AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
, let 
denote the number of its positive divisors (including 1 and itself). Determine all positive integers 
for which there exists a positive integer such that 
.
let 
be the foot of the altitude from 
to the side 
and 
, 
, 
be the incenter, -excenter, and 
-excenter, respectively. Denote by 
and 
the other intersection points of the circle 
with the lines 
and 
, respectively. Prove that 
.
such that![\[(a^2-b^2)(b^2-c^2) = abc.\]](http://latex.artofproblemsolving.com/7/1/8/7188261c40a45f5e15533100432d36abbf515efd.png)
. If this is still too hard assume in 
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. be an integer. Show that, if 
is an integer, then it is a perfect square.[/quote]
. If is an integer, then 
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 
, and that, for every such pair 
, the "next" such pair can be calculated as
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 is composite and for every divisor 
of at least one of 
and 
is also a divisor of with 
, such that 
and 
.
such that 
is prime, 
with 
and ?
be an acute triangle with orthocenter 
and circumcenter 
. Let 
and 
be points on the line segments 
and 
respectively such that 
is a parallelogram. Prove that 
.
that satisfy 
for all 
be an arbitary chord of circle 
,
be the midpoint of segment 
and 
be two other arbitary chords from 
intersect 
respectively,prove that 
.
is a point on 
intersect the circle at 
respectively therefore:
(I)
.
be the reflection of point 
wrt 
.
wrt point 
respectively thus:
(II)
be two other arbitary chords of 
be three points on the sides 
of triangle 
such that 
are concurrent at point 
intersect 
at 
intersect 
at 
,
and let 
intersect 
be a point on side 
of triangle 
and let 
intersect 
respectively at 
prove that6 
,
is the intersection of 
and 
is cyclic where 
is the intersection of 
.
and 
lies between 
.
be the second intersection of the circle with diameter 
(other than C).let 
intersect 
are colinear.
be three points on the sides 
,
then the pencil 
is harmonic 
,
.
and 
.
,
,
,
and 
,
.
then the division 
is harmonic,then using lemma we obtain that 
.
,
,
,
be the intersection of 
.
intersect line 
respectively,so 
,
i.e. 
we mean that segment 
has been devided internaly and externaly by 
we mean that segment 
be the circumcenter of triangle 
meet 
,
in the same way,prove that:
p
lie on a line 
.
,
,
in the same way,prove that 
be a point on 
now let 
be the line passing through 
,
in the same way,prove that 
are concurrent.
be the incircle of 
(
in similar way.Let 
be the tangent lines to 
at points 
respectively.Let 
be the points of their intesection,such that 
.
are collinear(similarly for other triplets of points) and it is easy to understand that 
(
,
is parallelogramm 
,
.
.
and 
,
to the collinearity of 
,
.
is the homothety center,which turns 
are collinear,where 
is Nagel point,and 
.
and 
be the 2 triangles such that 
are concurrent(at point 
)
are collinear.
.
through the points 
,
,
.
,
.
,
are concurrent,so we are done. 
and parallel to 
pass through its incenter 
and parallel to the line segments 
respectively, at one point, as the reflexion of the 
with respect to the incenter 
to 
and let 
intersect 
at 
,
.
is harmonic quadrilateral,hence 
makes a harmonic four,hence 
.
,
.
be points on 
are tangent to 
now denote 
.
intersect line 
respectively and also 
where 
but in the other hand we know that 
and 
i.e. 
so:
or in other words 
i.e. 
be a variant point on 
so 
line 
so 
is tangent to 
are concyclic and we write 
..then do we mean if they are projected about any point 
be points on circle 
intersect line 
now assume that 
i.e. 
.
(*)
and so on,so the equation (*) is equivalent to:
law in triangles 
we get that:
we have:
be two parallel chords of 
be the midpoint of 
with 
,
.
,
be the intersection point of 
,
and the lines 
intersects the circle at 
respectively and also the parallel line with 
:
so 
.
is tangent to 
are two tangents from point 
so 
is isosceles,but on the other hand 
thus:
.