[/quote]
,
.
, 
and 
.
.
such that 
is a divisor of 
.
,
are equal, and they intersect at 
intersect at point 
lying inside triangle 
,
intersect at point 
lying inside triangle 
.
.
denote the set of positive rational numbers. Find, with proof, all functions 
such that, for all positive rational numbers 
and 
we have ![\[f(x)=f(x+y)+f(x+x^2f(y)).\]](http://latex.artofproblemsolving.com/9/e/a/9ea71d2ee277dc3c689ba2d791b451e29353edb3.png)
.
and 
.
and 
and the point 
is given. Let 
be set of all 
segments of real line of type ![$[i, j]$](http://latex.artofproblemsolving.com/a/6/4/a644edd67caa59eaded765c417b6d9822f782ec2.png)
,
and 
are integers, 
.
is said to be valuable if the intersection of any two segments from 
is either empty, or is a segment of nonzero length belonging to 
satisfying 
,![\[ \frac{a^4+5b^4}{a(a+2b)} + \frac{b^4+5c^4}{b(b+2c)} + \frac{c^4+5a^4}{c(c+2a)} \geq 1- ab-bc-ca \]](http://latex.artofproblemsolving.com/f/e/8/fe85cf52b883c214806ab837a2eb25cebb511d7b.png)
such that there exists a unique integer 
such that 
with the following property:![\[ n!\mid a^n + 1
\]](http://latex.artofproblemsolving.com/e/2/3/e23d39aab685888993d355d8c93d4e6475f53487.png)
(use in a problem for junior) to practice. Can anyone share to me?
side length and with center 
.
is drawn where point 
.
Y by
Let 
be the incenter of triangle 
. Let 
and 
intersect at 
, and 
and 
intersect at 
. Suppose that 
is another intersection of 
and 
. Let 
be the midpoint of 
, and 
are the intersections of 
and 
, respectively. Show that 
are concyclic.
(ltf0501).
be the incenter of triangle 
. Let 
and 
intersect at 
, and 
and 
intersect at 
. Suppose that 
is another intersection of 
and 
. Let 
be the midpoint of 
, and 
are the intersections of 
and 
, respectively. Show that 
are concyclic.(ltf0501).
This post has been edited 3 times. Last edited by parmenides51, Dec 15, 2022, 7:27 PM
.
and 
.
Symmedain , 
are concurrent.
.
be 
.
.
are collinear we get
collinear, we need the following determinant to be 
.
____________________________________________________________________________________________
we have that 
is the 
and 
.
.
.
are pairs of an Involution 
and projecting 
we get that 
are pairs of an Involution. Hence, 
are Coaxial. 
are the power to each vertice w.r.t the circumcircle. We can easily compute 
by bash length. For 
,
and then substitute 
into equation of 
are coaxial 
.
.
.
ie 
.
where the 2nd last equality holds true due to angle bisector thm applied twice and last one due to 
to 
.
bisects 
.
is midpoint of arc 
and 
.
is cyclic as desired.
.
.
which suffices:
is 
.
with center 
is an involution. Project this involution onto 
is an involution. Since the reflections of 
over 
are 
we conclude that this involution is reflection over 
are isogonal lines on 
which yields the result
inversion and reflect over the angle bisector of 
.
are collinear. Let 
.
are collinear which is equavilent 
to be an involution since 
are concyclic. Project this onto 
is an involution. Let 
be the midpoint of arc 
.
are concyclic since 
are collinear. Also 
which is 
and it lies on 
.
are coaxial thus, 
at 
.
to 
and centered at 
.
and 
and ratio lemma to 
