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.
then 
.
+1 w
people who do not know each other. Prove that it is possible to introduce them in such a way that no three of them have the same number of acquaintances.
with circumcircle 
, excircle 
that is tangent to segment 
at 
. 
-mixtillinear incircle 
of 
is tangent to 
at 
. Suppose 
is the midpoint of the arc 
, and 
. Show that if the circles 
have the same radius, then tangents to at 
intersect on 
.
such that 
is a perfect square for all integers 
and 
.
such that for all 
the following holds:![\[f(x)+f(y)=f(xy)+f(f(x)+f(y))\]](http://latex.artofproblemsolving.com/a/c/a/aca098172ea7ce5663e9dd0dd10c2f1495630230.png)
be a convex hexagon, such that the triangles and 
are equilateral and the diagonals 
and 
are concurrent. Prove that 
or 
to the right; to the left.
squares north of the initial square after some finite number of moves if it is initially facing:
satisfying 
is a perfect square
, find all possible positive integer values of![\[
\frac{a}{b} + \frac{b}{c} + \frac{c}{a}.
\]](http://latex.artofproblemsolving.com/3/2/b/32b41635c159921e11245369e66185ec69b0d785.png)
be the set of the first 
perfect squares of nonzero integers. Suppose that 
for some 
. Here 
stands for the set 
. Prove that 
holds for every 
.
such that 
and 
be two chords of a circle with center 
such that they intersect at right angles inside the circle at the point 
. Suppose 
and 
are midpoints of the chords 
and 
respectively. Prove that 
is a parallelogram.
and 
is right angled, it follows that, 
Similarly, we can prove 
which follows that the opposite side in the quadrilateral 
are equal, hence, it's a parallelogram
meets 
.
and 
, we obtain 
. In the same way we have 
. Thus 
and 
.