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Đề thi và đáp án học sinh giỏi cấp tỉnh 2014 2015

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so GrAo DUC vA nAo rAo rAv NII.IH

xY rnr cHeN rrec sINH ctor Lop 12 THpr voNG
NAna Hoc zot4 - 2ors

rixn

NSi,y thit 24 th6ng 9 nf,m 2014

MOn: TOAN ru6i thi th* nh6t

Thli gian: 180 phft

pi

(khong k6 thdi gian giao dA)

of cnixn rrnlc
tnt

gim c6 0I trang, thi sinh khdng phdi chdp di vdo giAy thi)

Bii 1. 6 aiam)
Cho c6c s6 duong
Chimg minh
Beri 2.

6

x,y, zthoi mdn tli6u kiQn x + y * z=1.


L- * -=-l+
,ing
< 3.
e 3xz
- , +y
3z' + x + y
+z 3y' +z+x -;-]-

aieml

f thoe m6n di6u ki0n f(x) +2f (xy)= f(x +Zxy),Vx e
Chimgminhreng f(x+y)=f(x)+f(y),Vxe 1 ,Vye i

Cho hdm

BAri

3.

6

siS

i

,

Vy e

i


arcm1

Cho hai dudng trdn (O1), (O) cilt nhau tpi hai tti€m A, B vi PrP2 h mgt ti6p tuy6n
chung cira hai ttuimg trdn d6 (P1 thuQc (Or), Pz thuQc (Oz)). Gqi Qr vd Qz lAn luqt li hinh
chi6u vudng g6c cria Pr, Pz l6n dulng thing OrOz. Dulng theng AQr cdt (Or) tAi ili6m
thf hai M1, Dulng thing AQ2 cit (Oz) tai diem thri hai M2. Chimg minh reng ba cli6m M1,
M2, B thing hang.

Bifi 4.

6

aiaml

BCn trong hinh vudng c6 diQn tich bAng 6, d[t ba da gi6c c6 diQn tich cung b[ng 3.
Chimg minh ring vdi ba da gi6c dE d{t trong hinh vu6ng, ludn tim dugc hai da gi6c mi
di€n tfch phin chung cria chring kh6ng nh6 hon l.
.{

--- HET ---


so GIAo Drrc vADAo rAo rAyNIi\H

xi, rnr cHeN Hec srNH cror Lop tz rHpr voNG riNn
NAvr Hgc zot4 -zots
i\gny thi: 25 thfng 9 nIm 2AL4
Mdn thi: TOAX -Budi thi thfr hai
Thbi gian: 180 phrit (khdng kA thdi gian giao dA)


DE CHINHTHTIC
@A SOm cd

0l trang, thi sinh kh\ng phai chdp di vdo gidy

tht)

Bni 1. (5 diAm)
Giai h0 phuo-ng trinh

5)=64

I"'(3y+s
L"V (y'+ 3y + 3) = 12 +5 lx

Bei 2. (5 dihm)

(l
Cho day sO

(",) th6a mdn: J*'=5
Lt" +r) ' *n=

n2xn+r

* (+" * Z)XnXn*, Vn )

1, n e N


Tim limxn.

Bii

3. (5 diem)
Chring minh ring vdi n nguy6n ducrng
kh6ng c6 nghiQm nguy6n duong (x; y).

Bli

4.

vi

n > 3 thi phuong trinh 4xn + (x+1)2

=yz

6 adml

Cho tam gi6c ABC c6 ba g6c A, B, C deu nhgn vd AB > AC. Ggi D, E, F hn luqt ld
chdn c6c ducrng cao cta tam gifucABC vE ttr A, B, C xuiSng. c4nh di5i diQn.Ducrng thlng
EF cfu BC t4i P, dudng thang qua D vd song song vdi ef c6t c6c dudng tfrang AC va Ae
tuong img tpi Q vn R. Gqi M ld trung ei6m cpnh BC. Chr?ng minh ring b6n Oi6* P, Q, R,
M ctng thuQc mQt dudng trdn.

I

... H6t...



a
{/ri,

SO GIAO DUC VA DAO TAO TAY NINH

xi. Tm CHQN HQC SINH cror LOP 12 THPT voNG riNn
NAvt Hec 2or4 -zors

TTUONG

oAN

cnAu rm

vrON

roAN - su6r rnr rrrtl NnAr

trang

I


Bili

=

2


Chrfrng minh

rlng f(x

+ y) =

f(x)

+

f(y), Vx

f (0)

Cho v

Vdi

f(;

e R, Vy e IR '

f (Zxy), Yx e

v
x*0,dflt x-u; y=
2u

e lR'' VY € R


0

= f(2x)

Vai,

f(x +2xY)'Vx

thi f(u + v)

R,Vy e R

- f(u) + f(v), Vu * 0, Vv e R (1)

V6i x =0 thi f(x+y) =f(y)=f(0)+f(y) =f(x)+f(v), Vv e]R' (2)
(1) vd (2) suy ra: f(x + y) - f(x) + f(y), Vx e R, Vy e IR

i;

Bili 3
Q diam)

Cho hai tlulng trdn (O1), (Oz) cit nhau t4i hai Oi6m A, B vi PlP2li mQt fi6p
tuy6n chung cira hai dudng trdn d6 (P1 thuQc (Or), Pz thuQc (Or).Ggi Qr vi Qz
fa" foqt n trintr chi6u vu6ng g6c cria Pr, Pz l6n tlulng tneng OrOz.Dudng
tning AQr cit (Or) t+i aiOr" tfr1i hai M1, Dulngtneng AQz cit (O2) tli aiam
thrri hai M2. Chrfrng minh ring ba di6m Mr, Mz, B thing hing.

(hl)


(h2)

trang 2


r
1.,

(
I

trang 3


trang 4


so cIAo DUC YADAo rAo rAvNIi\H
xV rHr cHeN Hgc sINH cr6r Lop
NAna

12

rHpr voNG riNn

Hec zor4 -zots

i\giy thi: 25 thfng 9 nlm 20L4
M6; thi: ToAll - Buoi thi thfr hai
Thli gian: 180 ph fit (kh6rg kA thdi gian giao ai)

DE CHINHTHTTC
@A gom cd 0I trang, thi sinh kh1ng phai chdp di vdo gidy tht)

Bei L. (5 diem)
Giei h0 phuong trinh

[*'(3v
L*V

Bili2.

+

$'+

s

5):64

3y + 3)

: l2+5 lx

(5 diAm)

(1
Cho day sO

(*,)


th6a mdn:

J"'=5
Lr'+1) ' *n = r2Xn*r + ( 4n+2)XnXn+1 Vn > 1, n e N

Tim limxn.
Bei 3. (5 diem)
Chrlng minh rang vdi n nguy6n duong vd n > 3 thi phucrng trinh 4xn +
kh6ng c6 nghiQm nguy6n ducrng (x; y).

Bni 4.

6

(x+l)2 =y2

aidml

Cho tam girlc ABC c6 ba g6c A, B, C ddu nhgn vd AB > AC. Ggi D, B, F lan luqt ld
chdn c6c dudmg cao cta tam gifucABC ve tt A, B, C xuiSng c4nh e16i diQn.Duong thing
EF cat BC tai P, dudng tfring qua D vi song song v6i EF c6t c6c dudng thing AC vd AB
tuong img t4i Q vi R. Gqi M ld trung eiem cpnh BC. Chrlng minh ring b6n AiCm P, Q, R,
M ctng thu$c mQt dudng trdn.
,

oo.

HCt...



SO

GIAo DUC VA DAo TAo TAv NINH

KY THI CHQN HQC jsrNu cror Lop 12 THpr voNc
NAM HQC 2014 -201s
HrroNG nAN csAvr

riNu

rur vroN roAN - BUor rur rntruar
CACH GIAI

Bni 1
Q diem)

Giei h 0 phucrng trinh

I.'(ry
L*y

Vd'i x = 0,

h.6

+ ss) = 64

(y'+ 3y + 3) = lZ + 5lx

ph"""g


Dod 6 hQ tuong duong uoir

f_ __ 64
llv+5s-:'
*3

I

[r'+3
DAt t-

A
),thi

x

co

hQ:

y2 +3y

- 9+51
x

lZv+ 55 = t3
0,5

tr'+3y2+3y-3t+51


C0ngth.o;cffiipt,"ffic.iir,ethi.;6'
+6y+ 55 -

y3 +3y2

13

1,0

+3t+51

€> (y+1)3 +3(y +1)+ 51 - t3 +3t + 51 (*)

ii;=ffi;'e,

ffi;tio

Hdm'6
(Ho[c tii5n a6i (*) e (y+l-rl[fr+1)2 +(y+l)t

y+1=t)
Vdi

y=t-1

Vfly

hQ d5


*r,

0r5

"ry."

t

y+l

=T*'

] = o di5 suy ra

1r0

thi t3 -3t-52=0<>t-4<+x=1. Suy ray- 3.

cho c6 nghiQrn (1;3).

1r0

trang I


Bni

lr =5

2


Q diem)

Cho dfry s6

(*, ) thda miin: J*'

Lt,,+r) '*n-n2Xn*r +(4n+Z) xnXr+r Vnzl, ne N

Tim lirn;n.
Dtng phuong ph6p quy n4p chring minh duoc:
Bi6n ooi d6n.

DAt yn

=+

(n

(n + 1)'
Xn+l

Xn

)

0, Vn > l, n e N

+4n+2
-d

xn

-2 thi yr =3 va (r, * t)' (yn*, n2)- n'(y, +z)+ 4niz
-a
xn

* l)'yn+r - n'y n
n2

*Yn+r =-=,Yn

(n+l)
It'

(r,

Do d6

I

_

xn

v0y Iimxn

* l)'
I7
Y


./n

(" - 1)'
n'
(n- l)'"'22

Ft

rt-

+z:1+2-3+2n2 3
n-n

xn

J

\r

(rr*0-

Jn

)

n-

n'

- 3+2n2


0,5

I
1r0

2

Bei 3
Q diem)

chri'ng minh ring vrfi n nguyGn duong vi n > 3 thi phuorng trinh
4x' + (x +l)2 = y2 khdng c6 nghiQm nguyOn duons i*, v)
"

Gi6 st v6i n nguyCn ducmg vi n.) 3, phucmg trinh d6 cho c6 nghiQm
nguyOndu-(*+1)2 =(y_*_1)(V+x+t)

Vi

y-x-l

y+x+l

vd yj*+l cirngchEnho6c16;4x".hEnrren
cirngch5n

y_x_l


vd

y-x-l=2a thi y+x+l-2(a+x+l), suyra xn =a(a+x+l)
Do (a,a +I + 1) ; I n0n ton tai citc so nguyen u, v sao cho
Eat

a-un, a+x+1-vnrx-uv
Do n>3 n6n

uv+l = x+l = vn -un =(v_u)(vn-l *rn-2, +...+.rn-l) > l+uv +u2 .
v6 lf. vfly phucrng trinh dd cho kh6ng c6 nghiQm nguyEn ducrng.
trang 2


,?

^1

,'fi|)L i)

a(-

/

=

--Pd

t'>


f-

<--.

B+

cho tam gif,c ABC c6 ba g6c A, B, c tldu nhgn vi AB > AC. Gqi D, E, r B"
luqt lir chffn cfc dud'ng cao ciia tam gi{c ABC vE tri A, B, C xuiing-cirnh tISi
djgr.
thr_ng EF c_it tiC tgi P, d'u'd'ng thirng qua D vi song song vrii EF
Pf.ls
crt cic
dud'ng thrng AC vir AB tuo'ng ri.ng tqri e vn R. Ggi rvr re truig di6;
c?nh BC. chring minh rrng b6n tli6,r F, qln, Nr cnng thuQc mQt rludng trdn.

BAi 4
(5 di€rn)

I

..-l

I

Gqi M la trung di6m BC.

Do BEe = gFC = 900 non b6n di0m B, c, E, F cirng thuQc mQt dudrng
trdn. Suy ra PB.PC = PE.PF
BOn di€m D, E, F, ivi cung thuQc duong trdn Euler
PE.PF = PD.PM


cta tam gi6c ABC n€n

Do hai tarn giacAEF va ABC dong dang;
QR / IEF n6n
- - -^- -^
RBC - AEF = cQR. suy ratu gi6c ceBRnQi ti6p.
Do d6 DQ.DR

- DB.DC (2)

E[t MB=MC -a; MD-d; Mp=p
Khid6: PB=p+a; DB=a+d; pC-p-a;
Thay vdo (1) thi dugc: (p + a)(p

-

a) = (p

= (a+dXa-d):6;t;

-

CD

=a-d; Dp=p_d

d)p . Suy ra a2 = dp

+DB.DC _DP.DM (3)


Tt

(2) vA (3) suy ra: DQ.DR

VOy P, Q, R,

-

DP"DM

M cirng thuQc mOt dudng trdn.
,
......

HGt

eooooo

trang 3



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