[/quote]
,
.
be its 
be midpoint of arc 
,
is the midpoint of 
.
be the intersection of the line 
with 
,
be the intersection of the angle bisector of 
with the line 
,the intersection of line 
with 
and 
,the intersection of 
and 
such that 
.
are collinear. Who has a winning strategy?
be positive real numbers so that 
.![\[ \left( a - 1 + \frac 1b \right) \left( b - 1 + \frac 1c \right) \left( c - 1 + \frac 1a \right) \leq 1.
\]](http://latex.artofproblemsolving.com/7/6/e/76e794a62090c8c4b5f3a70ad4d036669418ca69.png)
,
.
.
.
the midpoint of arc 
the point where the 
meets 
and intersects 
,
.
or 
is tangent to 
,
be a point on the segment 
be the intersection point of 
with the diagonal 
.
at point 
.
such that 
.
and the segment 
.
and 
are concyclic.
that satisfy the following condition: for all 
and 
relatively prime to 
we have ![\[a \equiv b \pmod n\qquad\text{if and only if}\qquad ab\equiv 1 \pmod n.\]](http://latex.artofproblemsolving.com/e/8/5/e85c5b249ac8e75b45a525b8786414798a4921fe.png)
satisfying the following property: if 
,
and 
are integers such that 
,
![\[
\sqrt{x-3}+\sqrt{6-x} \leq p
\]](http://latex.artofproblemsolving.com/3/2/b/32b8bf5c8aa4d7561933e38ee0e2a0480e1a2f07.png)
In which 
,
.
be the set of positive integers less than 
relatively prime to it. Is there a nice pattern for 
?
or 
,
grid, the number 1 is written. A move consists of choosing a positive integer 
and a cell, and then multiplying the numbers in that cell and its neighbors by 
?
is a minuan pair if it satisfies the following two conditions:
are all the positive divisors of 
,
is precisely
circumcenter 
and orthocenter 
.
be a point on line 
.
are on lines 
respectively such that 
and 
.
is perpendicular to 
.
and 
meet at two distinct points 
and 
.
through 
(other than 
.
be the midpoint of the minor arc 
(i.e., the point on 
subtends the arc 
be the midpoint of the minor arc 
.![\[
MN \parallel \text{(radical axis of } \omega_1, \omega_2).
\]](http://latex.artofproblemsolving.com/6/6/4/6647f0f7182bdf5096ec5f2956a272803082f4f7.png)
Y by
In the triangle 
, from the incenter 
draw a perpendicular on the side 
, which intersects it at the point 
. From the foot of the bisector of the angle 
draw another perpendicular to the same side , which intersects it at the point 
. Prove that the segment 
is divisible by line 
in half, where 
is the point of intersection of the extension of the bisector of the angle with the circumcircle of the triangle .
(D. Basov)
, from the incenter 
draw a perpendicular on the side 
, which intersects it at the point 
. From the foot of the bisector of the angle 
draw another perpendicular to the same side , which intersects it at the point 
. Prove that the segment 
is divisible by line 
in half, where 
is the point of intersection of the extension of the bisector of the angle with the circumcircle of the triangle .(D. Basov)
