[/quote]
,
are defined inductively by 
,
for 
.
,
there exist some 
such that 
.
![\[x^5 + ax^4 + bx^3 + cx^2 + dx + e = 0\]](http://latex.artofproblemsolving.com/f/d/0/fd0ab11635ee44ad272a54c0f20a2bc47a3596de.png)
cannot all be real if 
.
be a sequence that consists of an initial block of 
positive distinct integers that then repeat periodically. This means that 
are 
for every positive integer 
.
prime numbers, find for which values of 
terms of the sequence.
and 
intersect at points 
and 
,
intersects 
,
intersects 
.
,
.
smallest positive solutions of 
that has an integer 
non-negative real numbers 
that satisfy 
,
such that![\[
x_i^2 + x_j \geqslant \frac{1}{2023}.
\]](http://latex.artofproblemsolving.com/2/3/c/23c312912bb0804a499fa33d1e254c3c1b554fad.png)
from the set of positive integers to the set of positive integers such that, for all positive integers 
and 
,![\[ a, f(b) \text{ and } f(b + f(a) - 1).\]](http://latex.artofproblemsolving.com/4/a/b/4ab97da27e05d0f9e7762f74d0d0bd2dc8b9842f.png)
are real numbers such that
,
.
.
.
.
.
we have 
.
we define the number of invariant subsets 
.
.
there exists a function 
.
and 
for 
.
there exists a positive integer 
.
satisfying following conditions.
for all 
.
with 
,
holds.
be an acute triangle with 
and let 
and 
be the circle 
.
and the circle of radius 
and 
,
and circle 
are concurrent.
of functions from the set of real numbers to itself that satisfy ![\[g(f(x+y)) = f(x) + (2x + y)g(y)\]](http://latex.artofproblemsolving.com/0/a/3/0a3bc25299a0bcbd3f3602a0e6881d68fce750c5.png)
for all real numbers 
.
Y by
Let 
be a square of center 
and let 
be the symmetric of the point 
with respect to point 
. Let 
be the intersection of 
and 
, and let 
be the intersection of 
and 
. Show that 
is the angle bisector of 
.
be a square of center 
and let 
be the symmetric of the point 
with respect to point 
. Let 
be the intersection of 
and 
, and let 
be the intersection of 
and 
. Show that 
is the angle bisector of 
.This post has been edited 2 times. Last edited by GorgonMathDota, Nov 8, 2020, 1:10 AM
be the vertices of square 
and 
,
:
is cyclic, as well as 
and 
and 
are similar, as well as 
and 
.
is midline in the 
,
is midpoint of 
.
are collinear.
.
,
.
,
be midpoint of 
.
is centroid of 
,
are collinear. Observe that 
.
.
,
is cyclic and consequently we have that 
.
as desired.
meet 
at 
.
so 
so 
.
so we have 
so 
so 
so 
.
where 
.
is cyclic. By cyclic, 
and 
.
,
.
and 
.
which implies 
.
giving 
as desired.
and 
.
and 
.
holds, implying 
is cyclic.
.
proving our claim. 
shows the second tangency as well.
.
.
.
giving the similarity between triangles.
implies 
is tangent to 
at 
.
is a circle with center 
and 
.
.
.
.
