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 the solid square region with vertices at![\[
(0, 0), \left(\frac{1}{2}, \frac{1}{2}\right), (1, 0), \left(\frac{1}{2}, -\frac{1}{2}\right).
\]](http://latex.artofproblemsolving.com/a/9/8/a9898f9237f07eaef5d17a9ac689d80ca631863f.png)
consists of two copies of 
: one copy which is rotated 
counterclockwise around the origin and scaled by a factor of 
, and another copy which is also rotated counterclockwise around the origin and scaled by a factor of , and then translated by 
. and 
below.
be the limiting region of the sequence 
. can be written as 
for relatively prime positive integers 
and 
. Find 
.
such that 
for all 
with parallel sides 
and 
, assume that there exist points 
on line 
outside segment , and 
inside segment 
such that 
. Denote by 
the point of intersection of and 
, and by 
the point of intersection of and . Let 
be the midpoint of segment , assume it does not lie on line . Prove that belongs to the circumcircle of 
if and only if belongs to the circumcircle of 
.
be prime numbers such that![\[
\frac{1}{pq}+\frac{1}{qr}+\frac{1}{pr}=\frac{1}{839}
\]](http://latex.artofproblemsolving.com/b/6/f/b6f882f3c93b0877419db625c2a427fc9561605b.png)
Hence find the numbers of ordered triples 
(i.e., the square on 
has as one of its sides and lies outside ). Show that the lines drawn from the vertices 
, 
, and 
to the centers of the opposite squares are concurrent.
with an area of 
, with 
. If 
, Hence find the value of 
.
be real numbers such that ![\[ \frac{1}{x^2-1} + \frac{1}{y^2-1} + \frac{1}{z^2-1} = 1. \]](http://latex.artofproblemsolving.com/8/f/5/8f5d49c9b44e3db3cd85fea7b6521e1dc02d4269.png)
Prove that ![\[ \frac{1}{x+1} + \frac{1}{y+1} + \frac{1}{z+1} \le 1. \]](http://latex.artofproblemsolving.com/c/0/4/c0438d6eba686d0a11fd4decb4afa9a14aa50bac.png)
? is about 
the intersection of 
's angle bisector and 
. We'll prove 
.
, we know that 
must be an isosceles trapezoid (the two opposite angle bisectors are symmetric) where 
and 
is cyclic, then 
, so 
.
is the perpendicular bisector of 
, so 
must be isosceles. By 
, 
.
, so 
is the inner angle bisector of 
. We conclude that 
.
.
Prove that
Let 
Prove that
.
.
.
such that 
.![$$2 \sqrt[4]{\frac{a}{b+c-a}}\ge 2 +\frac{2a^2-b^2-c^2}{(a+b)(a+c)}.$$](http://latex.artofproblemsolving.com/f/0/7/f0731fbe5c83aedc8474975f930f7194c08cc070.png)
is isosceles, so 
is a trapezoid.