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.
, the external bisectors of 
and 
meet at a point 
. Prove that the circumcentre of 
and the points 
lie on the same straight line.
be the circumcircle of acute triangle 
. Two tangents of from 
and 
intersect at 
, 
and 
intersect at . Point 
, 
are on 
and 
such that 
and 
.
are concyclic.
the centre of the circle passing through . 
, 
are difined similarly. Prove that 
, 
, 
are concurrent.
> 
that satisfy 
, for all 
€ .
be a fixed positive integer. Given a set 
of 
points in the plane such that no three are collinear and no four concyclic, let 
be the number of circles 
that contain 
in their interior, and let ![\[m(S)=a_1+a_2+\cdots + a_n.\]](http://latex.artofproblemsolving.com/4/9/b/49b928b1f20e2d1799a1234c7f25a2bc3d62d0dd.png)
Prove that there exists a positive integer 
depending only on 
such that the points of 
are the vertices of a convex polygon if and only if 
such that 
.
, then prove that the following inequality holds:![\[
\sum_{\text{cyc}} \sqrt{1 + \left(x\sqrt{1 + y^2} + y\sqrt{1 + x^2}\right)^2} \geq \sum_{\text{cyc}} xy + 2\sum_{\text{cyc}} x
\]](http://latex.artofproblemsolving.com/9/5/5/955863a0cf8747ae45b736b0631f243d3908eb84.png)
such that has exactly 2000 prime divisors and divides 
?
and 
is a subset of 
such that if 
then 
Find the smallest of 
that satisfy the following property: if 
, 
, 
, 
are all the positive divisors of 
with 
, then 
divides 
for every 
.
, in which 
: 
, but the proof is important.
be the longer side of the parallelogram. Let 
. One can angle chase to find that 
is a rectangle. Let bisector meet bisector at 
and 
be the intersection of 
bisector and bisector.
, 
. From 
, 
. With 
and 
. one can deduce is the midpoint of 
. Now one can compute side length of the rectangle as 
. Similarly, one can find the other one to be 
. By the problem condition![\[x\sin(\theta) \cdot x\cos(\theta) = 10 \iff \sin(2\theta) = \frac{20}{x^2}\]](http://latex.artofproblemsolving.com/e/6/7/e67f93e828e9e9e75ddce1bcca4216eef937f273.png)
Now ![$2 [\triangle BCD]$](http://latex.artofproblemsolving.com/0/6/6/066191c0daa8d613223cfdb6ddb9a40e9404c004.png)
is the area of parallelogram. One can just use the 
formula now to get![\[\text{Desired Area} = 2 \times \frac{1}{2}x\cdot 2x \cdot \sin(2\theta) = \fbox{40}\]](http://latex.artofproblemsolving.com/9/a/8/9a8f596599ff29a2ede380d674ebebbd09cd1f21.png)
and thus we are done. 
and 
Prove that
be real numbers such that 
Prove that
and 
.
,
,
