are positive real numbers, 
then
be the incenter of 
and 
be a point such that 
is perpendicular to 
is parallel to 
at points 
and 
respectively. Prove that 
.
,
,
,
and 
such that 
is a parallelogram and 
is a cyclic quadrilateral. Let 
be a line passing through 
at 
and intersects line 
at 
.
.
.
,
,
,
,
be the diametrically opposite point of 
and 
and 
,
,
,
are concurrent, i.e., the lines intersect at the same point.
with 
right and 
,
.
we take two points 
so that 
and 
.
and 
containing inside 
and 
,
and 
,
and 
,
and that 
,
and 
are collinear.
satisfying the conditions:
for all real numbers 
with ![$ t\in [0,1] . $](http://latex.artofproblemsolving.com/1/2/a/12ac756f5dc034618c1566c89e80dcf52761f105.png)
for any real numbers 
such that 
and 
and any 
real numbers 
the following inequality holds.
such that: 
.
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 
.
and 
.![\[
\angle I_KSC=180-\angle I_MSC= 180-\angle PSC=\angle SPC+\angle SCP=\frac{1}{2}(180-(\beta+\gamma))+\gamma=90+\frac{\gamma-\beta}{2}
\]](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.
