22
1
I = x - - th'i didu kiOn (*) tro thhnh
2
+(z-4)'1.
V6i viOc dat bidn phu trdn, bidu thrlc E d6\ xring ddi vdi t, y. Vi vAy c6 thrj iip dung ci{ch
gi6i trong cAu 2 cua Bhi to6n 1 vi ta s6 c6 mOt ldi girii dep.
. Criclr 2. Co thd, gi6i bang phuong phdp " dao hdm theo ttng bi€h" nhu sau:
Tit z = 3 - x -y vi do 0 < z < 2 n€,n 0<3- x*y<2el-y<x S3-y / l), . , vb, E =(x-l)2 +l v-- | +(l+x+ y)-. \" 2) It 3 l-- (l< -,0(y(2,0<z<2 )2 2 ls lt+v+z=- t2 vb. vqL-[I E =( , -!)'*l, ,-1)' 2) [" 2) = e -2t1- xxx - 4) ,2Jt+ I = -13x' - 1 8x + 331 = 2111 . xnaolat hdm so /r(x) tron [o ; 2], ta c6 og minft(x) = hQ\ = j ndn g>' . Ddns thuc rn )r a LL -.
xay ra khix= 2,! = z= 1. Ua, 2'2 minQ =2.
Nhdn xet. Cirng mdt hudng girii nhung c6 thd
trinh bhy gon hon bang ciich sii dung bat d&ng
thric ar + b: >(a + b):
nhu sau: 2
I
Xem _y Id tham so vdi 0 < y < 2, x6t him
" ( r\:
fl,r) = (.r-l)'+l v-- \." 2)| +ll+x+y)'v6i1-y<r<3-y. v6i1-y<r<3-y.
.f '(x) = 2(x -1) + 2(l + x + y) = 4x + 2y i
.f'(.r)=o<+ x =+=1-y (doo<y<2)."2 "2
Vi/(r) ddng bien tr6n (1 - y ;3 - y) ndn
min /(x) =f(t-y):y'*( y--!) ++=gtl)
il-r.rrl" r' ' [' 2)
I t\ r
.q'()') = 2y+21 y- -l:Q s y= . e (0:2).
" \" 2) 4
Lai do g'(y) drii ddu tir Am sang duong khi cli
I /r) 33
qua y=Z n€n
f1rtr):r[aJ=T ua,
min-E=Iou,rai x:f . r:!.,=2.
844
Nhdn .rit. D0 nhAn thay rang, cldi vdi cdu I cua Bdi todn 2 ndy thi cdch 2 vtia n6u tr6n
kh6ng hay vh c5 phAn dhi ddng. Nhung ddn
cAu 2 vai trd x, y, z khdng nhu nhau, thi c6c
phuong phSp gi6i o Bli to6n 1 dd kh6ng cdn hiOu luc, Ifc nhy c6ch2 vta nOu tr6n mdi ph6t
huy t6c dung.
Tro lai ket quil citu 2 Bdi todn 1, rning=2
2
ctatrai .r=2.!=z=l.r,r* 2 u(z,l ,i)( 2'2)
Ii hinh chieu cua A(4; l; 1) lOn canh AoA..
Trong cdu 1 Bdi toil,n 2, minE = I Uu, ,ui
8
3 1 /; \
)r=-, y=-, : = 2.DiemH[1.1.2 ] le
4 4 [44 )
hinh chieu vu6ng g6c cua diem S[,,] , O)
\2 )
l6n canh A,Arcia luc gi6c A1A2...46. Cdc ban hdy suy nghi xem, c6c nhAn x6t tren c6 1i
nghra hinh hoc gi kh6ng?l
Thay cho bdi tap, c6c ban h6y tim th6m nhidu c6ch giAi khric cho ciic bbi todn trong bhi viet
nhy vi hoin thlinh cAu 1 vi cAu 2 cira Bei to6n 2.
Chric ciic ban thhnh cong.
HU(rN(; nAn r;lAl ... eie'p rrans to)
PT duong thing qua F v) vuOng g5c vdi d ld
d, : -B(x-r'E) + Ay = 0,M = d e: tly. To4 dQ M li nghiOm cria h6:
l,lx+B)'=-C'
1
l-ar* Ay=-Jis.
Binh phuong hai ve cua tirng PT r6i cdng lai theo ve, ket hop vdi (1), suy ra x) + y) =9 .
2) Goi H l) truc tam tam gtdc ABC, chung
minh OH vudng g6c v6i mp(ABC). m'
mp(ABCt ' h ! +l 123+1 =l c>6x +3y +22 -6=0.
lx=6t
PT cludng thltng OH: ), = S, t e "R .
1", = z,
Suyra - Hf36,18,12).\4e 4e 4e)
CAu 7b. Goi A ld bidn cd "Ngoc, Thcio cd
phdn thtrtng giAng nhau".
NhAn thdy c6 2 hoc sinh nhAn s6ch (Tor{n,
Li); 3 hoc sinh nhAn silch (To6n, H6a) r,d
4 hoc sinh nhAn s6ch (Li, H6a).
56 phdn tir cua kh6ng gian m6u lh
,(o)= c;c;ci =1260.
IH1. Ngoc, Thio nhAn sdch (Todn, Li), so
khA nang ld CIC{ = 35.
fH2. Ngoc, Th6o nhAn sdch (Todn, Hoii), so
kh6 nang ld 7C?,Cl: 105 .
IH3. Ngoc, Thio nhAn s6ch (Lf, Ho6), sd khA nlng lh CiC.Ci = 210 .
Suy ra
"(A) = 35 + I 05 +210: 350.
n(.t\ 5
Dod5 n(A\=---.' n(a) 18
NGUYEN ANH DUNG
(Hd Nai)
TORN HQC
NGUYEN VAN MAU (DHKHTN, DHQG Hd NOi)
Ngdy 29.3.2009 t7i Gidng dttdng lon tiing 7 nha T5, Trudng Dai Hoc Khoa hgc Tu nhi6n Hit
Noi, Hoi Todn hoc He Noi dd td chut Ki thi Olympic Todn hoc Hd Noi md rong (HOMO) ldn
thOlV vdi st1 tham gia cia 50 em hoc sinh ldp I (Junior level, sinh sau ngdy 01 .01 .1995) vit
70 em hoc sinh ldp 10 (Senior level, sinh sau ngdy 01 .01 .1993) tUcdc trudng THCS vd THPT
thuoc Hd Noi, HAi Duong, Qudng Ninh, Bdc Giang, Bdc Ninh, Hoa Binh, Lao Cai, Ydn Bdi,
Phi Tho, Wnh Phic, Thiii Binh, Thiii Nguy6n.
Sau Ki thi ndy, Ban Td chdt sE chgn.ra 20 em bdc THCS vd 20 em bQ9 THPf tham du Ki thi
Olympic Singapore md r6ng vdo cudi thiing 5 nam 2009. Sau ddy h dd bei cia Ki thi HOMO:
el. What is the last two digits of the number Q9. Suppose that four real numbers a,b,c,d
1 000. I 001 + I 001. 1 002 + I 002. I 003 +. . .+ 2008.2009r satisfy the conditions
Important:
Attsv'er all l4 questions.
Enter your anst4.'ers on tlrc answer sheet provided. No calculotot's are allou'ed.
JUNIOR SECTION
Sundar.29 March 2009. 08h45 - I th45
(A) 25; (B) 41; (C) 36; (D) 54;
(E) None of the above.
Q2. Which is largest positive integer il
satislying the inequality
l I r I 6- ..._r-f -r.. f - /-') 1 .2 2.3 3.4 n(n + 1) 7 (A) 3; (B) a; (C) s; (D) 6;
(E) None of the above.
Q3. How many integral roots of the inequality
.-1
-l < ^ ' <2 are there in (-l0l l0) ?
.r+I
(A) 15; (B) 16; (C) 17; (D) 18;
(E) None of the above .
Q4. How many triples (a ; b : c) where a, b, c e
11,2,3,4,5,61 and a<b<c such that the
number abc + (7 - a)(7 * b)(l - c) is divisible by 7?
(A) 15; (B) 17; (C) le; (D)2r;
(E) None of the above.
Q5. Show that there is a natural number n such that the numbera = n! ends exacly in 2009 zeros.
a, b, c be positive integers with
on factor and satisfy the condition
. Prove that a+ b is a square.
Q7. Suppose that a - )tt * l9 , where 12 - )10n+t.
Prove that a is divisibleby 23 for any positive
integer rr .
Q8. Prove that m7 - m ts divisible by 42 for
any positive integer m.
ld+b2_cz ttl)=4
)
lac + brl =2.
Find the set of all possible values the number
M = ab + cd cantake.
Q10. Let a, b be positive integers such that
a t b = 99. Find the smallest and the greatest
values of the following product P = ab.
Q11. Find all integers x,y such that
rz + y2 = (2xy +1)2 .
Q12. Find all the pairs of the positive integers such that the product of the numbers of any pair
plus the half of one of the numbers plus one third of the other number is three times less
than 1 5.
Q13. Let be given LABC with area (LABC) =
60cm2. Let R, S lie in BC such that BR = RS =
SC and P, Q be midpoints of AB and AC ,
respectively. Suppose that PS intersects QR atT.
Evaluate area(A,PQT).
Q14. Let ABC be an acute-angled triangle with
AB=4 and CD bethealtitudethrough C with CD =3. Find the distance between the
midpoints of AD and BC.
(Xem ri€'p trang 26)
Q6. Let
no comm
111abc abc
3wd,; 70#?{*Q0a?%rt'l Wtfulafk*t f,** *d$arrtl, lidAn?i* fulafk*t f,** *d$arrtl, lidAn?i*
xuiT nix ri;' tt)at
So'382(4.2c09)
Toa soAn : 1878, Pho Gi6nq Uo' Ha N0i
0I Bl0n lail 04.351 21 607 0T " Fax Phat henh, Iri su : 04.35144272, 04.351 21 606
Emall: tapchlt0anh0c-lu0llre@yah00.c0m'Yn
ne.N cci vir't xttot. uoc
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GS, TSKH. NGLTYEN CANH TOAN
GS.TSKH. IRANvaN NHUNG
?S. NGTIYENVANVQNG
cs. DoAN QuiNn PGs. ?s. rnirvaN u4.o
cH{u raAcn xnrdut xuir nAl
Chfi tich HOi adng Quin tri ki6m
Tdng Gi6m doc NXB Gi6o duc Vi6L Nam NcO rnAN.tt
eha Td,rg Gi6m ddc ki6m
Tdng bi6n tQp N-XB Gi6o duc Vi6t Nam NGLI.EN Qrritneo
Phuong ph6p giai to6n - Math ProblemSolving Solving
LA Vd.n Lrtc - Stl dung tinh ldi, lom cria dd
thi hdm sdvdo chiing minh bdt ding thitc'
L6 ph6t d6ng cu6c thr truyOn ng5n, brit
ki, ph6ng sr: vd nginh Gi6o duc.
ru1r odNc BIEI\I raP
Tdng bian fi.p : PGS, rS. PHAN DOAN THOAI Ph6 Tdng bi€n td.p: IS. PHAM THI BACH NGQC
?S. TRAN OiNg CHAU, 7ftS. NGUYNN ANH OUNC S NAN NA.IT PUNC, TS. NCUYPN MINH DUC'
;. r;;fi* *r*, ;A, r;. NGUvEN vrsr uAt, pcs. ?s. r,s Qu6c uAN, ras P^HAM v+N HUN9
pcs.
"S. vU tHaNH'xntfr, GS. TSKH. NGUYEN VAN MAU, 6ng NGUYEN t<uA'C MINH' rs. rnAN llrju NAM, pcs.
"s.
NGtryEN DANc.puAr, ocs. ?s. TA DtrY PHUONG,
"lzs.
NGLIYEN THd THACH,
GS. TSKH. DANG HUNG THANG NS. VU KIM THUY, PGS. ?S. VU DIJONG THUY,
GS. TSKH. NCO Vtpr TRUNG, rhs. Hd QUANG VINH.
TRONG SO ruNY
@ Loi giii dd thi vdo 16p 10 chuvon To6n
tnrdng THPT chuY6n Hrtng YOn, ndmDirnh cho Trung hoc Co s6 - For Lower Dirnh cho Trung hoc Co s6 - For Lower
Secondary School
Hod,ngVd.n Chung - Di tim nhtng bdi to6n
thf vi tt m6t bdi to6n quen thu6c.
Dd thi vdo l6p 10 khdi THPT chuven DH Vinh, Ngh6 An, ndm hoc 2008 - 2009'
hoc 2008 - 2009.
Chudn bi thi vho tlai hgc - UniversitvEntrance Preparation Entrance Preparation
Trd.n Vd.n Xudn - C6ch lAp phttong trinh
dudng thing trong kh6ng gian.
@ ,O ra ki nhy - Problems in This Issue
Bia 3. Toin hoc mudn miru -