[/quote]
,
denote the set of all positive integers less than 
that are relatively prime to 
.
reducible over the rational numbers, given that it has integer coefficients?
is defined by 
and for all positive integers 
,
.
,
such that 
is a multiple of 
and incenter 
.
at 
respectively. Circumcircle of triangle 
intersects 
distinct from 
intersects 
distinct from 
points 
are concyclic.
,
on 
satisfies 
and 
.
be the point where the perpendicular from 
to line 
meets the angle bisector of 
.
holds.
be a positive integer, and let its positive divisors be![\[
d_1 < d_2 < \cdots < d_k.
\]](http://latex.artofproblemsolving.com/6/3/e/63ec82aba0d6c4eece89af6a1247a89498616281.png)
Define 
to be the number of ordered pairs 
with 
such that 
.
.![\[
n = p_1^{e_1} p_2^{e_2} \cdots p_k^{e_k},
\]](http://latex.artofproblemsolving.com/2/7/2/272dc2cc33ebbf695311cb61567fa0297e1299b6.png)
where the 
are distinct primes and the 
are positive integers.
such that 
.
that verify the following equation for all 
:
be a positive integer. Determine all sequences 
of positive integers such that
for all positive integers 
.
and 
be two circles with respective centres 
and 
intersecting in two distinct points 
and 
is an obtuse angle. Let the circumcircle of 
intersect 
and 
.
intersect 
; let the line 
intersect 
.
are concyclic.
.
be the feet of the altitudes from 
to 
respectively. Let 
be the circle with diameter 
.
be the intersection of 
and 
.
be the orthocenter of 
be the intersection of 
and 
be the lines through 
respectively. Let 
and 
.
and 
.
be the line through 
and let 
be the reflection of 
is tangent to 
.
where 
are integers in 
are each labeled with a positive integer less than or equal to 
such that every label is used exactly 
and 
such that the following property holds: For any two lattice points 
and 
that are both labeled 
,
and 
are both divisible by 
be an odd prime. Define 
over the positive integers as follows:
where 
such that 
.
such that 
and 
is a perfect square.
Y by cubres
Let 
be an arbitrary point on the side 
of the non-isosceles acute triangle 
. The circle with center and radius 
intersects the rays 
(after 
) and 
(after 
) at 
and 
. Prove that the orthocenter of triangle 
lies on a fixed line, independent of the choice of .
be an arbitrary point on the side 
of the non-isosceles acute triangle 
. The circle with center and radius 
intersects the rays 
(after 
) and 
(after 
) at 
and 
. Prove that the orthocenter of triangle 
lies on a fixed line, independent of the choice of .
be the reflection of 
be the reflections of 
and 
.
is the Steiner line of 
be the feet of the perpendiculars from 
to 
.
,
,
and ratio 
(or using midsegment arguments), it now suffices to prove that the orthocenter of triangle 
lies on a fixed line.
and 
be the feet of the altitudes from vertices 
- we claim that the desired fixed line is 
(one way to suspect this is by considering the cases 
and 
)
,
.
and the fact that 
is inscribed in the circle with diameter 
,![\[
AT = AD \cos \angle ACB.
\]](http://latex.artofproblemsolving.com/1/7/a/17aa0cd1b60cc7bd137b162e97a07a8cf96f7602.png)
![\[
\angle BAD = 90^\circ - \angle ADP = 90^\circ - \angle AQP = \angle TAB_1.
\]](http://latex.artofproblemsolving.com/1/0/6/106e53e7804ea5103c0137305f67fe784a807cd5.png)
Also, the quadrilateral 
is cyclic, so 
.
,
,![\[
AT = \frac{AD}{AB} \cdot AB_1 = AD \cdot \frac{AB_1}{AB} = AD \cos \angle ACB.
\]](http://latex.artofproblemsolving.com/7/a/5/7a526a24f8e97fcc5185a7bfdcfd3d65192f1064.png)
