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and 
such that![\[
1^n + 2^n + 3^n + \cdots + n^n = x!
\]](http://latex.artofproblemsolving.com/6/4/b/64b5f5919fd3710cc15e96724918eae216c8e4f0.png)
be a positive integer. Initially, a 
grid has 
black cells and the rest white cells. The following two operations are allowed :
square has exactly three black cells, the fourth is changed to a black cell;
square, the black cells are changed to white and white to black. such that for any configuration of the grid with black cells, all cells can be black after a finite number of operations.
touch sides 
, 
, and 
at points 
, 
, and 
, respectively. Let 
be the diametrically opposite point of with respect to the incircle. Let lines 
and 
intersect the incircle again at 
and 
, respectively. Prove that the lines 
, 
, and 
are concurrent, i.e., the lines intersect at the same point.
and 
have the property that![\[
\frac{a}{b} + \frac{b}{c} + \frac{c}{a}
\]](http://latex.artofproblemsolving.com/c/d/3/cd369fef55ed23cd01eef4b948cd1c155b0f78a0.png)
is an integer. Is it possible for![\[
\frac{a}{c} + \frac{c}{b} + \frac{b}{a}
\]](http://latex.artofproblemsolving.com/b/3/2/b3235637fdb32279ef39c1c072f340a4a6fd1d67.png)
to also be an integer except for the trivial solution? and have the property thatis an integer. Is it possible forto also be an integer except for the trivial solution?
is a triangle with circumcircle 
. Give a circle 
tangent , at 
, 
, resp, and tangent the circumcircle 
. Prove that 
bisects 
with 
is the centre of incircle 
be the midpoint of 
.
be the midpoint of 
, respectively .
be the B-excenter of 
and 
be the projection of 
on 
.
be the projection of on 
, respectively .
be the composition of Inversion 
and Reflection 
.
under ,
is the 9-point circle 
of under ,
is B-excircle 
under 
under ,
.
,
,
we get 
is the midpoint of 
, is the perpendicular bisector of .
to meet at 
we want to prove that is midline of 
note that are tangency points of mixtilinear circle in front of 
. apply an inversion with center and power 
we know that 
under the inversion and inverse of mixtilinear circle is excircle of 
in front of 
that touches at 
and since 
inverse of 
is a reflection of In the line 
call it 
let a line parallel to 
from intersect at 
so we get that 
is a nine point circle of 
also the inverse of 
is a circle that it is tangent to at 
and tangent to 
; call it 
but note that from a well known fact the nine point circle of a triangle is tangent to its excircle from this fact we get that is excircle of now because 
is midline of we get that 
are midpoints of 
so 
are midpoints of 
before the inversion and bisects .
