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 
, where 
are non-negative reals such that 
.
.
a tangencial quadrilateral that is not a rhombus, let 
be lengths of tangents from 
to the incircle of the quadrilateral which center is 
. Let 
be the midpoints of 
resp. Prove that![\[ \frac{MI}{IN}=\frac{a+c}{b+d} \]](http://latex.artofproblemsolving.com/e/7/0/e709acb8b17c0466a385ba8e7a89bb1bf16c72af.png)
is given. It turns out that for every positive integer number 
, for which the numbers 
and 
are positive, they have the same number of divisors. Is it necessarily true that 
?
with as its incenter. Let 
be the intersection of 
and 
and define 
in a similar way. Furthermore, let 
. Prove that if 
, then 
are concyclic.
positive integers. Product of any one of them with sum of remaining numbers increased by 
is divisible with sum of all numbers. Prove that sum of squares of all numbers is divisible with sum of all numbers
![$P \in \mathbb R[x,y]$](http://latex.artofproblemsolving.com/8/a/0/8a0576299e976c51607959fb68c7c742b2a49c6a.png)
with ![$P \not\in \mathbb R[x] \cup \mathbb R[y]$](http://latex.artofproblemsolving.com/a/f/7/af71ba8c643d5bd8a2928fa6714ccd91713bd641.png)
and 
homeomorphismes of 
, 
is an homoemorphisme too.
is inscribed in a circle centered at 
, and 
. Let 
be the second intersection point of the circumcircles of the triangles 
and 
.
, and are collinear.
and 
are the circumcenters of the triangles , and , respectively, prove that the points 
are concyclic.
where is a positive integer and 
is a prime.(Don't get mad at me,I've used the search function and did not see a correct and complete solution anywhere.)
![\[
2^a + 3^b + 1 = 6^c
\]](http://latex.artofproblemsolving.com/1/9/e/19eb5efa2ccdf54cd277339f8d3b99f75168779d.png)
is inscribed in circle 
. Let 
and denote its orthocenter and circumcenter, respectively. Let 
and 
be the midpoints of sides 
and 
, respectively. Rays 
and 
meet at 
and 
, respectively. Lines 
and 
meet at . Prove that 
.
satisfying both of the following properties:
.
.
and 
for each nonnegative integer 
. Prove that 
is the greatest integer less than or equal to 
, 
with the following property: for every positive divisor 
of we have that 
is a divisor of
. Prove that is a prime number.
with the following property: for every positive divisor of we have that is a divisor of. Prove that is a prime number.
be the smallest prime divisor of 
, where 
. satisfy the problem conditions then 
, so 
.
and 
, we have 
. isn't prime, , so 
must have a prime divisor. However, this prime factor is less than which contradicts the fact that is the smallest prime divisor of .
.
,
which is a contradiction. If instead 
,