ctn OQ vd PR thi b6n di6m A, D, P, K cingthuQc mQt iluong trdn. (Ban tlpc tU chimg minh) thuQc mQt iluong trdn. (Ban tlpc tU chimg minh)
Tro 14i gi6i bdi tohnTl2.
A
B6 qua truong hgp clon giin: AB = AC.
Ki hieu E = BH nAC;F = CH aAB.
Ggi (O), (Or), (Or) theo thri t.u ld cludmg trdn
chua cbc b0 b6n ditim (8, C, E, n; (8, C, Y, Q; (A, H, Y, 4; S = BC aEF;Z = SH nAO;
T =SAnoH. y\ frE=m:frfi =frF ncn
biln di6mB, C, E, F cingthuQc mQt duong trdn.
Tt d6, 6p dlrng UO AC t cho b6n di6m B, C, E,
F, suy ra AZH = ATH = 90o.
YQ.y Z, 7 ctrng thuQc clucrng trdn duong kinh AH. Ap dUng bO de Z cho b6n di6m B, C, E, F, suy ra b6n di6m A, B, C, 7 ctng thuQc mQt ducrng trdn.
Ydy 7 rr<o,> = sA.S7', = SB.SC :3 ,r(o,)
Di€u d6 c6 nghia li,S thuQc XI (1)
Vi AD. BE, CF tlong quy n€n
K(BCDS)= (BCDS):-1 (2)Tt (1) vd (2), chn y ring KDIKS, suy ra Tt (1) vd (2), chn y ring KDIKS, suy ra
BKD =CKD. A
Y NhQn xdt.Bdi to6n ndy kh6ng kh6, kh6 nhi6u ban
tham gia gi6i, xin n6u tdn m6t s6 ban c6 ldi gi6i tucnrg d6i t6t: Son La: Trdn Hodng Dqt, llAl,
THPT MQc Ly, M6c Chdu; Hir NQiz Hodng Le NhAt
Tilng,l2T2, THPT chuy6n KHTN - DHQG Hd NQi;
Nam Dinh: I/fr Thi DiQp, 11T2, THPT chuyCn LO
Hdng Phong, TP Nam Dinh; Thanh H6a: Dfing
Quang Anh, l0T, THPT chuy6n Lam Son, TP Thanh
H6a; NghQ An: Nguydn Dac Bdo, llAl, Nguydn
Hodng Quiic Khdnh,l2Al, THPT chuy6n Phan BQi
Chdu, TP Vinh; Ngh6 An: DSng Duy Bdo Khdnh,
llAl, THPT chuy6n DH Vinh; Hi finh: NguydnQu6c Khanh, Phan Th€ COng, Nguydn Qudc Khanh, Qu6c Khanh, Phan Th€ COng, Nguydn Qudc Khanh, 10T1, THPT chuy6n HA Tinh, TP He Tinh; Thira
Thi6n Hu6: LA I/iA Thanh Long, 11T1, THPT
chuyCn Qu6c Hgc Hui5; Quing Nam: Nguydn Pham Nam Dacrng, llT, THPT chuy6n I6 Th6nh T6ng,
TP HQi An; Long Air: Phsm Quiic Thdng, l2Tl,
THPT chuy6n Long An.
NGTIYEN MINH HA
TONN HOC,. -
Bni L1l470. Dat clidn dp , :400cosi00nt (I)
t,tia htti diur doan mach rndc ,Oi tAp gim cu6n
c,am thrtin r'6 dd tu' ctint 1,, di€n tro' R t;a fLt di€n to tJit''n tlung (' thut doi du'ot'.
I0' 1
Khi C-fl,='" ' 8n f hoiicC=:-C,thic\ng3
sudt u)o drtan mach co cilng gia tri. l() l
h'hi ('= ('_. = *- F ho€tc C : $fC2 thi did.n dp
l5n
hi&t c{rtng gi[i'o hai tldu tr,1 diQn r:o ciutg gia lri.
:. ^ ,:
Kli noi mot untpe ki xoalt chi2u fi u6'ng\ v(ti hai
diu t1t dien thi sd chi cua ampe kA h bao nhiAu? Ldi gffii. Khi Zc,=80Q,2s,=120Q thi c6ng
,,.
su6t cua mpch c6 cung giri tri nen
Z. Lt +2" L2 =2Zr=Zt=100{1,
I<hiZc,=t5Adl,Zr^ =300O thi diQn 6p higu
dpng hai diu tp tliQn c6 cing gi6 tri n€n: lll- ::-++:-= j-+ Zc"=200Cl. Zr, Zro zro P2 t72 Lai c6 Z" Lo =$= R = 100Q. zL
Khi C bi n6i tit bang ampe k6 li tucrng thi s6
chi cua n6 bing: ' t = ,^== =2(A). d
"tR'z+Zl
YNhQn xit, Cbc ben p6 loi gi6i dring: Hung YGn:
NSrryA" ViQt E*c, llA9, Tri€u Ninh Ngdn, 12A9,
THPT Ducvng Quing Hdm; Nam Dlnhz Doan Th!
Nhdi, 12 To6n 1, TTIPT chuy6n L6 H6ng Phong; B4c
Li6ul Dd Vdn Duong,l2cA, THPT Tr6n !5n BAy.
NGUYEN XUAN QUANG
Biti 1,21470. TrAn rndt khu dh rqng vd biirtg
,i
phdng, nnt em he nem di m6t vi€n sdi" Xac dinh
gdc nem di doan du'cing viAn soi bay di trang
kh6rtg khi td tbn nhiit. Bd cpra moi ma,sat, chiiu .'(tt) (nt he khong ttung ki:.
Ldi gi,rti. Phucmg tinh chuy6n tlQng cua vi6n s6i:
[x = (vcosa),
I ldx =vcosadt
I or2 3{
l.l = (vsina)t -T ldy = (vsincr - gt)dt
Thcri gian dti vi6n s6i i14t h^* ld: r= "ino,o
6
^ ..).
Suy ra chi6u ddi qu! cl4o vi6n soi:
vsin c g
t 0 vsino/g
k tanrl 0
LAy dpo hdm cria hdm Z theo o., ta dugc:
L'(a) =4r.oro.[t I L -rino.n(tin \coso/l" * I )l
L'(a) = o e 1 -sin cr.hf!Igl1) = o
\ cosc /
= C{,c
^, 5627'57'. D
YNhQn xd.Prit ti6c kh6ng c6 b4n niro c6 loi gi6i dring cho dA ra ki ndy.
DINH THI THAr QrriT.Ur
L=2. I@. tmq . tmq ) L=/v'cos'., oJ | 1*P11, 60 = arr**.[,tJi*p* rnlr + JrTp;i*" = 2v cos ct /\ PoiUl6n:k=[tano- 8' I ( ycoso/ - -vcoscr. -- ) dt =-dko 6
) L =4[*in., + cos2 .,,. tn( tin o * I )l
8L ( cosa /l OOi cpn: r*(ano- r' \' 0,. 1 vcoscr, / .,r / TORN HQC /.b - cftnfiUa sd 4?c (12-2o1G)
PROBLEMS .,. qfii;, rher.t tran,s l"i\
Prolrl*rm T'S/474. Given a tetrahedron SABC
such that SA, ,SB, and SC are mutuallyperpendicular. A moving plane (P) which perpendicular. A moving plane (P) which
contains the centroid of the given tehahedron
intersects SA, SB, and SC respectively at Ay,81,
l1t4
and C,. Prove that ': * '- * ^ 2*. where
sAi sB,' sci Rz'
R is the radius of the circumscribed sphere ofthe tetrahedron SABC. the tetrahedron SABC.
Problem 'f8fi74. Find the integral part of the
following number r =?.1.9.9...ry.
1 '3'5'7"' 2It5'
TOW' AIIDS MA'tHEh,I.4,'tI C A [-
O},YR,IPIAI}
Problem T101474. A positive integer is called a
"HV2015" number if in its decimal representation there are 2015 consecutive digits 9.
N{(}T' HU'#NG .". {'ti{St {fttc rrareg 7}
Let (an),n=1,2,3,... be a strictly incrppsing
sequence of positive integers such that tfrp
("\
sequence - I J l,n=1,2,3,... is bounded.
\n )
Prove that there are infinitely many "HY}A||"
numbers in the given sequence (a,),n=I7?,3,,,1 Probfem'flll474" Given a grid of the size lx n
(n > 2). Now we fill in each square with 0 or l.A way to fill the grid is called "good" if any A way to fill the grid is called "good" if any
three consecutive squares do not contain the same number.
Let an be the number of
"good" ways to fill the grid. Compute a,. I'roblem Tl2l4?4" Given a triangle ABC.Prove
that
o(.e B .c)- f.A :B f-
vz[s'rr7+srn7*t
1 ).l.*' *{rrZ *{*2,
Translated byNGUYEN PHU HOANq L.1N
(College of Science-I/ietnam National University, Hanoi)
A. m=0 B. m=l C. m:2 D. m=3
2. Cho hdm s6 ) = -na -2(m-1)x2 + m +l (m
ld tham sO;. VOI gi6 tri ndo c[ra m thi dO thi
hdm sO c6 ba dii5m cuc tri tao thdnh tam gi6c c6 di6n tich bdng32 ?
A. m:4
C. m=*4
3. Cho hdm sO !=xa+(3m+l)x2-3 (m lit
tham sd). V6i gi6 tri ndo cta m thi dO thi hdms6 c6 ba dii.lm cuc tri tao thdnh tam gi6c c6 dQ s6 c6 ba dii.lm cuc tri tao thdnh tam gi6c c6 dQ
a
ddi canh ddy bangi d9 dei canh b6n ? 5 55L A. trt=i B. nt:-; C. nt:-; D. m=-l 553 4. Cho hdm sd J = x4 +2(m'2)x2 +nf *5m+5
(mldtham sO;. VOi gi|trindo ctra m thi dO thihdm s6 c6 ba tli6m cgc tri tao thdnh tam giirc hdm s6 c6 ba tli6m cgc tri tao thdnh tam giirc ddu ? A. m=111 '' *=ry! B. *=4fi 3 ,.*=Y ham s6 ,=-z( * -4 -r\ rrqrrr rv ,---(" - 2 -' ) D4t '4\ o, =3T bro). Ta c6
AH =2aa,BC =2a, AB : AC = JA2 +4aB .Cf;rl lroi tr.Ta c6 AH.BC = 4- os =la cP =l Cf;rl lroi tr.Ta c6 AH.BC = 4- os =la cP =l
l
= m -;. Chon drip 5n A. J
{-jiu htli 2. Ta c6 m =}nC Z = Ja\4aB =3a
.)rF4{,
) q2 ='J2 = m =f . Chon clap an B.
III" ti,{r TiP TU'r.rrYt-H
1. Cho hdm s6 !=x4-2(m+l)x2+m2 (m lit
tham s6). V6i gi6 tri ndo oba m thi dO thl hdm sd
c6 ba di6m cgc tri t4o thdnh tam gi6c vu6ng ?
m=2+1,13 m=2-1lj B, m: -4 D. E6p iin kh5c, B. m= 2-{3 D. m = -trl1 C.
Sti aza (02-2o116) TOfiN HO( ^ * ^ * *l*ffiiffe/ /
0f;nrmno CIutfi TflonG Tnon ruilde nd
dnld cdc ban dd tung gap bdi to6n sau <16y
khi con ld hoc sinh tiilu hqc: "MQt chi€c kd
ddi 13 cm kh6ng c6 vqch khiic @tem chia) trAn
thabc. Ta phii phdn b6 + diA* chia vdo bAn
trong chiiii thadc do nha thii ndo dd vdi chi€c
thadc c6 vqch chia d6 ta cd the do daqc cdc
doqn thiing co chiiu aai mn fuqt ld I cm,2 cm,
3 cm,4 cm, 5 cm, 6 cm,l cm,8 cm,9 cm, l0
cm, ll cm, 12 cm, 13 cm?".
Ttr nay rd sa., ta sE goi c6c di6m chia nhu v6y
ld cac di€m chia trong. Ldi gi6i cira bii to6n
dugc minh hga qua hinh 1.
Hinh I
Bdi to6n dd dugc gi6i xong, nhtmg mQt c6u h6i dugc cl[t ra"LiQu 4 dd phdi ld sij diAm chia it nhdt chaa ?". CAu tr6loi ld dring.
Th{t vfly gi6 sir 3 h s6 di6m chia trong cua
chitic thudc diri 13cm, hi6n nhiOn hai ddu mritcua chi6c thu6c cflng lit circ di6m chia' Cing cua chi6c thu6c cflng lit circ di6m chia' Cing
v6i 3 di6m chia trong vd 2 dAu mrit cira thu6c ta
c6 5 di6m chia. V6i 2 di6m chia b6t ky ta x5c
dinh duqc 1 cloan thing c6 dO ddi ld s6 t.u nhi6n
< 13 cm. T6ng sd c6c do4n thing x6c dinh bdi 5
/< 1\ 5
cliem chia sd lir: #= l0 doan thang' Nhu
)
vfly n6u b6n trong
"rir6. ,no6" dei 13 cm chi c6
3 di6m chia thi v6i chi6c thu6c 6y ta chi do
duoc t6i da 10 dQ ddi kh6c nhau, nhrmg sO dopn cAn do lai ld 13 dopn. Boi th€ vdi 3 clitim chia
trong chi6c thu6c ddi 13 cm ld khdng thoa mdn
v{s
iNG DrIc rAx laal'1a4.,r , Ho' .,r , Ho'
diOu kiQn cua bdi to6n. Vd 4 ching ld s6 chia
, . i..
trong lt nlrat.
Xdt bdi to6n t6ng qudt: "Sd di€m chia trong {t
nhiit dAi vdi chidc thadc kd c6 d0 ddi n cm ldbao nhiAu dA vdi chi6c thadc c6 cdc didm chia bao nhiAu dA vdi chi6c thadc c6 cdc didm chia
: ' '' ; do duoc tdt cd cac dogn
trong aY la co tne
thting c6 dO ddi ti.r I cm d€n n cm?"
Bdi io6n t6ng quSt ndy dugc A.Savin <1A xuSt
tr6n Tap chi Kvant cta nu6c Nga tu ndm 1976,
vir trong srr6t 40 n6m qua vdn chua c6 ldi giii'
Ta xdt lcri giii bdi to6n AOi vOi mOt sl5 gi6 ta n
nh6 (<13). OOi vOi chi6c thu6c c6 d0 ddi 1 cm
thi khdng cAn c6 di6m chia trong ndo c6' Ddi
v6i chi6c thu6c c6 dO ddi 2 cmvd 3 cm thi chi
cdn c6 1 di6m chia trong ld dt. Vdi chii5c thu6c c6 d0 ddi 4,5,6 cm thi chi cin c6 3 di€m chia
trong ld dn. V6i chitic thu6c c6 d0 dni 10, 11,
12, 13 cm thi cAn ph6i c6 it nh6t 4 dir5m chia
trong. Todn bO c5c k6t qu6 nlry dugc minh hoa qua hinh 2. I r-..t 2ffi 3 r-r--r 4 =r----1-l s Er--1---r r. f-T---T--l 1 l-T-r----'---'f--1 3 r-t-I---:r-- q! | | I I ,,'- ,rffi .^ffi '.,G l1r t ! Hinh 2
Voi nhirng gi6[ n cdng l6n thi bAi to6n cdng trO nOn kh6 khdn hcn. DC giei chring rd rdng ld cAn
phii c6 mdt phuong phSp t6ng qu6t ndo d6 ? Sau
clAy chring ta sE tim c6ch dSnh gi5 s0 di6m chia
trong (x6c dinh cfln trOn vd c4n du6i)'
Gi6 sir tr€n chii5c thu6c c6 d0 dei n cmta ddt k
di6m chia trong. Tinh cd 2 ddu mft tr6i vi phii cira chi6c thu6c ta c6 (k+2) clitim chia' SO doan
thing tluoc t4o thdnh bbi (k+2) di6m chia d6 sE
^,. TORN HO(
olrrn "2(k+1)(k+2). Bdi vi theo tlidu kion cua
bdi to6n thi ta cAn phii nhQn dugc n tlopn c6 ilQ
(l,r1\(l,L1\
ddi kh6c nhau, nhrmg trong :1-l-:{11-2 <lo4n
2
d6 c6 th6 c6 mQt si5 do4n c6 d0 ddi bing nhau,
vi thr5 ta cAn phii c6bitding thric:
_ (k +l)(k+2)n< n< --- L^ ,__..( ,*s*Jsn+r)(o_-:+Jsr+t ), o -["-2J[^2)-" Jsn+1 -g
Tt d6 suy ra: Or- .--,,-Z (l)
Nhu v{y tr6n chi6c thudc c6 chi6u ddi n (cm) ta
c6n ph6i c6 it nh6t ,' JsT tl -: cti6m chia
Z
trong d,5 th6a mdn di6u kiQn cira bii to6n (n6u s5
trong bii5u thric ld kh6ng nguy6n thi ta phii l6y
t^
sd nguydn 16n hon vd gdn nhdt vcri n6, vi du gi5
d cria bii5u thric ld 6,25 chdng han thi taldy 7).
Ti6p theo ta sE phdn UO f AiCm chia trong tr6n thu6c nhu sau: DAu tiOn bit ddu tu d0u mrit b6n
trii ciathu6c ta <14t 1i6n tiiip tu tr6i sang phii m<lo4n thing bing nhau md m6i do4n c6 chiAu <lo4n thing bing nhau md m6i do4n c6 chiAu
dii 1 cm (m t k), ti6p theo tu di6m chia trong
thu m fiAnthudc ta clat li€n ti6p tu trSi sang ph6i
(k - *) clo4n thing bing nhau trong d6 m6i
tlopn thing 5y c6 chi6u dii (m + 1) (cm). B6y
gid ta x6t chi6c thu6c k6 c6 ft tli6m chia trong
nhu tr6n vd tlopn thing cu6i (c6 mQt dAu mirt
trung v6i dAu mrit b6n phii cira thudc vi tlAu
mirt cdn lai chinh ld di6m chia trong thu kmdta
vria thlrc hiQn) cfing c6 chidu ddi (lz+1) (cm).Tinh chi6u dii L cinchii5c thu6c tl6 theo mvdk, Tinh chi6u dii L cinchii5c thu6c tl6 theo mvdk,
tac6: