and 
.
and 
intersect the circumcircle of 
and 
,
and 
such that 
and 
are parallelograms (with 
,
)
be the point of intersection of lines 
and 
.
,
be an integer. In a configuration of an 
board, each of the 
cells contains an arrow, either pointing up, down, left, or right. Given a starting configuration, Turbo the snail starts in one of the cells of the board and travels from cell to cell. In each move, Turbo moves one square unit in the direction indicated by the arrow in her cell (possibly leaving the board). After each move, the arrows in all of the cells rotate 
counterclockwise. We call a cell good if, starting from that cell, Turbo visits each cell of the board exactly once, without leaving the board, and returns to her initial cell at the end. Determine, in terms of 
,
such that 
and 
are all perfect square.
Y by Adventure10, Mango247
The incircle of a triangle 
touches the sides 
, 
, 
at the points 
, 
, 
respectively. Let the line 
intersect this incircle of triangle at a point 
(apart from ). Assume that this point is the midpoint of the segment , this means, 
. Let the line 
meet the incircle of triangle at a point 
(apart from ), and let the line 
meet the incircle of triangle at a point 
(apart from ). Show that 
.
touches the sides 
, 
, 
at the points 
, 
, 
respectively. Let the line 
intersect this incircle of triangle at a point 
(apart from ). Assume that this point is the midpoint of the segment , this means, 
. Let the line 
meet the incircle of triangle at a point 
(apart from ), and let the line 
meet the incircle of triangle at a point 
(apart from ). Show that 
.
Power of the vertex A to the incircle (I) is 
Since 
we get 
BX is the B-median of the triangle 
Thus
In exactly the same way, we can show that 
As a result, EFYZ is a cyclic trapezoid with bases 
i.e., isosceles and with equal diagonals, hence 
,
.
,
.
,
should be parallel to 
should be parallal to 
where 
and 
.
,
,
and 
.![\[AX AD=AE AE\]](http://latex.artofproblemsolving.com/9/3/c/93caff3f2af3b6860646f8be2043925780283396.png)
![\[AD = (s-a)\sqrt{2}\]](http://latex.artofproblemsolving.com/5/a/9/5a909b8347358201ac5d614a89b494251e7d318e.png)
From triangle ADC we get![\[\frac{\sin \varphi}{\sin \gamma}= \frac{CD}{AD}\]](http://latex.artofproblemsolving.com/4/1/7/417e7cbf461aad0d1c90986ba0144a5ce4bf6055.png)
![\[\sin \varphi= \sqrt{2}\tan\frac{\alpha}{2}\cos^{2}\frac{\gamma}{2}\]](http://latex.artofproblemsolving.com/3/3/7/337d63702d62161786276a2f0a5dc02530d7d8ed.png)
![\[\sin \theta = \sqrt{2}\tan\frac{\alpha}{2}\cos^{2}\frac{\beta}{2}\]](http://latex.artofproblemsolving.com/7/6/5/7657af71cb7cc74df44169e0bf4a0ef530ade9f4.png)
In triangle AXB: ![\[\frac{\sin \theta}{\sin \xi}= \frac{BX}{AX}\]](http://latex.artofproblemsolving.com/7/b/b/7bbd38c28039a075bc6d8a9f5792db2442327b7a.png)
From triangle BFY: ![\[\frac{\sin \mu}{\sin \xi}= \frac{BY}{FY}\]](http://latex.artofproblemsolving.com/c/7/f/c7f11ff3938970fd8dedb1111b79dae3f6b676d5.png)
We have: ![\[BX BY = BD^{2}\]](http://latex.artofproblemsolving.com/a/0/7/a0794d8e643604053100d0467a00a01ee871b6da.png)
![\[\sin \xi = \sqrt{2}\sin \mu \sin \frac{\beta}{2}\]](http://latex.artofproblemsolving.com/a/2/5/a25e513203a495ff3ed1360f16ea0865f784c43c.png)
Also we have ![\[\sin \nu = \sqrt{2}\sin \eta \sin \frac{\gamma}{2}\]](http://latex.artofproblemsolving.com/6/7/0/6701c3c33033da33415e5decaa72b50d09748f6a.png)
From Ceva' s theorem we have![\[\frac{\sin \varphi}{\sin\theta}\frac{\sin \xi}{\sin \nu}=\frac{\sin(\beta-\xi)}{\sin(\gamma-\nu)}= \frac{CX}{BX}\]](http://latex.artofproblemsolving.com/8/6/9/869d298f4e7f0d9b79dc9b9c7d96ce98552e0abf.png)
but I don't know how to
and 
then using Pascal's theorem on hexagon 
we obtain 
and 
,
.
,
making 
an iscoceles trapezium. So the diagonals are congruent hence 
be the midpoint of 
.
and that 
is the 
s
,![\[
\angle BFY=\angle FXY=\angle MXD=\angle FAD
\]](http://latex.artofproblemsolving.com/8/0/f/80fd56fe2f42e13cf49eea31a510441dce9a6792.png)
meaning that 
