[/quote]
,
be real numbers such that 
.
Let 
.
and 
,
as follows:
where 
are all the positive divisors of 
.
Find all positive integers 
.
Find all positive integers 
.
be a convex quadrilateral. Suppose that the circles with diameters 
and 
intersect at points 
and 
.
and 
.
,
,
attaining all positive values and positive numbers 
are given. Numbers 
satisfy
for every 
.
.
?
be reals such that 
and 
.
and 
points on a plane is 
,
such that: in every set of 
points of a plane, no three of which are collinear, consisting of 
consisting of 
disting positive integers smaller than 
are given. Prove that there exists a positive integer 
that can be written in the form 
,
in at least 
different ways.
,
be the set of all positive integers not exceeding ![\[
P_n(x) = \sum_{k \in S_n} x^{k - 1}
\]](http://latex.artofproblemsolving.com/3/a/c/3ac60cb70baf4dfcf297e9af87480a8ee34ba70c.png)
and a polynomial 
with integer coefficients such that![\[
P_n(x) = (x^{r_n} + 1) Q_n(x).
\]](http://latex.artofproblemsolving.com/7/0/e/70e19a9d3e0a7e97abc7aa08d3f23126c14ea7d0.png)
is irreducible in ![$\mathbb{Z}[x]$](http://latex.artofproblemsolving.com/4/b/7/4b7d768fffbb7a1a6163af392ed348f2d49d8783.png)
.
integers are such that 
is prime and 
.
such that 
.
Y by nguyendangkhoa17112003, Adventure10, Mango247
Let 
be the incenter of an acute-angled triangle 
. Line 
cuts the circumcircle of 
again at 
. Let 
be the foot of the altitude from 
to 
, and let 
be the reflection of across . Show , and are collinear.
be the incenter of an acute-angled triangle 
. Line 
cuts the circumcircle of 
again at 
. Let 
be the foot of the altitude from 
to 
, and let 
be the reflection of across . Show , and are collinear.
![[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
import graph; size(8cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -3.290326064830163, xmax = 14.768502142812732, ymin = -8.51945754635313, ymax = 5.406188148977132; /* image dimensions */
/* draw figures */
draw((1.868264528394029,4.917345109873203)--(0.,0.));
draw((1.868264528394029,4.917345109873203)--(8.,0.));
draw((0.,0.)--(8.,0.));
draw(circle((4.,-2.9102038140509894), 4.9466439369856525));
draw((1.868264528394029,4.917345109873203)--(1.87,0.));
draw((xmin, -2.243614393209075*xmin + 4.195558915300971)--(xmax, -2.243614393209075*xmax + 4.195558915300971)); /* line */
/* dots and labels */
dot((0.,0.),linewidth(3.pt) + dotstyle);
label("$B$", (0.11160107991946724,0.1920194411365769), NE * labelscalefactor);
dot((8.,0.),dotstyle);
label("$C$", (8.123616411479343,0.3191942876692734), NE * labelscalefactor);
dot((1.868264528394029,4.917345109873203),dotstyle);
label("$A$", (1.98743006627674,5.024663609379043), NE * labelscalefactor);
dot((2.7001832797497842,1.8626111554481317),linewidth(3.pt) + dotstyle);
label("$I$", (2.8140665687392667,2.06784842749385), NE * labelscalefactor);
dot((5.299816720250215,-7.683018783550109),linewidth(3.pt) + dotstyle);
label("$E$", (5.421150922659544,-7.50205877409156), NE * labelscalefactor);
dot((1.87,0.),dotstyle);
label("$D$", (1.98743006627674,0.3191942876692734), NE * labelscalefactor);
dot((2.7001832797497842,-1.8626111554481317),linewidth(3.pt) + dotstyle);
label("$J$", (2.8140665687392667,-1.6838095452206963), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]](http://latex.artofproblemsolving.com/6/2/4/6244ce371822fd1912d2cdde360f3c96be1c919a.png)
,
.
,
will intersect at the center of spiral similarity that sends 
.
.
,
,
.
.
and 
be the projections of 
.
.
and 
,
.
,
and Fixed 
,
be the midpoint of 
.
and the ex-center are concyclic, so 
and 
because of Menelao on triangle 
.
,
is on the radical axis of the circumferences of 
.
is parallel to 
it satisfies that 
which means 
so we are done.
.
.
and also consider 
and 
.
and 
.
and we are done 
be the midpoint of 
;
and 
.
as the tangency point of the 
as the antipode of 
are collinear. Then 
to 
and 
are similar, from which we get 
.
and 
are homothetic, implying the collinearity. But this is clear by the homothety through 
.
and reflecting about the 
and 
(where 
since inversion preserves cross ratio (of course, you can prove this lemma by direct computation, but I think this is the fastest way).
gives 
which implies that 
and 
.
.
,
.
which is easy.
is harmonic and then 
is the angle bisector of 
.
bisects 
.
lie on 
.
and 
is the 
and 
are homothetic triangles
and 
is the reflection of 
if and only if 
if and only if(by angle bisector Theorem and by the fact that 
is external bisector) 
are collinear. As 
i
and 
be the 
be the midpoint of 
bisects 
that sends 
)
1, we have: 
)
= 2 
= 2 
= 
)
touches the perpendicular from 
)
.
are collinear. This follows from Thales theorem that 
,
,
.
are collinear you can use homothety with a parallel line to 
are collinear, which is analogue to note 1.
,
bisects 
thus 
are collinear
