AoPS Academy
, and 
are all polynomials such that ![\[ P(x^5)+xQ(x^5)+x^2R(x^5)=(x^4+x^3+x^2+x+1)S(x),\]](http://latex.artofproblemsolving.com/1/6/a/16a71abc110ece558d427a07d7be2d1f72148024.png)
prove that 
is a factor of 
.
such that the equality 
holds for all pairs of real numbers 
.
be the circumcircle of 
and let 
be a fixed point on 
, 
, 
. Let 
be a variable point on 
, 
. Let 
be the second intersection point of 
and . Prove that the circumcircle of 
passes through a fixed point.
. Let E the second intersection between the circumcircle of XYD and . DE meets on F. So 'cause DXYE is cyclic 
, so 
and then F is the symmetric of A w.r.t. the axis of BC. Then 'cause E is the intersection between FD and that are all fixed it's also fixed.
be the circumcircle of and let be a fixed point on , , . Let be a variable point on , . Let be the second intersection point of and . Prove that the circumcircle of passes through a fixed point.
be a fixed point on and 
be any point on (
are distinct from D). Let 
cut the circumcenter of the triangle at 
.
touch the circumcenter of the triangle at 
.
, therefore the point belongs to the circumcenter of the triangle 
, and since the points 
are mobile, the problem is solved.
. Let 
be the circle through , 
that is tangent to . Clearly is fixed. Let 
(so 
is fixed). Now ![\[ \angle XYE = \angle AYE = \pi - \angle AFE = \pi - \angle AFD = \pi - \angle EDX\]](http://latex.artofproblemsolving.com/c/3/b/c3ba6cd86b04b91129d14cecc45fb0cab031ecf0.png)
so 
is cyclic. Hence the circumcircle of 
passes through a fixed point, .
and 
is another variable point in 
and 
the intersection of 
with 
we have,
is cyclic and hence we are done
is an obvious solution. Now, let 
.
or 
.
.
,
.
.
and 
meet the circle at 
respectively. Allow the circumcircle of 
.
,
,
.
and 
,
.
,
is cyclic, which means that the circumcircle of 
passes through 
(which is fixed because 
passes through a fixed point, as desired.
