[/quote]
,
such that the equation![\[f(xf(x)+y) = f(y) + x^2\]](http://latex.artofproblemsolving.com/2/0/a/20a245460634f0455fc703d7ce41c475c1ae4970.png)
holds for all rational numbers 
and 
.
denotes the set of rational numbers.
with 
and 
.
and 
intersect at 
and 
intersect at 
be the midpoints of sides 
,
be points on segment 
and 
,
is the angle bisector of 
and 
is the angle bisector of 
.
is parallel to 
if and only if 
divides 
a point in the space and 
the centroid of a tetrahedron 
is divisible by 
for every positive integer 
,
.
is the number of integers not more than 
of positive integers with the following property: all of the terms of the sequence 
have a greatest common divisor 
,
,
be nonzero real numbers such that 
and 
.
.
be non-negative real numbers such that ![\[b+c\leqslant a+1,\quad c+a\leqslant b+1,\quad a+b\leqslant c+1.\]](http://latex.artofproblemsolving.com/b/9/e/b9e86e898c0536b7323a03611d5bdbf679caa710.png)
Prove that 
be a function satisfying
for all integers 
and 
for all integers 
such that ![\[f(xy-1)+f(x)f(y)=2xy-1,\]](http://latex.artofproblemsolving.com/8/8/8/888ca39f2b7f8cec6d6426bee28d40eade40a66e.png)
for any real numbers 
and 
![\[
f : \mathbb{R}^+ \to \mathbb{R}
\]](http://latex.artofproblemsolving.com/e/7/b/e7bfc5b236fb6f00ef328f850b1b14632cbf8416.png)
satisfying the condition that for all 
,![\[
f(x + y^2) \geq f(x) + y.
\]](http://latex.artofproblemsolving.com/3/a/a/3aa20083835682dbd81be692eba65cab19e923e5.png)
of positive integers with 
prime, 
,
,
is a multiple of 
.
be a 
makes a fixed angle with the tangent to 
be a rectangular hyperbola centered at the origin, scaled such that it is tangent to the logarithmic spiral at some point.
Y by sami1618, cubres
6. Circles 
and 
with equal radii are given. Let 
,
be the intersection of the circles.
points
and 
are on and such that they are inside and respectively.
Points
,
are on and respectively, such that 
and 
.Denote by 
as the other intersection of 
and 
. Prove that 
are concurrent.
and 
with equal radii are given. Let 
,
be the intersection of the circles.points
and 
are on and such that they are inside and respectively.Points
,
are on and respectively, such that 
and 
.Denote by 
as the other intersection of 
and 
. Prove that 
are concurrent.
are the intersections of 
with the lines through 
perpendicular to 
.
are 
in the direction of 
.
passes through the midpoint of 
.
be the midpoint of 
.
and 
are both parallelograms.
is the orthocenter of triangle 
.
is the orthocenter of triangle 
.
be the reflection of 
.
.![\[
\angle B'YQ = \angle B'PQ = \angle BPQ = \angle BYQ,
\]](http://latex.artofproblemsolving.com/0/2/9/02982c439ff0153cd7ee839d86887f45efd3c24a.png)
points 
are collinear. Because 
,
.
,
and 
be the centers of 
and 
be the midpoint of 
.
and![\[
BY = 2NO_2 = 2NO_1 = CX.
\]](http://latex.artofproblemsolving.com/a/5/5/a556730fd6e629c4e51d0810b9f6ed430b6aabd7.png)
and 
be the centers of 
and 
,
,
,
,
are parallelograms (specifically rhombi). We have![\[
\overrightarrow{Q} = \overrightarrow{O_1} + \overrightarrow{O_2} - \overrightarrow{P}, \quad
\overrightarrow{X} = \overrightarrow{O_4} + \overrightarrow{O_1} - \overrightarrow{P},
\]](http://latex.artofproblemsolving.com/4/e/9/4e992bebdbf91bab6ccbc6d90e00401765b6bd06.png)
![\[
\overrightarrow{S} = \overrightarrow{O_3} + \overrightarrow{O_4} - \overrightarrow{P}, \quad
\overrightarrow{Y} = \overrightarrow{O_2} + \overrightarrow{O_3} - \overrightarrow{P}.
\]](http://latex.artofproblemsolving.com/9/7/5/9756fa30376df9de62dd85db1875fcb27fe72404.png)
Thus, 
as desired. (This is known as the "AT" lemma. There is also a simple proof using congruent triangles.)
.
is equal to the radius of 
the orthocenter of 
.
the orthocenter of 
and 
is the orthocenter of 
is a parallelogram.
,
and 
circles are equal and 
(
degrees
is a parallelogram.
is a parallelogram.
and 
both 
and 
