[/quote]
,
and 
Prove that
Prove that
Prove that
satisfied 
Find minimum and maximum of 
such that:
with all 
is real number
be the set of integers. Determine all functions 
such that, for all integers 
and 
,
Proposed by Liam Baker, South Africa
be a positive integer. Find all real solutions 
to the system:![\[a_1^2 + a_1 - 1 = a_2\]](http://latex.artofproblemsolving.com/a/4/e/a4e26b83c929b82800576bac11e12de0c3aa1cf6.png)
![\[ a_2^2 + a_2 - 1 = a_3\]](http://latex.artofproblemsolving.com/9/9/5/995e8eec93828b82901b0639463f907f5ff95d08.png)
![\[\hspace*{3.3em} \vdots \]](http://latex.artofproblemsolving.com/e/e/1/ee15e1fefe2ad550d3c9b1c51207eaf66fbfa41f.png)
![\[a_{n}^2 + a_n - 1 = a_1\]](http://latex.artofproblemsolving.com/7/c/7/7c727d1f2044c1bf7c80981e1074bb3cde807b8f.png)
of positive integers is called central if for every positive integer 
terms of the sequence is equal to 
of positive integers such that for every central sequence 
there are infinitely many positive integers 
.
be the circumcircle of triangle 
,
be the second intersection of 
and 
,
be a transversal of 
be intersections of 
with 
be a tangent line of 
is cyclic
be positive integers that satisfy the inequality![\[
\frac{a}{b} < \frac{x}{y} < \frac{c}{d}
\]](http://latex.artofproblemsolving.com/8/d/9/8d982d52e655546ac29039fa6e440e68821bfe7f.png)
with 
.
,
.
and 
to 
to 
and 
as the reflections of 
and 
be the circumcircle of 
.
with 
the intersection of ray 
with 
is the circumcenter of 
can be inscribed within a circle.
Y by arturfetahu, Mango247
Let 
be the incenter, 
and 
midpoints of sides 
and 
of a triangle 
. Denote by 
and 
the midpoints of the arcs and of circumcircle of which do contain the other vertex of the triangle. If points , and are collinear prove that:
be the incenter, 
and 
midpoints of sides 
and 
of a triangle 
. Denote by 
and 
the midpoints of the arcs and of circumcircle of which do contain the other vertex of the triangle. If points , and are collinear prove that:
and 
.
and 
is the 
,
.
lie on a circle centered at point 
,
.
,
.
.
,
,
.
and 
,
,
.
,
.
and 
.
and 
.
touches 
.
.
and 
.
.
,
and we're done!Also note that 
is the 
.
.
,
is the external angle so 
because 
symmedian. So we get that 
is the angle bisector of 
and done!
regardless of the collinearity of 
