban c6 thd

bidu di6n tro

Ao

thuAt Luc

giac

ddi

mau.

MAu

hlnh

nayc6t6n

lir H6cxaphl6xag0n

(hexaflexagon) xu6t

phdttutiilng

Hylap:

hexld

sdu,

ftexld

gap,

tirdu6i

gon

chida

gidc. Trd choi

niry

do

mQt nghiOn

ct?u

sinh

todn

nguoi Anh ld Actho

X.

St6un, hqc tai

trudng

Dqi

hoc

M

Princiton,

tim

ra ndm

1939, sau

d6 lQp H6i

Phl|xagOn

dd

nghiOn

cfu

cdch ddi mau

c0a cdc hinh

pht?c

tap

hon.

DANH CHO

BAN

DQC

1)

Ldm

mQt

bing

gi6y

c6 cdc

nOp

g5p

tao thAnh

19

tam

gi6c

dbu vd t6

sdu miru

(ki

hiQu

mau

1,.2,g,4,

5,6)

nhrhinh5.

GAp

bdng

gi6y

nirytheo

n6p

g?p

ki6u xodn

5c sao cho.mit

phAi

dOu

nim

phia

ngoai dd

dugc bdng

2l6p

nhuhinh

1, sau d6

gAp

thanh

hinh

lr,rc

gi6c

d6u nhu

hinh

2

sao cho

mOt mit toAn

miu sO Z,

cbn

mdt

kia

toAn miu

s6

3.

2) Ban

hdy ldm

Ao

thuQt

bi6n ddi

lUc

giric

vira ldm

od

xuAt

nien

mQt

mat

todn

mAu

s6t ztoinmdu

s6qz

mau s6

5?

mau

s6

oz

H5y

chi

ra

cdch

bi6n

odi nnutno.

Ndm

phAn

thudng

dAnh

cho

cdc bqn

bi6n

ddi duoc

nhibu

mAu

nh5t.

PHI

PHI

(Xetn

tidp

trang

l3)

Hinh

5

7/17/2019 THTT So 279 Thang 09 Nam 2000


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ddng

bian

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:

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ttoAttc

cHUNG,

ttcO o4r

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r-E rHAc

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HUY

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ra

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vrEr

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ki,

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v0

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rsu

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di.en

yhia

Nam

:

TRAN

CHI

HIEU

23t

Nguy6n

Vdn

C*,

TP

H6

Chf

MiNh

DT

:

08.8323044

bri

hinh

hoc

du6i

r-E eA

ruc6

TRONG

SO

NAY

ffi

Oann cho

Trung

hoc co

sti

-

For

Lower

Secondary

Schools

Hb

COng Dang.-

Giei

to6n cuc

c6ch

nhin

dai s6

ffi

fidng

Anh

qua

cric

bii

tofn

-

English

through

Math

Problems

-

Ng6

ViAt Trung

ffi

f

ni

tuydn

sinh

viro

Dai hqrc

-

University

Entrance

Ex:rms

Dod.n

Tam

Hite

-

Db

thi

tuydn

sinh

m6n

To6n

vho

DH

XAy

dung

vi

DH

LuAt

He

N6i

2000

Gitii

thiQu

vd todn

hoc cao

cflp

-

Introduction

to

Higher

Mathematics

Hd Huy

Khod.i

-

B&y

bii

tor{n cria thi6n

ni6n

ki

Nhin

ra

thd

gitii

-

Around

the

World

Db

thi

Olympic

to6n

cria

Ddi Loan

Ban

dr2c

tim tbi

-

Reader's

Contributions

Pham

Hbng

Qud.n

-

Vii u6c

luong

trong trl

di6n

Di6n

tlin

d4y

hgc

toiin

-

Math

Teaching

lrorum

NguyZn

Huy

Doan

-

Yb

UO sdch

gi5o

khoa

To6n

THPT

chinh

li hop

nhAt

m

DO

ra

ki

niy

-

Problems

in this Issue

TLl279,

..., T101279,

LL,L2{279

Giei

bei

ki

trui'c

-

Solutions

to Previous

Problems

Giii

c6c

bii

cria

sd ZIS

ffi

Ca;

lac

bQ

-

Math

club

CLB

-

Gbp

nhau

qua

nghy

sinh

NGOC

MAI

-

Gi6i

vh

k6m

b6n

nhau

Sai

lim d tliu

-

Where's

the Mistakes

?

KIHIVI

-

Giai

d6p

"Tai

sao lai

th6

?"

NguyTn

Kim

Thanh

-

Phuong

trinh

v6 nghi6m

?

Bia

1; Nibm

vui n5m

hoc moi

b6n ng6i

truong

moi

cirathby

trb

truUng

THPT

chuy6n

Lam Son,

Thanh

Hoa

Bia

2

:

Toiin

hoc

mu6n

miu

*

Luc

gia

06i

m:u

Bia

3 :

Giai

tri

to6n

hoc

-

Math Recreation

Bia

4:

Toin tu6i

tho

-

nguUi

ban

moi

7/17/2019 THTT So 279 Thang 09 Nam 2000


.Ddnh

dna

SANC

t4o

cic

bh.i

todn

li

vigc ltrm

cdn rhidt

cria ngudi

hOc

to6n.

Qua

btri

vidt

ntry,

xin

dugc

trao

ddi vO vli

bli

todn cgc

tri hinh hgc

duoc

nhin

tri

bii

toin cgc

tri

dai

sd.

Tru6'c rl€t1,

x&vtri

vi

dg sau

:

Bii

todn

1.

Cho hinh

vuOng

ABCD

c4rlh

a.

M

lI

didm

di

dQng

trOn

c4nh

AB.

Dlmg

c6tc

hinh vuOng

c6

cAnh

MA,

MB

vd

bOn

trong

ABCD.

Xic

dinh

vi

tri

cria

M

dd diQn

rich

phin

cdn

lpi

S

cira

hinh

vu6ng

ABCD

le

lon

nhdt.

Ini

gifii.

Di,t

MA=x,

MB--yv1ix,y)0

th6a man

x*y

=

a.

Ggi

,tl

vL

52

lin

lugt

ld

diQn

tich hinh

vuOng

cA;nhMA

viMB

thi

Sr

=

*2 vI

52

-

y2.D6

thdy

S 16n

nhdt

<+

A

rM

)

)r\

GIAI

TOAN

CIIC

TRI

HINH

l.lOC

DU{II

CACH

NH|I\

oAI SO.

nd

cONc

DcrNG

(GV

tradng THPT

chuyAn

BinhThu|n)

tuorrg

trJ

nhu

trOn

suy ra

GTNN

cria S

bing

nRz

,,llf.c

d6

M

trung vot

tdm O.

Ldi

gi6i

c6c

bii to6n

trOn

ddu ddn ddn

viQc

x€t

gil,

tri nh6 nhdt

crla

bidu

thrlc dpng

*+yz

hopc.r'+y'....

trong

d6

x+y ld

hing

sd. Nhu vfly

viQc

gi6i

cdc bdi

to6n cgc

tri

hinh hoc

c6

thd

chuydn

vd

gif,i

bii

todn cr"rc

tri

dai

sd.

Ta

xdt

btri

to6n

cUc tri

dpi

sd

tdng

qu6t

hcrn.

Bii

todn

d4i sd

:

XAt

n

sd khAng

dm x1, x2,

...,

xnthda

mdn

: x1

+

x2

+

...

*

xn= a vhi

a ld

sd duong

cho traOc.

Tim

gid

tri nhd

nhdt

crta cdc

bidu thtic

:

a)

P

=

*r+*r+...+P,

b)Q=*?+4+...+f,

Ini

gini.

a) Ap

dung bdt ding

thric

Bunhiacffpxki

ta

c6

:

(x1+x2+...+xn)2

<

n(4+

4.+

...*

4la

a2

<

n(Py+

$*

...*

t)'

1

e

*1+ x,j+ ...

*

4>:

(l)

"n

Ddu

df,ng

thric xf,y ra

khi

I x1= x2= ...= xn

Vfly

GTNN

cfia

P

ra

d

r.fri

xl

=

x)= ...

=

n

a

Y

=-

"n

n

b)

DAt

tfxi=ti

>

0 vdi

i

=

1,2,

...,

n

tfi

xi=*vb,fi=1.fi.

Ap

dung

bdt

dhng thric

BunhiacOpxki

tac6:

$t4,+

b8*

...+

UPn)2

<

(Pr

+

?z+

...

+

h$f

+

4+

...

+

fi)

lnav

(Pr*

Pr+

...+

4)2

3

o(4,+ fr+

...t

4)

Q)

c&E

TRUNG

HOC

GO SO

S,

+

52

nh6

nhdt

<+

P

=

xz

+

yz

nh6

nhdt.

Tt

bdt

ding

th'uc

2(*

+

y\

>

(x+y)z

=

a2

suy

ra

gi6

tri

nh6

nhdt

(GTNN)

cria

^S1

+

S2

Uing

{,

2,

hic

d6

M

ltr

trung

didm AB.

Bii

to6n

2.

Cho

dulng

trdn

(O,

R).

M

lI

didm

di dQng

trOn

dudng

kinh AB.

X6c

dinh

vi

tri

ctn

M

dd

tdng

diQn

tich

c6c

hinh

trdn

c6

dudng

kllhMAvlMB

1I

nh6 nhdt.

Ldi

girti.

D{t

MA=?-r,

MB

=

Zy vu

x+y 2

0 th6a

x*

=

R

(khong

ddi).

Goi

A

. 1 vtr

^i2

ldn lugt ll

diQn

tich

hinh

trdn c6 dudng

ktfilMAvdMB.

Dd

thdy

S

=

.S,

+

S"

nh6

nhdt

e

P

=

n*

+

iyz

=

n(xz

+

y2)

ntr6

nhdt.

Lfp

lufln

7/17/2019 THTT So 279 Thang 09 Nam 2000


]\r

(1),

(2)

ta

c6

a(fi +

fr+ ... +

f;

>4

^n'

o

4+4+...

+

4.#

Ddu

ding

thrlc xtry

ra

khi

(1)

vl

(2)

ctng

xtry

ra ddu

ding

thric, khi

d6 i

x1= x2= .

..= xn.

Vfly

GTNN cira

Qrufixn

x =

xz--

...

=

xn

a

n

Bii

to6n

3.

Cho

dudng

trbn tam

O.

G6c

ZxMy

-

u

kh6ng

ddi

(90o

<

o(, <

1800)

quay

quanh

di€,m

M cd dinh

trOn

dudng

trdn.

Gqi

1

le hinh

chidu vu6ng

g6c

cria

M l€n

d0y

cung

'

r.,

ndi

2

giao

di€m cria

Mx, My

voi

dudng

trOn.

HXy x6c

dinh

vi tri

g6c

xMy

sao cho

gi6

tri

IA3

+

IB3la

nhd nhdt.

Ini

gini.

Gi6

str

Mx

vh, My

cdt

dudng trbn

ldn

luqt

6

A

vb B.

Yt ZAMB

=

u

khOng

ddi

nOn dQ

dei

A-B

bing

hing

sd a,

fi)

d6

IA+IB=

a.

Ap dung

kdt

qu6

btri to6n dai

sd thi

gi6

trf

IA3 + IB3 le

nh6 nhdt

khi

dO dei

IA

=

rB

-

I

lric

d6

g6c

xMy rhQn MO lL

dudng

ph0n gi6c.

Bii

todn

4.

Chia do4n

thing AB

cho trudc

b1i

cilc didm chia

theo thri'tU

A

=

My

M2, ...

,

Mn+1= B.

Goi

51, 52, ...,

,Stx

le

diQn

tich

cria

r

hinh

vuOng

c6

cdc

cpnh

ldn

luqt ltr M1M2,

M2M3,

...,

Mr\n+r.

Tim

gi6

tri nh6 nhdt

cria

tdngS=51

*Sr+...+Sr.

Bii

to6n

5. Chia

dudng kinh AB

cria

dudmg

trOn

(O,

.R)

cho trtrd,c

ldn luqt

thtrnh

n

doVn

x1,

x2,

..., xn

chi chung nhau

didm diu

mrit

sao

cho x1+x2

+

...

+

xn=

AB.

Tim

gi6

tri

nh6

nhdt

cria

tdng

diOn

tich

n

hinh

trbn

c6

dudng kinh

lI

c6c do4n

xy x2, ...,

xr.

Bii

todn

6.

Xet

n

hinh

lflp

phucrng

vfi

c\c

cpnh 11,

x2,

...;.r,? sao

cho tdng

x1

*

x2+ ...

+

x,

=

a

duong

khOng

ddi. Tim

gi6

tri

nh6

nhdt

cfia

tdng

thd tich cria chring.

Bii

todn

7.X$

n

hinh

cdu

vdi

c6c b6n

kinh

ty

t2t...t

rn

sao

cho

tdng

/1 * 12+...

+ fn=k

duong

khOng

ddi.

Tim

gi|tn

nh6 nhdt cria

tdng

c6c diQn

tich

vl

cira

tdng

c6c

thd

tich

cria chring.

Hi

vgng ring

gif,i

bli to6n bing c6c

c6ch

nhin kh6c nhau,

c6c ban

s0 s6ng

t4o'nhidu bli

toen

hay

hon,

hdp din hcrn.

rriNc

n

QUA

CAC

nA

NH

I

TOAN

BAI

56

33

Problem. Lel ABC...

Ebe a regular

polygon

of unit

side

1

. Considerthe triangles

at Athat are determined

by BEand

the diagonals at

A.

Then,

for each

of

these

triangles, the

length of one of the sides on

BC is equal

to the

product

of

the lengths of the other two sides.

Solution.

Lel AMN and APN be

two

adjacent

triangles, having sides

of the lengthes a, b,

c, x,

y,

as

marked

in

the

figure. Now, the

vertices

of a

regular

polygon

lie on a circle, and in this circumcircle

the

angles MAN

and NAP are subtended

by

equal sides

of

the

polygon.

Thus, these angles

are equal.

Therefore, ANbisects

angle MAP

and

we

get

xa

yc

From this

it

follows that

ab

bc

xyB

Now, denote

by

Q the

point

of

intersection of

AC

and BE,

By

symmetry

we

have IABQ

=

IBAQ, making

AA

--

BA.

From the above formula we can deduce

that

ab

-

AB'AQ

=AB=1

xBO

So we

have

in

general

that x

=

ab.

Td

mdi

:

regular

polygon

unit

triangle

determine

diagonal

side

adjacent

mark

f igure

vertex

circumcircle

angle

subtend

bisect

point

ol

intersection

symmetry

formula

deduce

in

general

=

d6u,

chinh

quy

(tinh

tir)

=

da

gi6c

=

don vi

=

tam

gi6c

=

xdc dinh

=

dudng ctr6o

=

cenh,

v6

=

kA, b6n.c4nh

(tinh

ttr)

=

d6nh

dAu

=

hinh

v6,

hinh

dang

=

dinh

=

durdng

trdn

ngoqi ti6p,

-

g6c

=

truong,

chin

=

chia

d6i

=

giao

didm

=

tinh

d6i

xfng

=

c6ng

thtlc

=

SUy

Io, suy

lufn

(d6ng

tLl)

=

n6i chung, th6ng

thtrdng

NGO VBT

TRLING

7/17/2019 THTT So 279 Thang 09 Nam 2000


I

ITH TITI

TUYBN

SINIT

IIION

TOAI{

VAO

DH XAY

DUT{G VA DH

I.UAT

IIA

ilU

NAM

TOOO

cAu

I.

cho hhm

sofl.r)

= :W.

2x'/+x-l

1.

Tim

tflp

x6c dinh

vtr

xdt

sg

bidn thien

cfia/(x)

;

2.T\mcictigm

cOn,

cli6mudn

vI

xdt tinh

ldi

l6m

ctra cld

thi"(.r)

;

3.

Chtmg

minhrIng

dno

htrmcfp

n

cira/(x)

bing.

,2n-12.

(-l\n

nl

\

12y_1yz+r

G+ty+t

1'

CAU

I'I.

1.

A

(PTTH

chua

PB) Giti

bdt

phrrng

trinh

ardg(tgx)

>

0

.

"r+1

B.

(THCB)

GiAi

b$t

phuong

trinh

1o

llf

-,>

r

J-.X

/'-3*+l'u:

2. Chrmg

minh

ring

vdi

2

sd tu

nhiOn m,

nkh6c

nhau:

lt

,l

ff

I

cosmx.cosnxdx

=

J

sinrux.sinn

xdx

=

0.

-fi

-lt

CAU

Iv.

1. Cho

4

didm

A,

B,

C,

D. Chimg

minh ring :

-++

a)

l,-s

L

cb

t*ivtr

chi

khi

AC

+ BD2

=

AD2

+

ao;

-)

--)

-)

--+

-+

-+

b)NduAB

L CDvd,AD

L BC,thiAC

L

BD.

2.Cho

4

di6mA(0;0;0),

B(3;0;0),

C(l;2;l),

D(2;

-l;2)

trong

hQ

tqa

dQ

Ddc6c

trgc

chudn Orye.

Vi6t

phuong

trinh

mlt

phtng

di

qua

3

did,m: C,

D

vd ttm

m{t ctu

nQi

tidp

hinh

ch6p

ABCD.

3. A.

(PTTH

chua

PB)

Tim

tflp

hgp

c6c didm

M(x,y)

trong

he

tqa

dd

D0cdc tnrc chudn

Oxy,

sao

cho

kholng

c6ch tri

M

d€n

didm

F(0;4)

bing

hai

Itn

khotrng

cichtir

M

d6n dudng

thfrng

y

=

1. Tap

hqp d6

lI

du0ng

gi

?

B.

(THCB)

Cho

l[ng tru

ddu

ABC.A'B'C',

c6

chi6u

cao bing

ft

vtr

2

dudng

thtng

AB

',

BC'

vuOng

g6c

vdi

nhau.

Tim

thd tich

llng

t4r

d6.

2.

Giei

ohtpns

tdnh

trl-sin2x

+

l+sin2x

=

4COS.r.

sin*

1

-,

r 3dx

cAU tll. 1.

Thh

I

-

i

t

+.f'

0

Ciu

1.

l. Hlm x6c dinh

vdi

nhrlng.x

kh6c

-1

vI 0.5.

jv2

-

1Crr + 7

f

'(x)

=

;/'(x;

=

g

-

(2x"

+x-l)'

5

-3^[,

5

+l"ll

t4tx,

=

vax2=-7-.

Hdmflx) ddng

bi6n trong cic

khotng

(-.o;

-1);

(-11

x1)

;

@2;

+a) vd

nghich bidn trong

c6c

khotng

(x1;

0,5)

;

(0,5;

x2)

;

dat cuc

dai tai

x1,

crJc

tidu

t?i

x2.

2.

TiQm

cfln

dtmg

i

x

=

-l vh r

=

0,5

;

ti9m

c0n

, ,

_4(-lf

+ l5xz

-

3x

+

2)

.

ngang

:y

=

u,)./

"(r)

=

'

ax,

.

r

D3

,

f

"(x)

=

0

tai

a

=

2.

Dd

thi

y

=/(r)

l6i

ten

ren

trong

cdc

khoflng

(-1;

0,5);

(2;

+*) vl

lom

(ldi

xudng

du6'i)

trong c6c

kholng

(-o;

-1);

(0'5;

2)

tqa dO

di6mudn

:

(2;

0).

4

tF

HUONG DAN

GIAI

3.

vdi n

=

t

c6:

(-1)r.1

|

(

*

-

*)=

(2x-l)/

(x+1)

-(x+l'f

+ 2(2x-l)2

7x2

-

l}x

+

t

?,,

-.\

Vfly cOng thrlc dring

vfi

n

=

1.

Gi[ sri

cOng

thrlc

dring

vdi

n,tficldc6

clao

htrm cdp n

bing :f")1x1=

\ -/

.'

\

12x-l)r*l

(x+ll+l

/

mot

lin

niia dugc

:

fn+t)1r1

-

t-t\nn

,

12"-1

.

(-2)'(n+t)

-

2(-1)'(n

+

1)

1

'

''

'-

'

\

(2x-l)n+2

(x+

l)"+z

)

=(-1),+1

@+t)tGh

;_Ol

Chung

minh

xong.

7/17/2019 THTT So 279 Thang 09 Nam 2000


CAu

II.

1. A. Tt

thric :

arctg(tgx)

lh

hhm

tuAn

hodn

chu

ki

m. Ttr thrlc duong khi

x

trong

khotrng

(at

;

nx+(nlD)

;

Am

khi

x

trong

khotrng

(nr

-

(nlZ)

;

nr) v1i ru

nguyOn.

Miu

thfc

ducmg

khi

-r

>

-1

vd 0m

khi

"r

<

-1.

Kdt hqp lpi

:

nghiQm

cira

b{t

phucing

trinh

li

+@

u

(nx-(rA);

tln)

W

w(nx;nrc+(tr,2))

n=0

n=O

U

(-nl2;-l).

B.

Tap x6c

dinh

:

-5

< x <

5.

Tt

ttlic

: /{l

duung

khi

0

<

r

<

5; Amkhi

-5

<x<

0.

Mdu thric c6

hai nghiQflrr

=

1

vi

x

=

3

;

dao

him

cria

m6u

thfc li 2' .

ln 2-3,

tri d6 suy

ra

ddu ctra

mdu

thfc

:

duong trong hai

khotng

(-5;

1) vI

(3;

5)

;

dm

trong

kholng

(1;

3). NghiQm

qia

bdt

phucrng

trinh

gdmhai

khotrng

(-5;

0) vd

(1;_3).

2. Phwng trinh tucrng

drrong

vdi

lcosx

-

sin-rl

+

lcosx

+

sinrl

=

4sin-x.cosx

vdi

-r

kh6ckn

(knguyOn)

<+

1

+

lcos2xl

=

Zsir?Zxvd'i

sin2x > 0

e2t2

*

t-

7

=0

vdi

/= lcos2xl

vtr

sin2r>

0

[cos2,r

=

-r

l/2

c+

{

-."

_-'

Q x

=

(nl6)

+

kn

ho{c

x

[sin2x

>

0

=(nl3)

+

kTE

(k

nguyOn).

CAU

III.1.

2.

Dtng

cOng

thric

luqng

gi6c

biOn

tich

thhnh

tdng

;

tr) d6 chimg

minh

clugc m6i tictr

phdn

bing

0.

CAU

ry.

ll

l.a) AD2 + BCz

=

nbz +

BC2

-+

-+^

-)

-+^

-

(AB

+

BD)t +

(BA

+

AQ/

-

=AB2+BD2+BA2+AC+

-

_+ _+

+2AB.BD+2AB.CA

=

-+ -) -+ -,

=

BD' + AC'

+2AB

.

(CA

+ A B

+

BD)

I f

=

BD2 + AC2

+

2

A'tJ .C'D

-+ -+

-+

-+

Trld6suyru:AB

L CD

a

AB

.

CD=O

eAD2+BO=BD2+AO

U)

Ap

dung

c0u a).

2.

Phirong

trinh

mlt

ph6ng

(ACD)

li

x-z

=

0;

phuorng

trinh mlt

phtng

(BCD)

ld

5x+3y+42

=

75;

mflt

phfrng

qua

3

didp C, D vd tAm

m{t

cdu

nQi

tidp

hinh

ch6p

lh mQt

$ong}m{tphtng

phdn gi6c

(P),

(Q)

cnanhi dipn CD.

Tri

cOng thric

khotng

c6ch

tri

cli6m ddn

m{t

ph6ng c6 phucrng

trinh

2

mqt

phing

d6 lI

:

(P)

:3y+92

=

15

vir

(Q):l0x+3y-7=15.

M[t

ph6ng

(Q)

cit

dudng thtng

AB

&

didm

E

(1,5;

0; 0). Vi

Enim

trong

do4nABftn(Q)

lb m[t

phfrng

phli

tim

((0)

di

qua

t6m

mfit

cdu

nQi

tidp

hinh

ch6p).

3.

A.

Kholng

c6ch

tirdidm

M(x,y)

ddndidmF(0;

Z1n:r[*

+

rj-+y

vh

ddnduong

th6ng

Alir

:

ly-11.

Vay

taa

dO didm

M thiba

mtrn

phuorg

trinh

:

4.(y1)2

=

x2

+

(y-02.

Do d6

tap

hsp

cdc

di6mM

,r2

Y2

Itr

dnlns

hvoebol

'.L

-

L

=

l.

412

B.

K6o

dA,i

hinh

l[ng

try

ABC.A'B'C' thOm

mQt

l6ng

tryA,B

rCr.ABCbilng

llng

trU dX

cho.

Khi

d6

tam

gilc

A.BC' vuOng

tai

B.

Gitr

srt cpnh

d6y

l[ng

trpbing

a.Tac6:

Trong tamgificABB':

AB"'2

=

a2

+

h2

;

A'

Trong

tamgiilcArBC

:

2ArBz

=

a2

+

(2h)2,

do d6

a2

=2h2.

n

frd tichlang

tru

r

F^13

Dang:

2

.

AI

DOANTAM

HOE

(Trudng

DHXdy

dwns

HdN1i)

|

(x2-x+t)

-

(x2-x-2)

=

I

-=:--

A.\

i

t+.C

-Lla*

x2-x+l

)

I

=rn(x+1)l;-;[ffi*.

2x-l

161

-

*{: i

{3

i1flf *1

\ 3

)

I

{,

3

3dx

+.r

=ir*-

=tn2-1

mr.r2-x+11I

2

',l

tv6

-l

=ln2+.'13

)dt

=

ln2

+

-rV6

Bl

1t/6

l.{r

j

-ltfi

fi

:

r/r

d(tst)

(tgt)z

+

|

C'

B

2.

TH

7/17/2019 THTT So 279 Thang 09 Nam 2000


1'trong

phiOn

hop

ngdy

241512ffi0

tai

paris,

Vi0n To6n

I

hsc

ClCi

(Clay)

b.

Kembritgio

(Cambridge)

(Massachusetts,

M )

cOng

b6

"Bey

bIi

to6n

cria rhiOn

ni0n

ki"

vi

I0p mQt

Qu

g6m

biy

rriOu

d0la

M

rtd

thuong cho

n\rmg nguoi gihi

dwc

bhy biu

tofur

d6

lmOitai

mOt

rt0u).

De gifp

ban doc

b6o To6n

hoc Tu0i

rr0 hidu

th0m

nhung

vdn

d0 to6n

hoc

crja thiOn

niOn

ki mdi,

tOi xin

gi6ilhi0u

van

t[r

n6i

dung

cia

biy bii

to6n.

pht{

n6i ngay

r}ing,

c6c

biri

to6n dOu

thu6c nhung

linh

vW

hiQn

dat

crja

to6n hoc,

0d

niilu

duoc

chfing,.

cin

phli

c6

nhtng

ki6n

thrrc

r6t

sdu

vd phong

phri

v0

nhiOu

ngdnh

kh6c

nhau

cfra to6n hoc

hi0n

dar. Vi

ttrC,

bdi

vi0t

ntr6

nhy

chi

c6 thd

gioi

thiOu

hOt

strc

so

luoc

nhrmg

f

tuong

chinh

crtachc

b[( to6n,

l.

P=

{P[1AY

P*ilP

Xin

ban

ch6 vOi vhng "gilrnr6c,'

p

0

hai

v6t O6

ctri

tI

c6ch

vi6t

mQt

bI(

to6n

nO'i

tii5ng

dflt

ra boi

St0phen

Cuc

(StephenCook)

nlm

1971.

Ta

thrl

hinh.

dung

6y

hai

sii

nguy0n

t6 dfr lon

(khoing

100

chft

si5; rOi

nnlo

u'6i nhuu.

V6i

mQt

m6y

tinh

diQn

tir, cilcban

lim

bli to6n

d6 kh6ng

Iau

lim. Bdi

roen n6i

tron

thuQc tdp

nhimg

bhi to6n mI

ton

tai mOt

thuat

toen

dd

giii

n6

v6,i

ttdt

giir

tI

mOt

da

thrn

cria

diu

vio

(input).

Ta

ki

hiOu P

li

top

cdc

b[{ to6n

nhu

vdy.

Biy

gid,

c6c ban

thri

gifr

bi

m4t hA

sd

da

dr)ng itd

nhdn

vdi

nhau

vi

cdng

b6

tich crla

chfing,

Khi

do, vdi miy

(nh

di0n

t&

vi

nhung

thuat

t0e0

da bi0t, nguol

ta c6 ttri

phii

mit

hhng

ti

nlm

m6i

tim lar

duus nai

st5 Uan

Oiul

Nhu

viy,,

bli

ben

ph0n

tich

mQt

si5 nguyOn

ra

thua

si5

nguy0n

t6

kh6ng

thuOc ldp

P.

Nguy0n nhdn

cria ditru

d6

lh,

thuflt.todn

xic

dinh nAt

sd

cd

phdi

ti

nguyAn

ui hay

kh1ng

bdng cich

ding

sing

Orafitxtenkh\ng

thuQc lop

p

(ta

sE

cdn

tro lar

di0u nly

trong

mfi

s6 bbi

to6n sau).

Tuy

nhi0n,

c6 nhing

thuQt

todn

xic

sudt

1nuil

to6n kh0ng

tdt

dinh)

liln

dwc viQc

d6 vdi

thdi

aian

da

thrrc

(c6c

ban

c6

thd

tim hidu

thOm vti vdn

Ci nhy,-vl

vc

c6c Udi

to6,n

s5 Itl

vI

VII

sE ninh

bdy sau.ddy,

trong cu6n

s6ch

cfra t6c

gii

blri

nly

:

NhQp

mdn

s6 h7c

thuilt

todn, NXB

Khoa

hgc

1997).

Bli

toan

P

vi

NP

c6 thd

ph6t

bidu

nhu

sau

:

phid

chdng npt

bii

toan cd the'giei

dw.c

bdng

nQt

thuilt

toan

khOng

Mt dlnh da thfu

(b\i

toan rhuQc l6rp

NP)

ki

cfing

gidi

dtntc

bdng

nQt

thuAt

bdn

AAt

dinh) da

thrrc

(tfu ti,

cflng thuQc ldp P)

?

Blri

toan

tr0n c6 vai

fd

h6t

sric

quan

trQng

kong khoa

hoc

m6y

tirh,

li

thuyi5t

dQ

phrrc

tap

tinh

to6n vI ti ttruy6t

mtt

mi, HAu

hOt

c6c

nhl

to6n hqc

tin

rlng

P *

NP.

t,

II.

GTA THIITET

POINCARE

Ta

thri 6y

mOt

quh

ctu

vi

ve tr0n

d6

mQt

duong

cong.

Dr) duong

cong

di

vd thO nio

thi ta cflng

c6 thO

"n6n

]i0n

6

ffiffiMffiMffiffiffi

HA

uuy

KHoAI

(ViAn

Todn

hpc)

tqc"

(kh6ng

l[rn

duung

cong

bi

dft)

cho dOn

khi

n6

chi

cdn lh

mOt

diOm

(trong

sui5t

qul

trinh

n6n,

duung

cong

lu6n

nlm

trCn mAt

cdu),

Vdi

duong

cong

v6

trOn chiOc

nh6n

(hinh

x,yy6n)

thi khOng phti

bio

gidlung

Iiln

duo.c

diOu d6.

Chlng han,

kh6,ne

th_i

n6n

liOn

tuc mQt

duong

cgng

ch4y bao

qualh

chiOc nhin

thinh m6t

diOm. Ta

n6i

r6ng,

m[t

cdu

don

]i\n,

cdn

m[t xuy6n

thi

kh\ng

don ]idn.

.

Nlm 1904

Hlngri

Po[ngcar0

(Henri

Poincar6) phit

biOu

gil

thuyi5t

sau

day : mqi

da tpp

ba chiiu

(tatam

triiiu

da.tap

3 chiOu Ih mOt

kh6i

hinh hoc

tron,

tuc li kI0ng

gO

ghO,

trong khOng

gian

3

chiOu) dun fiAn

compic vi kh6np

cd bi€n,

diu dlng

ph\ivdi

ngt

ciu ba chiiu'(tftc

ti

c6 th-d

6nh xa

mgt

OfSi

mQt tiOn

tuc hai

chibu da

t4p d6 lOn

m[t

ciu

ba

chibu).

Bei

t06n tr0n

duoc r6t

ntridu

nhi

to6n hqc

l6n

quan

tfun,

trong

d6 c6 nhibu

ngrroi

d6 tung duo.

c

Giii thuong

Phin

(Field$

(tuong

tu

nhu

giii

N0ben

(rlobel),

nhrmg

gianh

cho c6c nhi

to6n hoc) nhu

S. Sm0n

(S,

Smale),

W.p.

Thocston

(W.P.

Thurston),

S.Ndvicdp

(S.

Novikov).

Tuy

nhiOn,

cho d€n nay, ho

chi

gili.duo.c

bbi

toAn trong kh6ng

gian

vdi

chi0u

tri

4

trd lOn.

ThO

mdi

hay,

chc biii

ro6n dlr

cho khOng

gian

3 chidu

mi

chfng

ta dang s5ng

van

D

nr6

nhdt

IIT.

GIA

THW6TRTEMANN

Tolrn

tip

cOng

trinh

cria

nhi

to6n hqc Dric

Rioman

(B.

Riemann.1826-1866)

in

thdnh mQt

cu5n

sfuh

ch,i d9f

khofurg

gin

400 trang.

Vty

mi

tOn

cria

6ng

duo.

c

nh6c @n

hdu nhu

fong mgi

ng]rnh

cria

toan

hgc

hiQn

dVi.

Gii

thuyAt

Riemnn

duqc xem

lh mQt

trong nhimg

bI( to6n lon nh6t

cria to6n hoc.

Trudc

n6t

ta

xbt

UimLAta

Riemann

dtnh

nghia

b&i

ding thuc

sau ddy

:

(ro=i,'

Pl

It"

oC

miy

rhng,

chu6i n6i rOn hOi

tu vdi

c6c

s6

phrrc

s

c6

phdn

thuc ldn hon

l.

Ntru vfly,

hiln

z0ta

Riemann xdc

dinh

trOn

nua

mflt

ph[ng

n[m

bOn

phli

duung

thlng

Re(s)

=

I

(Re

lI

ki hiOu

ptr0n

ttrr;c

cria sd

phrrc).

Sau

O6 ntrd

phuong

ffinh

h[,rn

m]

hlq

zdta

Riemann

th6a

min,

ta

x6c

dinh drr.c n6

trOn

toln

mat

phing phtrc (rtl

ta

s

=

1,

vi

hkn

c6

gi6ri

h4n

v0 cirng khi

sddn dOn

l).

C6 thd chung

minh rlng, hd.ryn

zdta Riemann

blng 0 tai cic

gr|frr

s=

-2k

(vdi

mgi knguyOn

duung),

Gii thuy6t Riemann

n6i ring,

ngoii

cdc didm dd ra

noi

di€in khdc

tai dd

hara zAta Riemann

bins

0

diu c6

$en

nan

ting

lD,N6i

c6ch

kh6c, msi *nanf

oidmxntrc

ciia

hfuri zdta Riemann rl0u

nlm

trOn

dwng

th[ng

7/17/2019 THTT So 279 Thang 09 Nam 2000


Re(s)

=

1t2.

Bing.miy

tinh, nguoi ta

da

kidm

tra cluoc Ii

git.thuy0t

dring

d6i

vdi 1.500.000.000

(mQt

i

rutn) kh1ng

di€nctn

hi.m

z0ta Tuy nhi0n,

git

thuy0t

trOn vtn

lh

mdt

th6ch thuc

cho to6n

hoc

ciia thiOn

niOn

ki m6i.

C6

16

cung

can

gili

thich d6i dibu

di c6c ban hinh

dung dugc tai sao hfun zdta

Riemann lai

quan

ffgng

nhu

thO dOi vdi

to6n

hqc. Kh6ng

kh6 khln

gi,

ta chung

minh

dr-ru.

c

ding

thuc

sau

ddy,

gq1

h

fich

Oi

(Euler)

(c7c

b4n

hly tu llrn nhu la

hiii

ttp)

:

((s)

=

n-_]-

1-

1

f

trong d6 tich duoc 6y trOn tap hSp moi

si5

nguyOn

t5

p.

Chinh vi

c6ng thirc

[0n

mI

hIryn zdta Riemann

tim thdy

r6t ntriiiu

ung dung kh6c nhau

trong

li

ttruyiSt s6. Ctring

hal,

dr)ng

hlun zdta Riemann

ngudi

ta

ohung

minh

duo-c

Dinh li Dirichl0

(Dinchlet)

v0 su t6n tar v0

han

sd

nguyOn

t6 trong c6p si5 c-Ong

mI

sli hang

diu

vI c6ng

sai

nguyOn

tO cr)ng

nhau. NOu

gii

thuyOt

Riemann

lI

dring,.ta

sO c6

cOng

thrrc

kh6

chinh

xlc

d0 mo

tl

luit

phan

bO s0

nguyOn

t6.

vI tri

d6

d6nh

gi6.duo.c

thdi

gian

(90

pnrrc

t?p)

cria

thu|l to6n

phin

tich sd

nguyOn

ra thua sd nguyOn t6. Nhu

di

n6i

trong

phlnpidi

thi0u

bii

to6n

I,

diOu

nby r6t

quan

trong Eong

li thuyOt vd ung dung.

rv.

Li

THUY6T veuc.l,ru,r,s

NOu nhu

trong th6

gi6i

vi

md

c6

c6c drnh

lu|t

Nruton

(Nervton)

cta

co

troc

cO'Oidn

thi

trong th6

gi6i

c6c

hat co bfur

(*re

gi6l

vi md)

c6 6a d6ng cria

c6c

dinh

lutt

cria v6t

li

luong tu. C6ch ddy

gin

nrla

th6

t<i.

hai ntrl

to6n

hqc Yang

[ang)

virMin

(Mills)

phlt

hiOn ra rlng vit li

tusng

tu

chua

dgng

m6i

quan

h0 h6r

sfc

9ha cn6

giira

vit

li

c6c

hat

cs

bin vi

to6n

hoc

cila

mQt

sd dOi tuqng

hinh

hoc

(c6c

phdn

thd,,..).

Hg

dua

ra

phwng

trinh

nOi

tiOng

drri t0n.gqi Phuung trinh Yang-Mills, nhd

d6

tiOn

doan

duVc nhidu

hiQn

tuong trong

vit

li

c6c

hat co btn. Nhi0u

ti0n do6n trong sl5 d6

di

duo,c chrmg

minh

bing

thgc

nghiOm

tai

c6c

phdng

thi

nghiQm vir v4t li nlng luqng

cao

o

Anh.

M .

Ph6p, Nhnt... Tuy nhiOn, cho

dOn

nay, ngudi

ta

chua tim dWc c6ch

qili

phuong

trinh

Yang-Mills

sao cho

thda min duuc

hai

diOu

kiOn

: vua bto dlm

tinh

chinh

x6c

to6n

hgc, vua m0

ti

$uo.

q

c6c hat c6

khOi luong.(hat nang).

Ddy khOng

phii

lI

l'an

diu ti0n

md mQt

li

thuydt chua

thdl

"chlt

ch6"

v0

to6n hqc

lal

d

ung

dung trong

vflt

li. D'ic

biQt"

c6c

nhh

v0t

li

thuune

ding

eii

thi0t

"khe

khOi

luong"

diJ

giti

thich"ss

"khOng

i'9in tnily duut" cria cac

hat

quic

(quarks).

Didm

dac

biQt

cta

cichAt

"quarks" lh khi chfing

cing

xa

nhau, lurc

trrng t6c

giira

ch6ng cing m4nh,.vit cho

dOn

nay, nguui

ta

chua

thO

tao ra dri

nlng luqng ctn thiOt

O{ d.cir chring

rdi

nhau

(nhu

v6y,

c6c

h4t

"quarks'

Iudn o

r6t

gin

nhau.

v)

c6c

nhi

vft li

goi

lh

"h4t

quarks

cim

tr)"t.

Nhrmg

hiQn tuung

vnt li nly chua c6 duW

sU

m0 ti chinh

x6c tofur

hoc. D0 dat dr:qc tiOn

b0

nlo

tl6 trong

vin

d0

nly,

c6

10

phi,r

cfln d5n

nhirng

tu tumrg

mdi

ct trong to6n

hqc

vi

vflt ii.

v,

GIA

THUY6T

HoDGB

?d

nghiOn cuu mQt

Ci5i

tuqng

hinh hqc

ph0c

tnp,

ngay

tt thC

ki

19 nguoi

ta di

bi0t

c6ch

li.rn nhu

sau

:

"d6n"

thOm

vio

n6 mQt hinh

don

gihn

dti thu duo. c

mQt

d5i

firong hinh

noc

Oe

mio

s6t

hon

hinh

ban

ddu.

Tuy

nhi0n, trong nhiiru

trwng hqp. d0 thu durrc

m0t

d6i ftrong

hinh

hoc

dnn

gian

dOn

mrrc

ta c6 cOng cu

nghi0n

crm

chring, nhirng dOi tuong

phti

dr)ng Cii.Oan

tai

qu6 phfrc

tap

I

Gii thuy6t

Hotgio

(Hodge)

n6i r[ng, trong mQt sd tnrdng

hgp rdtquan trgng

cira to6n

hgc, c6c ddi tugng ding dC d6n c6 thO

duoc biOu

diOn

qua

,ei aortr:vng

kh7

don"gifur.

C6 thd

n6i

chi tiOt

hon mot

chft

nhu sau.

Xbt

cdc b0

r+1 sd

(x1,

...,

x,*1)

th6a

min hQ

phrrng

trinh

Pn(1,

,..,

t*t)

=

0, k=

1,.,,, fi,

trong

d6

PnlI c6c

da thfic thudn

nhdt, trlc

1)

c6c

da thric

mi m5i

tlon thrlc

cria

n6 c6 btc nhu

nhau. T|p hqp

circ

nghiOrn c'ua

h0

phuung

trinh

nhu

viy

lflp thrrnh.mdt. da tap dat

s6.xa'uh

trong

khAng

gian

xA

anh

n cht\u. ndu da tnp

dai-s6 d6

"khl

uon

tru" thi ta

n6i n6 Ih

da

r4p khOng

H

di.

DC nghiOn

cuu

c6c

da tap

dar sO

x4 lnh,

ta thuomg dr)ng

c6c chu

tinh

Hodge

dC

"de.'n'.

C6c chu uinh

Hodge dugc

th0.

hiQn

qua.mQt

dOi

hrvng

mh ta

goi

lI

ptrln

tri cria

nh6m Oi5r COng

diOu

hQ.s6

htu ti.

Gil

thuy6t

Hodge n6i rlng,

t€n

cdc

dg t4p dai.sil

xa

inh

khong M

dt,

ndi

chtl tri$

Hodge th tA'hqp tuy6n fnh

cfia chc chu trinb dai sO

(th1

hi0n

qua phin

tri

crja nh6m

Oi5i OOng

di0u

hq

s6

nguyOn,

tric

li

c6c

Oi5i tumg don

gihn

hsn

chu

trinh Hodge).

vr.

sV

rON

rm

vA

ri|ts rRoN

cte Ncslunr

PHUCTNG

TRINH

NAVIER.STOKS

Phuong

trinh

Navi0-St0co

(Navier^-Stokes)

II

phrrng

tinh

vi

phdn

d4o

hkn

riOng

dtng

d0

md

th

hiQn nrqng

s6ng. Chin^g

hpn,

ding

phwng trinh d6, ta c6

thO

m6

tt

dugc chuyOn dQng

c0a

nuoc

bao

quanh mOt con lbu dang

chay, chuyOn

dOng cr)a

lu6ng khi bao

quanh

mQt

m6y bay

dang bay.

Ph,rrung trinh d6

duol

viOt

nhu sau

(v6i

c6c

bnn

dang

hgc phO ttrOng, chua

llrn^quen

vor dao hfun

nOng thi

c6 tho

hieu

ki

hiou

0

ding

d0 chi

vi0c lly rlno

hirn cfra

mOt hhm nnidu ui6n tneo

irot bi6n

nbo

06,

uri

xem c6c

bi6n

kh6c

th c6

ointr):

a:dui

,

ui

+

Luiii

FL

dn

=vAui

-

i{

+

fdx,t)(xe

P,

f

>0)

(i)

dxi

la,

dtvri

=

). ia=O(xe

P,

t>0)

(2)

-.

dxj

Ftj

7/17/2019 THTT So 279 Thang 09 Nam 2000


Mlc

dr)

phr.rong

trinh

Navier-stokes

di

drnc

bi6t

dOn

tt the ki

XX,

cho d6n

nay, hidu

bi0t cira

con nguoi

v0

phuong.

trinh nhy

v6n

cdn r6t

han

cne. Trong

trudng

ho.p

hai

chiOu

(n

=

2), nguoi

ta

di

chrmg

minh

duoc

su tOn

tai

nghiOm

tron

(trlc

tir

c6

ttrd 6y

dao hdrn)

cira

phuong

trinh

Navier-Stokes.

Tuy nhiOn,_trong

truor.g

hop

3 chidu,

su

t6n

tai

nghi0m

nhu

v6y

v6n

cdn lb

v6n

dO

m&,

ngay

ci

trong

trwng

hop don gian

(v

=

0)

(phrong trinh

Euler).

C6c

ban

th6y d6y,

c6i

kh6

ntr6t

v[n

iat

n]im

trong khOng

gian

3 chiOu

I

'1'li.

i,{,

i"i{il1'tf

itii"{]t"l

vA

slviltl,l-ti {}},-

jla

. iJ

.

C6c ban

c6 thd xem

c6c didm cria

rlrrong

palabOn

y

=

I

nhu

c6c nghiOm

(x.

y)

cla

phw.ng

rrinh

y-

I

=

0.

N6i

chung,

ta xem mOt

da

thuc

hai

biOn

\x,

y)

xac

dinh mOt

dr.ung

cong C, N6u

c6c h0

s6 cria da

th(rc td

c6c sli triru

ti

thi ta c6 thd n6i

dOn

cdc nghiQm

huu

ti cria n6.

trrc

l)

cic

di(ln

hfru

li

cria

duong

cong.

Su tdn

tnr nghiOm huu

ti crja

mOl

da thrlc

phq

thudc

vIo

rn0t

dlc

trung hinh hoc

cua

dnong cong

tuong

ung,

goi

Ii

giting

cua

dudng cong.

TTtng

hoo gi6ng

cria

.duong

cong

bing 0 kh6

don

gian.

N6u

duong cong

c6

gi6ng

t6n

hon

hay

bhng 2

thi n6

chi

c6 huu han

didm huu

ri. D6

li ndi

dung

cria dinh li

Ph0nting

(Faltings)

nO'l

t6ng

(xem

bdi Dinh

ti Fermat

dd

dtruc

ghung

ntnh,

THTT,

1993). D6i

vu

cic duong

cong

c6

gi6ng

bing 1,

cho

d6n nay

chua c6

phrrong

ph6p

chung

nlo

dd

x6c

dinh mOt

drxrng

cong

d6 cho

li

co-tray ttrOng

d didm huu

ti. Khi

C

c6 didm huu

ri

rhi

C duur

gqi

tb

dwng

cong elliptic

vd,tap ho.

p

c6c didm

cira

n6

c6 cdu

trf c

cria nh6m

aben.

Nhiiru

tinh

ch6t

cria duong

cong

elliptic

d,ryg

ttii

hi0n qua

L

-

chu6i

LQ,

s) cira

nO

1CO

th mqt

chuOi x6c

dinh hodn

toin tuong

tunhuham

z0taRiemann.

Xem

NhQp m1n

sii hrlc

thuQt

todn).

Gie thuyOt

Boch

(Brch)

vI

Suynnoton-Daio

(Swinnerton-Dyer)

n6i ring

hpng

crta nh6m

cdc

di(ln huu

ti cila-dwng

cong

ellipttc hing

bLt

cila

kh1ng

rtiAm

cia

L

-

chudi

twng xng

t1i s

=

1.

D[c

biQt riuong

cong c6 vO han

di0m huu

ti

ttri

vii

chi

khi I

chu6i

tuong tmg bing

0

er

didm

s=

1.

Gii

thuy6t

Birch

vh

Swtnnerton-Dyer

quan

trqng ddi

vdi

to6n hoc

chinh vi vai

trd

cria duomg cong

elliptic

trong

ci*

vln

Od ffr6c

nhau. Ta

nhdc

lai

rling,

duong cong

elliptic

chinh

lh

c6ng

cu chri

yOu

dd

chung

minh

Dinh

li

ldn

Fermat.

Gan Oay,

duong

cong elhptic

ctng

tharn

gia

vio

vigc

thi0t

hp

c6c h0

mit

ml kh6a

c0ng khai. Mlt

khdc, n6u gil

thuyiSt

Birch

vi

Swinnerton- Dyer

dfng

thi

ta c[ng c6

thd

giii

duu.

c b}i to6n

m&

di

duo.

c d[t

ra

hlLng

[Im

nlrn

trvsc:

tin

tdt

ci

cdc sii

nguy\n

n

sao

cho

tin tat

tan

gihc

vuing

vdi s6

do cic canh

li

sd

nn

ti

vi si|

do

diQn tich

bing

n./.

**

thi

*lymPi*

?**m

cto

oAr LOAN

(3/1996)

Bni

1.

Gi6 sir c rc

goc

o,

0,

T

th6a

mln

0

< cr,

F,

T<

It

o

*0

+

y=

f,va,

tgu

=

,

-tt

tCP=i'tC^t:;

trong d6

a,b,

c

li

c6c

sdnguyOn

durng.

HXy

x6c

dinh c6c

gi

ti a, b, c.

BAi 2.

Gi6

str sd thUc

a

thda

rntrn 0 <

a

<

i vh

I

a

{ a,3'

vft

j

=

1,2,

...,

1996.

Chring

minh

ring

vdi

c6c sd thgc

khOng

0m

t1

(i

=

1,2, ...,1996) ma

I

t

=

t ,t i

6ssa

)(sqo

5

li,.,. llL.,.' I .1r,*1

t'

lu

t

t

ll-

t J

I

4\

a

)

li=r ltr=t

l

BniY.

e

v{ b

Ih hai

dldm

cd

dinh

rron dudng

trdn

d6 cho,

Gil

sfr

didm

P ch4y

tr0n

dr:Ong

trdn

ndy vd

di€m

M

tuong

img

sao cho :

hoflc M

thuQc

doan thf,ng

PA

vdi

AM

=

MP

+

PB,

holc M

tliuOc

doan

thing PB

v1i

AP + MP

=

PB. Tim

qui

tich

cr{c

didm

P nhu

thd.

Bni

4.

Ch(mg

minh

ring

vdi c6c

sd thuc

bdt

k\

a3,

a4,...,

o85

thi

cdc

nghiQm

cfia

phucrng

trinh

sau

khOng

phli

ddu

lir

sd

thqc :

assx85

+

ag4x84

+

...

+

a3x3

* 3x2+2x+7

=

O

Bni

5.

Tim tdt

c[

99

sd

nguyOn

a1,

a2, ..., oe,)

=

do

mA

br-t

-

aol2

1996

vdi

mgi

k

=

1,2,...,

99 sao cho

sd

m

=

maxll4l-t

-

a*li k

=

7, 2, ..., 99y

lA nh6 nh{t c6 thd

dr4rc,

vh xdc

dinh

gi6

tri

nh6

nh{t

m*

cita

m.

Bni

6. Gil

sir

eo,

Qt,

qz,

...\d

day

sd

nguyOn

th6a

mtrn

ddng

thdi

c6c diOu

kiQn:

(1)

vdi

mdi

m

>

n

thi m-nll,

udc

cta

Q*-

Qn

(2)

lq,l

<

n

1o

vdi

mgi

sd

nguyOn

n

>

0.

Chrmg

minh

ring

tdn

tni

da thrlc

Q(x)

sao

cho

Q(n)

=

qnYdimqi

n.

7/17/2019 THTT So 279 Thang 09 Nam 2000


VAI

UOC

LUONG

,:

TRONG

TU DIEN

PHAM

ndNc

euAN

(12

Todn,

THPT

Ngut6nTrdi,

HAi

Dwong)

MOr vii

uoc

luong

trong tri

diQn

d

bii

niy lI

sg

m0

ph6ng

c6c urvc

lugng

tuong tr; dx c6 trong

tam

gi6c

ma

theo tOi biOt

thi dAy 1I

nh*ng

ph6p

chimg

minh mdi

vd mQt mQt

vdi

kdt

qul mdi.

Ddi

vdi tam

gi6c

ABC

ta

ki

higu

S

ltr

dign

tich

vd,BC=a,CA=b,AB=c.

Ddi

vdi

tu dign

ABCD

ta

ki hieu

:

y

la thd tich

;

s(BCD)

lI

dign rich

cta

LBCD

ho{c

vidt

tdt So

=

S(BCD).

S,o=Sr+Sr+S.+S,

L=AB+AC+AD+BC+BD+CD.

P

=

AB.AC.AD.BC.BD.CD

(Xn

li sd do cfia ciic

nhi

diQn cnnh

XL

Xin

bdt

ddu vdi ud'c

ltrgng

liOn

quan

ddn

diQn

tich

totrn

phdn.

Bii

tor{n

1:

Cho tr?

diAn

ABCD.

Chftng minh

2

rdng:

,'

>

l2{t.tte

(1)

Dd

chftig

minh

ta srl

drJng

bd dd

quen

thuQc

sau trong tam

gi6c

:

Bd

de

1

:

Cho tam

gi6c

ABC, ta

c6 :

ab

+

bc

+

ca

>

+tE

S.

Dtng thric xly

ra

<+

tam

gil,c

ABC

dOu.

Tr0 lni

bdi to6n

1.

Ap

dung bd d0

1

cho

c6c

tam

gi6c

:

BCD,

CDA,

DAB,

ABC, ta dugc

bdn

bft

dtng

thrlc.

CQng

theo trmg

vd cira cdc b{t

ct6ng thlc

niry vtr

rut

ggn

ta

c6

:

(AB+CD)(AC+DB)

+

(AQ+BD)(AD+BA

+

+

(AD

+

BC)(AB+CD)

)

4{3

Sa

=

1, .2

i

(

<en*co)

+

(AB+GD)+ (AD+Bc;f

> 4r-3s*

J\

)

+

)

>12{3.SD.

Dtng thric xhy

ra

a Cdc tam

gi6c

BCD,

CDA,

DAB,

ABC

ddI

<+

Tii

dlQrtABCD ltr

tf

dien deu.

Bii

torin 2.

Cho

tO

di|n

ABCD.

Chtcng minh

rdng:s3a>216.iiVQ)

Dd

chr:ng

minh ta sir dgng bd

dd

quen

thuQc

sau

trong

tam

gi6c

:

Bd

d€

2.

Cho

tam

gi6c

ABC.

ta

c6

: a

+

b

+

c

2

l2{3 S.

Ding

thric

xtry ra <+

tam

gi6c

ABC

dd.u.

Tr0

lpi

bdi to6n

2.

Gqi

H

ld

hinh

chidu

cfia

D

trOn

m{t

ph6ng

@Bq.

Gqi

4

F,

Kla.hinh

chidu

cl0;a H

$et

cac duong

thlng

AB,

BC,

CA.

D[.t DH

=

h, HF

=

x,

HK= ,HE=Z

Theo dinh

li

Pitago

' r.*

"

Hinh

1

lac6:DF

="'l

h'+x',

OX=t[112

ayz, DE=^[h,

*

r'.

Vfly

:So

+

Su

+ S.

=

6c.or

+

GA.DK

+

AB.DE)

2'

=

g",[

rl* +

u'{r}+yz

+

r{rP**l

2'

=)<",trrnf

.@ *^[runf

<uyf

*

='^

2

>;^tt

"[Tsrh\4s:.,

(Theo

bd de 2)

=^{:

s;h,

*

g

Suy

ra

:

(so

+

s,

+

S.)2

>3"'[isonz

+

9o

=+

sr(sa

+

. ,

+

.lc

-

sD)

>-

z^[z sonz

=+

S,r(Sa

+

S,

+

sc

-.lD)2sD

> 5415'fr

_

,So

+S,

+

s.-

s,

+

2sr)2

>

s+r.6y,

5,p[

-z

)

-

9,e>2t6\[rrf,.

Df,ng

thrrc xhy

rakhi

vtr

chi

khi

ff=H=fr,

MBC d6u,

H nim trong

MBC,

Sa

+,SB

+. c

-

So

=

2So,

nghia

lI

khi

vi chi

khi

ch6p

D.ABC

ltr

ch6p

ddu

vX Sa

=

Sr

=

,lc

=

SD,

trlc

lb

tri

dllnABC

ltr

f0 diQn

ddu.

Cicbdt

cl8ng

thrlc

(1),

(2)

dtr

c6

trong

nhidu

ttri

ligu

to6n

so

efp,

ch&ng

han b6o

THTT sd

116

thtng

5

*6

n6m

1980 vdi bdi

"vhi

rrdc

luqng

hinh

hgc"

ola

GS

Phan

Drlc

Chinh.

Tuy

nhiOn

cdc

ph6p

chrtoig

minh

cria

(1)

va

(2)

theo

tOi

lI mdi.

Bdi

torin

3,

Cho

tu

diAn

ABCD. Chrhng

minh

rdng:P>-7XP

(3).

Bdt

tttng thric

(3)

m0i

xu{t

hipn

gin

d0y trOn

b6o

THTT

s6

2lt

thtrng

1 nlm

2000, chfmg

minh

c6

trong

ldi

gifti

bdi

T1,0/267.

BDT

(3)

cho

ph6p

ta

dodn

nhan

kdt

qutr

sau

:

(ch)2

+

(cz)z

7/17/2019 THTT So 279 Thang 09 Nam 2000


Bii

torin

4

: Cho

tri

dipn ABCD.

ChLmg

minh

rdng

:

(Sa

.

Sr

.

sc

.

SD)3

,# rr

@)

Trudc h€t

ta

ph6t

bidu

vtr chimg

minh kdt

qutr

sau.

Bd tI6

3. Cho

tu

diQn

ABCD,

ta

c6

:

II

sin(XI)

=

f

:)',

trong

tich

ndy

(XY)

ldy

dri

gi6

(xh

\

9/

tti

cdc g6c

nhi

diQn

cria

trl

dipn.

Chil:ng

minh.

Gqi.F1

ld

hinh

chidu

cta

D

tr}n

m{t

ph&ng

(ABC)

(xem

hinh

1). Ta

c6

:

cos2lBq

+

cos2lcA; +

cos21Atr;

/

S(HBC)

\2

/

S(HCA)

\2 i

.S(F1AB)

\2

=[

s,

,-[

s,

]"(.

s..]

Ap

dung

bft

dtng

thric

BunhiacOpxki

ta c6

:

cos2lBg +

cos2icA;

+ cos2lAB;

>

r

S(HBC)+S(FI

CA)

+

S

(H

A

Bt

rz

zt

si+s|+S.

/

=

"os21Bg

+

"or2icA;

+

"or2(AB)

,"S;\

>l-

-l

(*)

\Si+Sfi+S'cl

Ddng thrlc xiy

ra

<+

lS(HBC)

S(HCAt SIHABl

=

-=-

{

su .tB

.tc

I

[H

nim

trong

tam

gi6c

ABC

<+

(Bq

=

(CA)

=

(AB).

Tuong

tU

nhu

bdt

dtng

thric

(*)

ta

c6 4

b{t

dtng

thric

ddi vdi

c6c

m{t

vir

c6c

g6c

nhi dign

kh6c.

CQng theo timg

vd

cria

4

BDT n6i

trOn ta c6 :

q2

(xn

@.B,c,D

"B

',

"c

'

"D

trong

tdng

nly

(A,

B,

C,

D)l{y 4

gi6

tri

ho6n

vi

vdng

quanh.

Dettrdy)

,4

.r+

(A.

B.

c.D)

'B

+

s'z.

+

s|

-

3

Ttr

d6 suy

,u

:

)

"or'1xv1>

3

(x4

+

)tr

-

sin2(x4)

.? *)

rin21x4

<

U

(xn

3

l*o

3

*

o

{in

"nrxa

.*= r sin(XY).

(;l

6n

J

(Xt)

\

Ding

thric

xly

ra

<+

tdt cL

c\c

g6c

nhi

dign

bing

nhau c+ Tr1 diQn

ABCD

lI

tri

di-On

ddu.

10

Ta srl dgng

bd dd

quen

thuQc

sau

d0y :

Bd

dc

4.

Cho

tri dian

ABCD,

ta

c6 :

2

Sc.So .sin(AB)

3AB

Tr&

l4i

bdi

to6n

4. Ap

dung bd

d0

4

sdu

ldn

vd

nh6n

theo

tlng

vd

cria c6c d8ng thrlc

nh0n

dugc

tac6:

.)-6

(

i )

tso.s,

sc.sD)3,

rt^sin(xD

,-,6-\')

(xb

Sri

dung kdt

tac6:

to6n

3

vtr

bd

de

3

P

qufl

cria

bhi

(if

u,

,,

^tc

sD)3

rrt<

1^

vL

72

-

tl2

+f<f*tso.sr.s..sr)3

Dtng

thic

xly

ra

<+

Tri

dign

ABCD

ld

ttr

di0n

ddu.

C6c

kdt

qui nhdn

dugc trong bd d0

3

vI bdi

toin4

theo

tOi

li mdi.

Nhd

bdt dfrng thric

COsi, ta

th{y

(

)

li

su

m0

rQ"g^:p

(2).

Nhd

bat ding

thlc

quen

thuOc

53

a

+

o26212

ddi

vdi tam

64

gi6c

ABC bdt ki,

ta

th{y

(4)

li

su

m& rQng cria

(3).

Xin

mdi

c6c bpn

gitri

tidp mQt sd bii to6n

sau

:

Bii

todn

5. Cho

tf diQn

ABCD.

Ch(mg

minh

rlng:)sin(Xl)

<4'[r.

6n

Bii

torin 6.

Trong

cdc ttr

diQn

ngopi tidp

rnQt

mflt

cdu

cho trudc,

h[y

tim

tri diQn

c6

dign tich

todn

phhn

l6n nhdt.

Bii todn

7.

Cho

tri

diQn ABCD.

Cqi

R,

r theo

thrl

tW

ld b6n

kinh

cdc

mlt

chu

ngo4i tidp

vtr

nQi

tidp. Chrmg

minh

ring

:P>Zq\V.

r

Bii torln

8

:

Cho

tr1

dien

ABCD

nQi

tidp

mlt

cdu

(O,

R) vn

M

lh

mOt

didm

nim

trong

trl diQn.

Chtmg

minh

ring :

MA'V

(*rro)

+ M B.V

g6pty

+

+ MC.V,*oo4+

MD.Vludnq

<',[R24,W

.v..nco)

Cudi cing

xin ctm

on thdy

Nguy6n

Minh

Hi

di

dQng

vi0n vtr

girip

(lo

nhiou

dd

tOi

hoin

thinh

bdi

vidt

ndy.

7/17/2019 THTT So 279 Thang 09 Nam 2000


DAY

UA

T,f

.ra*

hqc

2000-2001

c6c

trudng

THPT

trong

ctr

ntrdrc

so ding

chung

1

bQ

s6ch

gi6o

khoa

To6n,

thay thd

cho

3

b0 sfch

gi6o

khoa

To6n

(SGKT)

dl srl

dung

tr)

n6m

1990.

Xin

gidi

thiQu

t6m t5t

mQt

sd

di6u chinh

ctra

b0

s6ch

gi6o khoa

chinh

li

hqp nhdt

(SGKTCLHN)

so

vdi

c6c

b0

SGK

trudc

day.

Trong cfc

cuQc

th6o

lufn vd

SGKT,

i

kidn

chung

dOu

cho

ring

nQi

dung

cira SGKT

phli

gdm nhung vfn d0 co

btrn

nhftctra

bQ

mOn To6n,

d6p

rmg dugc

nhtng dOi

h6i cira

khoa hgc,

cita

ddi

sdng

x6

hQi

vl

phtri

khOng

lac hflu nhi0u

so

v1i cic

mrdc

ti0n tidn.

Qua

10

nlm

sir

dgng,

SGKT

da bQc

l0

nhftrg

ur

khuydt didm cira

n6,

trong

d6 c6

mOt sd

vdn dd

bi

khai

th6c

qu6

stu

cho

mqc dich

thi

vtr

luyPn

thi.

BO SGKTCLHN

v6n

bao

gdm

nhung

kidn

thuc

co

bf,n

nhu

trong 3

b0

SGK

trudc

ddy,

nhung

c6

mOt

sd di6u chinh

nQi

dung

bing 3

biQn

ph6p

sau :

*

Lopi

b6

nhtng

ki6n thrlc

khOng

thflt

co

btrn.

x

Gi[m

nhtng

ydu

td c6

tffi

ch{t

kinh

viQn,

hqc

thuflt;

tlng

cutng c6c

ydu

td th\rc

hAnh.

Ching

hnn,

b0

nhtrng

chimg minh phrlc t3p,

tim

c6c

phuong ph6p

tidp

cfln don

gi6n

tuy c6

phli

hi

sinh

phin

ndo

tinh chinh

x6c

khoa hgc,

lga

chen

thOm c6c

vi dU

minh

hOa...

*

Dd

cao clc

ydu

t6

su

phpm

nhu

: thdng

nhdt

cic

kr hipu vh

thuQt

ng0

dtng

trong

s6ch,

chti

i

tinh

m6u mgc cta cdc

vi

du

hay

bdi

gili m6u,

sd

luqng

bli

Bp

ra

vta

phli

vtr vdi

nhrarg

yOu

ciu

thich

hgp, b6 c6c bli

Qp

qu6

kh6.

Sau

diy

trinh

btry

mQt sd

nQi

dung

cu thd :

vi

oAr

s6

D4i

sd

10.

HAu hdt

c\c

nQi

dung

ctra chuong

khoa

hec vI

ki thu|t

tinh

to6n

dvgc

chuydn sang

b0

m1n

khdc

thich

hqp

hon,

chi

gin

/4i

phhn

n6i

vd sai

sd

vI tinh

gin

dung

dd

girip

hgc

sinh

trong

thuc

hlnh

giti

to6n.

Bd

sung

thAm

mQt

sd

nQi

drurg

cin thi6t

vd

lOgic

vi

c6c

ki

hiQu

lOgic

vdi

mrlc dQ don

gitrn

dd

hoc

sinh

cd

tnd

hidu

vI

srl dung

dring

vd sau

nhy.

-a\

a ,

vE

B0

SeeH

6_[&0

KH0e

TOAI{ IHPI

C}II}IH

tI llOP

l{llAl

(Srt

dvng

trt

ndm

hOc

2000-2001)

NGUYEN

HUY

DOAN

(NhI

xudt

b6n Gi6o

duc)

C6c

nQi

dung

kh6c

n6i

chung khOng c6

gi

thay

d6i,

ngoni

trtr c6ch

trinh bdy

mQt

sd

vfn

d0

c6 don

gidn

hon

(nhu

htrm sd

vI

dd

thf,

li

thuydt

vd

phuong

trinh

vd bdt

phrrng

trinh,

bft

dtng

thLrc).

Dai

sd

vi

Gi6i tich

1.1

VA

fugng

gidc,

bO cdc

vdn dD :

htrm sd

luqng

gi6c

ngugc, bft

dtng thrlc

lugng

gi6c

vI

bft

phuong

trinh

lugng

gi6c.

Dd

chinh xdc

hda

thu\t

ngfr

vh,

tr6nh

nh[m

l6n, c6c thuflt

ng8

"h]m sd

lugng gi6c

cira

mQt

g6c hay

mQt

cung"

trudc

dty,

nay

ggi

lh

cdc

"gi6

tr

luqng

gi6c

cfia

g6c

hay

cung

d6";

Cdn thuQt

ngt

"hdm

sd

lurqng

gi6c"

chi

dtng

cho c6c

htrm sd

luqng

gi6c

v0i

bi6n

s6

thw.

VA

gidi

tich,

kh6i

niQm

gi6i

h4n

cfia

hdm sd

duqc dinh

nglia

th1ng

qu,a

khdi

niQm

gihi

hqn

ctra day

sd

chri

kh0ng dung

ngOn

ngir

delta-epsilon.

CLct.

ntry

llm

cho

nhi6u vfn

dd

tr&

nOn

don

gi6n

hon,

pht

hgp vdi

nhfln

thtrc

cria

da

sd

hoc sinh.

BOn cpnh

d6,

viQc

b0 bht

lrr}t vli

kh6i niQm

nhu c6c

khfi

niQm

vd

tdng,

hipu,

tich

thr.rong

ctra

hai day

sd

hoic

hai

htrm

sd,

khfi

nigm

htrm

sd

liOn

tuc

mQt

b0n,

cilng

lhm

cho

DSGTIl

nhg nhtrng

hon nhi0u so

vdi

tnrdc

dty.

Gi6i tfch

12. Trong

phdn

d4o

hdm vI

khto

s6t

hlm sd, c6c

cOng

thrlc tinh

dpo

hdm cria

hlm

sd

hqp v[ so

tld

khto

s6t

hdm sd cir.g vdt

cdc

yAu

cdu

cU

thd

khi khto

s6t ting

loai

hdm

sd

duqc nhfn

m4nh

hon,

girlp

cho

hQc

sinh

tr6ch

dugc

nhtrng

sai s6t

dE

mdc

phti khi

thuc

hlnh

gili

to6n.

Thufi ngn khdo

sdt hdm

sd

&wc

dr)ng

thay

cho cqmkhdo

sdt

vd

ve

dd

thi

crta

hdm

sd.

BO

bdi

tldny,hho

s6t

htrm sd d4ng

phtn

thrlc

mtr

cl

tir

thric

vi

m6u

thr.rc

ddu

ll

tam thrlc b0c

hai.

Clc

vi dtl

vd

tim

gid

tr

lhn

nhdt hay

nhl

nhdt

cira

mQt h)m

sd

ffAn

m|t doqn

hay mQt

khodng

duqc

xdc

dinh

cU

thd

khOng

nhiing

girip

hqc sinh

n{m

drryc

phuong

ph6p

gili

to6n

mi

cdn

tr6n}r

clrryc sU

l6n lQn

giira

kh6i

nipm

nhy

v6i khdi

ni€m

gid

tr

ctrc

tlqi

vd

gid

tri

clrc

tidu

ct0;a

mQt

hlm sd.

(Xem

tidp

trang

23)

11

7/17/2019 THTT So 279 Thang 09 Nam 2000


DC NO HI

ilNT

A

-^

\ffif,T

i;'):r'

ciic

sd thuc

th6a man

didu

/ /' ,^ \ )

rnANNauDuNG

t

(

NIJ |

<cv*nooTodnrrxdnsDHKHWrpHdchiMinht

r-Ac

r.cypr,Hcs\

\t/ f

Bdirstz79.

I*

giri

tri

ldnnhdtcria

bidu

thric

Biti 771279.

Tim

gi6

rri ltln

nhdt

cua

bidu

thric

P=-x

-+-1

.+

2.,

7+.f,

7+y'

7+2"

I

x2y

+

y2z

+

zzt

+

Px

-

xyz

-

yzz -

zP

-

tP

thi

sd circ

ulc

cfia

a

md"

nguyOn

td vdi

p

bang

(vinn

c6ng nghp th6ng

tin)

sd

ciic u6'c

cria a mh

khong

nguyon

td voi

p'

Bii rgl2lq

Tren

rnlt

phing

ch,

ba

iludng

NcuvENntrusiNc

trdn

d6ng tdm

o

viii

b6n

kinh

lAn

tuor

tir

(GVrrudngTHCSBdnThrty,vinh,Nsh€An)

rr=

l, rr"=.,{z,r:

=

lF. cnla, a, c

tiu"

oie^

Bdi

T2/27g.

Chring

minh

ring

ndu

c6c

sd

khOnq.mlrq

hing

ldn

lugt

nim

trel

ba dudng

thgc

.r,

y,

a,

b

th6a man

c6c di6-u

kiQn

r+y

-

trpn

d9

_Cpi I

li

diOn

tich MBC.

Chring

minh

a+bva'/

+y+

=a4

+b4lilii/.

*f

=a"iW

rdng

S<

3. Tinhd0

dAi c6c c4nh

AABCkhi

vrii

moi

so'nguyon

ducmg

n.

t

v

,s

=

3.

LEDUYMNH

uoANGHoATRAI

(Khoa

Todn

trudng

DHSP

Hd NQi

2)

-

9'

'::*t

THPT chuv2n

L0 Khi6t'

Qudng

Ni''ii)

Biti

T101279.

Cho

rri

diQn

ABCD

sao

cho

c6c

BitiT3l279.

Chring

minh

ring

:

cqnh AB,

BC,

CA

ddu nh6

hcm

c6c c?rth DA,

(P+yz))/nrny"

+

(fl-f)?

DB,

DC. Tim

giri

rri ldn ntrft

vi

gi6

tri

nh6

trong

d6

r,

1,rtr

c6c

sd

duomg

vi

n

ltr

sd

S#;pr=r;fT|iy.*i&.Pth6amandidu

guYOnduong.

Nv'rz

-rtr

atD

trL

nuyNs

rAN

csAu

nru

xuAN

riNit

(GV

r

rwmg

THPT chut,\tt

Lurmg

vdn

chdnh;

phil

y1n)

(GV

trutug

THPT

Inm scrr' Thanh

Hda)

Bdi

r4t27g.

cho

tam

gi6c

dou AIlc.

Tim

tap

cAc

oe

vAr LI

hqp

tdt

cti cdc

Ctdm U-nim

trong AABC

sab

_Biti

Lll279.

MOt

xe chd

c6t chiu

t6c

dsng

cho ndu

hinh

chidu

g|a

M

tr6n

c?c cVnl

BC,

th99

nfuong

ngang

b0i

mOt

luc

k6o f

khOng

ddi

cA, AB

ldn

lugt

ld

D,

E, F

thi

c6c

dulng

th6ng

c6 hu6ng

trung

vq

hugng

cria

vecto

vln

t6c

v

AD,

BE,

CF

ddng

quy.

"

cria

xe.

Do

mot

l5

thring

0 sln

xe,

cdt chtty

Ncuy6r*Htruplludc

xudrrg

vdi.

hnr

lugng-

kh^6ng

0di c

(fg/s).

Xac

(sv

trudng

DH

Bdch khoa,

Hd Noi)

fjnh

eia

tdc;n v0n

tdc

cria xe.&

thdi didm

r,

ndu

Iric

,

=

0 khdi luo. ng

cria

xe

bing

movd vQntdc

B.di

T51279.

Cho

tarh

gi6c

ddu

ABC

vd, M

lh,

cria

xe

bing

khOng.

86

qua

moi

ma s6t.

mQt

didm

nim

trong

tam

gi6c.

Gqi

X

Y,

Zldn

NGUvEN xuANquaNc

luqt

ltr

didm

ddi

xring

cira

M

qoa

BC,

CA,

AB.

(GVtrudngTHPTchuyonvinhPhfic)

:,1S9".T'*rinB

cfc

tam

gi6c

ABC

vt'

XYZ

c6

Bdi

L2tzlg.

eu6

cdu nh6

rich

diQn

treo

bing

cunS

rrong

ram.

NcuyENMrNrrHA

ddy

nhe, khOng

dtrn,

cdch diQn,

dii

I

=

|m,

(GVkhdipT,cttu,-anrcdntradnsDHSpHdNriit

[:ilr#'g:f[t:%'f{

dou

nim

nsans,

dav

reo

Sau

d6

ddi

dot

ngot

hutug

dien

trulng

cAcr.oprHpr

'#.1i

:urtF(5i1;?tli

it.fT#luT?'1,?l

B^ir6tzis.

rim

msi

sd nguyen

do*e

L_:T

ilXT?,"#.-.1%ttT#iLt#tiiit?i*';

cho n

(

In,

trong d6

,, la

sd c6c

udc

nguyOn

il;y'ffi;;a

ch4m

diQn trudng

6i

"gxi

iol

duong

cianz.

uliirayfenddndOcaontro

?

"

vO

DOc

soN

rn.ANuANuHt

NG

(SV

K 4l Khoa

Todn

Tin trudng

DHKHTN Hd

N1i)

(GV

khdi Chuycn Todn-Tin, DHSP

Vinh, N

gh€

An)

12

?1

7/17/2019 THTT So 279 Thang 09 Nam 2000


PROBLEMS

IN THIS

ISSUE

FOR

LOWER

SECONDARY

SCHOOLS

Tll279.

Find

all

natural

numbers

a

(a

>

l)

such

that

for

every

prime

divisor

p

of

a,

the

number

of

divisors

of a

which

are

relatively

prime

to

p

is

equal

to the number

of

divisors

of

a

which

are

not relatively prime

to

p.

TZl279. Prove

that if the real numbers

x,

y,

a, b

satssfy the

conditions,

+

I

=

a

+

b and

*u

*

yo

=

oo

+

ba

then

{ *

y"

=

an

+

b"

for

every

positive

integer

z.

T31279.

Prove

that

(x2+yz)>2nf

f

+

@-:r,n)2

where

x,

y

are

positive

numbers

and

n

is

a

positive

integer.

T41279.

Let

ABC be

an equilateral triangle.

Find

the

locus

of

points

M inside

AABC

such

that

if

the

orthogonal

projections

of

M on

the

lines

BC CA, AB

are

respectively

D, E,

F

then

the

lines

AD, BE,

CF are concurrent.

TSl279.

Let ABC

be

an equilateral

triangle

atd

M

be

a

point

inside

MBC. Let

X, Y, Zbe

respectively

the

mirror-images

of

M

through

the

lines

BC,

CA, AB.

Prove

that the

triangles

ABC

atdXYZhave

the

same

centroid.

FOR

UPPER

SECONDARY SCHOOLS

T61279. Find

all

positive

integers

n

such

that

fl

I tn

,

where /,

is

the

number of

positive

divisors of

n2.

T71279.

Find

the

greatest

value

of

the

Yyz

expressionP

=

"

.+

-"+

--'

l+x' l+y' l+z'

where

x,

l,

z

are

real

numbers satisfying

the

condition x

*

|

*

z=

t.

'f81279.

Find

the

greatest

value of the

exoression

'

xzy

+

y2:, +

zzt

+

tzx

-

*y'

-

yz'

-

at2

-

txZ

where

x,

,

z

are real

numbers

belonging

to

to;

1l

T91279.

In

plane,

let

be

given

ttuee

concentric circles

with

center

O and

radii

r|

=

l,

rz.=

.,[i,

r:

=

{i.

Let

A,

B,

C

be

t}ree

non

collinear

points

lying respectively

on these

circles and let S be

the

area of

MBC.

Prove

that

,S

<

3. Calculate the

measures

of the

sides

ot LABC when.

=

3.

T101279.

Let ABCD

be

a tetrahedron

such

that the

measures

of

the

sides

AB,

BC, CA

are

all

less

than

the

measures

of the

sides

DA,

DB,

DC.

Calculate the

greatest

value

and the

least

value

of

the

measure

of

PD

where P

is

a

point

satisfvins

the condition

'

"PDz

-

PAz

+

PBz

+

PCz.

fOAN

HQC

,ilUo\l

AIAU

Gidp

bia

2)

Gi6i

ddp

bdt

:MANDENBROT vA

uiuu rVDdNe

DAN9

1) Da

sd cdc

ban da fte

ldi

:

Ilinh

bOn tr6i

m0

tA csn

ldc xo6y, hinh

ben

phAi

m0

tA c6nh tAng

blng

trOn

bd bidn.

2) NAm

qng

phdm

dAnh cho c6c ban c6 t0n du6'i d0y da

v0 hinh E+

dtng

:

-

NguydnTidn

Hmg,69lLC Thanh Ngh ,

thlnh

phdHtri

Duong

-

PhqmTidn Dnng,

ThOn 12 khuydn NOng, TriQu

Sun,

Thanh H6a

-

Ngul1n

Phwong

Ngoc,

10A6, THPT

Holng

Qudc

ViQt,

Mao KhO, D6ng Tridu,

QuAng

Ninh

-

Phqrn Xudn

Huy,

11A1,

TIIPT

Th6i Phric, Th6i Thpy, Th6i

Binh

-

Traong Minh Nghla,

t0p thd co

khi

diOn

tir,

Thanh Xu0n Bdc,

Qu0n

Thanh Xu0n,

Hd

N0i.

13

7/17/2019 THTT So 279 Thang 09 Nam 2000


glnr

onr

ri

Tnuoc

T*

a =

ap-1ap*1.ta

k=2,3,...,n.

vay

ai+ a\

+

...+

fint

vdi

mgi

,ta

c6

an a1

a*l an+l

,

at-t

ak

co-

ak ak+l

Ar At An-l

An

DltS= -:--:=...=

'

42 43

An A*L

nnnn

d 41 A2 An-l

A"

o

--

a; ,A ai aX*r

Biti

Tll275.

Tim mqi

nghiQm

nguyAn

phuong

trinh

x+y

_1

i

-xY+Yz-

7

Ld'i

giii.

ciaD(ng

Thdnh

Long,gA,

THCS

YOn

Phong, Bfrc

Ninh.

Gi6sfr x, e

Zthbarrrn,

,--ry;=1

x-

-

x\+\)

7

hay

7(x+t)

=3(.x2-ry+y2;1t;

uOi

c6c x,y'*0.

DLt

p

=

x+ ,

Q

=

x-y ta c6

x

=

Y,,

=

P-q

"

2,r 2'

Thay

citc.r,

y

nly vno

(1)

ta

c6

:

28P

=

3gz

+

3qz)

(2)

Tt d6

suy

ra

28p:3hay

pt

3

+p=3k(keZ)

Thay

gi6

tri

cta

p

vlo

(2)

tac6

:

28k=3(3k2

+

q2)

(3)

Suyraki3=k=3m(meZ).

Thay

gi6

tri cria k

vlo

(3)

ta

duqc

28m=27m2 +

q2

=

m(27m-

28)

=

-q2

<

O

Ttd6suyra

O=*=#

VAym=0hoicm=1.

YOt

m= 0

thi

p

=

q

=0

+x

=

0,

)

=

0

(loai)

Ydtm=1+p=9,Q= 1

T6m

lai tacd

(r5,

y=4)

hoic

(x

=

4,

y

=

5)

Vfly

nghiQm nguyOn

cria

phuong

trinh

da

cho

ltr

(x

=

5,

y

=

4)hoac

(.r

=

4,

y

=

5).

Nhgn

x6t.

C6

rdt nhieu ban c6

ldi

gili

tdt.

.

T6NGUYE}.I

BiiliTZt275.Cho

n+I

(n>2)

sdth1rc

a,

a?

..., ar+t

khdc 0

thda mdn

azo=

ao-rao*,

vhi

mqi

k

=

2,

3, ...,

/t.7ir1,

j4 :

JL

theo a,

ai+

q+

...

+

dj*,

vd ao*r.

Ldri

giii.

(cria

ban Nguydn

Tidn

VieL

8B.,

THCS ThAi

Nguy0n, Nha

Trang, Khdnh

H0a

vl cria

nhi0u

ban

kh6c).

t4

nnt,

al

+

a2+

.,.1an

ai+d+..,+a ,*1

a, a-

an_l

Mat kh6c,f

=

j

'A2A3A,

4+oi+...+4

al

an+l

NhAn

x6t.

C6 rdt nhi6u ban

gitri

giOng

nhu

ldi

giti

ren,

nhung

tdt

cl

cdc ban

ddu

quOn

ring

c6 thd

xAy ra

truong

hqp

ai

+ a\ +

...

+

al)*,

=

0

(chtng

hqn vji n=2,

at= a3=

l,

a2

=

-1)

khi dd ti sd cdn tinh

khOng x6c

ttinh.

Hdu h6t

ldi

giti

cta c6c

ban tl6u

ngdn

gqn,

trt

mQt sd ban chring

minh

bing

phuong ph6p quy

n4p,

kh0ng dr:gc

ng{n

ggn

bing

ldi

gitri

trinh

btry uOn.

Cong

c6 ban uinh

biy ldi

gi6i qu6

tdt,

Ia

didu

nen

tr6nh

vT]DiNHHOA

B

i

T3t275.

Cho cdc sd

thqc x,

,

z

ndm

trong

[-2;

2].

Chftng

minh

rdng

:

21x6

+y6

+

z6

)

-

(*y2

+ya

zz

+

za* )

<

I

92.

,

Ld,i

girii.

Cdch

I. Tt

giA

thi6t

ta cd

t',f',

z'

e

[0;41

tq-f)(o

+

y2

-i).0

(4-*)(4+zz-f)-o

(4-*)(++*-t).0

/

*yo

-#."

(1)

42

y4+24-T<rc

Q)

42

za+/-T=tu

(3)

Cqng

trmg

vd cria

(1), (2), (3)

ta

cd

:

7/17/2019 THTT So 279 Thang 09 Nam 2000


GrAr BAr ri

rnudc

zqxa+ya+za)

-

)t*of

*f

z'*zo*')

.

4g

=

41xa+ya+r\

-l{"ul'+yoz2+zo*1<

96

(4)

Mlt

kh6c

:

Vi 0

<

l,

2,

z'

.

4 ncn

4xa >

*u

;

4yo

2

y6

;4za

,

,u.

Do dd

:

41xa+ya+za) > xu

+

y6

+

z6

(5)

Tti

(4)

ve

(5)

ta

c6

:

*6

+yu +

7u

-

){*o

t'

*lo

z'

*

zo

*')

.

96

+ z(x6

+y6+26)

-

1xay2+yaz2*zor')

< Lgz

Dtng

thric

xtry

ra

e

(x2;

y':

z')

e

l(4:

4;

4)

:

(4;

4:

o)

;

(4:0:4), (0;

a;

a))

<+

(.t;

y:

z)

e

l(x2;

+2;

t2)

;

(t2;

t2;

0)

;

(+2:0;

+2);

(0;

12:x2);

Nhgn x6L 1) Nhi0u

ban

tim c6c

kh6 nang

dd tling

thric

xtry ra

cdn thidu :

trong 3

gi6

tri

Lrl

;

lyl

;

ld c6

2

gi6

tri

ll

vl

1

gi6

td

la

0.

2)

MOt

sd ban d1 tdng

qu6t

h6a

btri to6n

vA cho

kdt

qui

dring.

3) C6c b4n cho

ldi

gili

tot

ld

:

Hi NQi:

NguyZn

Hodng

Thanh,9A, THCS

Nguy6n Tndng

TQ, Ddng

Da;Trdn

AnhTudn,gAl,

THCS

Luong Thd Vinh;

tlii

Phing:

Bili

Vdn

Tudn,9A, THCS

Tu

Cudng,

Ti0n

Lang;

NghQ

Anz

Trung

Tudn

Dfing,

Trwrtg

Binh

Nguy€n,98,

THCS

Dfug Thai

UA;

Tp

Yinh; Nguydn

Trgng

Chung,

9A,

TIICS

L,0

H6ng

Phong,

Hrmg

NguyOn;

Phri

Thq:

Hodng

Ngpc

Minh,

Trdn Thanh

Hdi,9C,

THCS Vier

Tri; Ninh Binh: Ngay|n Vdn

Dgo, 6F,, THCS

YOn Phong,

YOn

M6,

Phqm

Quang

Hny,9

To6n,

THCS

thi-trdn Ninh,

YOn

Kh6nh; Ddng

Thdp:

V0

Hftu

Trt,SA1

THCB

thi

xl

Cao Lanh; Hn

TAy:

Trlnh Xudn

Trt,

88,

THCS

Nguy6n

Thr:qng

Hidn;

Phqm Mtnh

Quydt,gA,

THCS Kim Dudng,

Ong

Hda; B4c Li6ut NgryEn Thdnh Nhdn,9A, TTIPT thUc

htrnh Su

pham,

thi

xt

Bgc

Li0u;

Vinh

Plclrdcz

Trdn

Vdn

Nam,8C, THCS D6ng ich,

LSp Thach; YGnB6* Trdn

Binh Minh,gK, THCS

If

Hdng Phong,

thi

xa

Y€n

B6i; Bdc Giang:

ThAn

Thi

HuA,

8C,

THCS

Ti6n

Phong,

YOn Dung;

Hii Duong:

Ngwy4n

Thdnh Nam,

9A,

THCS Nguy6n

Trai;

Tp Hd Chi

Minhz Nguy4n

Dinh Khudy,8A1,

THCS

NgO

Tdt Td,

Q.

Phf Nhufln;

Bdc Ninh:

D(ng

Thdnh

lnng,9A,

THCS

Y€n Phong;

Quing

Triz Phan

Qudc

Hung,9A,

THCS

thi trdn

Hai

Ltrng;

Thanh H6az L€ Khdc HuyAn,98,

THCS Thign

Vfln, ThiQn Y€n.

..

lEm6NcNuAr

B

i

T4t275.

Ch*ng minh rdng

MBC

vhi

BC

=

a, CA

=

b,

AB

=c

ld

tam

gidc

vuOng

khi

xdy

ra

m1t tron?

cdc

ddng

tht?c sau

:

lb-cl

Cdch

2.

Do

vai

trd binh

d8ng cria

x,

y,

z

nOn

gi6

sfr

lxl

<

lyl

<

lel

<

2.

Khi d6

:

l*o1r'-y'1<o

(r)

lx'1xo-zn1

<

o

(z)

ly4eyz-t)

.

yoz'

<

,u

(3)

=

2(x6

*yu+zu)

-

1fy2+yazz+zo*')

<

326

<

3.26

=

192

(*)

Ta

thdy

(*)

trO

thlnh ding

thric

<+

lzl

=

2 vd

(l), (2),

(3)

ddng

rhdi

trOthlnh

dfng

thric.

Vi

(3)

trd thanh

ding

thric

++

lyl

=

lzl

(z)

aothlnh df,ng ,h,i.

*l;Lot.t

(1)

trO tlrnnh ding

rhric <+

l5

o,

,

|xl=

lyl

Vfly

(*)

tr0

thenh

ding

thric

<+

lyl

=

lzl

=

2

vtr

l.rl

=

2

hogc

.r

=

0, ti d6 ta c6

kdt

qut

nhu

phAn

cuOi c6ch

l.

Aa

1)t5=;'

^A

2)

tg'1=

b+c

LO'i

gi6i.

Cdch

1.

Gqi

AD

(D

e

BC)

lI

dudng

phtn

gi6c

cria

g6c

A.

Tr0n

tia ddi cria

tta

AB

ldy

didm

E sao

choAE=AC=b.DE

thdy

IBAD

=

ZDAC

=

IACE

=

ZAEC

nOn

ADllEC.

KC

duOng

thfrng

BK

vuOng

g6c

vdi

BE,

cdt

dudng

thhng

EC

0K.

Ta

c6 :

,rt=rrr='fr=#

(1)

KtrOng

mdt tintr tdng

qu6t

git

s&

b

>

c

thi

ZABD

> IACD

+

IADB

<

IADC

+

IADC

vuong

ho[c

tr).

Tt

d6

vd,

tn

ADIIEC

suy

ra

ZDCE ltr

g6c

nhqn

(2).

1)

Tr) gi6

n&

te|

=

-n*v}

(1)

suy ra

BK

=

a

=

BC.

Tt

dd

hoflc

K tn)ng

vdi C

nghia ltr

ZABC

=

90o,

hoic

MCK

can

&

dinh

B nhung

di6u

nly

khOng

xty

ra do

(2).

2)

VOt

gitr

sil

b

>

c

gll

thidt

tro thanh

tez

=?.**hsp

vdi

(l)

c6

-2

b+c

l5

7/17/2019 THTT So 279 Thang 09 Nam 2000


GrAr BAr

xi rnuoc

- E-

-@-c)(btc)

=

nr(

=

bz

-

c2.

(b +

c)2

(b+c)z

uai

u,a.

AP

=

at{

+

cz

nen AI{

=

b2

=

eC. md6

hoac

K

trDng

vdi C

ngtria

ld IABC

=

90o,

hoac

MCK cln 0

dinh

A

nhtmg

didu

niy

khOng xtry ra

do (2).

Cdch

2.

Gil

sir dulng

trOn

t6m

l

b6n

kinh r

nQi tidp

MBC.

KC

1E]-ACthi

AIEr_pr_S

IO-=-=-

=

bZ

AE

p-a p(p-a)

p(p-a)

Aa

l)

Tri

tg;

=

-

vl

(3)

suy

ra

I D+C

a'

o'

-

(b-r)'

(b+c)z

(b+c)z

-

a2'

R[t

gon

ta

dx)c ao

=

(bz

-

,2)z

hay

az

=

lbz-czl.

-Iu

do

suy

ra

MBC

vu6ng 0

B hoac

C.

,A lb-c'

Drntgz,a='ffi

vl

(3)

suy

ra

'

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R.

Goi

Q

ld

tdm dwdng

tr\n bdng tidp

gdc

A

cfia

tam

gidc

ABC.

Chthng

minh rdng

:

1) IK

-

#g

,ronr

dd

IK

ld khodng cdch

ttt

I

ddn

B'C'.

(i)

2) IA' +

IB'

+

IC'>

6ril-5

,r,

'

llR+2r

'

a

Lcri

gitii.

(Dua

theo Hodng Ngpc

Minh,9C,

THCS

Viet

Tri,

Phri Thq)

1)

Gqi ,S vi ,S'

ldn

lugt

ltr

trung didm

cung

ACB

vb, cung

AB

khOng

chf'a

C, thd

thi

,S,S'

di

qua

O

vl

SO L

AB

0 trung didm

P

cfla

c4nh

AB,

CS'

lI du0ng

ph6n gi6c

ctn

ZACB nOn

di

qua

C'.

Tt)

d6 du0ng

phfln gi6c

ngoli

CQ

crla

g6c

ZACB di

qua,S.

Dpg

/"/

I AB

0 J

vd,

IK L

B'C'

0

K,

ta se

chfng

minh

ring

:

NJKcn

LOOS.

ThQt

vfly,

IJC'K

vd,

CC'PS

ltr

nhrrng

dr

gi6c

nQi

tidp

nOn

tac6:

ZIKI

=

ICC'B

=

IOSQ

.

Lai

dung

QE

L

AB

0

E,

QF

L

AC

0

Fvh

OM L

8E

0

M,

ON

L

QF

ONrhdthi

ta

c6

:

ZB'AC'=

ZCAB

=

INOM

(2)

DAt

BC

--

a, CA

=

b, AB

=

c,

a+b+c

=

2p

,

D6

thdy

:

AE

=

AF

=

p.Tt

d6

suy

ra

:

OM

=

PE

=

r'9

=o*b

.

'

z z'

vi:oN=o-L-c+a

-'-"

r

2 2'

Theo

tinh

chdt dtlng

ph0n gi6c,

dE

ding

tinh

dugc

:

AB'=b,

eC'=J\.

c+a'

a+b

Tt d6 ta

duoc

.

AB'

-a+b

AC'

c+a

(4)

Tri

(2), (3)

vI

(4)

ta du,gc

:

LAB'C'<r'

LOMN

(5)

Tt)

d6

suy

ra

:

IAC'B'

=

IONM

=

ZOQM

(do

OMQN Ii

tri

gitic

nQi

tidp)

(6)

M4t kh6c, EOCC'lI

hi

gi6c

nQi

tidp

nOn

ta

dtrgc

:

IAC'C

=

ZCQE=

ZSQM

(7)

Tt)

(6)

vI,(7)

suy

ra

:

ISQO

=

ZIC'K

=

ZIJK

(do

tti

gific

IJC'K

nQi

tidp)

(8)

Cudi

ctng, fi

(1)

vI

(8)

ta suy ra

NJKa LOOS.

Tt

d6 ta duo.c :

IK IJ

OS.IJ

Rr

-

1l\

=

=-

(li

o.t oQ

o8 o8

Gqi

IL,

ITldn

lrrgt la

khotng

cfch tr)

/

ddn

C'A',

A?.

Chring

minh

ttnmg

tU

nhu

ften

ta

cUng

thu

dugc

cdc

he

thtrc trrong ttJ nhu

(i).

Gqi do, ds,

dsldn

luqt ln khof,ng

cich

tr)

O d€n

t0m

dudmg

trOn

blng

tidp

cdc

g6c

A,

B,

C

cfia

tam

gi6c

ABC;th€

thi

ta

dugc

cdc

hQ

thric

sau

:

IK=A.L=9vh.IT=A.

dA' dB

dc

D

Ap

qrng

BDT

EcdOso,

trong tam

gi6c

A'B'C'ta

c6

:

IA'

+

IB' +

IC'>

2(IK

+ IL +

II)

=zR{+.+*+)

\d,

dB dr)

19

7/17/2019 THTT So 279 Thang 09 Nam 2000


crAr

BAr

xi

rnudc

Ddu ding thric

x1y ra

khi

vtr

chi

khi

tam

gilc

A'B'C'ld

ddu

nhfln 1ltrm'trgng

tim.

Theo BDT Bunhiac6pxki,

ta

duo.c

:

3(rtA

+

8,

*

h

>

(de

+

d,

+ dr)2

Tri

d6 :

do+

dr*

dc<lXfo* *r*

taOo>

Str

dqng c6c

cOng

thric Ole

dl

bi€t

vtr

hQ

thrlc

:

ra+ rb * r,

=

{ft+1

&=n2+2Rro

&=n2+2Rr6

j

d'L=fr+ZRr,

(trong

d6

ro,

16,

r,

ldn

lu$ le

bfln

kinh

dndng

tr0n beng

tt€p.ctc

g6c

A, A

vi C)duqc :

rto

*

8a

+

&r=3R2

+ZR(ro+16*rs)=

=

3R2

+

2R(4R+r)

=

11R2

+

2Rr

Tri

(10)

vI

(11)

ta

duu.c

:

do+

du

+

dc< tffiTznr

I

Mf,t

kh6c,l4i

c6

:

Vi

do

d6,

tri

(12)

ve

(13)

suy

ra

:

9

(14)

Cudi

ctng

ft

(9)

vI

(14)

ta thu duo.c

BDT

(ii)

cin

tim.

Ddu

ding

thric

xby ra

trong tdt

ce

c6c

hQ

thric tren khi

vtr

chi kni

ABCld,

tam

gi6c

oeu.

Nh{n

x6t

l) Trong

ldi gili

phdn

1)

cria btri todn,

nhi6u

ban sri dqng

cac

he

thfc

luqng

gi6c,

bi6n ddi

phrlc

tap v) cdng k8nh.

Duy

c6 ban

Minh

chi srl dqng

kidn

thrrc

hinh

hOc THCS

cUng thidt

Iflp

dugc

hP

thft

(i)

nhu

dt

neu

trong

ldi

giti

tren.

Mot sd it

ban

kh6c srl

dung

dinh

li

hlm

sO sin trong tam

gi6c

cung tlridt

lftp

duqc

hQ

thrlc

(0

mQt

c6ch

nhanh

ch6ng

(chfrng

hqn,

b4n

Nguy6n Thanh IIAi, 11A1,

TI{PT

chuyOn

Vinh

Phric)

\

Ini

gfttt phAn

2)

ddi

h6i

cdc ban

phtri

bi6t

sri

dung

BDT 6cd0so

trong tam

gi6c.

Tuy nhiOn da sd c6c

ban ddu srl dqng ctrc

h€

thrlc dl thu dtrgc

trong

bf,i todn

T9l27l, kdt

qu6

bidn ddi ttl:dng cdng

kdnh, khong

thtt

gQn'

3)

Ngoli

hai

ban

nOu trOn, c6c ban sau dty

c6

ldi

gitri

twng

ddi t0t :

Bdc

Ninh:

Nguyhn

Thd Thtiy, Tll,

THPTNK Hln

Thuy0n, Bdc

Ninh;

Nam

Dinh:

Hodng

Vdn Giang,

10

To6n,

TI{PT

I,0

Hdng Phong,

Nam

Dinh;

NghQ An:

LA

Xudn Hilng,lOA5,

NguyAn

Trpng

Tdi,

10A5,

Phgm

Vdn

Tudn,

10A2,

THPT

chuyOn

20

Phan BQi

Ch0u, thdnh

ph6

Vinh; Tp.Hd

Chi Minh:

Nguy€n Vdn Thdng, l0CT, THPT tE

Hdng Phong

NGUYENDANGPHAT

Blhi

TlAnTS.TrAn mfi

phdng

P

cho

dudng

trdn daMS

kinh

AB. Ldy didm

c ran

fia

AB

sao

cho AC

=

2AB. M\t

dadng

thdng

qua

C

cdt

dwdng

trdn tei

M

vd

N.

Dltng

didm

D

sao

cho DB

=

AB vd

DB

vu1ng

gdc

vhi

m(t

phdng

sinzlBDM

+

sinzBDN

=+

2

Ld'i

giii.

(cria

ban

T0

Minh

Hodng,

1lT,

THNK Htii

Duong)

(do+

d,

*

dr)

(;.

h.

*1r-,

(11)

(t2)

(13)

111

-+-+->

o' dn' dc-

Tac6:

CM.CN=CB.CA=2C82

=

CBZ+DBZ

=

CDz

CM

CD

CD

CN

MD

CD Drt

CDz

'

DN

CN

DI,F Clf

DtvP zC*

--

"

=

"

\r,,

DM

CN'

Ta

c6

:

LCMB

a

A,CAN

CB

BM

=

-

=- \21

CN

NA

Tir

(1); (2)

suy

ra

:

DI,f

zBki2

BIY(

:

-=

rn2

DN' AM

DM

AI\tz

2Drt

vay:

sinz1BDM

+

sinz

BDN

=

4

.

4

DM'

DM

A]'f

BI\P

f--

2D1\F, DI,P

Atf

+ 2atl

2Dfr

7/17/2019 THTT So 279 Thang 09 Nam 2000


erAr

sAr x'irnuoc

t*

+

ntf

zoxP

go?

+

alf

ot'f

1

zofi

zofi-

z

Nh{n x6t.

1)

Nhi6u

ban tham

gia gili

bii

to6n n)y

vdi

c6c c6ch

gitri kh6c

nhau, tdt

cA

d6u

giti

dring.

Z)

Ctrcban

sau dAy

c6

ldi

gili

tOt:

Yinh Phric

NgryAn

Xudn

Trudng,

10A, TIIPT

chuyOn

Vinh

Phric;

Hda Binh:

Nguy1n Thdi

Ngpc,

10T, TIIPTNK

Hohng

VIn

Thu;

Hii

Duung: D6

Thi

Ngpc

Qu)nh,

10T,

THPT

chuy0n Nguy6n

Trai; H:ii

Phdng: Ddng Phuong

Thdo,10T,

THPTNK

Trdn

Pht;

Hi

Tinh: 1?

Khdnh

Hung, llA, TIIPT

Minh

Khtri; Thanh

H6az

Hd

Xudn

Gidp,10T,

TIIPT

Lam

Son;

tli

Ndi:

V[

Qudc

My,gH,

THCS Tnmg

Vuong,

Nguy4n Hodng

Thgch,

10T,

TI{PT HA

NOi

Amsterdarn

NGUYftNMINHHA

Bdi L11275.

M|t

dogn

mqch

AB gdm

cd

ngudn

(e,

r)

trong

d6

e

=

36V, r

=

lC-,

cdc

di|n trd

Rr

=

8f); Rz

BQ mdc

theo so dd

nha

tAn

hinh.

{-r)

R2

Cd cdc

bdng

ddn

rdi

:

D; 10V-5W;

D2

:

l0V-4W

vd D3

:

BV-6W. Hdy chi

ra

cdc

phuong

dn

mdc

cdc bdng

d\n

tr€n vdo

cum

AB

dd chilng

sdng

binh

thwdng

(mdi

phaong

dn

phdi

cd drt cd ba bdng

d|n).

T{nh

cdc

diQn trl

phu

cd

mfit trong

cdc

phuong

dn d6.

Hrdng

d6n

gif,i.

Vd

nguy€n

tic c6 thd

dd

ra nhidu

phtrong

6n

kh6c nhau

vdi

sd

diQn

tr&

phg

du-o. c lga

chgn

mQt

c6ch tuong

img.

6

Cay

ta

chi

xdt

cic

phuong

dn tdi

uu

(vdi

sd

dign

trO

phU

it

nhdt,

cdc

dien

tr&

phu

ntry

c6

tr sd

hqp

li

dd dtrm btro

c@ng sudt tiou thu

0

c6c

dien

tro

ph\r

ltr

it

nhdt).

,

Ki hiQu

I,

ltr

cuOng

dQ

ddng

diQn

qua

mgch

gdm

cdc

dEn

vi

dien

trO

phu,

ta c6

(xem

hinh

v0)

:

/=.fr *

12;

Utp--

IzRz=

e

-

I(r

+

Rr).

Suy

ta;

(Jou=

24

-

6lr

Cudng d0 dinh mfc

cria c6c dOn :

P

Iat=

U

=

0,5A;

In=0,4Ai

Ia:,=0,75A.

Do

d6

d€

cdc

ddn

s6ng binh

thrdng,

khOng

thd

m6c

ndi

tidp

c6c

d0n. Suy

ra

phti

c6

1,

>

Iur+

I*=

0,9A

€

Utn= 24

-

6Ir

< 18,6V

€

Uea

=

10

+

8

=

18(V)

(phuong

trn

a),

hoirc

Uou= 10(V)

(phucrng

6n b),

)

Ito= 1A

vd Is= X.

3

Vdi

phuong

6n a, mdc

nhu

sau :

(D

rl

I

Dzl

I

Ri

nt

(Dr/

I

R.r)

;

_@I'?

+atf1+atf

-

zDp-

Uar

10

0,1

khi

d6

Ra

=

Ib- Iil- Id2

100f),

vd

R3=

=-

Iu- Iat

0,25

Vdi

phuong

6n

b,

mdc

nhu

sau :

(D

tl

I

D

zl

/

(D

3ntRr)

/ R

)

;

khi

d6

R,=

R3=

uat

=

14,6f);

Iw-

(Ia+

I*+

I*)

Nhf;n x6t.

C6c ban c6

ldi

gitri ggn

vd

dring :

NghQ An: Ddo

Vinh

Quang,

11A3,

THPT Phan

BQi

Ch6u,

Vinh;

Ti6n

Giang: Trdn

Tdn

L\c,

L2

Li,

THPT

chuy0n

Tidn

Giang; Nam

Dinh:

Ngwy1n Ngpc

Tdn,

llB, THPT

Duy

Ti0n

A;

Phri Tho:

Vtl

Qudc

Hny,

llB

(CL)

chuyOn

Hing

Vuong;

Ddng Nai:

Irdn

Hnu

Hiiu, l0 Li

L

TIIPT

chuyOn

Luong Thd

Vinh;

Vinh

Phric: THPT

chuy0n Vinh

Ph[c;

Nguy4n Thd

Anng,l2A3,

Nguydn Minh KiAn, l0AI

MAIANH

Bbi

L2l27S.

Cd

hai

thdu

kinh hQi tqt

L,

vd

4

ddt cilng trryc

chtnh

cdch

nhau

70cm.

VAt

sdng

AB

dSt

trudc

L,

(phta

khOng

cd

Lr)

ta

dwqc

dnh

A'B'

ndm sau

L,

ldn

gdp

6

ldn

vAt

vd

AA'

=

370cm.

(Hinh

vc).

Dfit

thAm

thdu

kinh

Lrtqi

Or(gifra Orvd O) cing

trqc

chtnh

vhi

hai

thdu

kinh

thn.

.

Vhi

OrOz= 36cm

thi

dnh A'B' khang

ddi.

o

Vhi OtOz

=

46 cm

thi

dnh

A'B'

ra

m va

cilng.

ud3

=

32Q

Uar- Un

8

^

Iil

3--'

21.

7/17/2019 THTT So 279 Thang 09 Nam 2000


crAr

eAr

ri rnuoc

HAi

Opz

bdng bao nhiAu

thi

d0 l6n

dnh

A'B'

khlng

ddi klti

AB

tinh

tidn trahc Lr.

Hutrng

din

gi6i

:

Khi

chua

df,t

tr,

so

d6

tao

inh :

Lr

L3

AB-Apo-4'3'

dr d'r h

d')

Theo

dd bhi

:

dr+d'r=

dr

*

P,

-

M'

-

otot=

3oo(cm);

'

dt-fi

d3=

O1o3

-

d't=70

-

i L'

ar-fr

-fifz

rt-=

^

--.

.

...]

=6

It-

dt

h-

dt

g4d,

g4d,

Suv ra :

f,

=

-----l-'

d'.

=

rr-

144*dr'

"

I

-

60+d,

4700- t4d,

h=

,o*

(l)

Khi

dft thdm L2,

so

dd t4o

6nh

Lt

L2

L3

AtBt-A282-A'B'

dr

d'r

d2

d'2

4

d,3

r

Theo

dC

bei,

khi

OrOz

=

36cm,

Lnh

A,B,

khong

ddi

=

t2

khong

c6 tdc

dung

trong hQ,

nghia

ltr A&o

=

AtBt

:

A2B,

+

d2

=

d',

=

g

.

khi

d6 d'

t

=

OrOz=

36cm

=+

(theo

(1))

dr= 45cm+fr=

20cm;fi

=

30cm.

r

Theo

dA

bei ldti

OrOz= 46cm

thi

d',

=

m

+

dt=fi

=

30cm,

=

fl'r=

OzOt-

dt=

70-46-30=-6(cm);

dy= O1O2-

d'r= 10

(cm)

dcd',

=fz=-:-.

^

=-15(cm)

;

02+A

2

.

Dtt

OtOz=

l,tac6

:

,.

dtfi

20dl

t7 .

=-

dr-fi

dt-20'

=

909

dr1t2-ao.-+lsoo)

+

2tooo

+

800/

-

zof

DC

A'B'

c6

d0

ldn

khong

ddi,

tric

ta

k

khong

ddi,

khi

vi tri

AB

thay

ddi

ph6i

c6 :

f-au+5oo=o

=+l=

10cm;hoac/=50cm.

Nh{n x6t.

C6c ban

c6

ldi

gili

ggn

vd

dring :

Y€n Bdi t Hodng

Anh

Tdi, llA3,

Ph(tm

Hdng

Chtnh,

LlAl,

TIIPT

chuyOn YOn

B6i;

Phri

Y€n:

1l

Nggc Thi1n, 11

Li, THPT

Luong

Vdn Chinh,

Tuy

HOa; Ngh€ An;

12 Ngpc Tudn,

l

lA3,

THPT

Phan BOi

Chdu,

Vinh; Tuy€n

Quang:

Nguy4n Trung

Hidu,

1182, THPT

chuy0n Tuy0n

Quang

;

Phri

Thq:

Dfng

Minh

Son, 114.7,

THPT

cOng nghiQp

Vier

Tri,

y,

Qudc

Huy, 118

(chuy€n

Ii),

chuyOn

Hnng

Vugng

;

Hn

Tinh:

NguyAn

Vdn Hidu,

12A,

TI{PT

H6ng Linh,

Hdng;

Linh;

Quing

Tri; Trdn

Viet

Anh,

11

To6n,

trudng chuy0n

te

Qui

DOn;

Khdnh

lIdat

Trdn Hodng

Thanh, 1l

Li, THPT

tE

Quy

DOn,

Nha

Trang;

euing

Ngdi:

Dfiig

Dinh

ThuQn,

11 To5.n,

chuy€n tf Iftidt;

Qu,ing

Namt Hodng

Anh

Vn,

117,

TIIPT

Trdn

Cao

V6n, Tam

Ki;

Quing

Binht

Duong

Dttc Anh,

l0 Li,

THPT

NK

Qutng

Binh;

Tidn

Giang:

Trdn Tdn

L\c,

12

Li,

THPT

chuy€n Tidn

Giang;

Th6i

Nguy0n:

Va

Qudc

ViQt, 10

Li,

THPT

chuyen Thdi

Nguy€n;

Son

Laz Nguy4n

Thdi Binh,

ll Tofn

3, TIIPT

NK

Son

La;

Tp

Hd, Chi

Minh:

Hu)nh

Thi€n Phfic,11T,

PTNK

;

Ddng

Nai: Trdn

Hftu

Hodng,1l

Li l,

THPT

chuyOn

Luong

Thd

Vinh;

Vinh

Phrd,ct

Nguydn

Thd

Cwng,

12A3,

TI{PT

chuyOn

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