[/quote]
,
satisfying: 
,
and the function 
defined by![\[f(n)=\left\lfloor \frac{an+b}{cn+d}\right\rfloor\text{ for all } n\in\mathbb{Z}_{\ge 0}.\]](http://latex.artofproblemsolving.com/8/f/b/8fb372df4b44c3d7c0690b310b12c684134fee74.png)
Prove that the following are equivalent:
is surjective;
,
and 
.
and 
be on sides 
and 
such that 
.
be the circumcenter of 
and let 
.
be the n-th prime number in increasing order for 
.
satisfying 
for all 
be the midpoint of side 
,
be the feet of the altitudes from 
and 
,
and 
intersect at a point 
,
is perpendicular to line 
such that ![\[f(f(x)) + xf(xy) = x + f(y)\]](http://latex.artofproblemsolving.com/a/6/f/a6f84dd53eec985b90bad3df6580eeaf29d73195.png)
for all positive real numbers 
and 
.
,
with center 
and radius 
intersects the circle 
and 
.
.
such that for any real value of 
we have:
and 
are points, such that 
,
= 
and 
and 
passes through the fix point.
Consider an 
-
in the plane. An eligible set is a non-empty set of the form 
where 
determine the smallest possible number of eligible subsets 
such that :
Y by Akatsuki1010, AlastorMoody, nikaaryu, jhu08, Adventure10, and 2 other users
Let 
be an acute triangle and let 
be a line in the plane of triangle 
We've drawn the reflection of the line over the sides 
and 
and they intersect in the points 
and 
Prove that the incenter of the triangle 
lies on the circumcircle of the triangle
be an acute triangle and let 
be a line in the plane of triangle 
We've drawn the reflection of the line over the sides 
and 
and they intersect in the points 
and 
Prove that the incenter of the triangle 
lies on the circumcircle of the triangle 
at 
and the extension of 
at 
become internal bisectors of 
and 
is the internal bisector of 
become external bisector and internal bisector of 
and 
respectively 
is the internal bisector of 
Therefore, 
is the incenter of 
From the same fact that 
it follows that 
and 
Then 
and the result follows.
,
,
denote the reflections of 
,
,
.
and define 
be the incenter of 
.
and 
denote the reflections of 
across 
,
,
bisects 
.
,
,
.![\[
\measuredangle (\ell_B, \ell_C)
= - \measuredangle (\ell_B, \ell) - \measuredangle (\ell, \ell_C)
= -2\measuredangle (\overline{AC}, \ell) - 2\measuredangle ( \ell, \overline{AB} )
= 2\measuredangle (\overline{AC}, \overline{AB}).
\]](http://latex.artofproblemsolving.com/2/b/8/2b8e9747056d37e2368d194b720f6e1beabed5c2.png)
It follows that 
.
.
is actually the pole of the Simson line which is parallel to 
. 
which is not true when 
to 
to 
.
meet 
,
and 
are the angle bisectors of 
and 
,
is the internal angle bisector of 
and 
.
is the internal angle bisector of 
and 
.
.
,
is cyclic, as desired.
and 
bisects 
Hence, 
and 
bisects 
Similarly, 
and 
bisect 
and 
respectively, meaning that 
and 
the incenter of 
We then see that 
