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be the set of all functions 
with the following properties: is defined for all real numbers and takes only real values.
the following equality holds: 
such that
,
.
, the diagonal 
bisects neither the angle 
nor the angle 
. The point 
lies inside and satisfies ![\[\angle PBC=\angle DBA\quad\text{and}\quad \angle PDC=\angle BDA.\]](http://latex.artofproblemsolving.com/c/2/0/c20761f3eadd054958f40259f3d1c05f26279783.png)
Prove that is a cyclic quadrilateral if and only if 
.
be a cyclic quadrilateral in which the lines 
and 
meet at a point . Let 
be the point of the line 
, different from 
, such that 
. We construct the parallelograms 
and 
. Prove that the points 
lie on the same circle.
defined on the set of all positive integers and taking non-negative integer values, satisfying the three conditions:
for at least one 
;
for every positive integers 
and 
;
there are infinitely many positive integers such that 
for all 
.
be positive real numbers such that 
. Find the minimum value of the following sum :
knowing that the denominators are positive real numbers.
point 
was chosen such that 
and 
. Prove that if perpendicular from to passes from the intersection of diagonals of , then 
.
balls to 
boxes as she wishes. After that, Aslı and Zehra make alternated moves which consists of taking a ball in any wanted box starting with Aslı. One who takes the last ball from any box takes that box to herself. What is the maximum number of boxes can Aslı guarantee to take herself regardless of Zehra's moves?
, 
. The internal angle bisector of 
and the external angle bisector of intersect the ray at points 
and , respectively. Given that 
, prove that 
.
![\[f(xf(y) + x) = xy + f(x)\]](http://latex.artofproblemsolving.com/9/9/f/99f580ebc50846e6bc2c004667559922749a4dfa.png)
, , 
and be points on the same circumference with 
. Let and be the projections of onto and 
, respectively. Prove that 
cuts the segment 
into equal parts.
, , and be points on the same circumference with . Let and be the projections of onto and , respectively. Prove that cuts the segment into equal parts.
such that:
therefore