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JTST, Greece, 2016
statements
Problem 1 (a) For every real number x  0 prove that: x 3  3x  2 .
(b) For all real numbers x, y, z  0 , prove that:
 y
x2 y y 2 z z 2 x
z
x 


 2     9 .
z
x
y
 xz xy yz 
(1)
When does equality hold?
Problem 2. Let c( O , R ) be a circle and A, B two diametrically opposite points. The bisector of the angle
AB̂O intersects the circle c( O , R ) at point C , the circumcircle of the triangle AOB (say ( c1 ) ) at
point
K
and the circumcircle of the triangle AOC (say
( c2 ) ) at point L . Prove that the point K
is
the circumcenter of the triangle AOC and the point L is the incenter of the triangle AOB .
Problem 3. The positive integer n is such that n 2  9 has exactly 6 positive divisors. Prove that
gcd  n  3, n  3  1 .
Problem 4. Vangelis has a box containing 2015 white and 2015 black balls. He follows the following
procedure: He chooses randomly two balls from the box. If both are black, then paints one of them white
and put it in the box, while drops the other out of the box. If both are white, then he keeps one of them in
the box and drops the other out of the box. If one ball is white and the other is black, then keeps the black
in the box and drops out the white ball. He follows the procedure, until only 3 balls remain in the box.
Then
he
realizes
that
in
the
box
there
exist
balls
of
both
colors.
Determine how many white and how many black balls exist finally in the box.
Solutions
Problem 1 (a) For every real number x  0 prove that: x 3  3x  2 .
(b) For all real numbers x, y, z  0 , prove that:
 y
x2 y y 2 z z 2 x
z
x 


 2     9 .
z
x
y
 xz xy yz 
(1)
When does the equality hold?
Solution
(a) We have
x3  3x  2  x3  3x  2  0  x3  x  2 x  2  0 
x( x  1)( x  1)  2( x  1)  0  ( x  1)( x 2  x  2)  0  ( x  2)( x  1) 2  0
which is valid, because x  0 .
(b) The inequality is equivalent to
z  2 2 x  2 2
 y     z    9 .
x 
y y
z
2
2
From question (a), by dividing both parts by x we obtain x   3 .
x
y  2 2
x  
z
x
(2)
Therefore it is enough to prove:
 y z x
y z x
3     9     3 .
z x y
 z x y
(3)
The latter is valid, as we can easily see, by applying the inequality of arithmetic and geometric mean as
follows:
y z x
yzx
   33
3
z x y
zxy
The equality holds if and only if x  y  z  1.
(4)
Problem 2. Let c( O , R ) be a circle and A, B two diametrically opposite points. The bisector of the angle
AB̂O intersects the circle c( O , R ) at point C , the circumcircle of the triangle AOB (say ( c1 ) ) at
point
K
and the circumcircle of the triangle AOC (say
( c2 ) ) at point L . Prove that the point K
is
the circumcenter of the triangle AOC and the point L is the incenter of the triangle AOB .
Solution
Segments OB ,OC are equal, as radii of the circle ( c ) , and so the triangle OBC is isosceles. Hence:
Bˆ1  Cˆ1  xˆ
BC
(1) .
Figure 1
is the bisector of the angle OB̂A , and hence
The angles
Bˆ1  Bˆ2  xˆ
B̂2
and
Ô1
(2) .
are inscribed in the same circle and correspond to same arc
( c1 ) . Hence
ˆ  xˆ
Bˆ2  
1
(3) .
OK
of the circle
Similarly we have KO  KC and the triangle KOC is isosceles and hence:
Oˆ 2  Cˆ1  xˆ
(4) .
From (1)  (4) we conclude that: ˆ 1  Ô2  x̂ , i. e. OK is the bisector, hence and perpendicular
bisector of the isosceles triangle OAC . The point K is the midpoint of the arc OK ( BK is the
OB̂A ). Hence the perpendicular bisector of the cord O of the circle ( c1 ) (which is also
side of the isosceles triangle OAC ), is passing through point K . It means that K is the circumcenter of
bisector of
the triangle OAC .
From (1), (2), we have: B̂2  Ĉ1  x̂
and since ˆ 1
Â1  Â2  Ô1  Ô2
and hence AB // OC . Therefore OÂB  AÔC , that is
 Ô2  x̂ , we conclude that:
Â1  Â2  2Ô1  2 x̂ .
The angles
Â1
and
Ĉ1
are inscribed in the circle
( c2 ) and correspond to the same arc OL . Hence
Â1  Ĉ1  x̂ .
From the last two equalities we have that Â1  Â2 , that is AL
is the bisector of the angle BÂO .
Problem 3. The positive integer n is such that n 2  9 has exactly 6 positive divisors. Prove that
gcd  n  3, n  3  1 .
Solution
In order for the positive integer n to have exactly 6 positive divisors, it must be of the form n 2  9  q5
n2  9  p 2 q , where p, q
n 2  9  q 5  (n  3)(n  3)  q 5
or
primes
mutually
different.
In
the
first
case
we
have
(1)
It gives that
n  3  q s and n  3  q t , with t  s and t  s  5 .
t
s
By subtracting we get q  q  6  q s (q t s  1)  2  3 and hence q 2,3.
For q  2 (1) gives n 2  41 , (it has no integer solutions), while for q  3 , (1) gives n 2  252 ,
(impossible).
In the second case n 2  9  p 2 q , it follows that (n  3)( n  3)  p 2 q , and so
2
( n  3  pq and n  3  p ) or ( n  3  q and n  3  p )
2
or ( n  3  p and n  3  q ).
From the first case we get p  q  1  6 , and hence ( p  3 and q  3 ), which must be rejected because
p  q.
From the last two cases we conclude that gcd  n  3, n  3  1 .
Problem 4. Vangelis has a box containing 2015 white and 2015 black balls. He follows the following
procedure: He chooses randomly two balls from the box. If both are black, then paints one of them white
and put it in the box, while drops the other out of the box. If both are white, then he keeps one of them in
the box and drops the other out of the box. If one ball is white and the other is black, then keeps the black
in the box and drops out the white ball. He follows the procedure, until only 3 balls remain in the box.
Then
he
realizes
that
in
the
box
there
exist
balls
of
both
colors.
Determine how many white and how many black balls exist finally in the box.
Solution
We will consider what happens at every step of the procedure for the number of white and black balls. In
the case of selection of two black balls, no ball is going back to the box and so the number of the black
balls is reducing by 2. In the second case of selection of two white balls, the number of the black balls
has not any change. In the third case of selection of balls of both colors, also the number of the black
balls in the box remains invariant. Therefore we observe that, in every step, the number of the black
balls in the box remains invariant or it is reducing by 2. Since their initial number is 2015 (odd) at every
step their number will remain odd. Therefore in the case of 3 balls in the box of both colors, we conclude
that one of them must be black and the other two must be white.