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.
players take part. Every two play exactly one match. The schedule is such that no two matches are played at the same time, and each player, after taking part in a match, is free in at least 
next (consecutive) matches. Prove that one of the players who play in the opening match will also play in the closing match.
be a fixed positive real number. Find the minimum possible value 
such that for any positive reals 
, 
, 
, 
, we have
and 
with no common divisor greater than 1 such that 
divides 
. (Professor Yongjin Song)
and 
. Prove that there always exists a positive integer 
such that 
. is a power of a prime number, but for arbitrary then well.....
, 
, with 
, 
. Prove that there exists 
such that for every 
![\[
f_n(x) = \frac{x}{1 + nx}, \quad \text{if and only if } f(x) = \frac{x}{1 + x}, \quad \forall x \geq 0.
\]](http://latex.artofproblemsolving.com/0/f/c/0fc36d9264eb7e103128c489aeae521a859c1fd4.png)
(For and , the notation 
represents 
. )
, , with , . Prove that there exists such that for every (For and , the notation represents . )
be the circumcircle of acute triangle 
. Points 
and 
are on segments 
and 
respectively such that 
. The perpendicular bisectors of 
and 
intersect minor arcs and of at points 
and 
respectively. Prove that lines 
and 
are either parallel or they are the same line.
, , with , . Prove that there exists such that for every (For and , the notation represents . )
is a positive real number, find the minimum of the following expression
. The angle bisectors of the angles and 
intersect the sides 
and 
at the points 
and 
, and intersect each other at the point 
. The line 
intersects the circumcircle of triangle at the points 
and 
. Prove that the circumradius of triangle 
is twice as long as the circumradius of triangle .
such that 
is also a prime number.
be a cyclic quadrilateral. Let 
, 
, 
be the feet of the perpendiculars from to the lines 
, 
, , respectively. Show that 
if and only if the bisectors of 
and 
are concurrent with .
such that for all 
,![\[
mn + f(n!) = f(f(n))! + n \cdot \gcd(f(m), m!).
\]](http://latex.artofproblemsolving.com/c/0/c/c0cdf1bcb8053f5dc26e9df6355207b5e479f861.png)
is a sum of two circles passing through 
without point . be the midpoint of and 
be the midpoint of . Let 
be the vertex of equilateral triangle 
.
that lies on circle 
. Let be the intersection of 
and 
on the other side of than . Obviously, 
is tangent to , therefore 
. As 
is equilateral, also 
is equilateral. So the altitude of from passes through . Hence, 
. This implies 
. So 
is the length of the altitude of 
from 
. But 
, so our point satisfies the necessary conditions. For a given altitude from there is only one "good" point for each side of line . Therefore, the requested set of points is a circle 
homothetic to with respect to in scale 
and a circle symmetric to with respect to (excluding point ).
![\[
\frac{5^{2000} - 1}{4}
\]](http://latex.artofproblemsolving.com/8/b/a/8ba5d1ed8bafd6653f9e6f351558a9976c43a57a.png)
is divided by 
.
