such that 
.
be a function such that 
.
.
be positive integers and 
be prime. We know that 
and 
.
is a perfect square.
be a set of 
residues 
.
of of 
contains at least half of all the residues 
such that for every polynomial 
with integer coefficients and for every integer 
the integer![\[p^{(j)}(k)=\left. \frac{d^j}{dx^j}p(x) \right|_{x=k}\]](http://latex.artofproblemsolving.com/a/2/7/a2793d7f46345bcba4ccaa3082cc47afcd35074b.png)
(the 
)
order, and if the constituents of its first, second, ..., 
figurate numbers of the first, second, ..., 
be expressed as a polynomial in 
with real coefficients?
holds inequality 
,
,
satisfying![\[
2a^n+3b^n=4c^n.
\]](http://latex.artofproblemsolving.com/0/2/9/0295405b01494edc05ce9867b2ec48a4e069bf26.png)
such that following equality holds
by 
,![\[ a_{n+1} = \left(n+1-\gcd(a_n,n) \right) \times a_n \]](http://latex.artofproblemsolving.com/1/3/1/1316c6496b6faec9bbaab8d87560bf995ad2141d.png)
for all integers 
.
if and only if 
.
denotes the greatest common divisor of 
and 
.
![\[ \dfrac {\text {gcd}(m, n)}{n} \dbinom {n}{m} \]](http://latex.artofproblemsolving.com/c/0/a/c0a6877cd49e904728b24e7743fe8021efd6d312.png)
is an integer for all pairs of integers 
.
Y by
Let 
be an acute-angled triangle with orthocenter 
. A line passing through intersects 
at points 
respectively so that 
. The bisector of 
intersects the circumcircle of the triangle 
at point 
. Show that 
bisects segment 
.
be an acute-angled triangle with orthocenter 
. A line passing through intersects 
at points 
respectively so that 
. The bisector of 
intersects the circumcircle of the triangle 
at point 
. Show that 
bisects segment 
.This post has been edited 1 time. Last edited by parmenides51, May 23, 2020, 7:37 AM
Reason: typo: replaced ''perimeter'' with ''the circumcircle of the triangle''
Reason: typo: replaced ''perimeter'' with ''the circumcircle of the triangle''
is the intersection of the 
bisector and 
.
is diameter of 
,
and 
.
and 
from the parallels we have that 
and 
and 
parallelogram. Now see that 
and since 
which means 
and so 
.
on 
then 
is a parallelogram and the conclusion follows.
