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 triangle with circumcenter 
. Point 
is the intersection of the parallel line from to 
with the perpendicular line to 
from 
. Let 
be the point where the external bisector of 
intersects with . Let 
be the projection of onto 
. Prove that the lines 
have a common point.
be an acute triangle with 
and let 
and be its orthocenter and circumcenter, respectively. Let 
be the circle 
. The line 
and the circle of radius centered at 
cross at 
and 
, respectively. Prove that , the circle on diameter 
and circle 
are concurrent.
for which the following statement holds: among any five positive real numbers 
(not necessarily distinct), one can always choose distinct subscripts 
such that![\[ \left| \frac{a_i}{a_j} - \frac {a_k}{a_l} \right| \le C. \]](http://latex.artofproblemsolving.com/a/e/e/aeeab8e846634c7922e9312ef139739ad3fbe6eb.png)
be a triangle with circumcircle 
, incenter 
, and -excenter 
. Let the incircle and the -excircle hit 
at 
and 
, respectively, and let 
be the midpoint of arc without . Consider the circle tangent to at and arc 
at 
. If 
intersects again at 
, prove that 
and 
meet on . be an acute triangle with 
. be the circumcircle circumscribed to the triangle and the midpoint of the smallest arc of this circle. Let and points of the segments and respectively such that 
. Let 
be the second intersection point of the circumcircle circumscribed to 
with . Let 
and be the intersections of lines 
and 
with other than 
, respectively. Let 
and be the intersection points of lines 
and 
with lines and respectively. Show that the 
line passes through the midpoint of
are real numbers, find the smallest possible value of the expression:
and 
be prime numbers. Prove that if 
is a perfect square, then 
is also a perfect square.
and are primes and is a perfect square, prove that 
is a perfect square.
be a positive integer. Aslı and Zehra are playing a game on an 
chessboard. Initially, 
stones are placed on the squares of the board. In each move, Aslı chooses a row or a column; Zehra chooses a row or a column. The number of stones in each square of the chosen row or column must change such that the difference between the number of stones in a square with the most stones and a square with the fewest stones in that same row or column is at most 1. For which values of can Aslı guarantee that after a finite number of moves, all squares on the board will have an equal number of stones, regardless of the initial distribution?
reals, and 
.
?
be positive real numbers such that 
. Find the minimum value of the following sum :
knowing that the denominators are positive real numbers.
such that: Among every vertexes of the polygon, there always exists 4 vertexes forming a convex quadrilateral such that 3 sides of the quadrilateral are also sides of the polygon.
and in a non-isosceles triangle with incenter intersect its circumcircle at 
and 
, respectively. The line through , parallel to , intersects 
at . Prove that 
is tangent to .
is a triangle with incenter ; 
the midpoint of arc not containing .Let the bissectors of 
cut respectively 
at 
then 
and 
are collinear and 
through cut 
at 
; it ' s clear that 
are isoceles besides 
we get 
thus 
is bisector of 
hence 
idem 
the midpoint of arc not containing 
; 
we need to show that 
then 
.
we get 
are collinear which closes the proof .
and 
- so 
is the perpendicular bisector of 
, whence 
. Hence 
(the latter is from the line parallel to ) and 
, which solves the problem.
.
.
, 
.
is tangent to 
, hence![\[PI^2 = PC_0 \times PA_0.\]](http://latex.artofproblemsolving.com/0/4/9/0495c60a93d89cb998267c141bd6dc870579922a.png)
But we notice that is the symmetric point of wrt 
. is the perpendicular bisector of , so 
.![\[PB^2 = PC_0 \times PA_0,\]](http://latex.artofproblemsolving.com/5/2/9/529c43329e92ef2c5e41d9f088cc82627aa3132c.png)
which implies that 
.
.
