[/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 
to the side 
and 
,
,
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 
be an acute triangle with orthocenter 
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 Amir Hossein, nguyendangkhoa17112003, Adventure10, Mango247, ItsBesi, Rounak_iitr
The incircle of triangle 
touches the sides 
, 
, 
at 
respectively. 
is a point inside triangle of such that the incircle of triangle 
touches at 
, and touches 
and 
at 
and 
respectively.
Show that
are concyclic.
touches the sides 
, 
, 
at 
respectively. 
is a point inside triangle of such that the incircle of triangle 
touches at 
, and touches 
and 
at 
and 
respectively.Show that
are concyclic.
and 
,
,
are concurrent,and let 
,
,
,
,
are concyclic.
.
.
,
(I mean polar wrt the incircle). Since 
and 
belongs to the polar of 
.
,
.
and 
to find that 
and 
coincide (they must both be equal to the harmonic conjugate of 
,
.
and 
.
,
is on the circle with center 
and radius ![$[BD]$](http://latex.artofproblemsolving.com/3/2/6/3261e689901bce018ecdef47b9bc60a78ead3746.png)
.
.
and 
.
.
are concyclic.
; 
; 
.
.
(or 
,
,
and 
are concurrent), and 
is such that 
is harmonic (because 
,
and 
concur), than 
.
are concurrent, you can use Menelaus' Theorem on 
,
.
gives
gives
we have
and we are done!!
is the radical center of 
and 
.
are tangent to their incircle and 
tangent to theirs. Then 
and 
.
so 
is cyclic as desired. 
is isosceles with 
!
.
are concurrent, and so 
.
are concurrent, and thus 
.
are concurrent.
and we are done.
and 
,
is a bisector of 
Because of the concurency of the lines 
which means that the cross ratio 
,
is harmonic bundle. A similar argument shows that, 
is harmonic bundle, where 
is the intersection of 
and 
Now, it is easy to see that, 
Therefore, we have
where 
denote the power of point 
This prove that the points 
,
are the angle bisectors of 
,
,
,
.
is cyclic iff bisectors of 
is circumscribed, what can be easily proven by calculating it's side lengths.
.
:
and 
.
for some real constants 
,
,
and so 
.
,
,
still holds, along with the relative positions of 
.
and 
intersecting on 
and 
,
by the circle centered at 
passing through 
)
satisfy the relation that 
is tangent to 
at 
,
are concylic.
centered at 
and a circle 
centered at 
.
and 
both become lines perpendicular to 
,
becomes a rectangle. In particular, it is cyclic, so 
is cyclic.
and 
.
.
.
concur at the Gregonne Point of 
.
.
.
concurs at a point 
.
and another circle centered at 
.
![[asy]
size(11cm);
defaultpen(fontsize(10pt));
pair A = dir(140), B = dir(210), C = dir(330), I = incenter(A,B,C), D = foot(I,B,C), E = foot(I,C,A), F = foot(I,A,B), P = extension(B,C,E,F), I1 = I+dir(I--D)*abs(I-D)*0.35;
path p = circle(I1,abs(D-I1));
pair Z = intersectionpoints(circle(B,abs(B-D)),p)[1], Y = intersectionpoints(circle(C,abs(C-D)),p)[0], X = extension(B,Z,C,Y);
dot("$A$", A, dir(120));
dot("$B$", B, dir(270));
dot("$C$", C, dir(330));
dot("$D$", D, dir(270));
dot("$E$", E, dir(15));
dot("$F$", F, dir(135));
dot("$P (Q)$", P, dir(210));
dot("$X$", X, dir(100));
dot("$Y$", Y, dir(260));
dot("$Z$", Z, dir(315));
draw(C--B--A--C, linewidth(1.2));
draw(B--X--C);
draw(Y--P--E^^P--B, dashed);
draw(incircle(A,B,C)^^p);
draw(circumcircle(E,F,Y), dotted);
[/asy]](http://latex.artofproblemsolving.com/a/9/4/a94da0dc9eabe4e02f53d769c83b8def501552df.png)
then 
and 
.
.
,
We know that 
,
,
,
,
implying 
by Power of a Point, and the result follows. 
concur at the Gergonne Point of 
concur at the Gergonne Point of 
or 
concur(say at a point 
,
,
and 
,
hence 
,
are concurrent 
are concurrent 
are concyclic.
and 
at 
and 
respectively. Since 
Hence, 
.
.
,
.
.
.
so 
are concyclic, as desired.
are concurrent at some point 
which clearly finishes. 
and 
.
the lines 
are concurrent at 
.
And 
are different
and 
meet 
i
but since 
and 
,
.
concur.
by Power of a Point, which implies the result. 
