such that for all real 
:
Prove that: 
is the sum of 
square numbers
Let 
and 
is the sum of 
is the sum of 
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)
and 
are the incenter and the 
-
,
be the midpoint of arc 
on the circumcircle, and 
be the foot of the
.
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 
.
Y by cubres
Given an acute triangle 
where 
. Point 
is the circumcenter of triangle , and 
is the projection of point 
onto line 
. The midpoints of , 
, and 
are 
, 
, and 
, respectively. The line 
intersects 
and 
at points 
and 
, respectively. Prove that is the incenter of triangle 
.
where 
. Point 
is the circumcenter of triangle , and 
is the projection of point 
onto line 
. The midpoints of , 
, and 
are 
, 
, and 
, respectively. The line 
intersects 
and 
at points 
and 
, respectively. Prove that is the incenter of triangle 
.
cyclic.
be the midpoint of 
where 
is the orthocenter, and let 
be the feet of the altitudes from 
.
is perpendicular to 
by symmetry, and as 
at 
by reims theorem, we have that 
.![\[\measuredangle PEQ=\measuredangle PED = \measuredangle PND=\measuredangle PAQ\]](http://latex.artofproblemsolving.com/2/8/5/285884435dc7a6bb542241f18bda104fe884c96e.png)
by parallel lines and nine point circle. This proves the claim.![\[\measuredangle PQD=\measuredangle PQE =\measuredangle PAE =\measuredangle EPA =\measuredangle EQA =\measuredangle DQH\]](http://latex.artofproblemsolving.com/7/e/5/7e5a8e580c89298a32268f32d79266e3a5b33c4d.png)
so 
bisects 
,
bisects 
as desired.
.
.
is concyclic.
.
bisects 
.
bisects 
.
,
.
