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.
is given, find the largest real number 
such that there exists a geometric sequence 
with common ratio satisfying
for all positive integers 
. Here, 
denotes the distance from the real number 
to the nearest integer.
, where the common ratio 
. Given that
.
is your favourite function, 
is your mother's favourite function, and 
is your father's favourite function. It was discovered that for any reals 
, 
Prove that you are boring.
is 
, then is called a perfect number. For example, the sum of the positive divisors of 6 is 
, hence 6 is a perfect number.
, where 
are positive integers, and 
are odd primes.
to be 
, where ![$[x]$](http://latex.artofproblemsolving.com/b/c/e/bceb7b14e55d33a8bca29b7863ad3cdae95afce4.png)
is the largest integer no greater than 
, and ![$\{x\}=x-[x]$](http://latex.artofproblemsolving.com/6/1/0/61036a03415081af535bd57c408b7be79aae9458.png)
. For each positive integer 
, find the maximum possible real number 
such as there exists an infinite geometric sequence with common ratio and isolation index of each term being at least .
and 
, with 
. Prove that there exists and only exists one point 
such that 
.( corresponds to , 
corresponds to )
, let 
be the 
-excenter. Points 
and 
are placed on line 
such that 
is between and , and is between and . Moreover, and are the contact points of with the -excircle of triangles 
and 
, respectively. Let 
be the -excenter of triangle 
, and let 
be the reflection of the orthocenter of triangle with respect to its circumcenter. Prove that , , and are collinear.
be positive real variables. Prove that
has no solutions in integers , 
, and 
.
be a 
board. Some of its cells are colored black in such a way that every 
board of has at most 
black cells. Find the maximum amount of black cells that the board may have.
intersect at , and there exist a circle 
tangent to the extensions of 
at 
respectively. Circle 
passes through points 
, and is externally tangent to circle at 
. Prove that 
.
, let 
intersect 
in 
such that 
. Upon inspection of similar right triangles 
and 
we have:
(i).
in 
can be expressed via the Law of Cosines:
;
(ii).
, then setting (i) equal to (ii) yields:
(iii).
.
and 
Prove that
and prove 
be a right triangle in 
.
and 
be points on 
in the order (A, D, E, C), and let 
and 
be the projections of 
and 
,
,
,
