[/quote]
,
![\[f(xf(y) + x) = xy + f(x)\]](http://latex.artofproblemsolving.com/9/9/f/99f580ebc50846e6bc2c004667559922749a4dfa.png)
be a sequence of positive real numbers such that for every 
,![\[
a_n = \max_{1 \leq i \leq 2025} a_{n-i} - \min_{1 \leq i \leq 2025} a_{n-i}
\]](http://latex.artofproblemsolving.com/e/3/0/e30e9e533eb9a41cf847484e676ac4522edda665.png)
Prove that there exists a natural number 
such that for all 
,![\[
a_n < \frac{1}{1404}
\]](http://latex.artofproblemsolving.com/0/1/f/01f1a81eed2230332d51a15501ef2a6ad3a7ec82.png)
be an acute triangle with 
and let 
be its orthocenter. Let 
such that 
and 
,
such that 
and 
a point on the perpendicular bisector of 
such that 
and 
lie on the opposite side of 
w.r.t 
.
and 
square is filled with numbers 
.
squares is summed,and arranged in an increasing order. Is it possible to obtain the following sequences as a result of this operation?
be a triangle with points 
lie on the perpendicular bisector of 
lie on a circle. Suppose 
are perpendicular to sides 
at points 
The tangent lines from points 
to the circumcircle of 
intersects at point 
Prove that: 
are parallel.
,
be four points in the plane, with 
and 
.![\[\frac{AB \cdot CD}{AC \cdot BD}, \]](http://latex.artofproblemsolving.com/f/3/c/f3cf9729b32b210faeea6e2d23da3582be2ae061.png)
and 
are orthogonal. (Intersecting circles are said to be orthogonal if at either common point their tangents are perpendicuar. Thus, proving that the circumcircles of the triangles 
such that for any positive real numbers 
and 
,
Proposed by Athanasios Kontogeorgis, Grecce, and Dorlir Ahmeti, Kosovo
of positive real numbers satisfies ![\[a_{k+1}\geq\frac{ka_k}{a_k^2+(k-1)}\]](http://latex.artofproblemsolving.com/0/3/9/03909a5f60d944063182ec97cf73f283178e8d4d.png)
for every positive integer 
.
for every 
.
,
and 
be the respective intersections with BC of the internal angle bisector, external angle bisector, and the median from A. The circumcircle of 
intersects 
a second time at point 
different from A. Define 
and 
analogously. Prove that the circumcenter of 
lies on the Euler line of ABC.
such that 
for all 
of the acute-angled triangle 
is tangent to its side 
at a point 
.
be an altitude of triangle 
be the midpoint of the segment 
is the common point of the circle 
(distinct from 
are tangent to each other at the point 
is called lucky if it has at least two distinct prime divisors and can be written in the form:![\[
n = p_1^{\alpha_1} + \cdots + p_k^{\alpha_k}
\]](http://latex.artofproblemsolving.com/7/4/4/744a5ccaeb9476ebd7d999c395762cb6e99a7a71.png)
where 
are distinct prime numbers that divide 
,
.
,
,
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.
we get that the reflections of 
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 
respectively. Prove that as the circle varies, the reflection 
of 
lies on a fixed line.
the line 
is fixed and passes through 
,![\[
\angle A_1QA_2=\angle A_1QP+\angle A_2QP=\angle A_1B_1P+\angle A_2B_2P=-\angle B_1CB_2=\angle A_1CA_2
\]](http://latex.artofproblemsolving.com/d/8/4/d8438c8205010cd65769746e9038a705be94d067.png)
so 
is cyclic with center 
,![\[
\angle O_1QO_2 = \angle O_1PO_2=-(\angle A_1PO_1+\angle A_2PO_2)=\angle A_1B_1P+\angle A_2B_2P=\angle A_1CA_2=\angle A_1QA_2=\frac{1}{2}\angle A_1OA_2=\angle O_1OO_2
\]](http://latex.artofproblemsolving.com/8/1/a/81a18a048ef676c8dafa6c37b0d0f9a67d5a000f.png)
and we're done.
