[/quote]
,
be a triangle and 
its circumcentre. A line tangent to the circumcircle of the triangle 
intersects sides 
and 
be the image of 
under 
.
is tangent to the circumcircle of triangle 
be a semicircle with diameter 
be a point on 
touches 
in a point 
.
touches 
and 
and 
with no parallel sides is inscribed in a circle 
.
are inscribed in triangles 
,
and 
,
,
,
.
and 
Prove that
,
,
,
are integers such that 
.
and 
of 
such that for each integer 
with 
,![\[ n\mid a_i - b_i - c_i
\]](http://latex.artofproblemsolving.com/1/a/e/1ae8d54f068b646ac7a1e43d75e7f26510cfdc95.png)
be an odd prime and 
be integers so that the integers 
are divisible by 
Prove that 
be real numbers such that 
Prove that 
Where 
.
and 
respectively denote the midpoints of the arcs 
denote the midpoint of arc 
(the arc 
including 
be the incenter of 
be the circle that is tangent to 
be the circle that is tangent to 
,
friends, 
people are seated around a circular table in a fixed order. Ali places 
.
lie on 
,
is tangent to circumcircle of 
.
is tangent to circumcircle of triangle 
Y by ImSh95, Adventure10, Rounak_iitr
Let 
be a trapezium inscribed in a circle 
with diameter 
. Let 
be the intersection point of the diagonals 
and 
. The circle with center 
and radius 
meets at the points 
and 
(where is on the same side of as 
). The line perpendicular to at intersects 
at 
. Prove that 
is perpendicular to 
.
Greece - Silouanos Brazitikos
be a trapezium inscribed in a circle 
with diameter 
. Let 
be the intersection point of the diagonals 
and 
. The circle with center 
and radius 
meets at the points 
and 
(where is on the same side of as 
). The line perpendicular to at intersects 
at 
. Prove that 
is perpendicular to 
.Greece - Silouanos Brazitikos
it is enough to prove that 
,
.
so 
.
.
.
.![\[ \cos{\widehat{KBL}}=1-\dfrac{KL^2}{2KB^2}=1-\dfrac{KL^2}{2BE^2}=1-\dfrac{2(AB^2-BE^2)}{AB^2}=2\dfrac{BE^2}{AB^2}-1 \]](http://latex.artofproblemsolving.com/c/4/7/c472344faf3c8081631f8208dcbce85fd7d3d337.png)
,
and with the basic trigonometry we get that both expression equal to 
.
and 
.
.
.
is cyclic and tangents to 
.
we get that 
are concurrent at one point 
and 
.
it gives 
and 
.
.
and 
It is enough to prove that 
.
.
parallel to 
.
;
.
.
.
.
is parallelogram.. 
.
.
.
.
as well as 
.
.
.
.
,
.
.
and 
are equivalent triangles. Thus 
so 
lie on the radical axis 
and 
.
.
,
intersect 
respectively.
are inverses with respect to 
.
,
,
be the circle with centre 
,
,
and 
.
,
so 
as they share 
and both have a right angle.
.
,
,
and 
so that 
.
is isosceles, so that as 
,
is isosceles; moreover, 
,
.
.
,
as required.
.
lies on the radical axis of the two circles,or in other words 
are collinear.From this it immediately follows that 
(
is cyclic.Thus 
or in other words 
.
not containing 
we see that 
at 
,
be the reflection of 
the orthocenter of 
are collinear and 
is a parallelogram. From 
and 
,
.
be the midpoint of 
,
is a rhombus so 
so 
,
and 
are concurrent at a point 
and 
;
.
;
is an isosceles trapezoid; hence, line 
,
,
.
it is well known that 
are concurrent...and its easy from here as shown by posts above 
.
and 
are reflections of each other in 
.
around 
and 
.
is tangent to 
lies on 
.
,
.
.
.
with 
.
hence 
,
.
in 
.
and 
hence 
is a diameter in 
unit circle...so that
Let us firstly note that
Since 
On the other hand 
Equalizing 
and 
yields
.
.
.
be 
and 
be 
.
is 
and so 
is 
.
is 
and so 
,
at 
is equal to 
, or equivalently, 
is isosceles. Since 
is the perpendicular bisector of 
.
.
=
=
=
is cyclic. We are done. 
and 
is tangent to 
.
is tangent to 
.
are concyclic , we get 
are concyclic
.
,
and 
,
.
as required. 
we deduce that 
and 
.
,
,
and 
so 
collinear. Note that the radical axis must be a line perpendicular to 
,
are reflections across 
