such that![\[
f(f(x)+y) = x+f(f(y))
\]](http://latex.artofproblemsolving.com/9/d/7/9d780faa5901646f7035e88400a9e3d8023f04ff.png)
for all 
.
points in a circle and number them arbitrarily with numbers from 
to 
pairwisely non-intersecting chords can be drawn for each of whom its endpoints differ in at most 
.
satisfies that for any positive integer 
and 
Suppose there is 
non-negative reals 
such that for 
Determine the least possible value of 
is chosen. Prove that there exists a prime number that divides infinitely many terms of the sequence 
,![\[
b_k = a^{k^k} \cdot 2^{2^k - k} + 1.
\]](http://latex.artofproblemsolving.com/1/7/5/1751a60482d729a36c71b77ac9c978e724f40da0.png)
be the polynomial with leading coefficent 1 and rational coefficents, such that ![\[p\left(\sqrt{3 + \sqrt{3 + \sqrt{3 + \ldots}}}\right) = 0,\]](http://latex.artofproblemsolving.com/0/f/4/0f472be9f2cede0221e1b4d970d65de94dba01b9.png)
and with the least degree among all such polynomials. Find 
.
such that 
is a positive integral power of a prime positive integer."
for positive integral values of 
be the largest perfect square that divides 
less than 
such that 
is divisible by 
but not 
.
be the product of all positive integers less than 
.
is divided by 
.
,
.
be a tetrahedron such that 
A sphere of radius 
is circumscribed about tetrahedron 
Given that 
,
,
,![$$r^2+\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \ge \frac{9\sqrt[3]{4}}{4}$$](http://latex.artofproblemsolving.com/6/c/6/6c665b4611f5fc8ebeba03ac649bfee4e2634950.png)
with equality at ![$a=b=c=\sqrt[3]{2}.$](http://latex.artofproblemsolving.com/b/b/9/bb9c9671b23cb45e5f85ae01bf25d566c2e43e91.png)
th and 
th toss, is
square such that all rows and columns sum to the same value. Please note that diagonals do not need to sum to the same value. How many ways can this be done?
,
.
.
be a dot on the 
side.
at point 
at 
.
A
.
is the intersection of 
and 
,
is a bisector of the angle 
Y by Davi-8191, Adventure10, Mango247, ehuseyinyigit, Rounak_iitr, cubres
A circle with center 
passes through the vertices 
and 
of the triangle 
and intersects the segments 
and 
again at distinct points 
and 
respectively. Let 
be the point of intersection of the circumcircles of triangles and 
(apart from 
). Prove that 
.
passes through the vertices 
and 
of the triangle 
and intersects the segments 
and 
again at distinct points 
and 
respectively. Let 
be the point of intersection of the circumcircles of triangles and 
(apart from 
). Prove that 
.
.
.
.
of triangle 
again at distinct points 
r
,
.
.
are concurrent.suppose 
intersects at 
.
.
is eaqual .and 
changes to 
respectively.and invert of M,should be colinear with them.
is tangent to circl
.
.
and 
,
.
and 
.
and 
is the polar of 
,
,
is a diameter.
,
then 
also belongs to the circumcircle.
is a hexagon inscribed in that circumcircle.
intersect 
are collinear, and we are done.
and 
.
and the point 
(the intersection between at least the circumcircles of the triangles 
it is well-known that the center 
(the quadrupoint 
is orthocentrically). In conclusion, 
,
.
and the points 
.
,
of the triangles 
,
cut again
,
respectively. Prove that
.
?
,
.
.
.
:
.
be the circle center 
that is orthogonal to 
is the polar of 
is the polar of 
wrt 
and 
is on the polar of 
is the polar of 
.
.
is the polar of 
are tangents to 
is orthogonal to 
,
is cyclic. Bypower of a point, 
.
and a simple triangle similarity, 
.
.
maps to line since 
maps to 
and 
maps to 
,
.
maps to 
.
is the intersection of these two circumcircles, the intersection of these two lines is the inversive image of 
.
are both on segment 
,
intersect at 
,
,
is cyclic.
,
,
,
is cyclic.
,
,
,
.
passes through the vertices 
and intersects the segments 
and 
again at distinct points 
(apart from 
.
because of the cyclic quadrilaterals 
and 
.![\[ \angle AMB = 180 - \angle ACB\implies \angle AMN = \angle AMB - \angle BMN = 180 - \angle ACB - \angle ACB = 180 - 2\angle ACB
\]](http://latex.artofproblemsolving.com/c/2/0/c200731beb5a0a950a1e8c6bc124e2cf6f865373.png)
Since 
,
,
is cyclic. This means that 
.
,
and 
be 
and 
respectively. Notice that 
and 
,
.
,
.
is cyclic. Yet, 
since 
is cyclic. Thus, 
is cyclic, so 
,
to be collinear and 
also collinear in such a way that 
forms a homothetic circle to 
.
where 
are tangents from 
be 
.
are concyclic, it implies 
are collinear . However, as 
.
.
.
degrees. (As tangents from an exterior point are equal).
and 
.
.
,
,
.
.
,
is cyclic 
.
are concurrents in 
)
centered at 
and 
.
be 
.
.
.
and the line 
.
the image of 
is 
.
are collinears
.
and 
concur by Radical Axis. We are done as 
where 
and 
and 
?
and denote the intersection of 
are concyclic we have 
and also since points 
are concylic we have 
now this gives 
.
and if 
(angle subtended by chord 
)
are concyclic points we have 
or 
which gives 
which gives 
which gives quadrilateral 
is a kite and hence 
and hence 
as desired 
and 
and 
meet at 
.
and 
.
,
,
,
Then, by Master Miquel Theorem, 
is collinear. By radical axis, 
are collinear. By Brocard's, 
