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.
and 
such that for all 
if 
then 
Find the smallest of sum of all elements of 
such that there exist positive integers 
for which the following holds 
for all 
.
where 
is a perfect cube
of a prime number 
and a positive integer 
for which 
is an integer.
the point 
is the centre of the excircle opposite the vertex 
This excircle is tangent to the side 
at 
, and to the lines 
and 
at 
and 
, respectively. The lines 
and 
meet at 
, and the lines 
and 
meet at 
Let 
be the point of intersection of the lines 
and , and let 
be the point of intersection of the lines 
and 
Prove that is the midpoint of 
opposite the vertex 
is the circle that is tangent to the line segment , to the ray beyond 
, and to the ray beyond 
.)
such that if 
and 
are distinct rational numbers satisfying 
or 
, then 
? Justify your answer.
be a positive integer. Consider the following equation: ![\[ \{x\}+\{2x\}+ \dots + \{nx\} = \lfloor x \rfloor + \lfloor 2x \rfloor + \dots + \lfloor 2nx \rfloor\]](http://latex.artofproblemsolving.com/8/2/9/8299b3ceb1650418c0a373536cb1d2931900efd9.png)
a) For 
, solve the given equation in 
., the equation has at most 
real solutions.
and 
intersect at two points 
and 
. Let 
be the line tangent to these circles at 
and 
, respectively, so that lies closer to than . Let 
be the line parallel to and passing through the point , with 
on and 
on . Lines 
and 
meet at 
; lines 
and meet at 
; lines 
and meet at 
. Show that 
.
is such that 
for any positive integers 
. Assume there exists a positive integer such that 
for all positive integers 
. Determine all possible values of 
.
satisfy that 
. Prove that: 
we are given :
and one side length of the triangle equals , then under these conditions find the maximum area of the triangle .
and since ![$\angle{C} \in (0, \pi) \Rightarrow \sin(\angle{C}) \in (0,1]$](http://latex.artofproblemsolving.com/d/e/2/de206f373aa89473291f595c7f75d57ab07980e1.png)
we require:
;
;![$4-4\sin(\angle{A})+\sin^{2}(\angle{A}) \le 3[1-\sin^{2}(\angle{A})]$](http://latex.artofproblemsolving.com/2/1/2/212de277aab5f9a21356699c113ef69aebad932e.png)
;
;![$[2\sin(\angle{A})-1]^2 \le 0$](http://latex.artofproblemsolving.com/9/2/c/92cd26045627e050025fb7d9c0030ec32881a787.png)
;
(i).
yields 
, or 
respectively, which only the former angle is admissible 
. If one of the sides of 
has length , then we have the following triplets 
:
(ii).
is the area, then (ii) yields: 
.
,![\[
\sum_{\text{cyc}} \sqrt{1 + \left(x\sqrt{1 + y^2} + y\sqrt{1 + x^2}\right)^2} \geq \sum_{\text{cyc}} xy + 2\sum_{\text{cyc}} x
\]](http://latex.artofproblemsolving.com/9/5/5/955863a0cf8747ae45b736b0631f243d3908eb84.png)
and 
.
