V2 cosa + sina
tr6n.
2)Tdt c'it cric ldi girii grri vd Tba soan ddu ddng. Ngoii
LtanTti, ciic ban sau cfing c6 ldi gitii t6t, ngin gon: Hi Ndi: Trdn Nhar Tdn, l2Al To6n, DHKHTN, DHQC HA Noi, Ddo Trottg Anh, llA4, THPT Nguydn Gia Thidu. Long Bi6n, Nguvttr So'nTing, l2A2' THFrf Ngo Quydn, Chau Son, Ba Vi; Vinh Phict Kint Ditit
San, llAl, THm chuyen Vrnh Phric; Hrii Duong: NinhVcin Qui,llA, THm Kim Thlnh, L€Vdn Huinh,
l2Tl, THP'| chuyen Nguyen TrIi, TP. Hrii Duong; Hi
Nam: Irdri Trung Kiin, l1 Torin, THPT chuyOn Hlr
Nam; Nghd An; Nguy€n Bd Hltu DL?c, I lA1, kh6i THPT chuy0n DH Vinh, Nguy€n AnTinh, llAl' THFrf
chuy€n Phan Boi Chau, TP. Vinh; Thira Thi6n - Hue:
Li Tltarrlr Pluic, I I To6n, THPT Qu6c hqc Hue, TP' Hue; Dnk Lnk: L€ Quang Hi?ir, llCT, THPT chuydn Nguy€n Dul An Giang: TrdnThatfi Pho', llAl, THFrI
Tdn ChAu; TP. H6 Chi Minh: Bii Vdn Minh, llCT'
THPT chuy€n L0 Hdng Phong.
ud euaNc vrNn
*tl:ii I-l 37fi. !,4Qr qtil bdng tlitr k)i rrii ri'tll i ttrt /r = l.0rn so t't)i iltL.il sL'ttt tttirtr rtqutr;1. tll i ttrt /r = l.0rn so t't)i iltL.il sL'ttt tttirtr rtqutr;1.
S(iit tn(;i \'(i (ltLu)t;r':'i stitt briig, tttcr't 19%: co'
n(iitq,.\Ltu tlti'ii trLtrt burt ldtt tlri lxiilg tlifitg
itril .l't'ott,? titr';r '<itur d(t bitrg tli lr tlrrt.rc
tlttLittli rltir)rr.! il,,'tiitl i)rttt ttltiatt iii,-;i 'l
Loi gidi. (Theo ldi gi(ii ctia mbt s0'ban). Co nang ban ddu ciia qua b6ng Ih E,t = mgh.
Theo d6 bii, sau m6i ldn va cham b6ng mat
l97o co nang, nghia lh n6 chi cbn 817o co nang so v6i lfc ngay tru6c va cham. Neu goi
Ei vd /i,lh co ndng vA dO cao b6ng dat duoc
sau ldn va cham thf i, ta c6 E, = ki.Er',h,= P.1r', (6 day k = 817o, i e N). h,= P.1r', (6 day k = 817o, i e N).
Thdi gian b6ng bay ngay tt sau va cham thf i E; I;
clen va cham tidp theo li ,,=Z^14 =2^14.,8 .
' !g \g
Thdi gian dd b6ng dung ln t = to tir, , ,tong
= -^P *2,P (L+ Jk * Jr ,...- Jl ) ) !g Vgt --E!--,@-fL-_4)Vt -!t [ ,-Ji ) lzt,It+Jt 2Jk"-'\ -!'I t-Ji ) m clo 1^ " - i-!g va cham len cham. Ta c6
lA thdi gian b6ng b6t ddu roi ddn
thf nhat v6i sirn, n lh so ldn va
* Bii L21378. lVl()r klfi rtr! tllit tr0tt mr")r mrit bitr
t'tip rtdm ngung.
l,[6t ti'm vtitr tltit
tt.tit t'io l;lt(t-i It'r.t
t't) trghi0,ng so t'i'i nitit pltiirtg itg,iln!
nrt.)t glc 60" (.rcnt
Ititrlt I L lJii't l;lrt,i ltt,ttti:
, = W.2W*,{r,'
!g Vgfr
Ilittlt I
','it clriirt tlii ctitt ltint
Tt}'Feq h*ffi(
si aaz (4-2009) * ,a *-; +r,,r,
po JE < I non khi n + +"o thi JL'" -r O.
6i(t*v'r\
Suvra ' 1= \ !---' s l' (r-Jk,":l=12s.
Qudng dudng b6ng di du-o. c Ih
s= h+rtr,=h+zhlk'= h + 2hlk t k' +...* k" ) = h + 2hlk t k' +...* k" ) = -h +zh(t + k + k2 +...* r') /' "';-r ) (t**-2t"-') = -h+2hl t-K l=hl-1. [r-k, I t-k )
Po JF<l ncnkhi n++co thi k"*t)o.
/r'r\
Suy " r.a S = rl (l-ki-+ l=19)m. A(Nnan x6t. Cdc ban sau clay c6 ldi giAi tot: (Nnan x6t. Cdc ban sau clay c6 ldi giAi tot:
Thanh Hoit L€ Tlnnh Minh, l2cl2, THPT L€ Van Huu, Thieu Ho'6, Mai Chi Dot, I lE, THFrf Hi Trung;
Quing Ngai: Ngu-yfu Tdh D6ng, llLi, Ngutin Ngoc
Duy, \ZLi, THPT chuyen Le Khiet; Bic Giang:
Nguyen Hoitng Hi€p, l2To6n, THPT chuyOn Blc
Giang; Bic Kan: Ngd Vi6t Hing, l2Todn, THPT
chuy0n Bic Kan; Vinh Phric: Le DL?c Di€p, l2Al0,
THPT chuy0n Vinh Phric; Nghe An NgLvin Duv
Manh, lOAl, THPT Qu!'nh Luu I, Ngrry€n Bd Hint
Dirc,llAl , Khdi THFrf chuy0n, DH Vinh, NguvdnVan
Linh, 12A3K35, HA Trong Hting,Vii Dinh Ht;i,10A3
K37, THFrf chuyOn Phan Boi Chau'
t'rin lirt lrrr.fi bt)tng kln'i lturg t'ci dtr'rng kinlt
cilu kh(t'i tt'r.r. Coi h0 sr)'mo stit tui ba dii'm ti0p ric li nltturhau.
tt) Chrilrg tr) rdng ltc nto sitt rt' ba didnt til'p .ric bing trhuu.
h) Hny tim lqiti tt'i nlrd rrlui't r'ilu h0 si'mn sit
rlc'dtim btio .sr.t ctirt btirrg c'riu h0.
Ldi'giii. a) Cdc luc t6c dung len m6i vAt nhu hlnh 2'
- 1)
Hinh 2
M II dicm tiep xric giffa thanh vdi tru. Theo bdi ra ta thay tam gi6c AMC d€,t. X6t ci he ta
thdy F, vh tr, lh cdc ngoai luc theo phuong
nging do c16 chring phii c6 dd l6n bang nhau.
X6t cAn bang m6men cfia tru doi vdi trucquay di qua O ta duoc F = Fr. quay di qua O ta duoc F = Fr.
NhuvAy F=Fr-F2.
b) C[ng nhAn thAy rang phan luc vu6ng g6c N nho hon N, vh N, , do d6 dd h€ can bang chi
cdn hO so ma sdt phai cho thoi man pr> L .
X6t can bang ciia thanh theo phuong ngang
Fcos60o * F2 =Ncos30o - F =+.l1
VJ
(Nna, x6t. Cric ban sau clay c6 ldi girii t6t:
Th6i Nguydn: Vii Quang Minh, l0 Li, THFrf chuydn Thrii Nguy0n; Vinh Phric: L€ Dic Di€p,12 Al0, THFrf
chuy€n Vinh Phric, Ltttt Cao CLtd'ng, Nguyirt Vdn Tfi,
Pham Ngoc Xtyin, llA2 THPT Ngo Gia Tq, LQp
Thach; Nghd Ln Hd Trong Hting, Hd Philc Dat, Nguydn Huy Hodng, Nguyin Trung Hung, Mai Trong
Dat, Nguy|n Bd Dting, l0A3 THm chuy€n Phan BOi Ch0u; Quring Ngei: Nguylrr Td'n Dbng, 1l Li, N guyen
Ngoc Duyt,l2 Li, THP| chuyOn LC Khiet.
NGUYEN XUAN QUANG TONN HOC
sd 582 ,4-zoog
Do vAy U, +
26 - 6[t1fii[E_ s0 eaz r+-zoOgt
KI THI... (Ti€p trang 3l)
SENIOR SE(ITION
Sttnda,-, 29 March 2009 08h45 - I th45
The problems from Ql to Q4 are the same as
these in the Junior Section.
Q5. Suppose that a =)b *19, where 17-)t}tt+t. Prove that a is divisible by 23 for any positive
integer n .
Q6. Determine all positive integral pairs (rr ; v) for
which 5rr2 + 6rrv + 1t'2 = 2009.
Q7. Prove that for every positive integet ri there
exists a positive integer rt such that the last rl
digists in decimal representation of rr3 are equal to 8.
Q8. Give an example of a triangle whose all sides
and altitudes at'e positive integers.
Q9. Given a triangle ABC with BC = 5, CA = 4,
AB = 3 and the points E, F, G tie on the sides
BC,CA,AB, respectively, so that EF is parallel to
AB and area (AEFG)=1. Find the minimum
value of the perimeter of tliangle EFG.
Ql0. Find a1l integers -r. y, z satisfying the system
[.r*t'-t: =8
l
['tt*Yt+z]-8'
Q11. Let be given three positive numbers a,p
and y. Suppose that four real numbers a,b,c,d
la2+F=a
salisly the conditiont
] .t * ,l: = 0lac+bd=y. lac+bd=y.
Find the set of all possible values the number
M = ab + cd canlake.
Q12. Let a, b, c, d be positive integers such that
a + b + c + tl = 99. Find the smallest and the greatest values of the following product P = obcd.
Q13. Given an acute-angled triangle ABC with area S , let points A', B'. C' be located as follows:
A' is the point where altitude from A on BC meets the outwards facing semicirle drawn on .BC
as diameter. Points B' , C' are located similarly.
Evaluate the sum
7 =(area MCA'12 +@tea LCAB')2 +(area MBC')2.
Q14. 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 7 times less than 2009.
DE R4... (Ti€p trang t6)
U,,,c'= lY vd 1 = lmA. Cho biet, neu gifr cho
(J o, luOn lu6n bang lV, tang tdn so dbngdi0n I6n qu6 1000H2 thi thdy dbng di6n trong di0n I6n qu6 1000H2 thi thdy dbng di6n trong mach giam. Hoi
1) Doan mach AB chrla linh kiOn gi? Doan
mach BC chria linh kiOn gi ? 2t Tinh L vh C.
NGUYEN QUANGHAU
(HA N6i)
Bii L21382. Mot he thong c6c bAn 16 noi c6c
dinh cua ba hinh thoi c6 ti le do ddi c6c canh
li,3 :2: I (nhu hinh ve).
Cho dinh A, chuydn d6ng v6i vAn toc l theo
phuong ngang. Tim vAn t6c cua cdc dinh A, ,
A2, Bt vi B, tai thdi didm c6c canh cua hinh thoi vu6ng gdc v6i nhau.
a ^.
DO TUAN
(GV THPT chuy,!nVrnh Philc)
pR0ffi.,,,,D,HFt$ l i TH$ l$$tlE
F()R I,OWER. SECONDARY SCHCOI-S
T1/382. (For 6th grade) Compare the sum
(consisting of 2010 terms)
12320092010
-+-+-+...+-+2009 20091 2009 20093 2009200e 200920t0
tl'tth ...:.
2008'
T21382. (For 7th grade) Find a root of the
polynomial P(x)=v3 +ar2 +bx+c given that it has at least one root and a + 2b + 4c : -! .
2
T 3 I 382. Let a r, a2, a1, a 4, a 5, a 6, a7 ) aB, agbe nonnegative real numbers whose sum equals 1. negative real numbers whose sum equals 1.
Put S* =ak+ak+tlak*z*ar*t (k =1,2,"',6)'Determine the smallest possible value of Determine the smallest possible value of
M =tndx{S,,S,S.,,So,S.,So }.
T41382. Let m, n, a, b, and c be real numbers
such that the following conditions hold:
m1000+n1000_a.
*sooo ansooo - c
*r.OOO an OOO
=?! .
36
Find a formula relating a, b and c which
does not involve m, n.
TSl382. Let AH be the altitude from A of a
triangle ABC. Choose a point D on the half-
plane created by BC which contains A such
AB
that DB = DC = !=. Prove rhar the lengrhs
Jz
of the line segments BD, DH and HA are the
side lengths of a right triangle.
FOR T]PPER SECONDARY SCH0OI,S T61382. Determine the maximum possible
value of x2 + y2 where -r and y are twointegers chosen arbitrarily within the interval integers chosen arbitrarily within the interval
l-2009;2009) such that
( r' -zr.t- y')' :4.
T71382. Consider two polynomials with real coefficients:
P (x) = v' * a,,-1x"-t +...+afi+ ao and
Q(x):x2 +x+2009.
Given that P(x) has n distinct real roots but P(Q(x)) does not have any real solution. Prove
I
rhar P0009) > ^
4"
T8/382 Let ABCDEF be a regular hexagon and let G be the midpoint of BF. Choose a
point 1on BC such that BI = BG. Let H be a
point on 1G such that difu = 45" , and K is a
point on EF such that 6FE = 45" . Prove that
DHK ts an equilateral triangle.
Translated by LE MINH HA
TO6N HOC
b) p=21-6x+g-212 > 2x2 _-6x+g -(y+-')'2 2 = 2x2 -6x+9 -Q -x)'z = !13r, -6x+9)=g(x). 22 Khio s6t hhm s6 g(x) tr6n ll ;2), ta c6 tlls(x) = g(1) = 3. Do d6 P > 3. Ding thric
x6y ra khix= | = z = 1 V4y minP = 3.
lf,lnAo xd. Ca,;u b) c5 the giii gon hon bang
c6ch sit dung bdt ding thirc Bunyakovsky
P = x2 + y2 + "3 r' >+(x+ y* z)2 :3.
o Cdch 2 (Stl dung phuong ph6p hinh hoc)
pffimilefuM$ffiuryffi
bh"g rrhd*r;a *6ekz
x
NGUYEN NGOC KHOA
(GV THPT chuy1n Le Khi6t, Qudng Ngai)
n di v6i m6i dang to6n, m5i cAu h6i khSc
l-l nha\ m5i bIi todn, ta cdn phrii chon loc
tlng c6ch giii thich ho. p, Muon c6 ki nang
nhy, ngudi ldm to6n cdn phAi suy nghr tim tdi
nhi6u phuong ph6p giii hay, nhi6u bii todn
mdi. Sau day ld mdt so bdi to6n minh hoa.
Cho x, y, z ld, c6c so thuc tDy y thoa min didu
0<x<2,0<y32,01212
x-t)t-lz=3 (*)
ki0n
,{... ,
ilttrtii! irr;i:: I "\'r''. t.,.r ..., t 'i, ' .i.. ti:.. .:.;
i:'i l/itt;rirrit i'ii ;4i.1 tli;ilr,'j r;'.::; r'tit: ctit; ,;i,',:rltfi'c sttri rltfi'c sttri
l) 1' = -t' + r,2 +:r:
2) Q =(r- 4)t + (.r, - 1)r + (: - i)r" Loi gidi.
l). Cdch I
a) Vai trd x, y, z nhu nhau, kh6ng mAt tinh tdng qu6t, gi6 sir,r > y > z suy ra I < x < 2.
P : (x + y + z)2 -2(y + z)x -2yz
= 9-2(3-x)x-2yz <2x2 -6x+9 = f (x).Kh6o sr{t him so /(x) trOn doan [1 ; 2) ta dc Kh6o sr{t him so /(x) trOn doan [1 ; 2) ta dc
ding tinh duoc max "f (x)- f(1)= f (2):5.
ttl2l
Do d6 P < 5. Ding th(rc xily ra khi x = 2, ! = l, z = 0. V4y maxP = 5.
Trong kh6ng gian Oxyz lay didm MQ ; y : z)
th6a mdn (*). Khi d6 OM2 =.( + y2 + z2.Tap c6c didm M ld rhi6t di0n cua khdi lap phuong
canh bang 2 vdi mdtphing (K): x * y * z = 3
(hinh vE) thiet diOn niy lb hinh luc gi6cAtA-,A3A4A.A6 (kd ci nhfrng didm thuOc mi6n AtA-,A3A4A.A6 (kd ci nhfrng didm thuOc mi6n
trong), ki hiOu le @) v6i At1;0;2) ,
Ar(0;l;2) , At(0;2)l) , Aq(l ;2;0) , A5(2;1;0) ,
AoQ;0;1).
Bdi to6n quy vd tim vi tri M trdn (H) sao cho
OMzl6n nhdt, b6 nhdt.
a) OMz dat gi6 rri l6n nhdt khi M trDng vdi A,
(i = 1, 2,3,4,5,6).
Ta c6 OA, = 5 . Vay maxP = 5.
b) Goi T(xu ; yu ; 2,.,) ld, hinh chidu vu6ng g6c
cia O l6n mat phing (IQ. Ta c6
?*firu qe.-1 ffi*{
- qo g6 ;,.at1 , . '1"
l*n lo zo]T= r =T ]T= r =T lro*lo tZu= u Y =v =7 :l' --0 J0 -0 -) suy ra I(1 ; 1; 1) e (11). VAy minP = 3.
2) VOi cdu 2 nhy, tinh doi xung cia x, y, z
kh6ng cdn, vi vdy ta kh6ng thd giii nhu ci{ch 1
trong cAu 1 dd tim maxQ. Su dung phuong
ph6p 'hinh hoc" & c6.ch 2 ta tim duoc maxQ kh6ng c6 gi thay deii.
max Q = max{AAl, ..., AA:} = 77,voiA(4;l;1). voiA(4;l;1).
Q = 17 tai (r; ! | z) =(0 ; 1 ; 2) holc (0 ; 2; 1).
Nhung vi6c tim mtnQ dd kh6c! Vi hinh chidu vu6ng g6c cta A(4 ; l; 1) l6n mat phing (riJ kh6ng thu6c (H) (ban doc hay kidm tra lai). NhAn x6t rang bli torin doi xrrng d6i y, z, v\
vAy ta c6 thd su dung cr4ch 1 cria Bhi torin 1 dd
tim minQ nhu sau.
Q =(r-4)'+ (y- 1)'+(z-I)2- (x + y + z - 6)' - 2(y + z - 2)(x - 4) - (x + y + z - 6)' - 2(y + z - 2)(x - 4) -2(y - 1Xz - 1) = 9 -2(l- xXx - 4) -2(y -l)(z -r) > 9-2(l-xXx-4)- (y + z -Z)') (, - 4)' + (y -l)2 + (z -l)'z ..r (v+z-272 2 (.r-4\'-" 2 , (1-x)l = (x - 4)- a:--:- = /(x) . 2
Kh6o srit hdm r(r) trOn [0 ,2) ta c6 kdt quri.
tlnai toiin 2. Vri'i ttiitt t:iatt ('t')
\)'{'im giti tt'ittlrri nlrti't r. rlt: ltii'tt l!tti't
/ rl
,4=(.t-l)t+l r -' I *t. -,ltt.
\- 2)
2) Tim ,qid tri lritt rtlut't t'ti liiti tt'i tiri rrltri'r criu
bii'tt iittc
F = ( r' - 1 )t - 2.r'' + 3t-- -4;r Loi gidi
l). Cdch /. R6 rhng bdi toiln kh6ng cdn d6i xrlng vdi ,r, y (hoac y, z hod,c z, x) n6n kh6ng thd 6p dung truc tiep crich girii c0u 2 cua Bdi
toitn l vio bli to6n nhy.