[/quote]
,
,
and 
.
is prime, 
with 
and 
be a function such that for every regular 
-
we have 
.
for all reals 
.
)
are the altitudes of the triangle. The bisector of 
intersects 
at 
.
be a point such that 
and 
.
passes through the midpoint of 
& 
.
with 
and at most one of them being zero, show that![\[\frac{1}{\max(a^2,b)}+\frac{1}{\max(b^2,c)}+\frac{1}{\max(c^2,a)}\geq\frac{27}{4}\]](http://latex.artofproblemsolving.com/9/e/1/9e14a9ef2d1a0a98006bcf716fa33c3d9ea29bd5.png)
be an integer. Find all sequences 
satisfying the following conditions:![\[ \text{ (a) } a_i \in \left\{0,1\right\} \text{ for all } 1 \leq i \leq n^2 + n;
\]](http://latex.artofproblemsolving.com/3/c/5/3c509ec2e9013e8d3be492c8eb44a7c33841b74e.png)
![\[ \text{ (b) } a_{i + 1} + a_{i + 2} + \ldots + a_{i + n} < a_{i + n + 1} + a_{i + n + 2} + \ldots + a_{i + 2n} \text{ for all } 0 \leq i \leq n^2 - n.
\]](http://latex.artofproblemsolving.com/9/7/d/97d2a467d1c0dc8594ec024c3bb9b8c87ee85b19.png)
and 
.
Equality holds when 
or 
and 
,
cookies from one of the piles, of which he eats 
and puts 
for which Marica has a winning strategy.
be non-negative real numbers such that 
Prove that
be the circumcenter, symmedian point of 
.
at 
,
at 
,
at 
are concurrent.
Y by Adventure10, Mango247
Let 
be a circumcenter of acute triangle 
and let 
and 
be circumcenters of triangles 
and 
, respectively. Circumcircles of triangles and intersect side 
in points 
(
) and 
(
), respectively. Perpendicular bisector of side intersects side 
in point 
(
). Prove that circumcenter of triangle 
lies on iff lies on line 
be a circumcenter of acute triangle 
and let 
and 
be circumcenters of triangles 
and 
, respectively. Circumcircles of triangles and intersect side 
in points 
(
) and 
(
), respectively. Perpendicular bisector of side intersects side 
in point 
(
). Prove that circumcenter of triangle 
lies on iff lies on line 
be the circumcentre of 
is isosceles and 
we have 
.
hence 
are concyclic.
.
hence 
are concyclic
.
hence 
are concyclic.
are concyclic.
so 
is parallel to the height through 
and 
are collinear 
is the perpendicular bisector of 
are collinear.
are collinear iff 
:
.
are concyclic iff 
be the midpoint of 
be the center of 
.
and 
,
,
.
.
is cyclic. Let the perpendicular bisector of 
.
,
.
.
are cyclic.
,
is cyclic. Similarly 
is cyclic. 
As desired.
