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 an even integer and let 
be real numbers satisfying 
.
and 
such that 
.
has a solution
and 
be different points on a circle 
such that 
is not a diameter. Let 
be the tangent line to at . Point 
is such that is the midpoint of the line segment 
. Point 
is chosen on the shorter arc of so that the circumcircle 
of triangle 
intersects at two distinct points. Let 
be the common point of and that is closer to . Line 
meets again at 
. Prove that the line 
is tangent to .
such that
table. It is permissible to reverse the signs of all the numbers in any row or column. Prove that after a number of these operations, we can make the sum of the numbers along each line (row or column) nonnegative.
, prove that 
is the circumcenter of 
, and 
are points on segments 
and 
respectively with 
. Let 
be a point such that 
and 
.
intersect and at points 
and respectively. Let the line passing through and perpendicular to 
intersect and at points and respectively. Prove that points 
, and are concyclic.
defined by 
and![\[
u_{n+1} = \frac{1}{2}u_n^2 - 4 \quad \text{for all } n \in \mathbb{N}.
\]](http://latex.artofproblemsolving.com/9/9/4/994aa754cc1288ce4f28a95a0276e64282fb5f66.png)
a) Prove that there exist infinitely many positive integers 
such that 
.![\[
\lim_{n \to \infty} \frac{2u_{n+1}}{u_0u_1\cdots u_n}.
\]](http://latex.artofproblemsolving.com/f/f/1/ff174e7431cbfbc17c650d109651241286756a1a.png)
and 
be altitudes in an acute triangle 
which meet at 
. Suppose that 
meets the circumcircle of at 
and 
such that lies on the shorter arc of 
and lies on the shorter arc of 
. Let 
and 
meet at 
. Show that the circumcircles of 
and 
and the line 
concur.
and 
are constructed in the extirior of a right-angled triangle 
with 
.Let 
be the midpoint of 
.Given that 
and 
.Find the length of .
, but I might be wrong with calculations.
are equilateral 
![$[CG] = [CA] = [GA]$](http://latex.artofproblemsolving.com/a/b/4/ab421e7805fb68ff78fa7ef170bc6193534bf750.png)
, ![$[FB] = [AB] = [AF]$](http://latex.artofproblemsolving.com/c/6/f/c6fe82d238ee035deb72c832bd709a8eab1b177f.png)
and 
by SAS ![$[GB] = [FC]$](http://latex.artofproblemsolving.com/4/d/4/4d49ac5d96d4e5de516d1ad4566e59c96f8ad94d.png)
.
are G-, F-medians of 
![$[GM]^2 = \tfrac{1}{4}\left(2[GB]^2 + 2[GC]^2 - [BC]^2\right)$](http://latex.artofproblemsolving.com/e/1/f/e1fb86879f913c0da9bf27f79fb4d5c977e1ac8f.png)
and ![$[FM]^2 = \tfrac{1}{4}\left(2[FB]^2 + 2[FC]^2 - [BC]^2\right)$](http://latex.artofproblemsolving.com/9/b/e/9bed683d4e3a6886c553e24ef14d12ad349dbb38.png)
![$[FM]^2 - [GM]^2 = \tfrac{1}{2}\left([FB]^2 - [GC]^2\right) = \tfrac{1}{2}\left([AB]^2 - [CA]^2\right) = \tfrac{1}{2} [BC]^2 =$](http://latex.artofproblemsolving.com/a/e/4/ae4bfe990229b05c41464d1bba8ed0f5941eeae3.png)
![$[BC]^2 = 144$](http://latex.artofproblemsolving.com/1/d/7/1d729c3d6702a814cca0ff398eb2165df65bd570.png)
![$[BC] = 12$](http://latex.artofproblemsolving.com/f/c/a/fca981613c8213607683ec4f336bfad3402be878.png)
.
and are constructed in the extirior of a right-angled triangle with .Let be the midpoint of .Given that and .Find the length of .
,
.
be the midpoint of ![$[AB]$](http://latex.artofproblemsolving.com/a/d/a/ada6f54288b7a2cdd299eba0055f8c8d19916b4b.png)
and ![$[AC]$](http://latex.artofproblemsolving.com/0/9/3/0936990e6625d65357ca51006c08c9fe3e04ba0c.png)
, 
and 
.
is right at its vertex 
hence
, 
formula (cosine formula ) gives
, so we have
we get
we get
