be a triangle, and let 
be a point on side 
.
at 
at 
intersects the circumcircle 
of triangle 
and 
intersect 
and 
respectively.
such that 
for all real number 
and 
and 
on the plane such that 
is an equilateral . Let 
and 
at 
and 
at 
,
on 
satisfies 
and 
.
be the point where the perpendicular from 
to line 
meets the angle bisector of 
.
holds.
over 
.
and 
Prove that
![\[\lfloor x \rfloor\]](http://latex.artofproblemsolving.com/d/3/f/d3f8facf62413e1f17c89b9c183a8735c2defe8b.png)
denote the greatest integer less than or equal to x . What is the sum of the squares of the real numbers x for which ![\[ x^2 - 20\lfloor x \rfloor + 19 = 0 \]](http://latex.artofproblemsolving.com/f/a/9/fa96f2d6217a16029e5f03499e0a83deffdfa28a.png)
Prove that
of the pizza, where 
is a positive integer. Find 
![\[\mathcal{P}(x) = x^{108}+x^{102}+x^{96}+2x^{54}+3x^{36}+4x^{24}+5x^{18}+6\]](http://latex.artofproblemsolving.com/e/b/2/eb28918359848dd07784a3c2c6a34b934743c016.png)
be 
.![\[\dfrac{r_1^6+r_2^6+\dots+r_{108}^6}{r_1^6r_2^6+r_1^6r_3^6+\dots+r_{107}^6r_{108}^6}\]](http://latex.artofproblemsolving.com/3/1/5/315341a7adf35bc04e40ee0685a011c9f7b76115.png)
without Newton Sums.
Is there an elegant way to prove this?
Y by Adventure10, Mango247
Two circles 
and 
intersect at points 
and 
(with 
). Let 
be the centre of and let 
be the centre of . The exterior angle bisector of 
intersects again at 
and again at 
. Prove that the interior angle bisector of 
passes through the circumcenter of 
.
For points
that lie on a line 
in that order, and a point 
not on , the interior angle bisector of 
is the line that divides into two equal angles, while the exterior angle bisector of is the line that divides 
into two equal angles.
and 
intersect at points 
and 
(with 
). Let 
be the centre of and let 
be the centre of . The exterior angle bisector of 
intersects again at 
and again at 
. Prove that the interior angle bisector of 
passes through the circumcenter of 
.For points
that lie on a line 
in that order, and a point 
not on , the interior angle bisector of 
is the line that divides into two equal angles, while the exterior angle bisector of is the line that divides 
into two equal angles.
intersects 
,
respectively. Obviously 
and 
are collinear. Let 
.
be the center of 
.
is cyclic.
.
is cyclic. We known from the miquel configuration that the miquel point 
is cyclic. Notice that 
is a parallelogram. So 
,
is cyclic. Which means 
is cyclic. 
,
is isosceles. And 
bisects 
,
.
to 
)
is the internal 
b![\[90 + \measuredangle AZX = \measuredangle BAX = \measuredangle YAC = 90 + \measuredangle YZA\]](http://latex.artofproblemsolving.com/0/0/c/00c4506238f9fc37a024b954b4b9fac61870dd9c.png)
from which it indeed clearly follows that 
as desired.
-
which implies that 
d
.
