[/quote]
,
be a triangle with incenter 
.
intersect the incircle of triangle 
.
is perpendicular to line 
.
.
is tangent to line 
.
lies on line 
such that 
.
.
,
be its circumcentre, and let 
.
and 
at 
and 
respectively. The circumcircles of triangles 
and 
.
and 
then 
is tangent to the circle 
,
,
,
be distinct positive integers, 
.
and 
such that 
does not divide any of the numbers 
,
,
.
written on the first page and 
empty pages. Suppose in each of the next 
be the largest possible number such that for any 
,
on the last page of her grimoire. Prove that there exists a positive integer 
such that for all 
,![\[n^{0.99n}\leqslant F(n)\leqslant n^{1.01n}.\]](http://latex.artofproblemsolving.com/5/7/e/57e2b7e3d97564665d8de28a3afd74bdcf74ba5b.png)
,
and 
be a pair of isogonal conjugate points. The line 
intersects 
at 
,
intersects 
.
and the circumcircle of 
intersect again at 
(other than 
bisects 
.
is right angled (for clarity, 
)
as a point on line 
is parallel to 
,
meaning 
,
,
.
,
meaning 
.
,
or 
and we are done.
from the reals to the reals such that![\[f\left(f(x)+y\right)=2x+f\left(f(y)-x\right)\]](http://latex.artofproblemsolving.com/e/b/4/eb4e0d6580d9d3d6a627e39837edb3d1943d4596.png)
.
is called 
if
is a perfect square, and
,
is the smallest positive integer such that 
is a perfect square.
for all integers 
.
.
-
meets 
and intersects 
.
or 
is tangent to 
,
,
are
with 
,
are all coloured blue.
of the set ![$[n]=\{1,2, \dots, n\}$](http://latex.artofproblemsolving.com/8/f/e/8fea25e7cf72aabc60f48c3b30f4d4818a084ae3.png)
and keeps it hidden from Bob. In a move, Bob tells Alice a permutation 
of ![$[n]$](http://latex.artofproblemsolving.com/0/d/e/0deff9086fd7840ef23860f5bc924f1d4d2d2310.png)
,![$i \in [n]$](http://latex.artofproblemsolving.com/e/6/7/e67206e47d5d5875105b339f177a3ab8f5151a5e.png)
such that 
.
moves.
denote the set of all real numbers. Find all functions 
such that
for all real numbers 
,
.
for which there exist infinitely many positive integers 
such that ![\[ \frac{a^m+a-1}{a^n+a^2-1} \]](http://latex.artofproblemsolving.com/2/8/a/28a68e409451bbdcaee7c1286341f6aeea9ffdad.png)
is itself an integer.
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 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 
and 
and let 
.
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 
intersects 
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.
