[/quote]
,
such that for all real 
:
be positive reals with 
.![\[\sum_{i=1}^n \frac{x_i}{2-x_i} \ge \frac{n}{2n-1}\]](http://latex.artofproblemsolving.com/f/6/5/f65f6fce4465125dfacf002f0659c898df7ae150.png)
Prove that: 
is the sum of 
square numbers
Let 
and 
is the sum of 
is the sum of 
and 
are the incenter and the 
-
,
be the midpoint of arc 
on the circumcircle, and 
intersects the circumcircle again at 
. For
on the circumcircle of triangle 
intersect 
. Prove
are concyclic.
squares, we cover the area with several S-Tetrominos (=Z-Tetrominos) along the square so that in every square, there are two or fewer tiles covering that (tiles can be overlap). Find the maximum possible number of squares covered by at least one tile.
satisfies
for any non-negative integers 
and 
.
.
be a function from the Euclidean plane to the real numbers such that 
for any acute triangle 
with circumcenter 
and orthocenter 
.
be the positive real number. Prove that
,
grid. Alice goes first, and they take turns drawing a black point from the coordinate set![\[\{(i, j) \mid i, j \in \mathbb{N}, 1 \leq i, j \leq n\}\]](http://latex.artofproblemsolving.com/4/3/f/43f7493d1dff43079eb5481ee17bc8f16d7b9553.png)
There is a constraint that the distance between any two black points cannot be an integer. The player who cannot draw a black point loses. Find all integers 
.
unit squares depicted in the drawing below.
Y by
The angle bisector of 
intersects the circumcircle of the acute-angled 
at point 
. Let the perpendicular bisectors of 
and 
intersect sides 
and 
at points 
and 
, respectively. If 
is the circumcenter of , prove that the points 
, and are concyclic.
Proposed by Petar Filipovski
intersects the circumcircle of the acute-angled 
at point 
. Let the perpendicular bisectors of 
and 
intersect sides 
and 
at points 
and 
, respectively. If 
is the circumcenter of , prove that the points 
, and are concyclic.Proposed by Petar Filipovski
This post has been edited 1 time. Last edited by Lukaluce, Apr 14, 2025, 12:57 PM
. Let 
we can furthermore prove that 
![$$\angle OFD +\angle OED =[(180-2\angle BAD)-(90-\angle BAD)]+\frac{[360-(180-2\angle BCD)]}{2}=90-\angle BAD +90 +\angle BCD=180,$$](http://latex.artofproblemsolving.com/5/9/8/5982526233fb8c7c21b636531dbf96da5af63759.png)
done.
