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.
, such that 
and 
.
such that 
is prime, 
with 
and ?
be a function such that for every regular 
-gon 
we have 
. Prove that 
for all reals 
.
be a triangle (
). 
are the altitudes of the triangle. The bisector of 
intersects 
at 
. Let 
be a point such that 
and 
. Prove that 
passes through the midpoint of 
.
with 
and at most one of them being zero, show that![\[\frac{1}{\max(a^2,b)}+\frac{1}{\max(b^2,c)}+\frac{1}{\max(c^2,a)}\geq\frac{27}{4}\]](http://latex.artofproblemsolving.com/9/e/1/9e14a9ef2d1a0a98006bcf716fa33c3d9ea29bd5.png)
be an integer. Find all sequences 
satisfying the following conditions:![\[ \text{ (a) } a_i \in \left\{0,1\right\} \text{ for all } 1 \leq i \leq n^2 + n;
\]](http://latex.artofproblemsolving.com/3/c/5/3c509ec2e9013e8d3be492c8eb44a7c33841b74e.png)
![\[ \text{ (b) } a_{i + 1} + a_{i + 2} + \ldots + a_{i + n} < a_{i + n + 1} + a_{i + n + 2} + \ldots + a_{i + 2n} \text{ for all } 0 \leq i \leq n^2 - n.
\]](http://latex.artofproblemsolving.com/9/7/d/97d2a467d1c0dc8594ec024c3bb9b8c87ee85b19.png)
and 
. Prove that
Equality holds when 
or 
and 
, Ivica and Marica play the following game: Given two piles of cookies, on each turn a player takes 
cookies from one of the piles, of which he eats 
and puts of is arbitrary in every move. Players take turns alternatively, with Ivica going
for which Marica has a winning strategy.
be non-negative real numbers such that 
Prove that
be the circumcenter, symmedian point of 
. Show that the tangent of 
at 
,the tangent of 
at 
, the tangent of 
at 
are concurrent.
be a triangle such that 
. Find two ways of dissecting the triangle into three isosceles triangles by straight cuts.
the bisector of 
with 
on 
Let 
a point on so that 
Then the triangles 
and 
are isosceles.
the bisector of 
with 
on 
Let 
the perpendicular from to 
with 
on Then the triangles 
and 
are isosceles.
then method #2 above can be applied from both vertices and 
& 
.
