[/quote]
,
?
and 
.![\[ \frac{a^2+1}{a+1}+\frac{b^2+1}{b+1}+\frac{c^2+1}{c+1} \geq 3\]](http://latex.artofproblemsolving.com/7/d/2/7d2e34116452912cd3da2925312ad500bb976f15.png)
be positive real numbers such that 
.![\[ \frac{x^2-yz}{x^2+x}+\frac{y^2-zx}{y^2+y}+\frac{z^2-xy}{z^2+z}\leq 0. \]](http://latex.artofproblemsolving.com/e/a/1/ea1f56cc594862673040411c3ad322cb00cfe904.png)
with 
being the midpoint of 
.
of 
.
be the projection of 
.
that is not on 
intersects 
at 
.
and 
intersect at 
.
is an isosceles triangle.
be the set of all positive integers. We say that a function 
is Georgian if 
and, for every positive integer 
,
such that![\[
f^{(k)}(n) = 1, \quad \text{where } f^{(k)} = f \circ f \cdots \circ f \quad \text{(applied } k \text{ times)}.
\]](http://latex.artofproblemsolving.com/5/3/c/53ca7c663237d939a7a049aa3569f8d595b259cf.png)
If 
is a Georgian function, we define, for each positive integer 
as the smallest positive integer 
such that 
.
for which there exists a Georgian function such that, for every positive integer 
,
.
prove that
denotes the sum of the integers, 
denotes the sum of the square of the integers and 
denotes the sum of cubes of the integers(that is, if the three integers are 
,
,
,
)
and 
,
.
and 
that satisfy the system of equations:
and 
be the sets of integers and rationals respectively.
such that the sets 
are disjoint?
denotes the set 
,
and for 
.
grid 
)
black cells.Detemine the number of balanced colorings.
be a quadrilateral inscribed in a circle with center 
and 
.
be the circumcircle of 
and 
be the circumcircle of 
.
.
.
.
,
denote the number of 
s![\[\sum_{i=1}^{n}(-1)^{s(3i)} > 0.\]](http://latex.artofproblemsolving.com/d/e/9/de975cc2632685d7a3f71c023115535d0aeb5dbb.png)
Holden Mui
for which there exist real numbers 
satisfying 
,
and
for 
.
Y by narutomath96, swamih, Gaussian_cyber, Adventure10, swynca
We are given triangles 
and 
such that 
, 
. Let the circumcenter of be 
, and let the circumcircle of intersect 
again at 
respectively. Prove that the perpendiculars to through 
respectively intersect at a point 
, and the lines 
intersect at a point 
, such that 
are collinear.
Proposed by Sammy Luo
and 
such that 
, 
. Let the circumcenter of be 
, and let the circumcircle of intersect 
again at 
respectively. Prove that the perpendiculars to through 
respectively intersect at a point 
, and the lines 
intersect at a point 
, such that 
are collinear.Proposed by Sammy Luo
concur at the orthocenter of 
.
denote the isogonal conjugate of 
.
are collinear and 
is the pedal triangle of 
lies on line 
,
and we have 
are collinear. Let 
again and since 
,
are collinear and hence 
are collinear and we are done.
and perpendicular to 
are concurrent at the orthocenter of 
,
and 
are orthologic 
the perpendiculars through 
to 
are concurrent at 
.
are concurrent ,
are concurrent at 
.
is the pedal triangle of 
are concurrent at 
and 
) we get 
are collinear.
are collinear
which is the orthocenter of 
as the circumcenter of 
.
,
is the foot of the perpendicular from 
(define 
similarly), we get that 
are concurrent and therefore 
is a harmonic pencil, and so 
is a harmonic quadrilateral. Thus if 
is the reflection of 
is symmedian of 
we get that 
is a median of 
lies on 
,
(since 
is a diameter).
,
that passes through 
.
.
denote the inverse of 
for any point 
.
invert to themselves. Notice that the circle 
has a diameter 
,
)
are the feet of the altitudes in a triangle, and that 
lie on the sides. Since 
are concyclic, so are their inverses and so 
are the midpoints of said triangle, and their circumcenter is the 9-point circle. Finally, since 
,
are cyclic, so 
are collinear. From this, 
.
.
.
have the Euler line of 
be the barycenter of 
be the intersection of 
and the 9-point circle. Then 
since 
is cyclic because 
.
are concyclic. But 
.
of 
.
of 
.
are Orthologic, so the perpendiculars to 
which is the Isogonal Conjugate of 
are concurrent at 
are Orthologic with centers of Orthology 
.
but 
.
.
