Đề thi và đáp án học sinh giỏi tỉnh 2013

1,
sO crAo DUC vA DAo TAo

rAv Nrr\H

Ky rHr cHeN Hec srNH Gror Lo'p 12 THpr voNG
NAtu

Hoc zot3 -zot4

rixu

I\giy thi: 26 thftng 9 nf,m 2AI3

MOn thi: TOAN BtrOi ttri thir hai
TIrd'i gian: 180 phut (khong ke tho,i gian giao dA)

or csixn rsrlc

@a thi gom co

0l trang, thi sinh khong phai clt6p di

vdo giay thi)

Bai 1. (4 diem)
Giai phrrong trinh:

(Z*-

3)

+2x+2

= x2

-2x+

4

Bi i 2. (4 diLrn)
Giai

Bni 3.

hQ

phucmg trinh:

I*' - 6x + y + 1

1

lv'-v-x-3

g arcm1

* t I vol rnoi n nguyOn clucrng.
=1[r,
z\ un)
Chimg minh ring v6i moi sO nguyOn ducrng n Z 2 thi ,n , .,6. Tim limu,,.

Cho ddy s6 (r,.,) xac dinh

boi: ur=2 vi

un*r

Bei 4. & aiam)
Tim t6t cA cdc s6 tq nhi0n m sao cho v6i n ld mOt s5 tU nhiOn nAo rlri, ta
mn =1(modn)

Biri 5.

cci

thi m=1(modn).

g aie4

Cho tam gi6c dOu ABC canh a. M

AC sao .ho

AM

* AN = l.

MB

vi N ld hai cli6m di dQng lAn lucrt tren hai canh AB vd

NC

ring MBCN la tf gi6c ngoai ti6p dugc m6t dudng trdn.
b) Tim gi6tri lcyn nh6t cria diQn tich tam gi6c AMN theo a.
a) chring minh

IIeI --He ve ten thi s inir:
So bao danh:


s0crAo

DUC

vAEAo rAo rAy

NrNH

rY rnr cHeN Hec jsrNH cr6r lop

12

Trrpr voNG riNn

NAM HQC 2013 -2014

uudxc

oAN cnAna

rnr

nnoN

roaN

(Bu6i thi rhrl hai)

CACH GIAI
Bei

I

Gidi phwong

@ diem)

Di6ukien:
EAt

3)W=

trinh (z* -

- 2x + 4

xrl2

t- W.t>o

Phucrng

x2

thi x'=tz -2x-2.

trinh dd cho tucrng ituong v6i t2 -

Giai phucrng trinh tr6n tim dugc t

Vdi t - 2x- I thi dugc

W:2x-1<+

(Z*-:)t - 4x + 2 = 0

- -2 (lopi) , t = 2x: l

phucrng trinh

3x2-6x-1-o (do 2x-1>o)

Giai phuong trinh trOn tim dugc nghiQm
lopi
BAi 2
@ diem)

[*' =6x+y+11

Girti h0 pltwnng trinlt

1,,
He tuong ducnrg

-y-x-3

v6i I(": L(v'+ 8)-(y+2)=3-x

27)-6(x-3)-y+Z

I ti;:;;i*r*;;.;r;
+2)(y2
-3
-Zy+3)

Lfv

;i;;*:"*;;;;";; -ri. 3>0 n6n x>3=+y> -2=)x<3.V6 lli.
Iyg$-ju
Suy

'a

cho trudrne hqp x

ne;;;hio*

<3.

ilt;h at (3;-2)

.

trang

I


L

BAi 3

G diem)

Cho ttdysd (u" ) xdc illnh bdi: vr

=2vri

un*, =

clwong. Chfing minh rdng vd,i mgi s6 nguyAn

t

1[r"
2[ " * un ) vdi mgi n nguy\n

)
dwong n> 2 thi un, Jj.

n*

limun.

*

i

u2=

I( s) s
;1",.i ):;'^6

0r5

Gia sri b6t ddng thfc dring vdi

fl-k,tac6

1(
:r["k
Khid,6u
k+r

.2

+

s) ,,

1

Uk

,.16.
0r5

",1

Vay un ,.6, Vn e N* ,n) 2.
Un+l

::[",.*) .E =;(.,,-.8)*][i-.8)

-

1/
:r('"
- .6) *
Suy ra: 0 ( rn

Vi
Bni

4

@ diem)

*(6

-,.,6

-.6),nr,

Tim tdt cd ctic sd 4r nhihn m sao cho
mn = I (modn) thi m: I (modn).

N6u m

>

vdri

I

3

=o non lim(u, -..6)

NCU m = 1 thi bai to6n

0r5

- o. Suy ra limun - .,6

0r5

n

>t.Z

Gi6 sti p ld u6c nguy0n t6 ntrO nhdt cria n, khi dO Z":1 (modp) (*)

Tt (*) ta c6 p ld s0 16, suy ra n li

0,5

n ld mQt sa qt nhihn ndo itri, ta cd

thoi mdn.

=2,tac6 2" =1(modn) yoi -oi

VO,y t6n

1

- u. ) . 1(", - .6)

.#(r,

rim#(,,-.,6)

=

sO

le, do d6 (n,

p-1)

tai cdc so nguy0n d,b sao cho an + b(p - l)

-1

0,5
.

= 1.

0,5

.

0r5

Theo dinh lf Fermat nh6 thi: 2t -- 2nu.2@-r)b : I (modp).V6 lf.

N6u m > 2,ta"6 *(m-l)' =[1+(m-1;1('-rt2 = 1 (mod(m-t)r)

0r5
0r5
.....................-..-.....

MAt khSc
T6m lpi

m/l

(mod(m

-t)').

Bdi to6n kh6ng tho6 man.

ffr:1 ld c6c gi|tri cAn tim

0r5
0r5

trang 2


Bni

5

ft diem)

Cho tsm gidc it1u ABC cgnh o. M vd N ld hai iti6m di ctQng ldn lwgt trhn
cgnh AB vd AC sao

hui

N*4N 1.
"6 MB NC =

a) Ch*ng minh rdng MBCN ld tir gtdc ngogi ti6p itwgc mQt itwdng trbn.
b) Ttm gid tr! td,n nhdt crta diQn tich tam gidc AMN theo a.

,A
XA,

B

A\

C

a) Dat AM = x, AN = y . Ggi E, F l6fl luqt ld trung rti6m cria AB, AC.

ri,

1 suy r" {Y:Y.
**g
MB NC=
LANI < NC

vay M, N ran ruqt thuQc c6c do4n

AE vd AF.
Ttr gi6thi6t

0,5

ta.o,

u

YB *

=

u,

3
IC = I =+ =L*j=
NC
MB NC =

MB

:i1YP:XE3P_I9:_

I
>

f

MN2 = AM2 + A}-.{2 _ AI\4.ANI
= (a - MB)2 + (a - NC)'

-

(a

- MB)(a - NC)

0,5

= a2 +MB2 + NC2

- 2a(MB + NC) + a(MB + NC) - MB.NC

= a2 +MB' +NC2

-2a(MB

+ NC) + 3MB.NC

- MB.NC

- a' +MB2 +NC2 -Za(MB + NC) + 2MB.NC
- (MB+NC -a)'

0,5

+ MN = MB +NC - a )MN

+BC = MB +NC + MBCN ld trl gidc
ngoqi.ti6p dugc. Dudng trdn nQi ti€p tti gi6c MBCN ctng ld nQi tiep tam gi6c ABC.
b) DiQn tich tam giac AMN la So*, = lapl.AN.sin600 _
2

M4t khric a = x + y + MN = x +

xy.\6

0r5

0r5

4

y+.ffiy' -t >Z,t"y+.ffi = 3r&y

0r5

trang 3


=+ xy

a2 Sor*
=;3
'u'Ji
36
-

v6y maxsor*

0,5

:U=f ex-y= a \- ANd ANI
3 AB AC 3.
36
/'--A

1

0,5

f

..

.... HOt o. o. . .

trang 4