GAi ccc D GDo B GBcc

Một phần của tài liệu Tạp chí toán học và tuổi trẻ tháng 8 năm 2007 số 362 (Trang 27 - 30)

)

Sau d6 nit gon ta thu duo. c h9 thtlc (i) cdn tim.

Tidp tuc chung minh tuong tu bang hodn vi vdng

quanh c6c b0 chft {A, B, C , Dl , Ai , Bi, C; , D; ta duo. c 3 h0 thri'c nfra:

GB GC GD GA-:-= -- + -- + --- (ii) -:-= -- + -- + --- (ii) GBz GC1 GDt GAr GC GD GA GB -:-= -- + --- + -- (iii) GCz GDr G& GBt GD GA GB GC -::-= --- + -- + -- (iv) GDZ GAT G& GC1

Cudi ctng, cdng theo vd b6n ding thrlc (i), (ii),

(iii), vi (iv) ta thu duo. c h0 thrlc (*), dpcm. O v(GCrDAz) =GA, v(GBCD) GA' v(GDr&Az) =GA, v(GBCD) GA Ddn d6y cong (3), (4) chidu vdi (2), ta duoc

G&.GCr.GDt GB.GC.GD GCT,GDI GC.GD GDT,GBT GD.GB vd (5) theo vd rdi (4) (5) (*) Loi gitii. (Dqa theo Duong Til, 11T, THPT chuyOn LC Quf D6n, Binh Dinh). hdt, ta l4p he (1) (2) Ldi gidi 2. (Theo NSO Thi Mj Li€n, THPT LC Quy D6n, Binh Dinh) Xet t(r di6n GBCD, gqi B Gt le trong tam cira n6 C6 hai trudng hgp phAi x6t

oTradng hop l: Ar* Gt. Gqi (a) li mat phing

di qua G, vit (a) ll mp(B1Cp1); @) ldn luot

(3) c/r GB, GC, GD theo thrl tu 6 8., C3, D.,. Khi, G& GCt GDt GAz ,, \ , G& GCt GDt GAz ,, \ dOtaCO -=-= ili G& GCI GDZ GGT 3ffi0{ Fto( ' qu.6i$e sd 362 (s-2007) ilP"*t 25

1 ,-

rud6 Gd=;@*cc*a) Q)

Mat khric, vi G, Ii trong tAm trl di6n GBCD nen GG +crE+ci+Gfi=d

Ph( Tho: Nguydn Ngoc Trung,9Al, THCS Lam

Thao; Hrmg YCn; Hodng Dttc Tdm, 12 To6n, THPT

chuydn Hung YOn; Hii Phdng: Trdn Hodng Bd, ll

To6n, THPT NK Trdn Phri; Thanh H6az L€ Duy

Todn, llA11, THPT Luong Dic Bing, Hoing H6a;

Dir Ning: I'lguydn Nhu Dfic Trung, llAl, THPT

chuy0n LC Quf Ddn; Binh Dinh: BDi Hfru Nhdn,10

To6n, THPT chuyOn Lc QuI DOn; Bi Ria - Vtng Tiru; DrongVdn An,l1Al, THFrI Chdu Thinh, TX.

Bi Ria.

NGUYEN DANG PHAT

*gai L1l358. C6 mot mach di'n nhtr hinh v{.

Ngudn dian llY; hai ru 1,0pF vd 4,01tF : mQt

di€n trd 50O vd ba di€n trd 25Q. Bo'n ngat

di|n,lilc ddu tdi cd ddu md.

a) Hdy tinh di€n tich cila m6i tu sau mdt thdi gian ddi trong mdi trudng hop sau:

1) Khi Kn, K,, K2 d6ng; K, md ,

lt) Khi Ko, K,, Kr. K, d6ng.

lii) Khi Kovd K., md ; K, vd K, d6ng.

b) Khi K, vd K. md; Ko vd K, d6ng. Sau m1t

thdi gian ddi, ta cho vdo trong tu L,0 pF mbt

tdin di€n mdi c6 hdng sd diAn m|i ld 3,0 ld'p ddy kh1ng gian giria hai bdn ctrc. Tinh cdng

todn phdn duoc thtc hi€n khi dtto rd't chdm tdm di€n mbi vdo trong tu (tftc ld cdng duoc

thuc hidn bdi ta vd bdi ngudn).

Ldi gitii. a) (Theo ban Trdn Vdn Giai, A3,

K34; THPT chuy0n Phan BOi Chdu, Ngh€ An). a) i) Khi Kr, Kr, K2 d6ng; K, m6. Sau m6t thdi

gian dii, kh6ng c6 dbng di6n chay trong

mach, hiOu diOn thd gifia hai bin cria m6i tu bing sudt diOn d6ng cira ngu6n.

Di0n tich fiAn c6c tu 1,0 pF vi 4,0 pF ldn luot lh:

Vi c6c didm 8., Cr, Dt theo thrl tu nam trong

c6c doan GB, GC, GD n€,n ta duo. c

Gd =!( 1LcF, * rc cc * GD

Go,. l.

' 4\G& - GCt ' GDt )

Lai vi bdn didm G1, 83, Cr, D, ddng phing ndn

ta c6 h€ thrlc 1(g*g.g)=r, hay4[GB3 GCt GDt ) 4[GB3 GCt GDt ) GDt GDt (3) GA GAZ GAz GGr la rdo GA = 4) IGG,) GB GC GD GA T- G& GC3 GDt GGl GB G& GC GCI GD v -.-T T-

G& G& GCI GCI GDT

Ddi chidu (3) vd (1) ta thu duoc he thrlc (i)

nhu Ldi giii I dd chi ra & tr0n:

GA GB GC GD

-::-=--+--+- (i)

GAZ G& GCT GDT

t Trudng hop 2: Az = Gl, khi d6 mp(a) =

mp(&CrDr). Bang cdch chrlng minh tuong tu

nhu 6 trudng hqp 1 ta cfing thu duo. c (i).

Cu6i cing, bang cdch ho6n vi vdng quanh ta

thu duoc c5c hO thrt'c (ii), (iii), (iv) nhu Ldi

gi6i 1 rdi sau d6 thu duoc h0 thrlc (*) li dpcm.

(Nnan x6t. l) N6i chung c6 hai phuong ph6p co

bAn duoc s& dung dd gini bdi to6n trcn day. D6 Ie

phuong ph6p thd tich vh phuong ph6p v6cto du-o. c thd

hiOn ldn luqt b&i c6c ldi giii 1 vi 2 nhu di n0u lOn &

tren. M6i ldi giii ddu c6 m didm ndi bat vi s6c thdi ri0ng biOt r,hvng ttu chung ld v8n dung nhfing kidn thrlc co bin nhu c\ng thtc vd ti sd thd ttch cila hai

hinh ch6p tam gidc (c6 chung nhau m6t g6c tam

di€n) vi didu ki€n dd bdn didm ddng phdng (lil.n quart ddn vecto) dd thidt lfp h€ thtic (i) rdi tt d6 mi thidt

lAp hQ tfltc (*) nOu trong bhi todn.

2) Ngoni hai ban di dugc n€u ldi giAi trcn dby, cdc

ban sau dAy c6 ldi girii dfng vi gon ging hon ci:

TGffi !.|QCsd 362 {8-2007) & ctuditi6, sd 362 {8-2007) & ctuditi6,

l.0prF

Qr = cyE = 10-6.10 = lo-s (c) = 1o (pc)

Qz= cz.E = 4.10-6.10 = 4.10-5 (c) = 40 (pC)

ii) Khi Ku, Kr, Kr, K, d6ng.

Sau m6t thdi gian dii, di0n tich trOn ci{c tu dd

dn dinh, cudng dO dbng di6n trong mach li:

l= =-3:==-]L: =0,1(A)

Rr +R: +& 50+25+25

Hi0u di6n thd gifra hai bin cdc tu 1,0 pF vi

4,0 1rF ldn luot li:

Ur=Rl=50.0,1 =5(V)

Uz= Rr,l =25.0,1=2,5 (Y)Di€n t{ch tuong rlng trdn c6c tu: Di€n t{ch tuong rlng trdn c6c tu:

Qr= clt= 1o-6.5 = 5.10-6 (c) = 5 (pC)

ez= cz(Jz= 4.10-6.2,5 = 10-5 (c) = 10 (pc)

iii) Khi Ko, K, m&; K1, Krd6ng

C6c tu kh6ng tich diOn: Qr = Qz= 0.

b) Khi Kr, K, m&; Ko, K, d6ng, sau mOt thdi gian

dli hi€u diQn thd trOn tu 1,0 FF bang sudt di6n

ddng cria ngudn, tu 4,0 pF kh6ng tich diOn.

Ning luong cta tu 1,0 pF li:

w, = c,E' - l0-6'l02 = 5.lo-5 (J).

'22"

Sau khi cho tdm diQn m6i vho tu di0n, di0n dung

ctra tu tang l6n s= 3 ldn; C'r= {r= 3.10-6 G).

Nang luong cia tu hic niy:

W,, = C't E2 _ 3.10-6.102 = 15.10-5 (J)

'22"

Cdng toin phdn duo. c thuc hiOn dd dua tdm

diOn m6i vdo trong tu bang dQ bidn thiOn nang luong cira tu didn:

A = LW =W', - W r= 15.10-5 - 5.10-5 = 1004(D. D (Nnan x6t. Ngodi banTrdnVdn Giai, cdc ban sau

c6 ldi giAi tdt:

Vinh Phric: Ta Dtc Manh, 11A3, THP| chuydn;

Bdc Ninh: BiiVdn Hdi,llAl, THPT Qud V5 s6 l;

Hir Tny: Nguydn Ngqc Ddng, 11A1, THPT Phri

Xuy0n A. Htmg YGn: Pham Dtc Linh,ll L)r, THPI

chuy€n Hung Y€n; Thanh H6az L€ Bd Son, I lF, TFIPI

chuyOn Lam Son; Quing Nam: Nguydn Bd Hu)nh

Quang,l l/1, THPT Nguy6n Duy Hi€u, Di0n Bin.

NGUYEN XUAN QUANG

*Bii LZl358. Ngttdi ta quay m\t xilanh

thdnh mdng bdn kinh R cho d€'n vdn tdc gdc a6

rdi ddt viio gifia hai mdt ph'dng nghidng d= 45"

so voi phtong ngang (hinh l). H€ s6' ma sdt

gifia mdt phdng nghi€ng vd xi lanh lit 1t. Xdc dinh sA'vdng quay cia xilanh cho d€'n khi dtng

lai. Coi truc cila xi lanh dftng y€n khi bi hdm.

Hinh l

Ldi gitii. C6c luc tr{c dung lOn xilanh duoc bidu di6n nhu hinh 2.

Hinh 2 Dd truc cira xilanh drlng ydn thi

F * lr, + frz + F^1+ F^,2 =d

Chidu (1) lcn ffqc Ox:

-Psincl+N2+fl,,l=0

Chidu (1) lcn truc Oy:

-Pcosd+N1 -F^z=0

Mat khSc F^t = ltNr

Frrz = llNz

Tit c6c phuong trinh (2), (3), (4), (5) vi

a = 45o ta tim duoc:

n Pms(t+ P)

rmst=W.t\

(Xem ti{p trang 9)

(1) (z) (3) (4) (5) thay (6) reffid$ffi{- cltudiue s6 36218-20071 27

LING DUNG... (Tidp trang 2)

chinh gifa cira nfta dudng trbn di cho) (h' 6).

a.g;, <1M.AB= l-.r =l a ag-63!224 224

nuv-! 24a2 Dhngthtlc xiy ra khi vi chi khi

' l-a2

M=M^ oo'=L."2

Hodn rodn tuong tu ta cflng ,5

*7 4b2;1 1 = 4c2 dod6M=-f *-] - * ' l-c2 - '- 1-a2 1-b2 l-c2 > 4(a' + b2 + ,'1= 4.1 = 6. 2

Y4y M dat 916 tri nh6 nhdt bang 6 khi a2 = bz

=c2=!oo=b=c=Q.o22

Dd kdt thric bhi vidt chring t6i xin mdi c6c ban ldm quen v6i phuong ph6p dd trinh bhy 6 tr0n

qua c6c bdi tdp sau:

Bdi l.Cho a > b > c > 0. Chfng minh rang

J;-J-b < J;i .

Bdi 2. Cho a > c; b > c > 0. Chrlng minh rang

,lr@ - r) + JGO -.1 .Jou .

Bdi 3. Cho x, y, z, t ld ci{c sd thuc duong thu6c khoiing (0 ; 1). Chrlng minh rlng

x(l - y) + y(l - z) + z(l - t) + t(l - x) <2 .

Bdi 4. Chring minh rang vdi moi sd thuc

duong a, b, c, d ta ludn c5

{A . *W . O *'K* * a'l<u' * A

Bdi 5. Chung minh rang vdi moi sd thuc duong

a, b, c ta lu6n c5

lJt, +uz -Ja, +Cl<tb-ct.

Bdi 6. C}to a, b, c ld. c6c s6 duong thudc khoing ,. ftt

(0 ; 1) vd a + b +, ='u=' . Chung minh rang

2"

M= 7 *-l - *--l- >-6J;.

a(l-a2) b(l-bz) c(l-cz)

28 rSru,trroonTffiffi

PROBLEMS... QiaP ffang t7)

T61362. O is a point chosen arbitrarily inside

a triangle ABC. AO, BO and CO meet BC, CA

and AB at M, N and P respectively. Prove that

( ottp\( on.au\ ( cr.cN\thevalueofratiol- -rr thevalueofratiol- -rr

-rr-

i

, -bP )\ oM )l ot{ )

does not depend on the position of the point O'

T71362. Let M be a point inside a circle (O)

with center at O and radius R. Draw twochords CD and EF through M but not passing chords CD and EF through M but not passing

through O. The tangent lines with (O) at C

and D intersects at A, and the tangent lines at

E and F meets at B. Prove that OM and AB

are orthogonal.

FOR UPPER SECONDARY SCHOOLS

T81362. Let p > 2007 be a prime nqmber and

n is an integer exceeding 2006p. Prove that

q2oobn -Cfooo is a multiple of p, where o =l+l

Lp)

is the largest integer which is not larger thanL .

T91362. Le:t (u,) be a seque n". giu"n bypu, and the formula u,*t=Yl!-, ' l-u, where r > 0, where n = 1,2, ... . Given that ur. = ar, find

the value of t.

TlOt362. Suppose a, b, and c are the three lengths ofsides ofa triangle. Prove that

a b c - 3

Ja2 +3bc ..lb'+3"o Jc2 +3ab 2

T111362. Let P be a arbitrary pgnt in a

quadrilateral ABCD such that PAB , PBA,fda , F1B, fjb, fDa , FZD , and FD) fda , F1B, fjb, fDa , FZD , and FD)

are all acute angles.Let l+4, N, K, L denote the

feet of the altitude from P on AB, BC, CD and

Dl respectively. Find the smallest value of

AB BC CD DAthe sum the sum

-+-+-+-.PM PN PK PL

T121362, Let ABC.A'B'C' be a triangular prism.

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