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Problem 2. Let be given polynomial P (x ) = rx 3 + qx 2 + px + 1
where p, q, r are real number, r > 0 . Consider the sequence (a n ) , n = 0, 1, 2,
… defined by
a0  1, a1   p, a2  p2  q
an3   pan2  qan1  ran (n  0).
Prove that if P (x ) has only one real root and has not multiple root then sequence
(an ) contains infinitely many negative terms.
Solution. From the conditions of prolem follows that characteristic equation of
recurrent relation x3 + px2 + qx + r = 0 has one real root and two conjugate
complex roots. Let these roots be -a, R(cos + isin), R(cos - isin) với a > 0, R
> 0, 0 <  <  then
an = C1(-a)n + C2Rn(cos + isin)n + C3Rn(cos - isin)n
where C1, C2, C3 some constants, C2, C3 are conjugate complex numbers. We put
C2 = R*(cos+isin) với   [0, 2), then
an = C1(-a)n + Rn(R*(cos+isin)(cosn + isinn) + R*(cos-sin)(cosn
- isinn)) = C1(-a)n + 2RnR*(cos(n + ))
Let by contrary that there is natural number n0 such that an  0 for all n  n0. Then
we have
0  an+1 + aan = 2Rn+1R*(cos((n+1) + )) + a2RnR*(cos(n + ))
= 2RnR*(Rcos((n+1) + ) + acos(n + ))
= 2RnR*.C.cos(n + *) (C > 0, *  [0, 2))
for all n  n0.
This is impossible because of the condition 0 <  <  and there are infinitely
many n such that n + *  (/2+k2, 3/2+2k).
Problem 3. Let a, b be positive integers. a, b and ab are not perfect squares. Prove
that at most one of following equations
ax2 – by2 = 1 và ax2 – by2 = –1.
has solutions in positive integers.
Solution.
Firstly we prove following lemma
Lemma. Let be given the equation Ax2 – By2 = 1
(1)
where A and AB are not perfect squares. Let (a, b) are minimal positive roots of
associated Pell equation x2 – ABy2 = 1
(2)
Assume that (1) has positive integral solutions and (x0; y0) be its minimal positive
roots. Then (x0; y0) are the only roots of system of equations
a = Ax2 + By2
b = 2xy
Proof. Let (x0; y0) be minimal roots of (1). Let u = Ax02 + By02, v = 2x0y0 then we
have
u2 – ABv2 = (Ax02 + By02)2 – AB(2x0y0)2 = (Ax02 - By02)2 = 1.
This shows that (u; v) are roots of (2). But (a; b) are minimal roots of this equation
so u  a, v  b.
We will prove that u = a, v = b.
Indeed, assume that u > a ; v > b. We have
a  b AB  (a  b AB )( a  b AB )  a 2  ABb 2  1
 (a  b AB )( A x0  B y 0 )  ( A x0  B y 0 )
 (ax0  Bby 0 ) A  (ay 0  Abx0 ) B  ( A x0  B y 0 )
In other hand
(a  b AB )  (u  v AB ) 

A x0  B y 0

2
 (ax0  Bby 0 ) A  (ay 0  Abx0 ) B  (a  b AB )( A x0  B y 0 )
 ( A x0  B y 0 ) 2 ( A x0  B y 0 )  ( A x0  B y 0 )
Put s = ax0 – Bby0, t = ay0 – Abx0 then the last inequalities can be rewritten as
s A  t B  x0 A  y 0 B
(3)
s A  t B  x0 A  y 0 B
( 4)
Now, we have (As – Bt ) = A(ax0 – Bby0)2 – B(ay0 – Abx0)2
= (a2 – ABb2)(Ax02 – By02) = 1.1 = 1.
We see that s > 0 since
s > 0 ax0 > Bby0 a2x02 > B2b2y02  a2x02 > Bb2(Ax02-1)
 (a2-ABb2)x02 > - Bb2  x02 > - Bb2.
The last inequality holds, so s > 0.
We also have that t  0 since t = 0  ay0 = Abx0  a2y02 = A2b2x02
(ABb2+1)y02 = Ab2(By02+1)  y02 = Ab2. But the last one is impossible
because A is not perfect square
.
If t > 0 then (s; t) are positive roots of (1), due the minimality of (x0 ; y0) we have
(1) s  x0 ; t  y0, thus
s A  t B  Ax0  B y0 , that contradicts (3).
Similarly, if t < 0 then (s ; -t) are positive roots of (1) and we come to a inequality
that contradics (4).
Hence u = a, v = b or in other words, (x0 ; y0) are roots of abovementioned system.
2
2
Let return to the problem. Assume that bot equations
ax2 – by2 = 1 (5)
and
bx2 – ay2 = 1 (6)
have solutions
Let (m ; n) be minimal roots of x2 – aby2 = 1, (x1; y1) – minimal roots of (5) and
(x2, y2) – minimal roots of (6). Apply the lemma, we have
m = ax12 + by12
n = 2x1y1
and
m = bx22 + ay22
n = 2x2y2
Since ax12 = by12 + 1 and ay22 = bx22 – 1 it follows that
ax12 + by12 = bx22 + ay22
2by12 + 1 = 2bx22 – 1
 b(x22-y12) = 1.
But it is impossible since b > 1.
Problem 4. Find all real number r such that for all positive real numbers a, b, c
the following inequality holds
a 
b 
c  
1

r 
 r 
 r 
  r  
b  c 
c  a 
ab 
2

3
Solution 1. Due the homogenusness of our inequality, we can assume that a + b +
c = 1.
Our inequality can be written in the form
r2
ra a
(b  c)   )  0
2
4 8
2
r
ra a r 2 1
Put S a  (b  c)     (4r 2  2r  1)a , and similarly for Sb, Sc we write
2
4 8 2 8
 (b  c) 2 (
the inequality as
 (b  c)
2
Sa  0
+ Take b = c then the inequality equivalent to Sb = Sc  0  4r2 – (4r2-2r+1)b  0.
Let a  0, b  1/2 we have 4r2 – (4r2-2r+1)(1/2)  0  4r2 + 2r – 1  0. So, it is
the necessary condition.
We prove that it is also the sufficient condition for which the inequality hols for all
a, b, c. Indeed, assume that a  b  c. Since 4r2 – 2r + 1 > 0 we have Sa  Sb  Sc.
Furthermore, in this case b  1/2 thus
Sb = r2/2 – (4r2-2r+1)b/8  r2/2 – (4r2-2r+1)/16 = (4r2 + 2r – 1)/16  0
Sb + Sa = r2 - – (4r2-2r+1)(a+b)/8  (4r2 + 2r – 1)/8  0
Hence
Sa(b-c)2 + Sb(c-a)2 + Sc(a-b)2 = Sa(b-c)2 + Sb((a-b)+(b-c))2 + Sc(a-b)2
= (Sa+Sb)(b-c)2 + (Sb+Sc)(a-b)2 + 2Sb(a-b)(b-c)
 0.
Solution 2. (official)
Let k be real number for which the inequality
a 
b 
c  
1

k 
 k 
 k 
  k  
b  c 
c  a 
ab 
2

3
(1)
holds for all a, b, c > 0. In this inequality take a = 1, b = c = n and let n   we
have
k(k+1)2  (k+1/2)3

8k(k2+2k+1)  8k3 + 12k2 + 6 + 1

4k2 + 2k – 1  0.
(2)

k
1 5
1 5
k 
.
4
4
Conversely, we prove that if k fulfils conditon (2) then inequality (1) holds for all
a, b, c.
Indeed, expanding (1), we have
(1) 
b
c
3 
ab
ca
bc
3
 a
k2


   k 


  
 b  c c  a a  b 2   (b  c)(c  a ) (a  b)(c  b) (a  b)( a  c) 4 
abc
1
 0
(a  b)(b  c)(c  a ) 8
 4k 2 [2a (a  c)( a  b)  2b(b  a )(b  c)  2c(c  a )(c  b)  3(a  b)(b  c)(c  a )] 
2k[4ab(a  b)  4bc(b  c)  4ca (c  a )  3(a  b)(b  c)(c  a )]  8abc  (a  b)(b  c)(c  a )  0
Let S = a3 + b3 + c3, T = a2(b+c) + b2(c+a) + c2(a+b), P = abc then the last
inequality can be rewritten as
4k2(2S – T) + 2k(T – 6P) + 6P – T  0

8k2.S + (6 – 12k)P  (4k2 – 2k + 1)T.
(3)
Now we will use two wellknown facts to prove (3):
1) S + 3P  T (Schur’s inequality)
2) 2S  T  6P (Simple AM-GM for (a, a, b); (a, a, c) …)
+ If k  1/2. Apply 2), we have
8k2.S  4k2T = (4k2 – 2k + 1)T + (2k –1)T  (4k2 – 2k + 1)T + 6(2k-1)P so
(3) is true
+ If k < 1/2. Apply 1), we have
2(1-2k)(S + 3P)  2(1-2k)T
Apply 2), we have
(8k2 + 4k – 2)S  (4k2 + 2k -1)T
Add two these inequalities side by side, we get (3). Problem is solved completely.
So, all possible value of k is: k 
1 5
1 5
k 
.
4
4
Problem 5. Let a circle (O) with diameter AB. A point M move inside (O).
Internal bisector of AMB cut (O) at N, external bisector AMB cut NA, NB at
P, Q. AM, BM cut cut circle with diameter NP, NQ at R, S. Prove that: median
from N of triangle NRS pass over a fix point.
Solution. Let I = NS  PQ, J = NR  PQ. Since M, N, Q, R lie on a circle, we
have AMP = RMQ = JBM = JNQ. It follows that M, N, B, J lie on a circle.
Similarly, M, N, A, I lie on a circle. Thus JBN = IAN = 900.
From A and B draw the lines perpendicular to MN that meet NJ and NI at U and V
respectively. We have
AU NA
NA

 AU  JP.
,
JP
NP
NP
BV
NB
NB

 BV  IQ.
.
IQ NQ
NQ
Hence
AU JP NA NQ NA NQ S NJP NA NQ NJ .NP. sin PNJ NA NJ

.
.

.
.

.
.

.
 1.
BV
IQ NB NP NB NB S NIQ NB NP NI .NQ. sin QNI NB NI
=> AU = BV. But AU // BV so AUBV is parallelogram => O is midpoint of UV.
On other hand,  NAI ~  NSP ~  NBJ ~  NRQ =>
NA NS NB NR
NA NS NB NR
NU NS NV NR
NU NV




.

.

.

.


NI NP NJ NB
NI NP NJ NQ
NJ NI
NI NJ
NR NS
=> UV//RS. It follows that median from N of triagle N pass throu midpoint O of
UV (Q.E.D).