AoPS Academy
board and made 
moves so that it visited all squares and returned to the starting square. At every moment, the distance from the center of the square the king was on to the center of the board was calculated. A move is called 
if this distance becomes smaller after the move. Find the maximum possible number of pleasant moves. (The chess king moves to a square adjacent either by side or by corner.)
be an acute-angled, non-isosceles triangle with circumcenter 
and incenter 
, such that![\[
\prod_{\text{cyc}} \left( \frac{1}{a+b-c} + \frac{1}{a+c-b} - \frac{2}{b+c-a} \right) \neq 0,
\]](http://latex.artofproblemsolving.com/7/8/1/78125759bf5c73149e4a56538e9b0c44309704a7.png)
, 
, and 
.
, 
, and 
be the excenters opposite to vertices 
, 
, and 
, respectively, and let 
, 
, and 
be the Lemoine points of triangles 
, 
, and 
, respectively.
, 
, and 
all pass through a common point 
. Moreover, the isogonal conjugate of with respect to lies on the line 
.
intersect at point 
. The points of tangency of the circumcircles of triangles 
and 
with their common external tangents lie on a circle 
. The points of tangency of the circumcircles of triangles 
and 
with their common external tangents lie on a circle 
. Prove that the centers of circles and coincide.
cut 
at 
and 
cut 
at 
respectively. Line passing through 
and 
(lying at infinity) is the polar of 
WRT 
Then if 
touches at 
We have 
is the midpoint of 
Similarly, we have 
is the midpoint of 
As a result, triangles 
and 
are both isosceles 
, 
, 
, 
; it is easy to see that, due to the fact that the triangles are homothetic at 
, 
is the contact of the incircle of the triangle 
with 
and 
. From 
we get 
and, with we get 
, so 
is the reflection of 
in 
.
and 
, so, by difference, we get the required equality.
is denoted as 
, etc. Let 
denote the tangency points of the incircle to 
. Let touch the incircle at 
. It is clear that 
are not moved and that 
are the midpoints of 
. Also note that 
is cyclic because those points lie on the circle with diameter 
. Consequently, 
lies on the image of that circle which must be a line through 
. Also 
are collinear. Further, it is obvious that 
are collinear, so 
.
and 
, 
are midpoints of 
and 
, so we can easily conclude that 
, so 
.
, so subtracting from above, we get 
. 
. By symmetry, we also have 
. Subtracting these relations gives us
where 
is the center of inversion and 
denote the images of 
, so using that above, we get
.
is an excircle to , we have 
or 
is isosceles .
is isosceles and we are done.
be a triangle, 
a point such that 
and 
,
such that 
and 
.