AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
the equation: 
, 
. Points 
lie on segments 
respectively. Points 
are such that 
and 
are equlateral, 
and 
lie on opposite sides of 
, 
and 
lie on the opposite siedes of 
. Prove that
+1 w
is a parallelogram. Take 
on the line 
so that 
and lies between and . Let the line through 
perpendicular to 
and the line through perpendicular to meet at 
. Show that 
.
and 
be two isogonal conjugate with respect to triangle . Let 
and 
be two points lying on the circle 
such that 
and 
are perpendicular and parallel to bisector of 
, respectively. Prove that 
and 
bisect two arcs 
containing and not containing , respectively, of 
.
be given positive integer. Suppose that there is only one positive integer 
which satisfies the equation
for some positive integers 
.Then find all possible values of the positive integer .
be positive real numbers such that 
. Find the minimum value of the following sum :
knowing that the denominators are positive real numbers.
such that 
, denote the sum of digits of in its decimal representation by 
. Find all polynomials 
with integer coefficients such that for any positive integer 
, the integer 
is positive and 
and 
as the orthocenter and incenter, respectively, of triangle 
. Let 
be the midpoint of 
. Prove that 
if and only if 
.
be an acute, scalene triangle with circumcenter 
and symmedian point . Let 
be the point on the circumcircle of triangle 
such that 
. Assume that 
. The hyperbola passing through , , , , and intersects the circumcircle of triangle at points 
and 
, distinct from and . Prove that 
is the perpendicular bisector of 
. is the intersection of the reflections of -median and -median across the angle bisectors of 
and 
, respectively. be a triangle with 
, incenter , and excenter 
. The incircle meets at 
. Define 
, 
. Show that the circumcircle of 
and 
are tangent to each other
prove that
such that
are being replaced with 
. Find the biggest number of steps, after which the number 
will appear on the board.
, because last four digits would have to be divisible by 
, but in each of this cases numbers 
are not. Similarly, we can see that 
leads to contradiction.
is coprime to 
. We know that 
is coprime with 
and 
for all , so function 
is bijective on integers relatively prime to 
(we count modulo ). To see that this function is bijective, we can also see that 
iff 
divides 
, but 
cannot divide neither 
, nor 
if 
is not divisible by , so we would have to have 
. The same argument works modulo 
. Therefore, we can have 
. Obviously, we can also have 
, so our answer is 
.
such that for any positive integer 
there exists a positive integer 
are equal to 
