[/quote]
,
satisfy:
![$[(x-2)^2+(y-4)^2][ (x-1)^2 + (y-2)^2] = 36$](http://latex.artofproblemsolving.com/b/1/0/b108dc2cac5b7491fb83418eb36c7184f6d5d30c.png)
such that
holds for all 
.
such that:
with all 
be a triangle, and let 
,
,
be collinear points such that 
,
,
.
,
,
are the reflections of 
,
,
,
is a nondegenerate triangle, then its circumcenter lies on the circumcircle of 
be three positive real numbers satisfying 
Prove that
There is a very elegant proof :-D Could anyone think of it?
such that 
and has exactly 
then
.
on 
on 
and 
,
lies on the circumcircle of triangle 
is the midpoint of 
and 
,
be a transversal of 
be intersections of 
with 
be a tangent line of 
is cyclic
be positive integers such that 
and 
.
.
and 
Prove that
at 
,
perpendicular to 
be the intersections of 
with 
,
be the projections of 
onto line 
be the second intersections of 
with the incircle 
be the intersection of 
and 
.
bisects segment 
.
Y by Adventure10, Mango247
Let 
be a cyclic quadrilateral with circumcircle 
. Let 
and 
be the incentres of the triangles 
and 
respectively. Let 
be the midpoint of the arc 
of circle containing the point 
. The line 
intersects again the circle at point 
. Prove that the points 
lie on a circle.
be a cyclic quadrilateral with circumcircle 
. Let 
and 
be the incentres of the triangles 
and 
respectively. Let 
be the midpoint of the arc 
of circle containing the point 
. The line 
intersects again the circle at point 
. Prove that the points 
lie on a circle.
be the intersections of 
with 
intersect 
at 
.
and 
.
is isosceles at vertex 
,
.
.
are harmonic. Projecting from 
are also harmonic.
,
.
.
.
m
.
.
.
and perpendicular to ![\[ -1=(E,F;D,B)\stackrel{K}{=}(T,A;Q,P)\implies AP\cdot TQ=TP\cdot AQ \]](http://latex.artofproblemsolving.com/f/3/7/f37455b07f4739d800a9e3bdf762fcd1d926fbcd.png)
By fact 5 we know 
.![\[ \frac{AP}{TP}=\frac{AQ}{TQ}\implies \frac{JP}{TP}=\frac{IQ}{TQ} \]](http://latex.artofproblemsolving.com/f/a/1/fa1efff7824faabd3e7be4abe79b473ee1525913.png)
And We know 
which means 
.
.
