[/quote]
,
have,where 
and 
are natural numbers.
inscribed in 
with 
being the midpoint of 
.
of 
.
be the projection of 
onto 
.
that is not on 
intersects 
at 
.
and 
intersect at 
.
is an isosceles triangle.
and let 
,
,
be a sequence of non-negative real numbers. Find the maximum value of 
in terms of 
.
,
as follows: write 
,
are non-negative integers and 
,
.
where 
and 
is the twist of 
for each positive integer 
.
if and only if the remainder when 
is either 
.
every next number (from the fifth number) is the unit number of the sum of the four numbers preceding it. Is there any cases that we get the numbers 
and 
in this series?
,
and 
meet at the point 
.
and 
.
prove that 
.
![$P \in \mathbb Q[x,y]$](http://latex.artofproblemsolving.com/a/3/8/a384cbf38fd3a7f812b22baaeb0860e7f6eb26c0.png)
with 
and ![$\forall (a,b) \in \mathbb Z^2 \cap [0,d]^2, P(a,b) \in \mathbb Z$](http://latex.artofproblemsolving.com/a/6/5/a65ecb4bb2d9b5e6096f322c2de8bd11b945ed6a.png)
.
?
.
be nonnegative real numbers such that 
.![\[
\frac{a}{4 - b} + \frac{b}{4 - c} + \frac{c}{4 - a} + \frac{1}{16}(1 - a)^2(1 - b)^2(1 - c)^2 \leq 1,
\]](http://latex.artofproblemsolving.com/5/b/d/5bd3349071e075519bd986c845c500125b7d46f8.png)
and determine all such triples 
where the equality holds.
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 
Y by Lsway, Adventure10, Mango247
Dear Mathlinkers,
1. ABC a triangle
2. P a point
3. A’B’C’ the P-cevian triangle
4. Q the pivot point wrt ABC and A’, B’, C’
5. (1), (2), (3) the circumcircles wrt to AA’Q, BB’Q, CC’Q.
Prove : (1), (2) and (3) are coaxal.
Sincerely
Jean-Louis
1. ABC a triangle
2. P a point
3. A’B’C’ the P-cevian triangle
4. Q the pivot point wrt ABC and A’, B’, C’
5. (1), (2), (3) the circumcircles wrt to AA’Q, BB’Q, CC’Q.
Prove : (1), (2) and (3) are coaxal.
Sincerely
Jean-Louis
denote the radical axes of the circumcircle 
with 
resp. 
and 
are pairwise radical axes of 
and 
concurring at their radical center 
and similarly we have 
and 
Hence if 
cut 
we have 
and similarly 
and 
concur at 
concur at 
then by Cevian Nest Theorem, 
and 
concur at a point 
Thus, 
and 
are coaxal with common radical axis 
and a point 
.
be the cevian triangle of 
be the Miquel point of 
WRT 
are coaxial and the second intersection of these three circles is the isogonal conjugate (WRT 
be the Miquel point of the complete quadrilateral 
(define 
similarly). Since 
and 
,
and 
share the same angle bisector 
,
is the Inversion with center 
is the image of each other under 
and 
we get 
and 
,
.
,
under 
,
.
we get 
,
.
and 
,
be the Inversion with center 
is the cevian triangle of 
WRT 
.
and 
,
lies on 
and 
,
(similar discussion for 
)
are concurrent (at the isogonal conjugate of the complement (WRT 
)
WRT 
are coaxial, hence notice 
is the image of 
,
.
and 
and similarly we get 
we conclude that 
Let 
Then 
meet at a 2nd point 
and 
again at 
resp. Thus we have 
Hence since 
and 
But if 
the isogonal conjugate of 
we have 
and similarly 
and 
Ceva circle”.Permit me to write down my solution here:
be the isogonal center of 
,
and 
n
then 
then 
.
are concyclic 
and a point 
be the cevian triangle of 
is the Miquel point of 
with respect to 
,
.
.
are coaxial and the second intersection point of these three circles is 
