Hinh 2
(Ban doc tu chrlng
minh bang cdch
dUng hinh binhhdnh BCDE (h.1), hdnh BCDE (h.1),
sau d6 sr? dr,rng
dinh li vd so s6nh
canh vh g6c ciratam gi6c ABE). tam gi6c ABE).
Btt6c2.D4t q = y (0 < y < n), AiBi = c,B;C;= at, Ci Ai = bi, (i e {1. 2,3].). B;C;= at, Ci Ai = bi, (i e {1. 2,3].).
Khi d5 ddy didu ki6n (*) duoc vidt lai lh:
ar+ b2-- azt bt= at*br (1)Ta sE chrlng minh rang ddy ding thtic (1) chi Ta sE chrlng minh rang ddy ding thtic (1) chi
duoc th6a m6n vdi at= a2= atYdbl= br= Sr.
a) Trudc hdt, gii sfr rang a1, a2, a,. d6i m6t khdc nhau vh do d6, theo (1) thi br, brvd b. cflng d6i m6t kh6c nhau. Kh6ng mat tinh tdng qu6t gii
dinh rdng ar 1 az< a., vd do d6 b, < b, < b, (2).
l, Sau d6, ta dung
mil g6c3y = y
rdi ldy c6c didm A'r, A'r, A'., tren
Cx vd. B'r, B'r,
B'., tr€n Cy sao
cho CA',= fi, vdCB',=4.
(i e {1, 2,3]l). Thd thi ta du-o. c
L A'i B'iC = L \BiC (c.g.c.), (i e { I, 2, 3l),
(h. 2). Theo c6ch dmg vh gii thi6t c6: A'1 B'y = A'zB'z = A'38'3 = AiBi= C (3). Ta sE chrlng t6 rang: A'18'1 I A'2 B'2 hodc
A'rB'r 1 A't B't.
Thdt vdy, tir (1) ta duoc: br- bt= a2- atvd
bt- br= a1 arnAn
A't A'z = B't B'2, A't A'i = B'z B'3 (4) X6t rf gi6c ldi A;EzBtA\ c6 A\A'2 = B'rB'z vd
tfi gi6,c ldi A! B'tB'zA'1 c6 A'1A'l.= B'rB'.r vir hai
cdp tia (A'rA'r, B'rB',)i (A'A'r, B'rB'r) ddu cat nhau 6 C. Vi vAy, theo bd dti 6 trcn thi ta c6:
A'rB'r) A'..8'l (5) vd A'.,8'., ) A'rB', (6)
X6t hai trudng hqp c6 thd xiy ra:
1) Ndu ar I btthi ADi < eiln suy ra
A', A\ g, .d{F, < 9oo < E\ A\1 vi dod6, trong AAi Bi A'3 tac6 A'z B'1 ) A'18'1 nOn d6, trong AAi Bi A'3 tac6 A'z B'1 ) A'18'1 nOn
ddi chidu v6i (5) ta duo. c A'1 B'1 1 A'z B'z Q). 2) Ndu ar) bt thi cflng ldp luAn tuong tu (ban doc tu vE hinh) ra duo. c f t B', gt < 90" <
T, Y, B\ vi do d5, trong A. A'1 B'1 B'2 ta c6A't B'z ) A'18'1 nOn ddi chidu vdi BDT (6) ta A't B'z ) A'18'1 nOn ddi chidu vdi BDT (6) ta
duoc A\ B'1 1 A'3 B'z (8).
Cric BDT (7) ho4c (8) thu duoc ttr gil dinh c6
c6c BDT nghiOm ngat (2), mAu thuSn vdi gii
rhidr cta bii to6n ld c6 d6y ding thrlc (3). Mau
thuSn d6 chri'ng t6 hodc ar = a2 = at yd do d6,
bt= bz= bt,hotrc it nhdt hai trong ba sd a; bang nhau (hay hai trong ba sd b, bang nhau),
(i e {1, 2,3lr).
b) Ta chung minh rang: Ndu hai trong ba s6 a,
(i e { 1, 2, 3 }) bing nhau thi ci ba sd a, d5 bang
nhau, do d6 ci ba sd b; cfing bdng nhau vi ba
tam gi6c AB,C, bang nhau. ThAt viy, ching han
gii sit a2= a3; thd thi tit (1) suy ra br= fi, vi
bay gid thi d6y ding thrlc (1) thu vd chi cbn mdt
ding thrlc d5li a, + b2= az* br (e).
Bdi todn quy vd chi cbn hai tam gii{c trong d6
didu kidn (*) dugc thay boi (9). Sau d6 cflng
chung minh tuong tu nhu phdn a) trOn co sd sit dung bd dd trcn thi chung minh duo. c rang
ar = a2 = at yd do d6 b, = b, = b., ta d16qc
dpcm.0
( Nnan x6t. l) Mot sd ban dua ra c6ch girli dai sd
theo hu6ng rau' Ap dinh li cosin cho M;B;C; (i e
ll,2,3l ta xem sdu sd duong a,, b,ld c6c nghi€m ciia he (fl gdm srlu PT sau: ar + bz= az t b, = q, * b, = i
vd al +fi *kaP1 = ,E+8+t"rrh = cfi,+$+kq\=c2
trong d6 k = -2cosf. Su dung tinh horin vi vbng quanh
ciia chi sd 1 -+ 2 -+ 3 -+ I vi tfnh chdt cta diy ti sd
bang nhau d6n ddn x6t tam thrlc bAc hai cua,t dung
f(k) = ak! + Bk + y, lronl d6 a, B, T ldn luot li cric da thrlc ddi xrlng ciia c6c dfii sd a1, a2, ar. Didu kiQn cdn vi dn dd t - -Zcosy ld nghi€m duy nhdt ciia h0 phuong
trinh (H) lh A = 02 - 4o'y - 0 <+ a, = e2= at (k6o theo br = bz- b.).
2) Bii to6n nhy thuQc loai kh6. Chi cd 22ban tham giagirii, trong d6 c6 m6t b4n ldp 7 vd m6t ban ldp 9. Dring girii, trong d6 c6 m6t b4n ldp 7 vd m6t ban ldp 9. Dring tidc li hdu hdt ddu giAi sai, hoac do v6i vhng ngQ nhdn, hoqc do tinh to6n nhdm ddu, hoac chi x6t trOn mQt cung trdn ctng d6n dOn ng6 nhAn.
NGUYEN DANG PHAT
*Bni T12B44. Xdt m\t lqtc gidc tii nei uep
ABCDEF. Eadng chdo BF cdt AE, AC ld.n luqt
tqi M, N. Euong chio BD cd.t CA, CE ti.n luqt tqi P, Q Duong chdo DF ciit EC, EA lin fuqt tai R, S.
Chi'ng minh rdng MQ, NR vd PS d1ng quy.
Loi gitii. Tru6c hdt, xin phrit bidu vd chrlng
minh hai bd dd.
Bd da' 1. Cho c6c sd duong a, B, a', B' th6a mdn di6u kidn a+ P = a'+ P < n; +g = tl1,
.
' slu:'B sinB'
Khi d6: a= a'vd B + p' .
Chfne minh.
Daty=n- (a+ f)=x- (a'+ fr') >0. X6thai
tam gi6c ABC, A'B'C' c6: A = o; h = p; e = y
vi .i'' = d, E,= f, ; e, = yra rhdy: e =e'
(cirng bang 7) (1)
Theo dinh lf sin vi theo gii thidt CA _sin0 _ sinf '
=7 4' (Z)
CB sina sina' C'B'
Tt (1), (2) suy ra: MBC <r> A,A'B'C' s a = di
0= F.
Bd dd 2. (dinh li Pascal) Cho luc gi6c ABCDEF
nQi tidp. AE cit DF tai S, BE cit CF tai X AC ci* BD tai P. Khi d6 S, X P thing hing.
Chune minh. Ap dqng dinh li C6va dang sin cho
cdctam gi6c)GF,BC (h.1), ta c6
r,
lsinXl sinEr sinfi _ ,
lsin& sinEz sinE
l'
lsin& sinG sinBr _ ,
.-- I
l.sin.li sinCa sin&
Mat kh6c, ta c6
le =e;A =e
i^ ^:
ffi=& i fr=Bq
(g6c n6i tidp cing chdn mOt cung)
T[ d6 suy ' ra: sin& sinXz - sin& sin& . Theo bd dd t, ta c5: fr =fr,f,, =fr * S,X,Pthinghing.
n. TOfrr) koa vH'rqol yR8
zo ""__,-_-__-"__ --*"""=.-**--*
348 (6-2006)
T16l4i viOc giAi bdi toiin.
Trudng hqp 1: AD, BE, CF khdng ddng quy
(h. 1).
Hinh I
Gii srl BE cit CF taiX BE cit AD taiY, AD cat CF IAiZ.
Theo bd dd 2, PS di qua X NR di qua Y, MQ di quaZ.
Ap dung dinh li C6va dang sin ldn luot cho cdc tam gi6c re,F, YAB, ZC D, ta c6:
sinX sinEr sin4
-l
sinXz sin& sinr9
sin)'i sin,4r sinBl
sinlz sinA sin&sinZ sinCr sin4 sinZ sinCr sin4
sinT+ sinG sin,D
Tt d6, vdi chri f rang:
E =D i F-z=&; A=fr ; 4=\; Q1 = 82;e:fr (4) (g6c ndi tidp cirng ch6n m6t cung) e:fr (4) (g6c ndi tidp cirng ch6n m6t cung)
Tt (3), (4) suy ra:
sin& sinli
sinXz sinY2
VAy, theo dinh li C6va d6o dang sin cho LWZ
thi PS, NR, MQ ddng quy.
Trudng ho.p 2: AD, BE, CF ddng quy tai didm I.
(3)
-1-1 -1
sinZ sinT+
Theo bd dc 2, PS di qua 7, NR di qua T, MQ di
qua 7. D
( Nnan x6t. l) C6 35 ban tham gia giAi vi ddu giAi
dfng.
2) Bii todn ndy l) mQt 6p dung hay ciia dinh li C6vadang sin. dang sin.
3) Xin nOu t€n m6t sd ban c6 ldi giAi tdt:
Hlr NQi; Nhfi Anh Tud'n, llA2, Hodng Vdn C*ong,
l0A1,Ti Ditc Phong,10T2, khdi PT chuy€n, DHSP Ha
NQi; Hi Tiryz Nguyan Ti€h Thdnh, 10T2, THP|
Nguy€n Trii, Hd Dong; Bdc Ninh: Ngti QuA'c Duy,
llAl, THFrf Li Th6i Td, TU Son; Nghd An: BiiVdn
Hodng, THCS Y€n Thlnh; Hh finh: Li NamTrurdng,
THPT chuy6n Hd finh; Di Ning: Nguydn Nhu Qu6'c Trung,THFT chuy0n LC Qu! DOn; Phri Yi,nt Nguyiin
Tudn Dfing, llA, THFrf Phan Chu Trinh, SOng Cdu;
\tnh Long: Chuong Cdng Danh,l0T1, TH[rf Nguyen Binh Khiem NGUYEN MINH HA Bni LU344. E Cho h€ co nlur hinh vd. Thanh AB cu'rtg rilt nh,e, ddu 4 A gdn vdo ban l€
sao cha thonh
v6i h =2.1(l - cosp) =
Visin8 -B"-lnen
I
cria vAt ld Er= LQx'f =2mx'2. 2'
Co nang ctra h€ b6o todn (b6 qua ma sdt) ncn:
2m'fi +
^c+= "l const . Ldy daohim hai vd hO
*lt
. DOng nang
thrlc niy ta c6: x" * $.
dao d6ng didu hba v6i
= 0; vit (trlc li thanh)
r;
tdn so g6c at =
l*
( Nnan x6t. Circ ban c6 ddp s6 dring (nhung lAp luAn
chua that chat chc):
Thdi Nguy6nl Trinh Quang Htrng, l2A, THPT chuyOn
Thrii Nguy0n; Bdc Ninh: Trdn Thdi Hd, Trdn Dtc
Trung, 12 Li, THm chuy€n Bic Ninh; Hi NOi: Nguydn
Ngoc Ldm, l0B chuy€n Li, DHKHTN - DHQG Ha NOi; Hrmg Yin; Trdn Anh Vfi, Pham Ditc Linh, l0 Li,
THPT chuyOn Hung YOn; Hii Duong: Vfi Tudn Anh,
1l Li, THPI. Nguy6n Tr6i; Thanh H6a: L€ Bd Son,
10F, THm Lam Son; Ngh€ An: Nguydn H6 Nhu f ,
Nguydn Anh Minh, Hodng Manh Hd, Nguyin Tdi Nghla,1043, THm chuyOn Phan B6i Chau; Ei Ning:
NguydnThi Qu6'c Todn, 10A3, THPT chuy€n LC Quf
DOn; Binh Dinh: Vd Hdo Nhon, Nguydn Hftu Qu6'c Dat, Nguydn ViAt Duy, 11 Li, THF'| chuyen Lc Quf
DOn, Quy Nhon;Vinh Phirc; Ta Duc Manh, 10A3, THPT chuyOn Vinh Phric; Bh Ria - Viing Tdut N guyin
Phic Hang,llAl, THFrf V6 Thi S6u, Dxt D6; Tiin Giang: Tr*ong Mai Thanh Td.m, l0 Li, THP| chuy0n Ti6n Giang'
MAI ANH
* gai L21344. Cho mach diAn nha hinh vd.
Bi€t R1 : lA,
Upg: 2l/,fu: 0,5Q. fu: 0,5Q.
Khi Ra: 0 {)
tki Ie: 1A; khi R4: 6Q thi Ie: 0 A;
khi R-t + @ thi U : ! A. Tlnlt R2, R3, R5.
7-
Loi gitii. (Theo ban Trinh Vdn Vuong, 11T,
THPT Lam Son, Thanh H6a)
rqflknggJgJggrmg 11 34S (6-2006) a t +trinr(P-) \2) Y2 E=*g I
Chu ki dao ddng ctra h€ li I = 2"ff O
cti thd quay theo moi htrdng, ddtt B gdn vdt cd
khdi laqng m. Ddy CD cr5 d0 ddi l, khong din
vo'i D id trung diAm cua AB. Kdo thanh AB lech
mQt gdc nho theo phtrong vu6ng gdc vdi AE rdi
tha ihe. Bo qua moi mo. sat, chang minh ring
thanh AB doo dQng di€u hda t,d t[nh chu ki dao
d|ng cua h€.
Loi gitii. Khi thanh lOch g6c a thi d6y OD llch
g6c F, D lOch kh6i vi tri cAn bang doan x ddn D'
vh B l0ch doan2x ddn B' (hinh vE).
Y\ a, p,,r li nh6 n6n thd nang cira vat lA (chon mdc tinh thd nang tai vi tri cdn bang): E, = m?h,
, -Ut -lr - lr - - - ,& , -ure - ,r- n, =&=&& , Upu lt - - - ,&
Khi & = 0C), mach diOn c5 dang nhu hinh 1, R2 Hinh I (Jue=tn(no+&)=0,5+R. 0,5 + Rs Ir=la*//2= Ta cflng c6 : R, . 0,5+& Rz z-(o,s+R) =1,5-& = && hay Rrft3 = 6 (2) Suyra:0,5+& =(O,S-R,)& (l) Khi R4 = 6 C), Ie = 0 A, mach di6n tr& thinh mach cdu cdn bang :
&
R3
a)&=10;R.=6f)tRs=00;
b) R, = 3 f) ; R,, =2 Qi Rs = 0,25 O. O
( Nna, x6t: Ngoii ban Vuong, c6c ban sau ddy c6
ldi giAi tdt vh ng6n gon :
Hrmg Ydn: Pham Dtc Linh, l0li,Trdn AnhVfi, l0 Li, THPT chuyOn Hung YOn; Tidn Giang: Trudng Mai Thanh Tdm, 10 li, THft chuyen Tidn Giang; Hii
Phdng: Vd Si C*dng,10 li, THm nang khidu Trdn Phri;
Binh Dinh: Nguy€nVi€t Duy,l l Li, THPf chuydn L€
Quf DOn; Phri Tho: PhingThi YAh Chi, THPT chuyOn
Hing Vuong; Khdnh Hoh: Ngd Hd Thdo, ll Li,
Nguydn Ngoc Ki Nam, 12 Li, THP| chuyOn Le Quf
DOn; Hii Duong: Vfi Tudn Anh, ll Li, THF f chuyen
Nguydn Thi Kim Khuy€n, ll Lf, THP'I' chuyen L€
Khidt; Son La LA Vfi Hdi, l0 Li, THm chuydn Son La;
Hlr NOi: Phan Dinh Dttc, 10A2, THPT Cao Bd Qu6t,
Gia Ldm; Nghd An: Nguydn Tdt Nghla, 10A3, Trdn
VdnThdng, l0A3, THm chuyen Phan B6i ChAu; Bdc
Ninh: Irdn Thdi Hd, 12 L(, THfrf chuy€n Bic Ninh;
Vinh Phtic: Vfi Ngoc Quang, 12A3, Nguydn Minh
Khtong, 11A3, THtvf chuyOn Vinh Phtic, D6 Duy
Ching, I lAl, THm Ng6 Gia Tu; Thanh Ho6; Trdn Hodttg Dai, llB3, THPT Tricu Son I, Ll Bd Son, l}F,
THPT Lam Son.
NGUYEN VAN THUA}.I
HINH THANH ... Qie:p ftans t2)
BDT niy cdn ilugc vit5t o d4ng sau:
ab(a - b12 + bclb - c)2 + ca(c--a)z >
> a21a-b11a--c) + b2@--c)(b-a) + c2(c--a)(c-b). I(6t hqp c6c BDT (10) vd (13), ta c6
(a+b+c)(a3+Ut+c3)
> 4(bcmo2 + acm62 * abm"2)2, (o' + b2 + c212 1t41
-Chgn d: p= y: I thay vdo (3*) ta.luoc:
(bc.m,)z +(ca.m)2 +(ab.m")z Z