[/quote]
,
,
denote the number of its positive divisors (including 1 and itself). Determine all positive integers 
for which there exists a positive integer 
.
let 
be the foot of the altitude from 
,
,
be the incenter, 
-
and 
the other intersection points of the circle 
with the lines 
and 
,
.
such that![\[(a^2-b^2)(b^2-c^2) = abc.\]](http://latex.artofproblemsolving.com/7/1/8/7188261c40a45f5e15533100432d36abbf515efd.png)
.
that 
is a prime.
is divided into rectangles with two cells. Two rectangles are called adjacent if they share a segment of length 1 or 2. In each rectangle the amount of adjacent with it rectangles is written.
is an integer, then it is a perfect square.[/quote]
.
is at least rational, so that 
must be a perfect square then. Using Conway's topograph method, I have found out that the smallest non-negative pairs 
for which this happens are 
and 
,
,
The eigenvalues of that matrix are irrational, however, so that any calculation which uses powers of that matrix is a little cumbersome. There must be an easier way, but I cannot find it. Can you?
interesting if 
of 
and 
is also a divisor of 
,
and 
.
is prime, 
with 
and 
and circumcenter 
and 
be points on the line segments 
and 
respectively such that 
is a parallelogram. Prove that 
.
that satisfy 
for all 
and 
satisfies the following recursion formulas: 
,
.
and 
be points on segment 
denote the circle passing through 
denote the circle passing through 
)
.
Y by Davi-8191, Adventure10, Mango247, Funcshun840
Let 
be the circumcircle of the triangle 
. Two circles 
are tangent to each of the circle and the rays 
, with 
interior to , 
exterior to . The common tangent of 
and the common tangent of 
intersect at the point 
. Let 
be the midpoint of the arc 
(not containing the point 
) on the circle , and the segment 
be the diameter of . Prove that 
, and 
are collinear.
be the circumcircle of the triangle 
. Two circles 
are tangent to each of the circle and the rays 
, with 
interior to , 
exterior to . The common tangent of 
and the common tangent of 
intersect at the point 
. Let 
be the midpoint of the arc 
(not containing the point 
) on the circle , and the segment 
be the diameter of . Prove that 
, and 
are collinear.
be the tangency points of 
and 
with 
It's well-known that 
and 
go through the exsimilicenter 
and insimilicenter 
of 
and the incircle 
(for this use Monge & d'Alembert theorem for 
and similarly for 
)
then the pencil 
is harmonic 
quadrilateral 
is harmonic 
be the Incenter, A-excenter of 
,
and 
be the projection of 
on 
.
be the tangent point of 
with 
be the tangent point of 
with 
,
are isogonal conjugate of 
(well-known) ,
,
is a harmonic quadrilateral 
and 
are collinear .
is radical axis of 
be incenter and A-excenter of 
.
be the points of tangency of 
.
be the inversion followed by reflection about 
such that 
.
,
.
,
is the midpoint of 
,
.
are on the radical axis of 
,
are all on this radical axis.
in 
with radius 
composed with reflection 
be the touch point of incircle,excircle,the foot of A-altitude,the foot of 
than it's well-known that 
are collinear.Projecting 
on 
be points on 
respectively. Let 
be points on 
,
.
,
are concyclic.
be the midpoint of arc 
in 
and 
be the tangency points between 
and 
,
and 
are collinear, where 
,
is the point at infinity along 
is harmonic and we are done.
,
the points of tangency of the excirlce and incircle 
(homothety) and 
and 
are isogonals wrt angle 
reflecting in the 
,
.
we get 
,
,
.
and 
be the 
be the 
.
is cyclic.
,
are collinear. From 
.
be the point at infinity on line 
.
are internal and external angle bisectors of 
,
.
as the touchpoints of 
resp. with 
resp. as the incenter and 
excenter and 
the midpoint of 
Then 
.
is the symmedian of 
which is obvious. Darn
the intouch,extouch point of inMixtilinear and exMixtilinear with 
be the incenter A-excenter and let 
be on 
, 
, 
are the midpoints of 
are on the A-mixtilinear circle which is 
also on the exterior A-mixtilinear which is 
is the radical axis of 
are collinear so 
are colinear
and 
are collinear
![[asy]
size(10cm);
defaultpen(fontsize(10pt));
defaultpen(linewidth(0.35));
dotfactor *= 1.5;
pair A = dir(130), B = dir(200), C = dir(340), I = incenter(A,B,C), D = foot(I,B,C), E = extension(A,I,B,C), M = (B+C)/2, F = 2M-D, G = foot(A,B,C), X = foot(E,A,M);
draw(A--B--C--A);
draw(incircle(A,B,C), gray);
draw(circumcircle(A,D,X)^^arc(circumcenter(A,F,X),circumradius(A,F,X),150,315), heavygreen);
draw(A--G^^A--E^^A--M^^A--F, blue);
draw(circumcircle(A,G,E), purple+dashed);
dot("$A$", A, dir(135));
dot("$B$", B, dir(200));
dot("$C$", C, dir(340));
dot("$D$", D, dir(270));
dot("$E$", E, dir(270));
dot("$F$", F, dir(270));
dot("$M$", M, dir(270));
dot("$G$", G, dir(270));
dot("$X$", X, dir(90));
[/asy]](http://latex.artofproblemsolving.com/1/1/7/117ff36ba92e521c053242551bddf0e70316e785.png)
,
,
.
at 
is cyclic.
,
.
be a function of point 
such that ![\[ f(\bullet)=P(\bullet, O_1)-P(\bullet, O_2). \]](http://latex.artofproblemsolving.com/4/2/2/4222e494302b8bcce2d6d66edece9065305760ef.png)
It is known 
,
and 
,
be the tangency point of 
,
respectively. We want 
.![\[ 4f(M)=2\left(f(I)+f(I_A)\right)=f(P)+f(Q)+f(R)+f(S)=0. \]](http://latex.artofproblemsolving.com/a/2/d/a2dc0d0dc937b917998ae302dea75183a1bd489b.png)
Done.
-
and 
be the circles passing through 
.
,
is tangent to 
being tangent is similar. It suffices to show that 
.
.
and
as desired. 
