[/quote]
,
be an acute triangle, 
the center of its nine point circle. Let 
be a point on the parallel through 
to 
such that 
and 
and 
a point on the parallel through 
to 
such that 
and 
and 
are the reflections of 
and 
are the intersections of 
and 
respectively with the circumcircle of 
and 
lies on 
.
undergoes the following procedure in order to obtain the number 
:
;
;
.
.
,
,
,
.
.
show that there exists a prime 
such that one can choose distinct integers 
such that p divides 
for all 
.
and a positive integer 
is given.
and for each 
![\[f_n(k)=\frac{a_n+\dots+a_{k+1}}{n-k}-\frac{a_k+\dots+a_1}{k}\]](http://latex.artofproblemsolving.com/0/4/1/041f80cf63f17c0f62c166021af474a5d305eb87.png)
Then for every integer 
the condition![\[a_{n+1}=\max_{0\leq k<n}(f_n(k))\]](http://latex.artofproblemsolving.com/b/1/f/b1f5aff787de8fc9c113cf8441a6b2900a19f24f.png)
is satisfied. Prove that this sequence must be eventually constant.
Prove that
of positive integers is called central if for every positive integer 
, the arithmetic mean of the first 
terms of the sequence is equal to 
of positive integers such that for every central sequence 
there are infinitely many positive integers 
.
![\[
\text{Find all functions } f : \mathbb{R} \to \mathbb{R} \text{ such that:} \\
f(a^4 + a^2b^2 + b^4) = f\left((a^2 - f(ab) + b^2)(a^2 + f(ab) + b^2)\right)
\]](http://latex.artofproblemsolving.com/2/a/7/2a73a5e58eab4a994fbd83160fb79a0b00152951.png)
be real numbers such that 
.
and 
.
.
of subsets of 
having the following properties:
distinct sets of 
where 
are nine distinct integers. Prove that there exists an integer 
such that for all integers 
the number 
is divisible by a prime number greater than 20.
,
.
Y by rkm0959, anantmudgal09, baopbc, ThE-dArK-lOrD, guangzhou-2015, buratinogigle, nguyendangkhoa17112003, Ya_pank, Adventure10, Rounak_iitr
Let 
be a non-cyclic convex quadrilateral. The feet of perpendiculars from 
to 
are 
respectively, where 
lie on segments 
and 
lies on 
extended. The feet of perpendiculars from 
to 
are 
respectively, where 
lie on segments 
and 
lies on 
extended. Let the orthocenter of 
be 
. Prove that the common chord of circumcircles of 
and 
bisects 
.
be a non-cyclic convex quadrilateral. The feet of perpendiculars from 
to 
are 
respectively, where 
lie on segments 
and 
lies on 
extended. The feet of perpendiculars from 
to 
are 
respectively, where 
lie on segments 
and 
lies on 
extended. Let the orthocenter of 
be 
. Prove that the common chord of circumcircles of 
and 
bisects 
.
be the isogonal conjugate of 
WRT 
respectively. Clearly, 
and 
so one of the intersection 
of 
and 
lies on 
Let 
be the second intersection of these two circles, then note that 
and 
are concyclic we get 
hence 
Let 
be the midpoint of 
Since 
are concyclic (the 9-point circle of 
)
be the orthocenter of 
.
concur.
.
is cyclic on circle 
.
so 
.
.
and 
,
,
,
are collinear. But since 
is the 
altitude of 
,
as desired, and the lemma is proved. 
be the midpoints of 
.
,
are collinear. It remains to prove that this line passes through the other intersection of 
.
be the unique right circumhyperbola of 
of 
as well. Thus, 
w.r.t 
,
so 
are collinear. Thus, 
bisects 
be a non-cyclic quadrilateral and 
be the orthocenter of triangle 
.
be the pedal circle of 
be the pedal circle of 
.
passes through the mid-point of 
.
be the projections of 
respectively. Let 
meet the sides 
respectively.
be the midpoint of 
.
.
is the common chord of 
.
be the isogonal conjugate of 
;
be the isogonal conjugate of 
are symmetric in 
.
lies on the pedal circle 
and 
hence 
all lies on the circle with diameter 
.
.
be the midpoint of 
and 
meet at 
,
and 
meet at 
.
,
and 
hence 
.
and 
are both tangent to 
and 
meet at 
.
by Brokard's theorem.
,
.
,
,
.
is cyclic, and similarly, 
is cyclic. Let 
,
are tangent to circle 
w.r.t. 
,
and 
.
be the orthocenter of 
.
and 
are concurrent one of the intersections of 
,
.
,
.
and 
are perspective and the concurrency follows. Now it is left to show that this point lies on both circles. Since 
and 
are cyclic, by Miquel's theorem it follows that 
it follows by homothety that 
passes through 
.
with radical axis of circles 
,
.
the circle passing through 
.
,
lies on line 
is one of the intersections of circles 
lies on the radical axis of 
.
,
with respect to 
,
