[/quote]
,
such that for all 
,
.
such that, for all integers 
that satisfy 
,![\[f(a)^2+f(b)^2+f(c)^2=2f(a)f(b)+2f(b)f(c)+2f(c)f(a).\]](http://latex.artofproblemsolving.com/a/9/e/a9edf1a3207d29facc970b6ee288d4fabdd43a4b.png)
denotes the set of integers.)
board. Some of its cells are colored black in such a way that every 
board of 
black cells. Find the maximum amount of black cells that the board may have.
such that 
is unbounded and![\[2f(m)f(n)-f(n-m)-1\]](http://latex.artofproblemsolving.com/0/f/7/0f77b5addbf8f5e603eb1b6923cdac795cd132e9.png)
is a perfect square for all integer 
,
be the 
-
and 
are placed on line 
such that 
is between 
,
and 
,
be the 
,
be the reflection of the orthocenter of triangle 
,
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.
Y by parola, Adventure10, Mango247
Circles 
and 
intersect at points 
and 
. Distinct points 
and 
(not at or ) are selected on . The lines 
and 
meet again at 
and 
respectively, and the lines 
and 
meet at 
. Prove that, as and vary, the circumcentres of triangles 
all lie on one fixed circle.
and 
intersect at points 
and 
. Distinct points 
and 
(not at or ) are selected on . The lines 
and 
meet again at 
and 
respectively, and the lines 
and 
meet at 
. Prove that, as and vary, the circumcentres of triangles 
all lie on one fixed circle.Attachments:
This post has been edited 1 time. Last edited by orl, Oct 25, 2004, 12:16 AM
is cyclic, meaning that we can forget about the 
'
.
are constant angles, it means that the triangles 
are all similar, where 
is the circumcenter of 
,
is the center of 
.
as a complete quadrilateral the problem becomes trivial once we know that the circumcenters of the 
"small" triangles of the complete quadrilateral and the miquel point of this complete quadrilateral (which in this case is point 
)
and 
.![\[\angle{CA_1 Q} + \angle{CA_2 Q} = \angle{QPB_2} + \angle{CA_2 A_1} +\angle{PA_2 Q} = \angle{PQB_2}+\angle{PB_2 Q} +\angle{QPB_2}=180\]](http://latex.artofproblemsolving.com/b/3/6/b363e3aaec6e4edad494abd8ef8af178f75ce814.png)
is cyclic. Let 
and 
denote the perpendicular bisectors of 
and 
and let 
.
is the circumcenter of 
and 
,
.
.
is cyclic and since 
are the centres of 
and 
.
.
and 
are perpendicular to 
and 
respectively as by definition all of 
are isosceles. Let the midpoints of 
be 
respectively. To show that the quadrilateral 
is cyclic it suffices to prove 
as then it would follow 
if we recall the right angles 
.
.
(at least on this diagram) so 
so 
is cyclic.
has fixed angles. So if we consider the diagram, but with another 
diametrically opposite 
,
lie on 
respectively. So this rotation must leave an equal against the common 
,
.
.
.
be the circumcenters of 
.
is constant. Since 
and 
,
is constant; thus, 
as desired. 
are concyclic,where 
be the centers of 
respectively.
and 
are cyclic.
are cyclic.
We have 
then we know 
As desired.
are concyclic.
,
,
,
.
w.r.t 
to the 
,
be 
resp.
is cyclic.
which doesn't depend on 
,
.
.
which is a fixed circle as desired.
.
is cyclic, because 
(because: 
)
.
are fixed 
So the Circumcenter of 
is same as the Circumcenter of 
.
from the Diagram. So, the problem can now be restated as
and 
are two circles and 
.
respectively. Then Prove that the locus of the Circumcenters of such triangles 
is a circle.
.
where 
and 
are the Inverted Image of 
intersecting 
respectively, then the reflections of 
are collinear.
are also collinear, and the reflections of 
map!
.
lies on the miquel circle 
which is fixed
is the Miquel point of 
.
form a cyclic quadrilateral with 
so 
is cyclic and the problem reduces to showing that the circumcenter of 
lie on a fixed circle (the 
.
respectively. Then the circumcenter of 
and 
to 
and this easily implies the result.
and so circumcenter of 
which is fixed.
be the centers of 
so 
is cyclic. 
Let 
We claim that 
is cyclic. First, we claim that 
This implies the claim.
Note that 
is the perpendicular bisector of 
the perpendicular bisector of 
Thus, 
and 
These sum to 
so we are done.
which means the circumcenter of 
where 
,
remain invariant over all 
,
remains similar as 
as a complete quadrilateral, 
is directly similar to a fixed triangle as well, where 
is directly similar to a fixed triangle as 
.
such that 
always holds, and the rest is clear. 
From here, we see that 
is the same as the circumcenter of 
and 
,
is a cyclic quadrilateral, which means that 
.
.
Thus, 
,
,
.
,
.
,
,
,
are concyclic.
also passes through 
so we just need to show that as 
lies on a fixed circle.
two fixed lines in the plane which intersect at 
An arbitrary circle passing through 
intersects 
respectively. Prove that as the circle varies, the reflection 
of 
lies on a fixed line.
the line 
is fixed and passes through 
