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 a cyclic quadrilateral. A circle centered at 
passes through 
and 
and meets lines 
and 
again at points 
and 
(distinct from 
). Let 
denote the orthocenter of triangle 
Prove that if lines 
are concurrent, then triangle 
and 
are similar.
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:
.
.
, which for all nonnegative 
satisfy 
Proposed by Igor Voronovich 
be the proposition that 
.
.
Substituting 
into the last statement gives 
, but 
, hence 
.
In the last statement, if 
,
In other words, if 
, 
for all nonnegative 
. Now consider some 
such that 
.
Which means for any 
, 
. Hence there exists at most such .
Hence 
. Also, since 
, 
.
, or there exists some 
where 
, and otherwise.
into the last statement gives , but , hence .
shouldn`t we prove function is injective first?
, which for all nonnegative satisfy Proposed by Igor Voronovich 
such that 
Here we get non-trivial solution 
. Is it true that there are no more solutions?
be the proposition that ..Substituting into the last statement gives , but , hence .In the last statement, if ,In other words, if , for all nonnegative . Now consider some such that .Which means for any , . Hence there exists at most such .Hence . Also, since , . Hence either , or there exists some where , and otherwise. It can be checked that these functions work.
we will get that 
so 
and this is contradiction with assumption that exists a positive value of 
, so its constantly zero.
