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 
-
on the circumcircle, and 
be the foot of the
.
intersects the circumcircle again at 
. For
on the circumcircle of triangle 
intersect 
. Prove
are concyclic.
Y by 62861, Adventure10
Let 
be an acute triangle with circumcenter 
, orthocenter 
, and circumcircle 
. Let 
be the midpoint of 
and 
the midpoint of 
. Assume the points , , , are distinct and lie on a circle 
. Prove that the circles and are internally tangent to each other.
Dhroova Aiylam and Evan Chen
be an acute triangle with circumcenter 
, orthocenter 
, and circumcircle 
. Let 
be the midpoint of 
and 
the midpoint of 
. Assume the points , , , are distinct and lie on a circle 
. Prove that the circles and are internally tangent to each other.Dhroova Aiylam and Evan Chen
is well-known to be 
,
,
,
be the reflections of 
in 
,
be center of 
,
.
,
be the deameter of 
(
)
is rectangle 
and 
and 
.
.
.
is a rectangle 
and now homothety 
ends it.
be a point on 
with center 
it sends 
to 
the reflection of 
is on 
But the reflection of 
.
so 
And 
is on both 
and 
Since we have the equality case, 
is a positive integer. But 
is 
and 
,
.
over 
Circumradius of 
.
cenered at 
.
.
)
)
and 
be the centers of 
and 
and 
.
.
,
be the reflection of 
has circumradius 
has circumradius 
has circumradius 
has the same radius as the circumcircle of 
.
.
.
is therefore 
,
.
and 
is equal to the radius of 
at 
)
.
denote the 
p
and thus 
.
and thus again, 
is cyclic, thus the intersections of the perpendicular bisectors of 
is the center of 
which is our 
.
are collinear, we have 
and thus the orthocenter 
.
,
,
.
So the circumradius of 
must have radius 
.
is homothetic to 
has radius ![[asy]
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[/asy]](http://latex.artofproblemsolving.com/5/f/0/5f00261c51d2aedc6f4971d0ebaf131e2afcfe75.png)
instead.
is the same as 
,
are the midpoints of 
let this value be 
Thus taking a homothety centered at 
but since the circle also passes through 
Let 
be the reflection of 
By 
homothety about 
is cyclic, with half the radius as the circumcircle. However, since 
,
