. Nd uH tring v6 iA thi kdt ludn tr0n van
D6n tdi (R 2 oM2) 2Ol4.M A MA)
98. THCS Trdn Mai Ninh.
NCUYEN MINH DIJC
*nai T10/361. Ct'htlrng minh rting vdi moi
tam gidc ABC ta dilr co
-1 < 5cos,4 + 3cosi] + ZcasC <J (l)
Ldi gitiị (Theo ban Mac'Th{Trtrdng, 10A1,
THPT chuy€n Vrnh Phric vi nhidu ban khr{c). Nhdn x6t rang vdi moi tam gi6rc ABC. ta dtiu c6
cosA+cos B = 2rorA r B.o, A -B
= ọ
22
cosA+cosC>0, cosA > -1.V4y 6cosA+3cosB+2cosC V4y 6cosA+3cosB+2cosC
= 3(cos A + cos B) + 2(cos A +cos C) + cos A > -1 .
ding thirc xtiy La khi ta xem xdt bdt ding thric trong Itlp chc tam gi6c suy rong (kd cii tam giaic suy bien).
NGU\'EN VAN MAU
*nai Tll/361. Cho tctm giac ARC noi tiip
dudng trdn (O , R). Goi E lit trtútg cliAnt cLtu AB. TrAn conJ't Ắ ltiy diim f- scrct clto {.
^ -!
Ặ 3
Dtrng hinh binh hdnh AEMF-. Ch[rng rninh rdng l,4A t MB r 4/(' < ,/t t(Ri OM: ) .
Gia sr (() R) ld dm)ng. rrin c6 dinh. LIit1,dqrng tarn gidc ABC n6i ti)p (O , l?) sao cho dqrng tarn gidc ABC n6i ti)p (O , l?) sao cho
MA + MB + MC = .[( R-' -(r],11 .
Sd -l 6 r,e tr6i cua (1) ln tdi uu vi ta chi cdn
chonB=C=F,A=x-ZBth\
lim(6 cos A + 3 cos B + 2 cos C) 0+o
= lim(6cos(zr - 2p) +3 cos p +2 cos p) = -1 .
P+l)
Mat khdc
6cosA+3cosB+2cosC
= 3(2cosA + cosB) + ZcosC
I
< ; (9 + (2 ms A + cos 8.12) - 2(ms A cos B-sin Asin B) )
;
= - ')'(9 + 4cos2A + cos'B + 4sin4sinB) l
< :(9 + 4cosrA + cos2B + 4sin2A + sin2B; = 7. 2
Sd 7 d vd phii cia (1) li toi uu vi ta chi cdn
chon A=B=d,C=tr-Zath\
lim(6 cos A + 3 cos B + 2 cos C) ư0
= lim(6 cos s + 3 cos a + 2 cos(r *2a)) =7.
a-0
T* d6 suy ra BDT cdn chfng minh. C
(Nnan x6t. DAy li mot bii toiin chfng minh bdt ding thfc trong tam gi6c thuOc loai trung binh. M6t so ban cdn su dung ciich chrlng minh quen thu6c theo phuong phdp vectd. Tuy nhiOn, rdt it ban chi duoc r[ng c6c udc luong hai phia cua (1) lir tdi uu v) đu
Loi gitiị o Til gii thiet bii toiin, ta c6
't _ l_ AM-AE+AF =!AB+!eC.a1 Z) l__1_. nen AM -: .)\ ( eU +MBlt : ( aru -MC\ , -j\ I noy '6 I .^;+ 2 ME *! 3 MC =o (r)
Mar khiic R: =OA2 =(OA4 + \/ U,q\' = oM2 +z.blrt.twe+ MA2 .
D6n tdi (R2 - oM2) -2Ol4.MA - MA)I I
-I
I
e l( R2 - OM:\ _2OM._ MA : ! hIẠ
666
Ldp Iudn tudng tu, cfrng c6
I _-_ I I _(Rr _ OM: ) - 20M._ MB = _ MBz 222 1 - r- - r j (R, _ OM: )_ZOM._ MC = _ MC? aaa JJJ C6ng (2), (3), (4) theo ve vn đ f đn thu duoc ffiHS(* ;H+*th"* Sd 365 (t I -2002) (2) (3) (4) (1) ta 25
= (MA + MB + MC)2 .Suy ra
MA+MB+MC<lT:(n'-ont'l
Ding thrlc xriy ra khi vh chi khi MA=3MB=ZMC
Tr€n mat phang, dung tam gi6c đLr AtBCt, ram llr M,. Xdc dinh ctic didm Br, C, sao cho
r.-l-
M ,8. - ; M,g, . M,C: - '^ M,C, . DUng duong
,ro, ,^*3o , bd,n kinh n,zrgoul tiep tam gi6c
A1B2C2. Dtrng tam gi6c ÁtB'2C'2 lir inh cira tam gi6c AyB2C2qua ph6p vi tu tAm O,, ti s6
Ạ Cuoi cirng, dung tam gidc ABC ld'anh cua
Rr