[/quote]
,
be a cyclic quadrilateral. A circle centered at 
passes through 
and meets lines 
and 
again at points 
and 
(distinct from 
)
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 
.
.
be lengths of tangents from 
to the incircle of the quadrilateral which center is 
.
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 
,
and 
are positive, they have the same number of divisors. Is it necessarily true that 
?
with 
and 
in a similar way. Furthermore, let 
.
,
are concyclic.
positive integers. Product of any one of them with sum of remaining numbers increased by 
is divisible with sum of all 
![$P \in \mathbb R[x,y]$](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.
.
be the second intersection point of the circumcircles of the triangles 
and 
.
,
and 
are the circumcenters of the triangles 
are concyclic.
where 
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)
.
and 
be the midpoints of sides 
and 
,
and 
meet 
and 
,
and 
meet at 
.
satisfying both of the following properties:
.
.
Y by centslordm
Let 
and 
be two non-empty subsets of 
with 
. Let
be the product of all elements of and let 
be the product of all elements of .
Find the minimum and maximum possible value of
and find all possible equality
cases.
Proposed by Greece
and 
be two non-empty subsets of 
with 
. Let
be the product of all elements of and let 
be the product of all elements of .Find the minimum and maximum possible value of
and find all possible equalitycases.
Proposed by Greece
This post has been edited 1 time. Last edited by Steve12345, Sep 12, 2020, 10:59 AM
So the closer 
and 
we have minimum 
.
and 
,
Now onto the nontrivial part, that is the minimum of 
.
,
and 
,
we would have 
and 
,
.
,
and 
,
.
,
and 
is minimal, now we should just find 
and 
.
,
.
,
.
as if we switch 
,
and has only prime divisors 
.
:
Now we only have to check the two possibilities: 
and 
,
.
