,
and 
,
is the least common multiple of the numbers 
.
for every positive integer 
be positive integers with 
.
,
,
,
,
,
positions to the right. Andrés and Beto alternate turns, starting with Andrés. The winner is the player who moves a token into the rightmost cell. Determine who has a winning strategy in terms of 
.
,
passing through 
at point 
.
Y by Adventure10, Mango247
Consider an isosceles triangle 
with 
, and a circle 
which is tangent to the sides 
and 
of this triangle and intersects the side 
at the points 
and 
. The segment 
intersects the circle at a point 
(apart from ). Let 
and 
be the reflections of the point in the points 
and 
, respectively. Show that the circumcircle of triangle 
is tangent to the circle .
with 
, and a circle 
which is tangent to the sides 
and 
of this triangle and intersects the side 
at the points 
and 
. The segment 
intersects the circle at a point 
(apart from ). Let 
and 
be the reflections of the point in the points 
and 
, respectively. Show that the circumcircle of triangle 
is tangent to the circle .
.
be the points where 
respectively. The quadrilateral 
is harmonic, so 
form a harmonic division. This means that the same holds for the intersection points between these four lines and 
,
.
,
are homothetic with center 
,
,
,
,
.
- harmonical pencil ; 
- harmonical division. From the well-known property we have 
,
a.s.o.
,
and 
.
and 
are the tangents to 
is the symmedian of triangle 
.![\[\frac{XK}{XM}=\frac{YK}{YM}\]](http://latex.artofproblemsolving.com/a/b/5/ab57a19986fc9c71b725d93550f0fdccd8c5707c.png)
is isosceles, we have that 
.
.
.
.
.![\[\frac{XM}{XK}=\frac{YM}{YK}=\frac{XK}{P'X} \quad \Rightarrow \quad \frac{P'X}{XM}=\frac{XK^2}{XM^2}\]](http://latex.artofproblemsolving.com/a/8/2/a821a435f50c67a1efb1be9aac8f79424ed497b1.png)
.![\[\frac{AX}{AK}=\frac{AM}{AX}=\frac{XM}{XK} \quad \Rightarrow \quad \frac{AM}{AK}=\frac{XM^2}{XK^2}\]](http://latex.artofproblemsolving.com/c/d/3/cd3faac883922079c40cd0e41b1dde561363595d.png)
combined with the collinearity of 
,![\[\frac{P'B}{BK}=\frac{P'X}{XM} \cdot \frac{AM}{AK} = \frac{XK^2}{XM^2} \cdot \frac{XM^2}{XK^2}=1\]](http://latex.artofproblemsolving.com/7/a/a/7aadc076927ed0cbcdcfeb54817de548705954dd.png)
and 
.
.
to the circumcircle of 
which is 
and 
intersect 
and 
,
,
.
(since AY is a tangent
,
,
.
since 
.
,
.
and 
,
.
are collinear. By power of a point, 
,
,
.
,
.
,
.
,
,
are concyclic and not collinear, We have 
by one of those similarities, 
,
and 
pass through the tangency points 
and 
,
,
at 
.
Let 
be vampire points in 
such that 
and 
are colinear. Obviously the quadrilateral 
is harmonic.
is homothetic to 
,
to 
,
be the touchpoints of 
be the infinity point on 
.
be a point on 
tangent to 
.
hence it suffices to show 
.
and 
.
.
hence 
and we're done. 
are the points where the circle touches 
.
and 
respectively. We first show that 
are collinear. As 
and the tangents at 
.
from the circle to ![\[-1 = (MK;T_1T_2) \overset{T_1}{=} ((T_1M \cap BC)K;B\infty_{BC})\]](http://latex.artofproblemsolving.com/c/5/2/c524f3cabbd923ffdcc39dbaaf0048b72eb2359a.png)
implying that 
.
are collinear. Let the tangent to 
meet ![\[-1 = (MK;T_1T_2) \overset{M}{=} (XK;PQ)\]](http://latex.artofproblemsolving.com/8/c/9/8c9aa6fd6088ec54aa33318950f1d6edac152d5c.png)
As 
.
is tangent to 
as well, finishing the problem as 
