AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
be a semicircle with diameter 
and midpoint 
. Let 
be a point on different from 
and 
.
touches in a point 
, the segment 
in a point 
, and additionally the segment 
. The circle 
touches in a point 
and additionally the segments 
and .
and 
are perpendicular.
with no parallel sides is inscribed in a circle 
. Circles 
are inscribed in triangles 
, respectively. Common external tangents are drawn between 
and 
, and 
, and 
, and and , not containing any sides of quadrilateral . A quadrilateral whose consecutive sides lie on these four lines is inscribed in a circle 
. Prove that the lines joining the centers of and , and , and the centers of and all intersect at one point.
, 
, 
, 
are integers such that 
.
and 
of 
such that for each integer 
with 
, we have![\[ n\mid a_i - b_i - c_i
\]](http://latex.artofproblemsolving.com/1/a/e/1ae8d54f068b646ac7a1e43d75e7f26510cfdc95.png)
be an odd prime and 
be integers so that the integers 
are divisible by . divides each of .
be real numbers such that 
Prove that 
Where 
be a triangle with circumcircle 
. Let 
and 
respectively denote the midpoints of the arcs 
and that do not contain the third vertex. Let 
denote the midpoint of arc 
(the arc 
including ). Let 
be the incenter of . Let 
be the circle that is tangent to and internally tangent to at , and let 
be the circle that is tangent to and internally tangent to at . Show that the line 
, and the lines through the intersections of and , meet on .
friends, 
people are seated around a circular table in a fixed order. Ali places apples for serving directly in front of himself and wants to distribute them among everyone. Since Ali and his friends dislike eating alone and won't start unless everyone receives an apple at the same time, in each step, each person who has at least one apple passes one apple to the first person to their right who doesn't have an apple (in the clockwise direction). such that after some number of steps, the situation reaches a point where each person has exactly one apple.
(i dont remember exactly if it was 8032038 or some other number, so correct me on this)
(i posted this at 12:29) lol
?
and 
Prove that
Prove that 
. Any help would be nice, thanks!
,
the foot of 
,
so 
