[/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 
.
be on sides 
and 
respectively in 
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 
intersects the circle 
and 
.
.
such that for any real value of 
we have:
,
= 
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 Adventure10, ehuseyinyigit
Let 
be a isosceles triangle with 
an 
be the foot of perpendicular of 
. 
be an interior point of triangle 
such that 
and 
.
and 
intersects at 
, 
and intersects at 
. Let 
be a point on ![$[AB]$](//latex.artofproblemsolving.com/a/d/a/ada6f54288b7a2cdd299eba0055f8c8d19916b4b.png)
and 
be a point on 
and not belongs to ![$[AP]$](//latex.artofproblemsolving.com/f/6/2/f6235de9bde9e47652ee5f3cfbad1a2666873220.png)
satisfying 
and 
. Show that 
be a isosceles triangle with 
an 
be the foot of perpendicular of 
. 
be an interior point of triangle 
such that 
and 
.
and 
intersects at 
, 
and intersects at 
. Let 
be a point on ![$[AB]$](http://latex.artofproblemsolving.com/a/d/a/ada6f54288b7a2cdd299eba0055f8c8d19916b4b.png)
and 
be a point on 
and not belongs to ![$[AP]$](http://latex.artofproblemsolving.com/f/6/2/f6235de9bde9e47652ee5f3cfbad1a2666873220.png)
satisfying 
and 
. Show that 
,
,
.![\[\frac{\sin (\theta + \alpha)}{\sin (\theta - \alpha)} \cdot \frac{\cos (\alpha + \beta + \theta)}{\sin (\alpha - \beta)} \cdot \frac{\sin \beta}{\cos (\beta + \alpha)} = 1 \quad (..1..)\]](http://latex.artofproblemsolving.com/2/1/a/21afc45eee8643c344f779b9c68696477b09ee62.png)
.
.
,![\[\frac{\sin (\theta + \alpha)}{\sin (\theta - \alpha)} \cdot \frac{\sin (\alpha + 2\beta + \theta) - \sin (\alpha + \theta)}{\sin (\alpha + 2\beta + \theta) - \sin (\alpha - \theta)} = 1 \\ \\ \Rightarrow (\sin (\theta + \alpha) - \sin(\theta - \alpha))\sin(\alpha + 2 \beta + \theta) = \sin^2(\theta + \alpha) - \sin^2(\theta - \alpha) \\ \\ \Rightarrow \sin (\alpha + 2\beta + \theta ) = \sin (\theta + \alpha) + \sin(\theta - \alpha) = 2\sin\theta \cos \alpha \quad (..2..)\]](http://latex.artofproblemsolving.com/2/f/7/2f7dbf6a5d29c7b350f092d79b79d0eabe37aa89.png)
,![\[\frac{\sin (\alpha + 2\beta + \theta)}{\sin \theta} \cdot \frac{\sin \alpha}{ \sin (2\alpha)}\cdot \frac{\sin \angle RST}{\sin \angle RTS} = 1 \quad (..3..)\]](http://latex.artofproblemsolving.com/9/1/6/916e29fe7c0856cf5f2af4a0d812f119d5449449.png)
and 
(by 
)
will be![\[\frac{2\sin\theta \cos \alpha}{\sin \theta} \cdot \frac {1}{2\cos \alpha} \cdot \frac{\sin \angle RST}{\sin \angle RTS} = 1 \Rightarrow \sin \angle RST = \sin \angle RTS \Rightarrow RT = RS\]](http://latex.artofproblemsolving.com/f/0/a/f0aa718e3950acb39a36077a3b5e43d60b308469.png)
.
,
,
.
,
the circumcircle of 
,
the intersections of 
with 
yields that 
are collinear. Since 
,
,
.
and 
,
and 
are congruent.
