be a scalene triangle, 
be the median through 
,
and segment 
meet 
.
be the triangle formed by lines 
be real numbers such that 
Prove that
Where 
Where 
are are the lengths of the sides of a 
g
then![$$(n-2)\left[\sum_{i=1}^{n}{\frac{a_i^2}{(1-a_i)^2}}-\frac n{(n-1)^2}\right]\geq(2n-1)\left(\sum_{i=1}^{n}{\frac{a_i}{1-a_i}}-\frac n{n-1}\right)^2.$$](http://latex.artofproblemsolving.com/9/3/e/93e5529518807b92cd9e0382f5fb07a293128b04.png)
and 
intersect at point 
and 
be a foot of the altitudes from 
and let 
be the midpoint of 
.
cuts 
in half.
and 
.
be an irrational number. Show that for every integer 
,
such that 
.
be positive integers with 
of the set of 
”
”
Y by PikaPika999
Convex quadrilateral 
is described on a circle 
, and is not a trapezius inscribed in a circle. Let the tangency points of and 
be 
respectively. A circle with a center 
, different from is tangent to the segement 
and lines 
. A circle with center 
, different from is tangent to segment 
and lines 
. A circle with center 
, different from is tangent to segment 
and lines . A circle with center 
, different from is tangent to segment 
and lines . Prove that the lines 
are concurrent.
is described on a circle 
, and is not a trapezius inscribed in a circle. Let the tangency points of and 
be 
respectively. A circle with a center 
, different from is tangent to the segement 
and lines 
. A circle with center 
, different from is tangent to segment 
and lines 
. A circle with center 
, different from is tangent to segment 
and lines . A circle with center 
, different from is tangent to segment 
and lines . Prove that the lines 
are concurrent.
and 
are homothetical, that is, there is a homothety which sends one into the other. Then lines 
would all pass through the center of homothety.
because both are perpendicular to the angle bisector of angle 
.
is perpendicular to the angle bisector of angle formed by 
is the angle bisector of angle formed by 
.
are collinear as all of them lie on the angle bisector of 
.
.![\[
\angle I_KSC=180-\angle I_MSC= 180-\angle PSC=\angle SPC+\angle SCP=\frac{1}{2}(180-(\beta+\gamma))+\gamma=90+\frac{\gamma-\beta}{2}
\]](http://latex.artofproblemsolving.com/2/f/0/2f0a53dfa8cb06240301eef2f5132d5c56d79930.png)
and![\[
\angle KMC = \angle KNM = 180-(\angle DNM+\angle ANK)=180-\frac{1}{2}(\angle NDM+\angle NAK)=180-\frac{1}{2}(180-\beta+\gamma)=90+\frac{\gamma-\beta}{2}
\]](http://latex.artofproblemsolving.com/e/8/d/e8d9613bb6a3edbbdf25c33572024622a6f01f92.png)
thus lines 
are parallel.
all lie on the external angle bisector of 
so they're collinear and 
,
.
so pairs of triangles 
and 
are homothetic and we're done.
