[/quote]
,
be an acute triangle with 
and let 
and 
be its orthocenter and circumcenter, respectively. Let 
be the circle 
.
and the circle of radius 
and 
,
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)
,
,
.
at 
and 
,
be the midpoint of arc 
at 
intersects 
,
and 
meet on 
.
respectively such that 
.
be the second intersection point of the circumcircle circumscribed to 
with 
and 
and 
with 
and 
and 
with lines 
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.
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 
reals, and 
.
?
be positive real numbers such that 
.
knowing that the denominators are positive real numbers.
such that: Among every 
equals to a positive integer power of 
.
Y by Adventure10, Mango247, Rounak_iitr
Let 
be a point out of circle 
. Let 
and 
be the tangents to the circle drawn from . Choose a point 
on 
. Suppose that the circumcircle of triangle 
intersects again at 
. Let 
be the reflection of with respect to 
. Prove that
![\[ \angle PBT = \angle {P}'KA \]](//latex.artofproblemsolving.com/8/e/8/8e8e84079b26b0674c3e68f5d333e94fbebe2328.png)
be a point out of circle 
. Let 
and 
be the tangents to the circle drawn from . Choose a point 
on 
. Suppose that the circumcircle of triangle 
intersects again at 
. Let 
be the reflection of with respect to 
. Prove that![\[ \angle PBT = \angle {P}'KA \]](http://latex.artofproblemsolving.com/8/e/8/8e8e84079b26b0674c3e68f5d333e94fbebe2328.png)
This post has been edited 5 times. Last edited by dan23, May 25, 2013, 5:07 PM
so 
so 
done.
is parallel to 
and because 
then we need to prove that 
then we have 
so now 
and we are done
and 
and 
for the point S and we will prove that SP=PK.
is distinct from 
.
is a 
.
is bisected at 
which (again after an angle chase) yields us 
intersects 
be the reflection of ![\[ \angle DBP = \angle QCA \]](http://latex.artofproblemsolving.com/5/9/5/595529a23011b193fbc4bac7c1f81d014562ff89.png)
and 
.
bisecting 
,
.
