be positive reals. Prove that there is a cyclic permutation 
of 
such that the inequality:![\[
\frac{a}{xa + yb + zc} + \frac{b}{xb + yc + za} + \frac{c}{xc + ya + zb} \geq \frac{3}{x + y + z}
\]](http://latex.artofproblemsolving.com/5/5/3/553527fc40a1b7d29dbdfb4b5e63635308f46c85.png)
and 
.
and circumcenter 
.
,
the midpoints of 
,
.
with diameter 
,
at a point 
.
meets line 
at 
.
and 
intersect at a point 
on 
.
Prove that: 
is the sum of 
square numbers
Let 
and 
is the sum of 
is the sum of 
Y by doxuanlong15052000, Amir Hossein, hydo2332, HamstPan38825, Adventure10, Mango247, Funcshun840
Let 
be a triangle inscribed in circle 
, and let the medians from 
and 
intersect at 
and 
respectively. Let 
be the center of the circle through tangent to 
at , and let 
be the center of the circle through tangent to 
at . Prove that , , and the nine-point center of are collinear.
Proposed by Michael Kural
be a triangle inscribed in circle 
, and let the medians from 
and 
intersect at 
and 
respectively. Let 
be the center of the circle through tangent to 
at , and let 
be the center of the circle through tangent to 
at . Prove that , , and the nine-point center of are collinear.Proposed by Michael Kural
be the midpoints of 
is the centroid of 
If 
cuts 
again at 
then 
i.e. the ratio of the powers of 
and 
thus the center 
of 
coaxal with 
verifies 
(H is the orthocenter of ABC) 
but 
Similarly 
is a parallelogram 
and 
bisect each other through the 9-point center.
is a parallelogram. Indeed, let 
.
,
,
by the Ratio Lemma. Similarly, 
.
and 
,
yields 
.
and 
,
yields 
.
,
.
,
.
is a parallelogram. It follows that 
passes through the midpoint of 
,
be the orthocenter of 
.
is a parallelogram, which will immediately imply the desired result. We proceed with a complex number bash. Assume WLOG that the circumcircle of 
by 
respectively.
by 
.
we have that 
so 
.
by 
.
-
so 
and so, after plugging in, we find that 
.
also is a root of this quadratic, by Vieta's formulas we find that 
.
denote the midpoint of segment 
and let 
be its complex coordinate. Then we easily find that 
.
we have that 
.
we have that 
.
.
so 
and so 
.
and solving for 
we obtain 
.
have complex coordinate 
we similarly find that 
.
has complex coordinate 
and 
it suffices to show that 
.
as desired so we are done.
be the radii of 
,
and let 
If 
cuts 
for a second time at ![\[MD \cdot MU = MC^2 = -MC \cdot MA = -MD \cdot MB,\]](http://latex.artofproblemsolving.com/0/f/7/0f7507574ba685143e856848f7fe3406eb87cdeb.png)
implying that 
Then since 
is the image of 
,
of 
Thus, it is 
If 
,
It is well-known that 
is a diameter of the nine-point circle. Therefore, 
It follows that 
is a parallelogram. Therefore, 
is a parallelogram as well, whence the desired result follows. 
is the nine-point center of 
is the perpendicular bisector of 
,![\[\measuredangle UO_1O' = \measuredangle UDC = \measuredangle BDC = \measuredangle BAC \quad \text{and} \quad \measuredangle UO'O_1 = \measuredangle UAC = \measuredangle BCA,\]](http://latex.artofproblemsolving.com/3/2/2/3225af9b6c6b06e573733680387945f2f6732f80.png)
where the last step follows since 
is a parallelogram. Therefore, 
Then since 
(which follows from the aforementioned homothety), we obtain 
Analagously, we find that 
,
Consequently, 
and meanwhile 
,
Therefore, 
is a parallelogram, and thus the common midpoint of 
as the unit circle. Applying Theorem 6.17 (Complex Intersection), we have
Let 
Then we must have 
and from Lemma 6.5 (Perpendicularity Criterion) we have
We also have 
which gives
Similarly, we have
now observe that
then 
is the midpoint of 
so we are done.
represent points 
and let 
,
Noting that 
:
In other words
Thus, the sum of the roots are: 
which after subtracting 
:
We know that the segment from midpoint of 
to 
Note that since 
is perpendicular to 
which means that:
Note that:
so
from which we have that:
and by symmetry:
The midpoint of 
which is precisely the nine point center, as desired.
be the midpoints of 
,
,
,
and 
and
using the equation 
.
is the midpoint, we have 
)
,
,
From the first equation, we have 
from the second 
and from the third 
Substituting from the second to the first, we have 
or 
Similarly, 
Thus, 
so 
are collinear.
and 
.
denote an arbitrary complex number 
for which 
.
intersects the unit circle at 
in some order. Thus, 
Thus, 
Similarly, 
We wish to show that these are collinear with 
Multiplying by 
it suffices to show that 
are collinear. Expanding this out gives 
We subtract 
from each to get 
The first one is exactly twice the last one (after simplifying of course), so they are collinear and thus we are done. In fact, this shows that 
,
.
where 
.
Computing the symmetric expression for 
,
hence the midpoint of 
parallelograms and 
midpoints of 
then 
so 
similarly 
.
meet at 
and let 
antipode be 
,
parallelogram. Midpoint of 
is the same as NPC when projected onto 
,
