be an acute triangle with 
.
the circumcircle circumscribed to the triangle 
the midpoint of the smallest arc 
of this circle. Let 
and 
points of the segments 
and 
respectively such that 
.
be the second intersection point of the circumcircle circumscribed to 
with 
and 
be the intersections of lines 
and 
with 
,
and 
be the intersection points of lines 
and 
with lines 
line passes through the midpoint of 
are real numbers, find the smallest possible value of the expression:
and 
be prime numbers. Prove that if 
is a perfect square, then 
is also a perfect square.
such that 
Edit: Proposed by Marian Vasile (congrats!).
.![$f : [0,1] \longrightarrow \mathbb{R}$](http://latex.artofproblemsolving.com/9/a/4/9a4b7997c35990f839742d29866122791ade8b7c.png)
is differentiable with 
.
for all ![$x \in [0,1]$](http://latex.artofproblemsolving.com/d/3/4/d3423a3200d0ad7660fc4265d59d5598a53d551d.png)
,
for all 
.
increasing with
?
?
and the distribution function is 
then
and find the expression of 
over all triples 
of distinct real numbers such that
Y by Dtong08_math
A table 
is filled with positive integers. Specifically, the first row is filled with numbers from the set 
. It turned out that for any two columns the difference of numbers from the first row is divisible by the difference of numbers from the second row, while all numbers in the second row are pairwise different. Is it true for sure that the numbers in the first row are equal?
Ivan Kukharchuk
is filled with positive integers. Specifically, the first row is filled with numbers from the set 
. It turned out that for any two columns the difference of numbers from the first row is divisible by the difference of numbers from the second row, while all numbers in the second row are pairwise different. Is it true for sure that the numbers in the first row are equal?Ivan Kukharchuk
table. By shifting all numbers by integer constant in the second row, we may assume that the smallest number in the second row is 1. Let 
be the biggest number; we have 
since all numbers in the second row are pairwise different. Let 
be all the numbers appearing in the second row.
be a function from 
such that for a column such that the second entry is 
,
as the entry in the first row of the same column. Now we have 
for all 
,
.
while 
,
.
the number of positive divisors of 
with 
where 
.
is divisible by both 
and 
.
,
,
divides 
,
such 
takes value 
for at least 
such 
(including 1 and 
)
,
.
,
,
is a positive divisor of 
such 
where the first inequality is using 
,
.
since half of the divisors are at most 
.
,
.
this does not hold, hence showing that our assumption cannot happen.
;
.
