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.
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)
be positive integers. Prove that it is impossible to have all of the three numbers 
to be perfect squares.
such that for all real numbers and 
,
be a quadrilateral inscribed in a circle of center 
. Let M and N be the midpoints of diagonals 
and 
, respectively and let 
be the intersection point of the diagonals and of the given quadrilateral .It is known that the points 
are distinct. Prove that the points 
are concyclic if and only if the points 
are concyclic.
such that there exists a pair 
of positive integers, such that 
is not divisible by the cube of any prime, and 
such that has exactly 2000 prime divisors and divides 
?
be triangle with 
and 
on the segment such that 
. Consider point 
on 
such that 
. Point 
is the intersection of tangent line through point to circle 
respectively . If the tangent through point to 
intersect at 
. Line 
intersect at . Prove that 
bisect 
.
be a positive integer. Determine all sequences 
of positive integers such that
for all positive integers 
.
.
such that 
, then by induction it follows that 
, which forms a valid sequence.
. Then, we have:
Hence, we obtain
, where 
, it follows that 
. For sufficiently large 
, the left-hand side grows linearly while the right-hand side grows quadratically:
which is a contradiction.
.
![\[
\frac{5^{2000} - 1}{4}
\]](http://latex.artofproblemsolving.com/8/b/a/8ba5d1ed8bafd6653f9e6f351558a9976c43a57a.png)
is divided by 
.
.