[/quote]
,
is chosen inside a convex quadrilateral 
.
and 
,
[list=a]
and 
,
satisfying this equation.
denotes the greatest common divisor of positive integers 
and 
.
,
a right-angled triangle can be formed?
has side lengths 
,
,
and denote them as 
,
,
.
?
.
and 
and 
and 
satisfying the following conditions:
for all 
Define the set ![\[S = \left\{ \left( \sum_{i=1}^{10} a_i x_i, \sum_{i=1}^{10} a_i y_i \right) : a_1, a_2, \cdots, a_{10} \in \{0, 1\} \right\},\]](http://latex.artofproblemsolving.com/d/e/b/deb50830b9fe10c86ea169c044c1b26f515b2514.png)
then 
，
be non-negative real numbers such that 
Prove that
.
of the written numbers were erased, and it turned out that any two of the remaining numbers are coprime. Find the largest possible value of 
and 
are equal, and they intersect at 
.
and 
intersect at point 
,
and 
intersect at point 
lying inside triangle 
.
.
be a positive integer. Egor has 
cards with the number 
”
”
.
be a point on side 
be a point on the extension of 
such that 
.
be the circumcenter of 
,
satisfying 
.
children standing in a circle. For each girl, it turns out that among the five people following her clockwise, there are more boys than girls. Find the greatest number of girls that can stand in a circle.
,
Y by Adventure10, Mango247
Let 
be a triangle with circum-circle 
. Let 
be a point in the interior of triangle which is also on the bisector of 
. Let 
meet in 
respectively. Suppose 
is the point of intersection of 
with 
; and 
is the point of intersection of 
with 
. Prove that 
is parallel to 
.
be a triangle with circum-circle 
. Let 
be a point in the interior of triangle which is also on the bisector of 
. Let 
meet in 
respectively. Suppose 
is the point of intersection of 
with 
; and 
is the point of intersection of 
with 
. Prove that 
is parallel to 
.
to obtain 
collinear.
.
is cyclic.
.
and we are done.
and 
.
so 
is a cyclic quadrilateral.
.
and hence 
and 
both of which are parallel to 
.
.
,
are cyclic.The first because of the previous angle equality the last one because of 
,
.
are collinear and since 
we have 
is incenter. This problem is Azerbaijan JBMO 2016 TST.
and 
are cyclic and hence evaluate 
and 
.
are collinear points.
and hence conclude that 
.
and 
,
gives, 
are collinear and hence, 
and 
are concyclic with radical axis 
cyclic and thus by Reim's Theorem 
.
we have 
.
.
HFAE, AEIG, AFDG all quadrilaterals are cyclic so we have 
means H,E,I are concurrent
hence done.
