Đề thi và đáp án học sinh giỏi lớp 12 cấp tỉnh

SO GIAO DUC VA DAO TAO

TAY NII\H

Ky rrrr cHoN Hec srNH Gror Lop 12 THpr voNG riNn
NAvl Hec 2ot3 - zot4
Ngay thi: 25 thing 9 nlnr Z0l3
thi: ToAx - Btroi ttri thri,ntrdt
Tho'i gian: 180 phut (khong kA thdi gian giao

Mdn

DE CHfI\H THTIC
thi
gont
co 0 t trang, thi ,sinh khong phai chep
@A

dA)

di vao giay thi)

Bii 1. (4 dient)
Cho ba s6 ducvnga,b, c tho6 mdn cli6u kiqn

Chirngminhr6ng:

a+b+c

* b * t ,3
-L

l+bc l+ca l+ab 2

= 3.

Ri i 2. (4 client)
Cho harn s6 f(x) xac clinh vcyi rnQi gia tri x > 0 rra
thoi m6n
+ y) = f(x).f(y), Vx > 0. vy> o
Irfx

Lrf28) - Q
Chu'ng mi nh r6ng: f(x) = 0, Vx > 0.

Ilii 3, (4 diem)
cho lrinh ch0'nh$t ABCD. Tr0n c6c cluong thang BC va
cD, l6y lAn lugt c6c di6m di
= e00. Gqi H rd hinh chi6u vu6ng g6c cria A tr6n MN.
rirn qu!
fl:il*"x,,il;i?r:n"

ffir

Bni 4.

g

aiam)

cho cludng thing d cri clua trtrc tam H cria ta,r gi6o
ABC. Gei ;,. dz, d3 lan lucyt la oac

dudng ttring dtii xung vdi d qua BC, CA, AB.
Chring rninh rang ba dtrong th[ng
dr, d2, d-, d6ng quy.
Bei s. G aiiim)

tno't minh rf,ng trong I I s5 thuc khirc nhau thu6c croan 10001 o6
the chon cluoc hai
[l;
,
so x vdy sao cho: 0 < x _y va t6n thf
So bao clarrh:
I-.1q,

Her


-

socrAo DUC vAoAo TAo rAy

NINH

xi, rnr cHeN Hec

srNH crOr Lop 12 THpr
NApr Hec zots -2014

nucrrvii ;iiii'&iM

ffi

rilbil

voNc riNn

idfi{''i;;il il ;il

nn6t)

CACH GIAI
BAi 1
ft diem)

cho ba so dwong o, b, c thod mdn rtiiu kipn a + b + c = 3. chftng minh

abc3

1+bc 1+ca 1+ab

a
l+bc

ring

2

abc

l+bc

)4 -i(ub+ac)

Bii 2 clto hdm saf@) xdc itlnh vdi mryi gid tri x ) 0 vd thod mdn itiiu
ft diem) ki€n:ltf* + y) - f(x).f(y), vx 2 o, vy > o
. Ch*ng minh rdng f(x) = 0, Vx > 0.
' Lf Q8): o
X6t xo >28 thi f(*o)=f(xo -28+28):f(xo-28).f(28)=0.
t(28) = f(l 4 +14): f(l4).f(14) = 0 =+ f(l4) = 0 .

Lim tuong tU nhu trOn thi f(x) LAp lupn tucrng tp thi co:

Vay f(x)

f(x) -

0, Vx > 14

0,

vx >7; f(x)

= 0,

vx

a!

- 0, Vx > 0
trang

I


rlei

3

g diem)

Cho hinh chfr nhQt ABCD. TrAn cdc itudng thhng BC vd CD, ldy tdn lw,qt cdc
ifiAm di itQng M, N sao c/ro
= 900. Ggi H ld hinh chi6u vuAng gdc ctia A
tr€n MN. Tiim qu! tich ctic iti6m H.

ffi

M(a; ma)

B(a; 0)

DUng hQ trpc toa d0 Oxy nhu hinh ve

Gi6 sri B(a;0), D(0; b),
y = mx thi M(a; ma).

l)Ni5u

&

)

v6i A = O(0;0).

0, b > 0 vd phuong trinh duong

th[ng AM ld

m*0:

Phuong trinh cludrng thdng AN
Phucmg trinh dudrng thing

h

y=

-l*.
m

Suy ra N(-mb;

b).

MN lA (b - ma)x + (a + mb)y - ab(l + m2) : g

(1)
Phuong trinh dulne th6ng OH le (a + mb)x

GillhA;il""s;'i"h

TU d6 thu dugc

*

*
aoI

(1,

=

;irfiil d;;;;
l.

-

(b

- ma)y

= 0 (2).

Phusng trinh ndy phuong trinh cfia dudng

thing BD.
2) NOu rD = 0 : Khi d6 M = B, N

Vfly quy tich

cdc

- D. Suy ra H eBD.

di6m H la dudrng thang BD.

Gtti chfi: Ndu thl sinh s* dung ki6n ththc itudng thiing Simson
gidc CMN vd di€m A dd gidi thi:
- Phdn thudn: 2il
- PhAn ddo: 2d

diSi

vdi tam

trang 2


Bni 4
G dieln)

Cho itudng *dng d iti,qua trgc tdm H cfia tam gidc ABC. Ggi d1 , dz, dj lhn lwgt
ld gdc itudng thdng ddi x*ng: vt6,i d qua BC, CA; AB. Ch*ng minh rdng ba dwd'ng
thdng & , dz, dj cl6ng quy.

Ggi A', B', I' lAn luqt ld c6c A', B', C' nim trdn du&ng trdn (O) ngoai ti-6p tu- gia. AnC.
vi d cft it nh6t mQt tr"ong cic dudng theng AB, BC, cA nOn gi& sir d c[t
BC t4i E. Gqi I ld gi,ro eti0m cira d1 vd d3.

N6uI=A'thi Ie(tr).
Xit I

I'hdp dOi xftng tqrc IIC hi6n

trpc AI3 bi6n

tr

Id

thanh

iJi,

phip

dOi xring

ilrdnh IrCt

DAt (BC,BA)_ er. O6c tam gi6c A'IJC', IBI2 cfin tai B vil
A'BC' - IBIz =Zrx II#n ARA'I: ABC'fr.
-ts-.+

Suy ra

-G-

* BI,C'

t'l6u I nim trong g6e
^

-+

-#

vA

-.

BA'I

^{B'
+

thi

A'IC': A'IB + RIC': BIzc'+ BIC'= lB00 -"2cr, cdn n6u I nim ngoii g6c
^
A'BC'
thi A'IC'=2a.
Suy ra b6n ditim

0,5

A', B, C', I dAu thuQc dulng trdn (O).

vi B'HI, - BI-IF. = BC'I = BB'I n6n duimg thlng c16i xrrng v6i d qua AC

,lTLll P'I.VAv {f: it,d, cliing quy t4i I.

l0sl
lo,ri

tl

t_:

l

trnng 3


Bni

ft

Chftng minh rdng trong 1I sd thgc khtic nhuu thuQc ttogn [1;1000], cd thA chgn
itugc hoi s6 x vd y sao cho: 0 < x - y < 1+ 3{/xy

5

diem)

) ,
xl,,xzt
tla SIu 11 Sto oac holldxl
G
JlA

ta

A

?

..Lp

,

Trtu gita ,n]le)trthi
\

./l\

l,

A

sO
SO

)A t,
t-,
S( oo

io

110 (k

=j.tr",:"
\

v' netnlth
6,. ffi

{{;
{\

0,5

,,,:,:.: ,11)

r0i do4n c6
rha&U, mo
h0n
[nbr&nEgrnt
thinh11t00 phi

ta I t;;1 0 l
C
lhia

Ill

1

Xlr.1.

.)

"r

r13

t

c o2'

s[,

inl i Dirichl

rguryrCn
le(o)n[

0,5

?

,\

Gia6sru hai sO do la
0t do anrnnh6i. G
thhl,u,a Qrcung mr6t
l

'x j

X,

vi
vd xi

Khid6

0
-fi
0,5

=0.(f-,--{E)'<1

0,5

=+0(Xi-xj -3!

0,5

=+0
0,5

1....--...-..--

I =o
-xj <1+3ff- (vi o<1f;.; -{tr<1)

0r5

I

Chgn

)(=xi,Y=xj thi 0...... H6t .....

0r5

o

trang 4