Hay tim trong tam giac ABC mot diem M sao cho MA + MB + MC dat gia tri be nhat.. HUdng ddn gidi Cdch 1: Lci giai cua Toricelli , , Durng ba tam giac dcu ABC,, BCA,, ACBi ra phia ngoai
Trang 1Chuy6n dg BDHSG To^n gia tr| lan nha't g\& tr| nhd nhSt - Phan Huy Kh5i
S + S, 1 +
•S + S, =s \, S p 1 + — - + -^
s s^
c2
= S + 2 S , + ^ = > S , - 2 S , + ^
Lap luan tiTdng tiT, ta c6: S2 = ZSj + —
,-s
S3 = 2S3 +
Tir (2) (3) (4) suy ra:
P= S , + S 2 + S 3 = 2 ( S , + S 2 + S 3 ) + l(sf+S^+S^)
Vi: ^ + ^ + ^ = l , n c n l a c 6 :
3
(2) (3) (4) (5) (6)
(difa vao nhan xet hicn nhien sau: Ne'u a + b + c = 1, thi a^ +b^ >
-3' Dau bang xay ra <=> a = b = c = ^)
S S S 1 Da'u bang trong (6) xay ra <x> ^ = ^ - ^ = - o O la trong tarn AABC S S S 3
Vict lai (5) du'di dang: P = 2S +
Tilf (6) (7) suy ra: P > 2S + - - - S
3 3 7S
P = — o CO dau b^ng trong (6) o O la trong tarn AABC
7S Vay min P = — o O la trong tarn AABC
Bai 2 Cho tarn giac dcu ABC canh a M la diem tily y n^m ben trong AABC
Xet tam giac c6 3 canh la MA, MB, MC Goi ^ la dien tich tam giac ay
Tim gia trj Idn nha't ciaa <S
Hudng ddn gidi
Goi S la dien tich tam giac dcu ABC canh a, thi S = (1)
Qua M ve ba doan thang A|B,, B 2 C 2 , A 3 C 3 tifdng iJng song song vdi AB, BC
CA (xcm hinh ve)
,atBB| = x ; B , C 3 = y ; C 3 C = z
De lhay A 3 B B, M , CCjMA,, A A 3 M C 2
cac hinh thang can , ; tfdoco:
•iB = A 3 B 1 ; A M = A 3 C 2 ; CM = C 2 B , „
Vay lam giac co ba canh MA, MB, MC cung chinh la A A 3 B 1 C 2
Do cac tam giac MB.A, MA.C va M B 1 C 3 la ^ cac tam giac deu vdi canh lan liTdt la x, y, z, nen neu goi S,, Sj, S3 tiTdng ufng la dien tich cua chung thi:
S1+S2+S3 ^S; ^S2 ^S3 ^ fy^ ^ (z^
(do a = X + y + z)
• x^+v^+z^
Tirddtacd: S , + S 2 + S 3 = ^ S
x^+y^+z^
(x + y + z)^
(1) Tac6:<^= SAA,B|C| ^ S A ^ A J C J +S^\MB,C2 +SAMA,B,
2 ( S M A , A A , +SMC2CC3 +SMB2BB1
=irs-(s,+s2+s3)
r
Hien nhien ta c6: x^ + y^ + z^ > - ( x + y + z ) ^ nen tir (1) suy ra:
s'l +S2 + S3 >is
\ a Da'u b^ng trong (3) xay ra x = y = z = -
T C f ( 2 ) ( 3 ) s u y r a : ^ < | (4) Da'u b^ng trong (4) xay ra o c6 da'u bang trong (3) •
o x = y = z o M l a tam cua A deu ABC
(2)
(3)
DoS = a'V3 minS = ^ ^ o M la tam cua AABC
4 12
357
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Trang 2Bai 3 Cho lam giac ABC c6 AB = c, AC = b va A=n Xet lap hdp cac du, • " " ^ « n i
thang A qua A va khong U-ung vc'ti hai canh AB, AC Goi P la tich khoang each if
B va C tdi A Tim gia tri Idn nhat ciia P yi^ \, Hii(fng dan gidi
DiTcJng thang A qua A chia thiinh hai loai:
Nhom I:
Dirftng A cii doan BC Dal BAH = 3
Kc B H l A, C K I A
Khido: ' ' '
P = BH.CK = bcsinpsin(a - p) ' ;
2 cos(a-2P)-cosa
cua gc)c
Dau bilng irong (2) xay ra '
<=> cos(a - 2P) = 1 <=> P = Y o A lii phan giac Irong cua goc A
Vay trong cac di/ctng thdng thuoc nhom 1, thi du'clng phan giac trong
A la dai li^ctng P dcU gia tri \&n nhat va gia tri do bSng: P| = be sin^ y
Nlwm 2:
Difilng A qua A va citt doan BC, & day C la diem dol xuTng cua C qua A Kc
BH va CK cimg vuong goc A
Kc C'K" 1 A =^ CK = C'K' =:> P = BH.CK = BH.CK'
Ap dung li luan phan 1 vao lam giac
ABC", la lhay tich BH.CK'(cung la tich
BH.CK) li'm nha'l khi A la phan giac ,
trong ciia goc BAC' (luTc A la phiin giac
ngoai ciia goc A) va gia trj Idn nhat do
hang p = be sin = bccos —
2 2
Theo nguycn li phan ra, thi:
maxP = max{Pi; P:} = max|bcsin^y;
u 2 «
bccos —
bccos^ —, ne'u A la goc nhon (o < a < 90") bcsin^ ^ , neu A la goc iD (90" < a < 180")
^bc, ne'u A la goc vuong (a = 90") NhiT vay ne'u A la goc nhon, ihi maxP = bccos^ y khi A la phan giac ngoai cua goc A va ne'u A la goc tu, thi maxP = bcsin^ y, khi A la phan giac trong cua g6c A va ne'u A 1^ goc vuong thi maxP = - b e khi A la phan giac trong
Nhqn xet: Trong bSi tren da stf dung nguyen li phan ra cua bai loan tim gia tri
Idn nhat va nho nhat, di nhien ke't help vdi cac lap luan ve hinh hoc phang c6 iJng dung lifdng giac!
Bai 4 (Bai toanToricelli) Cho tarn giac ABC c6 max {A, B, C} < 120" (tiJc la mpi goc cija tarn giac deube hdn 120")
Hay tim trong tam giac ABC mot diem M sao cho MA + MB + MC dat gia tri
be nhat
HUdng ddn gidi Cdch 1: (Lc(i giai cua Toricelli) , ,
Durng ba tam giac dcu ABC,, BCA,, ACBi ra phia ngoai tam giac ABC De tha'y ba dUcJng iron ngoai ttep cua ba tam giac deu ay dong quy tai diem T,
va T la 3 diem nhin 3 canh cua tam giac BAC dUdi ba goc b^ng nhau va
hlng 120" Do CJTA = ABCJ = 60",
nen Afc = 120"=^ C,, T, C th^ng hang Li luan tiTdng tir c6 B, T, B, thing hang va A, T, A| cung thang hang La'y diem I tren TB|, sao cho
TA = TI i=> AAB,I = AACT (c.g.c)
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Trang 3=> B|I = C T => T B + T A + TC = B T + T I + I B , = BB| (1)
Lay diem M baft k i thuoc mien tam gidc A B C Ta se chiJng minh rkng:
M A + M B + M C > T A + T B + TC , „: (2)
That vay diCng lam giac A M E sao cho E cf nufa m a t phang hci A M khong chuTa B
Ro rang A A M C = A A R B , => M C = EB, => M A + M B + M C > T A + T B + TC
:=> (2) diing Dau bang xay ra M = T
Vay d i e m M can t i m chinh la d i e m T n 6 i tren D i e m 66 thi^dng g o i 1^ diem
" T o r i c e l l i " • '
Cdch 2: Lay M la diem tijy y trong mien lam giac A B C J^'i ,
Thirc hicn phep quay R" ( B , 6 0 " )
K h i d o : R - ( B , 6 0 " ) > ' 1 ' V V , , , ,
C - y A , *
M ^ M '
Then tinh cha'l cua phep quay suy ra M B M ' la tam giac d c u => M B = M M '
N g o a i r a d o : R - ( B , 6 0 " )
M C ^ M ' A , , * vfiy M C = M ' A ,
V a y M A + M B + M C = M A + M M ' + M ' A , (1)
'' 7jy (*) jjnj^ diTcfng gap khiic, ta c6:
M A + M M ' + M ' A , < A A , (2)
Nhir vay tir (1) (2) ta da chi^ng minh difcfc rhng v d i m o i vj tri diem M thuoc
mien A A B C , ta luon c6: M A + M B + M C < AA,
Da'u bang xay ra o A, M , A2 thang hang
Gia sur Mo la vj tri cua M ma A, M„, A, thang hang
(3)
D o B M o M o = 6 0 " A M „ B = I 2 0 "
Theo tinh chat cua
phep quay ta c6 goc
tao bdi M,)C va M„A|
bkng 60"
=j^MoMoC = 6 0 "
B M o C = 120"
160
Civ !NMM M IV riWII Kh.il"! Viet_
Vay M(i la diem nhin ba canh tam giac durdi ba goc bling nhau = 120", tuTc Mo
la d i e m Toricelli phai t i m
C^ch3: ^ ^ :r m- y
R6 rang ta c6 bd de h i c n nhien sau day (vi the chung toi bo qua chtfng minh)
N c u QNP la tam giac deu, Ihi v d i m o i vj t r i cua d i e m M nam trong mien tam giac QNP, ta luon c6 long khoang each tif M tdi ba canh cua tam giac QNP
la hang so' (c6 the tinh di/dc ngay hang so nay bSng chinh chicu cao cua tam
^ giac d c u QNP)
R A p dung bo de ay ta giai bai toin Toricelli nhifsau:
• Do A A B C CO max(A, B, C) < 120", nen ton tai duy nhat d i e m T nhln ba canh
B c u a tam giac dvtdi ba goc bang nhau, tuTc la: A T B = B T C = A T C = 120"
• Q u a A, B, C dirng ba
Hdi/dng vuong goc
• T A , T B , TC Ba
Hdirdng nay c^t nhau
^ t a i Q, N, P De thay
I QNP la tam gi^c deu
L a y M la d i e m tiiy y trong A A B C G o i h , ,
h2, hi la khoang each
tijr M tdi ba canh NP,
PQ, Q N cua AQNP
M Ro rang ta c6:
m. M A + M B + M C > h| + h2 + hj
H Theo bo de neu tren (ap dung vao tam giac deu QNP), ta c6:
• h, + h2 + h j = T A + T B + T C
• Tir do suy ra: M A + M B + M C > T A + T B + T C (*)
• Da'u b^ng xay ra trong (*)<=> M s T Do chinh la dpcm
m:(ich4:
Sis dung b6 &G hien nhien sau day (ChuTng minh ra't ddn gian va x i n danh cho
cac ban doc)
B6 de: Gia si^ A M B = B M C = C M A = 120" va M A = M B = M C th i:
M A + M B + M C = 0 - :
Do gia thie't max(A B, C) < 120", nen t6n tai duy nhaft d i l m T trong tam giac sao cho A T B = B T C = C T A = 120"
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Trang 4ChuySn de BDHSG Toan g i i tri Ion nhat va g i i Iri nh6 nha't - Phan Huy KhAi
' , TA TB TC
An dung bo de suy ra: 1 1 = 0
' • ^ <i T A TB TC
(*)
TA TB TC
Do do dai cua ba vccUl , — , — dcu bang 1
TA TB TC Lay M 1^ diem bat ky trong mien tarn giac ABC, ta c6:
K^ro y^Ar^ MA.TA MB.TB MC.TC
M A + MB + MC - + +
TA TB Su" dung cac bat dang thtfc hien nhien sau:
TC
1 a|+a2+ + a„ < + «2 + +
2 a.p <|a| p ; ta c6;
MA.TA MB.TB MC.TC MA.TA MB.TB MC.TCi,
• + + > + + •
T A TB TC TA TB TC (2)
Delhay: VF(2) = ( M T + T A ) T A ( M f + T B ) T B { M T + T C ) T C
T A
= M T T A TB TC
TA TB TC
TB ^ TC + T A + T B + T C (3)
(4)
T i i f ( * ) v a ( 3 ) s u y r a : VF(2) = TA + TB + TC 'i
Ttr (1) (2) (3) (4) suy ra: M A + MB + MC > TA + TB + TC
Nfuln xet: Da'u bang xay ra <=> M s T Do chinh la dpcm
Cung gio'ng nhu" trong cac bai loan dai so', giai tich, v6'\c bai toan tim giii
trj Idn nha't, nho nha't trong cac biii toan khac noi chung (trong hinh hoc
phang noi rieng) cflng c6 rat nhicu each gitii khac nhau Bai toan tren la mot
vi du dien hinh minh chufng cho tinh da dang ciia cac phiTdng phap tim gia tri
Idn nha't va nho nha't ciia mot dai lufdng cho tri/dc
B a i 5 Cho tam giac ABC vdi a > b > e, d day qua a, b, c ki hicu do dai cac canh
BC, CA, AB ti/dng uTng Gia suT M la mot diem di dpng tren diTctng tron ngoai
tie'p tam giac ABC Goi x, y, z Ian liTdt la khoang each tif M den cac canh
BC.CA, A B
Tim gia tri be nha't cua dai ii/dng S = — + — + -
X y z
HUdngddngidi
Goi (~(f) la dircJng tron ngoai tie'p AABC
K h i d 6 ( ^ ) = B C U C A U A B (1)
62
Cty TIMHH MTV DVVH Khang Vi^t
Gia sur M e BC
Goi H, I , J ti/(tng i?ng la hinh chic'u cua M xuong AB, BC, CA
Khi do ta eo: M I = x; MJ = y; M H = k
Gia siJ K e AC sao cho AB = CK (do a > b > c) *
G o i L = M K n BC „
De thay A B L M - AACM ^ - = —
y X
p a cung CO ACLM - A A B M => - = —
z X
T i r ( l ) ( 2 ) = ^ ^ + ^ = - ^ l i l ^ = i
r^ - o a b c a a 2a
Do vay S = — + — + - = — + — = —
X y z X X X
(1)
(2)'
Nhuf vay min S = min
MeBC MeBC
^2a^ 2a 2a
max M l M,)Io
MeBC
d day Mo la Irung diem ciia BC
Ta c6: M,,!,, = BI„tan MoBI,, = t a n ^
-Tiif do suy ra: min S = —
MeBC a
2a tan
2 2
, A
= : 4 C 0 t —
A 2 Lap luan ttfdng tu", co:
i B C ' min S = 4cot— va min S - 4 c o t —
MeAC 2 MeAB 2 Theo nguyen l i phan rii, ta co:
min S = min min S, min S, min S MeBC M6AC MeAB
Tif (3) (4) (5) suy ra:
min S = min 4col—,4cot—,4cot— A B C
2 2 2
Do a > b > c =^ 180" > A > B > C > 0
9 0 0 > ^ > - ^ > | > 0 ^ c o t - < c o t - < e o t -
(3)
(4)
(5)
(6)
(7)
Vav lir (6) (7) di den: min S = 4cot— 44 M la trung diem cua BC
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Trang 5Ctiuygn de B D H S G Tcrin gia tf| I6n nha't g\i trj nh6 nliil Phan Huy KhJi
Bai 6 Cho hlnh tron ban kinh r Xet ta't ca cac tvt giac ABCD ngoai tiep dtfcj^
tron Tim gia trj nho nhat cua dai luTOng P = AB + CD ,
Hitdng ddn gidi
Goi M la tam dtfSng tron npi tiep ti? giac => M
giao diem cua cac diTdng phan gi^c trong cua cac
goc A, B, C, D cua ti? giac
Ve di/dng tron ngoai tiep tam giac ABM va gpi
ABN la tam giac can npi tiep c6 dinh la N sao
cho ANB = AMB
Gpi h la khoang each tif N xuong AB, con h, la
khoang each tir M xuong AB Khi do ta c6:
h > hi va hi = r
Ta CO AB = 2htan = 2htan
Dau bang trong (1) xay ra o h = hi
o M each deu A va B
Dau h\ng trong (2) xay ra o M each deu C va D
2 2 Dau bkng trong (3) xay ra o dong thcfi c6 dau bkng trong (1) (2)
AMB CMD
tan CMD
- Vi theo bat dang thiJc Cosi, suy ra:
^ AMB CMD
tan 1-tan
1
364
AMB Dau bang trong (4) xay ra <=> - ^ — = 4 3 Tir (3) (4) di den: P > 4r - (5) Dau bang trong (5) xay ra <=> dong thcti c6 dau bang trong (3), (4)
M each deu A, B; M each deu C, D va AMB = CMD = 90"
o ABCD la hinh vuong ngoai tiep dufdng tion ban kinh r da cho
Tom lai min P = 4r Gia tri nho nhat dat du-dc khi v^ ehi khi ABCD la hinh vuong ngoai tiep diTdng tron ban kinh r cho trifdc
Bai 7 Xet tat ea cae ti? giac ABCD ehi c6 duy nhat mot canh Idn hdn 1 Tim gia tri Idn nhat cua S, d day S la dien tich tu* giac ABCD
Hiidng ddn giai ^
Gia siJ AD > 1 Khi do ta c6: AB < 1, BC < 1, CD < 1
Dat AC = X va gpi M la trung diem cua AC
Ta eo: A B ' + B C ' = 2BM^ +
2
AC^
=^2BM^+ — < 2 = > B M < - V 4 - x ^ 2 ~ 2 (1)
Do AC < AB + BC < 2 0 < X < 2 Dau b^ng trong (1) xay ra o AB = BC = 1 Ke BH 1 AC Ta c6:
Dau b^ng trong (2) xay ra o H s M o AB = BC
Dau bang trong (3) xay ra o dong thcfi c6 dau bang trong (I) (2)
o A B = BC=I
T a c 6 : S A C D = -CA.CDsinACD<-x
Dau b^ng trong (4) xay ra o CD = 1 va AC 1 CD
(4)
Tir (3) (4) ta c6: S = SASBCD = SABC + SACD < ixVTI x^+ix, 2
Dau b^ng trong (5) xay ra o dong thdi c6 dau b^ng trong (3), (4)
o AB = BC = CD = 1 va ACD = 90"
365
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Trang 6ChuySn gg BDHSG Toan g\i trj Idn nhS't va g\i tr| nh6 nha't - Phan Huy KhSi
_ , 2x + x V 4 - x ^ x/, , , r f\
Ta co: = + 1 + V4 - x"^ j
4 4
A p dung ba't dang thiJc Bunhiascopski, ta c6:
(l + l + V 4 ^ ) < 3 ( 2 + 4 - x ^ ) hay 2 + V 4 ^ < 7 3 7 6 ^
T v / - / ; \ / ' 7 \x + x \ / 4 - x ^ 73 fZ T
Tir(6) (7) suyra: <J!_.xV6-x^ ' ,
4 4 Da'u bang trong (8) xay ra <=> c6 dau bang trong (7)
o 7 4- x ^ = l o x - V 3
Lai iheo ba't dang thiJc Cosi, ta c6;
(6)
(7) (8)
cVri-x^ < = 3 (9)
Da'u bang trong (9) xay ra o x' = 6 - x^ <=> x = N/3
/3
Tir do ket hcJp vdi (5) co: S < — 7 3 = - (10)
4 4
4i Ket hcfp lai ta c6: S = - o
A B = B C = C D = : 1
A C = N/3
A C I C D
o A B C D la nufa luc giac deu canh 1
T6m lai max S = — o A B C D la nii-a luc giac deu canh 1
Bai 8 Tren mSt ph^ng cho hai diem A, B v6i A B = d Xet tap hdp tS't ca cac
hinh vuong sao cho A va B nam tren cac canh cua hinh vuong ay Goi / la tong
cac khoang each tiT A den cac dinh cua hinh vuong T i m gii tri nho nha't ciia /
HUdng ddn gidi
Gia sijf CDEF la mot hinh vuong tuy y canh a va gia suT A e CD
T a c o : A C + A D = a
A E > a , A F > a
Do do canh \dn nha't trong cac doan AC, A D , AE,
A F la A E hoSc A F (thi du la AF)
Ro rang trong cac diem tren bien cua hinh vuong
thi diem each xa A nha't phai la mot trong cac
dinh
Tir do suy ra:
A B < m a x { A C , A D , AE, AF) = A F (1)
Cty TNIiH MTV DV'VH Khang Vigt
Hicn nhicn ta c6: A B < CE = aV2
TiS do suy ra: AC + A D = a > -A B 72 I 72 do A B <a
A E > a > A B
7^-TiJf (1) ta c6: A F > A B , nen ket hdp lai suy ra / = AC + A D + A E + A F >4^ + ^ + A B = A B ( I + 72) = d ( l + 72)
72 72
Nhir vay ta di den: / > ( l + 72)d ' "(2)' Da'u bang trong (2) xay ra o A B la diTcJng chco cua hinh vuong
Tom hii: m i n / = ( l + 72)d M
Gici trj nho nha't dat diTdc khi A B la diTdng cheo cua hinh vuong
i 9 Cho hinh vuong A B C D canh bang 1 Diem M va N Ian liTdt di dpng tren
AB va CD sao cho M B N -=45*' Gia sijf S la di?n tich tarn giac B M N Tim gia tri Idn nhat va nho nha't ciia S
Huding ddn gidi
Dat A M = X, CN = y, nhiT vay x, y e [0; 1 ]
Do M B N = 45" =i> a + 3 = 45" tan((v + 3) = 1 : A *
tan (x + t a n 3 , x + y
= 1=^ ^ = 1 1 - x y
l - t a n a + tan3
> x + y = 1 - xy ••.yy.A, (1)
Ta c6: S = SBMN = SABCB - SACM - SBCN - SMDN
A x M
_ ^ X y ( l - x ) ( l - y ) ^ 1 - x y
~ 2 2 2 2
D a t t = xy Khi do ta c6: S = 1 - t (2)
T a c o t C f d ) : [x + y 1 - t
[xy = t nen theo dinh li Viet x, y la hai nghi^m thoa man dieu kien 0 < x, < X2 < 1 cua phiTdng trinh:
De (3) CO nghiem nhiT vay, ta can c6:
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Trang 7ChuySn dg BDHSG Jo&n gia tr| Idn nha^t vi g\i tr| nh6 nhS't - Phan i h r y i.ii
f A > 0
( X , - 1 ) ( X 2 - 1 ) > 0
X| +X2 <2
X|X2 >0
X, +X2 >0
<=>
- 6 t + l>()
X | X 2 - ( X | + X 2 ) + 1>0 (*)
X| +X2 <2 ',
X | X 2> 0
Ap dung dinh li Viet vdi (3) ta c6: x, + X2 = I - t; x,X2 = t nen
t^ - 6 l + l > 0
t - ( l - t ) + l > 0 t
l - t < 2 <=>0<t<3-2V2 (3)
t > 0 , , „
l - t > 0
( * ) o
Tir(2) (3) suyra: V 2 1 < S <
S = - o t = 0<=>
2
X = 0; y = 1
x = l ; y = 0
M = A ; N = D
M = D ; N s C
o M , N tiTdng tfng la chan diTdng phan giac cua ABD, DEC
1 Vsiy maxS = - <=> M = A,N = D
minS= V ^ - I o M = M„;NsNo,
cl day BM,, va BN,, Ian liTdt la phan giac cua cdc goc ABD va DBC
Nhgn xet: Ta c6 each giai khac bai toan tren nhiTsau: (diTa vao li/cfng giac)
1
Ta c6: S = SBMN = - B M BNsinMBN
2
2 cosa cosp 2
4i
>/^ + 2cos(a-(3) (do a + p = 45")
Vi a, p e [0; 45"], nen - 4 5 " < a - p < 45"
V2 ^' ^
=> — < cos(a - p) < I
2 cos(a + 3)-cos(a-P)
(5)
(6)
368
Tir (5) (6) suy ra: 2 + V2 242 2
hay V 2 - 1 < S < ^ ' (7>
Lai c6: S = - o cos(a - p ) = l < = > a = P = 22"30'
2
42 :
2
"M = D ; N = C
M = A; N = D
S = V 2 - 1 « c o s ( a - p ) =
a - p = 45"
a - p = -45" <=>
a = 45";P = 0
a = 0;p = 45"
Ta thu lai ket qua tren! :S M
Nhdn xet: Mot Ian ni^a ta thafy du'dc tinh da dang cua cic phtfcJng phap d6ng de
giai bai toan lim gia tri Idn nha't va nho nha't trong cac linh vifc khac nhau:
dai so, giai tich, so" hoc, hinh hoc, liTdng giac
ai 10 Cho nufa di^cfng Iron bdn kinh R, di/dng kinh AOB C la mot diem tily y
tren nufa di/dng tron khong trung vdi A va B Trong hai hinh quat BOC va AOC ve hai di/dng tron npi tiep Gpi M v^ N 1^ hai tiep diem cua hai diTdng tron ay vdi diTcJng kinh AB cua nuTa diTcJng tron da cho DSt 1 = MN Tim gia
tri nho nha't cua 1 • >
HUdngddngiai
Gpi Oi, O21^ tarn ciaa hai di/cJng tron
Dat CON = 2a (nhir vay 0 < a < 90")
D 3 t M 0 , =R; NO2 = R2
De thay O ^ = ^ C O N - a ,
O.OM = - C O M = 90" - a
2 A
Ta CO / = M N = OM + ON :p R,cot(90" - a) + R2Cota
= Rjtana + R2Cota (1)
Trong tarn gidc vuong 0|M0, ta c6 R) = OO|Sin(90" - a) = (R - Ri)cosa
Rcosa
=> Ri(l + cosa) = Rcosa =:> R| = Hoan to^n ti/dng tiT, ta c6 R2 =
1 + cosa Rsina
1 + sina
(2) (3)
T i r ( l ) ( 2 ) (3) suy ra/ = Rcosa sina Rsina cosa
1 + cosa cosa 1 + sina sina
369
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Trang 8R s i n a R c o s a sina + coscx + l
• H =
K-l + cosa 1 + sina (1 + sina)(K-l + cosa)
2 cos
-= R
a a sin - + COS ~
2cos2^
2R
a a
sin + C O S
2 2
a ( a
cos 2 sin + cos
- (3)
sina + cosa + 1
Do sina + cosa = yfl cos(a — 4 5 " ) , ma 0 < a < 90"
=> 0 < sina + cosa < yfl ,\\e tiT (3) c6 :
2R
• = 2 R ( N / 2 - 1 )
N / 2 + 1
/ = 2 R ( N ^ - 1 ) <!=^ cos(cY - 4 5 " ) = 1
< ^ a = 4 5 "
* <?>C\A triing diem ciia A B
Vay min/ = 2 R ( N/2 - 1), khi C la diem chinh giffa cua A B
E Diem qua rngt so bai todn tim gid tri Idn nhd't, nho nhd't trong
luang gidc
Bai 1 Clio tarn giac A B C T i m gia trj nho nhat cua dai liTdng sau:
P = tan^ A + tan^ ^ + tan^ ^ - tan^ ^ t a n ^ ^ t a n ^ ^
2 2 2 2 2 2
HuHng ddn giai
Thco bat dang ihu-c Cosi, ta c6 tan^ — + tan^ — > 2 t a n — t a n —
2 2 2 2
tan^ — + tan''— > 2 tan—tan—
tan
Tir do suy ra:
-2 A 2 B
2
B B
2
2 A
2
^ +tan^ — > 2 t a n —tan —
2
C
I —
2
2
A
2
tan ^ + tan^ ^ + tan'' ^ > tan—tan — + tan — t a n — + tan — t a n —
Trong moi tarn giac A B C , thi VP ( 1 ) = 1, nhiT vay ta c6 :
(2) tan^ ~ + tan^ — + tan^ — > 1
2 2 2
Cty TNHH MTV DVVH Khang Vigt Dau bSng trong (2) xay ra <=> A = B = C
L a i Iheo ba't dang thuTc Cosi, thi
A B B C , C , A ^ J 2 A , 2 B 2 C
1 = tan—tan — + tan—tan — + tan —tan —>3:Vtan —tan —tan —
2 A 2 B 2 C 1
tan'' — tan —tan — < —
2 2 2 27
26
Da'u bang trong (3) xay ra o A = B = C Tif (2) (3) c6 P > —
Da'u b^ng trong ( 4 ) xay ra o dong thcJi c6 da'u bang trong (2), (3)
o A = B = C
26 Nhi/vay minP = — o A B C la tarn giac deu
Bai 2 T i m gia trj Kin nha'l cua bicu thuTc P = '^"'^ cos —cos— d day A , B, C B C
2 2
la ba goc cua mot tarn giac A B C
Hitfing ddn giai
Chu y rhng P > 0 vdi moi tarn giac ABC Tif do ta c6 maxP = VmaxP^
, ^2 2 A B C I
T a c o P = cos —cos—cos—= —
2
A
2
2 2
, 2 A
1 - sin —
2)
cos-B + C
• + cos -B - C ^
\ 2 A V A
1- s i n — sin — + cos B - C ^ l
2 2
D o c o s ^ — ^ < 1 , n e n t i r ( 2 ) c 6 P ^ < | l s i n ^
, A ^
1 + sin —
Da'u bang trong (3) xsiy ra o cos
Ta CO —
2 1- s i n ^ - ^ 2
, • A l
1 + sin —
2 j
B - C
2
1 ^
= 1 o B = C
A \
2 2 s i n
-2 A
, • A
1 + sin —
2
I • A
1 + sin — Tijr do theo ba't dang thuTc Cosi, c6
1 - s i n ^ A
1 - s i n ^ A ^
2) A;
2)
, A)
1 + sin —
2)
^ • A ' ( • A ^ (, • A ^
2 - 2 s n - + 1 + sin + 1 + sin
1 I 2> I 2 ; I 2)
n3
1 + sin —
2
4
,116
Da'u b^ng trong ( 4 ) xay ra o 2 - 2 s i n — = 1 + s i n — <=> s i u y = -
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Trang 9Chuyen dg BDHSG Join gii trj I6n nha't vh gia tri nhd nhat - Phan Huy KhAi
Tfif (3) (4) di den 27
P ^ = - ^ < : > B = C v a s i n ^ = i
27 2 3
Vay tiTCl) ta c6 m a x P = 16 4N/3
27
<=> A B C la tarn giac can dinh A vdi A = 2arcsin ^ \, 'ii
Bai 3 Xet cac tarn giac A B C thoa man he thtfc tanA + tanC = 2tanB
T i m gia tri Idn nhat cua dai liTdng S = cosA + cosC
HUdng ddn gidi
T „ ' A , \ r sin(A + C) 2sinB
Ta CO tanA + tanC = 2tanB — =
cos A cos C cosB
=> cosB = 2cosAcosC (do sinB = sin(A + C)^Q)
=> cosB = cos(A + C) + cos(A - C)
=> 2C0SB = C O S ( A - C) ,
L a i CO S = cosA + cosC
- A + C A - C - B l + c o s ( A - C )
= 2cos cos = 2 s i n — J
2 2 2 \
T i r ( l ) ( 2 ) s u y r a S = 2 s i n | ^ i ± ^
Tur (3) suy ra S > 0, nen maxS = V m a x S ^
T a c o S ^ = 4 s i n 2 - i i ^ ^ = ^?::ii^^i?Ki±l£^
2 2 2
V i the theo bat dang thiJc Co si di den
(1)
(2)
(3)
(4)
• ( 2 - 2 c o s B ) + (l + 2cosB) n2
8 •
= ^ c : > 2 - 2 c o s B = l + 2 c o s B o c o s B = -
8
TiT do ta CO max S = <=>
]_
4
c o s ( A - C ) = 2cosB cosB = —
4
<=> 3
B = arccos—
4
Cty TNHH MIV DVVH Khang Vijt
Bai 4 T i m gia tri nho nhat cua dai lifdng S = tan'' — + tan^ — + tan'' —, d day A,
B, C la ba goc ciaa mot tarn giac ' * ' *' ' '
Hiidng ddn gidi
- , T A ^ B , B i C T C ^ A
De lhay S > tan — t a n — +tan — t a n ' — + tan- —tan" —
2 2 2 2 2 2
I Da'u b^ng trong (1) xay ra <:5> A = B = C
A B B C C A Dat X = t a n y t a n - j , y = tan — t a u y , z = t a n y t a n y thi x + y + z = 1
( c h i i y x > 0 , y > ( ) , 7 > 0 )
Apdungba'tdangthuTcBunhiacopskichohaiday Vx, ^ y
^^ - r - ;
I ta CO ( X + y + z)(x' + y ' + z') > (x^ + y^ + v?)^
(1)
x^ + + z' > (x^ + y^ + 7})'
2
.»1,
, 2 2 2 ( X + y + z) 1
L a i CO x-" + y + z > =^ = -
TCf do suy ra x ' + y^ + z^ > ^
Dau bang trong (4) xay ra o dong thcfi c6 dau b i n g trong (2) (3)
o A = B = C
ViU
(2)
(3)
(4)
T i r ( l ) ( 4 ) CO 9
S = i o A = B = C
9
1 Vay minS = ^ la tarn giac deu
Bai 5 X c t cac tarn giac ABC vdi A la goc Idn nha't
T i m gia t r i Idn nhat cua dai lifdng S = sin2B + sin2C + - r
Hiidng ddn gidi
2
Ta CO S = 2sin(B + C)cos(B - C) + — — • , , j
sin A f
sin A
= 2sinAcos(B C) +
-sin A
Do sinA > 0 va cos(B - C) < 1, nen lif (1) cd:
2
S < 2sinA +
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Trang 10Chuyfin 66 BDHSG Toan gii trj Idn nha't vi giA trj nh6 nhat - Phan Huy Kh&\
Da'u bang trong (2) xay ra <=> cos(B - C) = 1 <=> B = C
Do A la goc l(^n nhii't trong lam giac => A > ^ => ~ < sin A < 1
X c l ham so i\x) = 2\ - wYi —<x< \ :
X 2 • • ,;
2 2 x ' - 2
Ta CO f'(x) = 2 — - = — , ncn c6 bang bicn thicn sau
, ,Vay ^ l a x i(\) = {
<x<l
yf3] 773
v 2
X
-f(x)
1
TCr (16 suy ra (kcl hctp vdi (2))
7V3
S<
3
S=:——-oB = C v a s i n A =
• 2
o A = B = C Vay maxS = <=> ABC la tarn giac dcu
Bai 6 X e l tarn giac ABC vdi A > 90"
Tim gia trj Idn nha't ciia dai li/iing P = tanBtanC + 2colA
Hiidng dan gidi
Tr\idc hct ta c6 tanBtanC < cot" A ^ , j,
Da'u bhng trong (1) xay ra o B = C
T i f d 6 P < c o t ' A +2cotA .;,
2
A
, i - i a n ^
Do colA =
, 1-lan^
tan A A •
2 tan
Thay (3) vao (2) ta di den P < c o l ' - + c o l - L
2 2 A ' col
2
Cty TIMHH MTV DWH Khang ViSt
Dau bang trong (4) xay ra ci> B = C ' >3 ; i i b i f
Xct ham so f(x) = x ' + x - — vdi 0 < x < 1 < ^
X
(do A > 90" ^ — > 45" 0 < cot — < 1) *
Ta CO f'(x) = 2x +1 + -J- > 0 Vx e (0; 1], ncn c6 bang bicn thicn sau
r ( x ) f(x) 1 + 1
Vay max r(x) = 1(1) = 1 Tif (4) suy ra
0<x<l
P < 1
P - l o B - C v a c o t — = 1
« ABC la tarn giac vuong can tai A
TCr do ta CO maxP = 1 o ABC la lam giac vuong can tai A
''hiiy: {I) chiJng minh nhU'sau 'f
Do A > 90" =o B + C < 90" cos(B + C) > 0
Ta CO cosBcosC = ^[cos(B + C) + cos(B -C)J
> -[cos(B + C)cos(B-C) + c o s ( B - C ) ]
P =>cosBcosC > - c o s ( B - C ) l l + cos(B + C)] = c o s ( B - C ) c o s ^ - ^ i ^
2
2 A
=> cosBcosC > sin^ —cos(B - C )
2
=> cosBcosC > sin^—(cosBcosC + sinBsinC)
2
=>(1 - sin" — )cosBcosC > sin^ — sinBsinC
2 2
Do B, C nhon => cosBcosC > 0
„ 2 A ,
cos Tir do ta CO 9 sin B sin C „ 2 A
^ > =>tanBtanC<cot'' — 2 A cosBcosC
sm
2
Da'u bang xay ra <=> cos(B - C) <=> B = C => (1) di/cJc chiJng minh! ' www.facebook.com/groups/TaiLieuOnThiDaiHoc01/