[/quote]
,
,![\[
\frac{a}{b} + \frac{b}{c} + \frac{c}{a}.
\]](http://latex.artofproblemsolving.com/3/2/b/32b41635c159921e11245369e66185ec69b0d785.png)
be the set of the first 
perfect squares of nonzero integers. Suppose that 
for some 
.
stands for the set 
.
holds for every 
.
such that 
,
satisfy that: 
satisfying 
is a perfect square
be a polynomial with integer coefficients. Suppose there exist infinitely many integer pairs 
such that 
.
![\[\sqrt x - \frac 1y - 2w + 3z = 1,\]](http://latex.artofproblemsolving.com/d/0/f/d0ff4819db851e6c12f738dcdb6b81bbd2ed0ca7.png)
![\[x + \frac{1}{y^2} - 4w^2 - 9z^2 = 3,\]](http://latex.artofproblemsolving.com/6/5/3/653c1547620f85ff761817a5cc333f16bce9c4c6.png)
![\[x \sqrt x - \frac{1}{y^3} - 8w^3 + 27z^3 = -5,\]](http://latex.artofproblemsolving.com/1/b/9/1b9c2ed383acc931aa3607ba82031e182e18f3d1.png)
![\[x^2 + \frac{1}{y^4} - 16w^4 - 81z^4 = 15.\]](http://latex.artofproblemsolving.com/e/0/3/e032ce109cd74bbae47dc6bca04dcab2d8f9ef29.png)
![\[
|MA|^2 + |MB|^2 + |MC|^2 = |GA|^2 + |GB|^2 + |GC|^2 + 3|MG|^2
\]](http://latex.artofproblemsolving.com/0/7/0/070ff15920d19d474defb77ceb606b2811fa85b4.png)
such that the numbers ![\[ (a_1)^{2018}+a_2, (a_2)^{2018}+a_3, \dots, (a_{2018})^{2018}+a_1 \]](http://latex.artofproblemsolving.com/6/1/a/61ad92b49bd9b4b8839e6f6c4e81c758a11835a3.png)
are all powers of 
?
of positive integers satisfying 
and 
for all positive integers 
and 
For a given positive integer 
find the maximum possible value of 
such for any 
:
.
Y by kun1417, K.titu, Davi-8191, Adventure10
Let 
be an arbitrary triangle. Let 
be two points on 
respectively such that their distance to the midpoint of 
is equal. Let 
be the second intersection of the triangles 
circumcircles . The tangents from 
to the circumcircle of 
intersect each other at 
. Prove that : 
be an arbitrary triangle. Let 
be two points on 
respectively such that their distance to the midpoint of 
is equal. Let 
be the second intersection of the triangles 
circumcircles . The tangents from 
to the circumcircle of 
intersect each other at 
. Prove that : 
and consider three arbitrary points 
then let 
be the projections of 
on 
be the midpoint of it. we define 
similarly then 
are collinear if and if only 
.
intersect 
be the orthocenter of 
,
the midpoint of 
and 
be the antipode of 
respectively. note that 
and 
are collinear (well known). Let 
cut 
again at 
since 
we get that 
now from the lemmma for two lines 
we deduce that 
hence 
.
then 
.
(For starters, see that sine rule will work nicely after this)
be a circle
is diametrically opposite to 
denote midpoints of segments 
where 
and 
are congruent, it follows that 
and from here it follows that 
But recall that 
hence 
On account of these 2 similarities it follows that 
Hence quadrilateral 
is harmonic, so it follows that 
are collinear, as desired.
with median 
.
such that 
.
.
.
.
be the midpoint of 
then 
are collinear. The tangents at 
of 
.
to 
,
so 
is cyclic. Combine with property of symmedian we get angles chasing 
We deduce 
is a point on perpendicular bisector of 
.
.
is circumcircle of triangle 
interesect at 
and 
cuts 
.
be inversion centered at 
(equality follows by spiral similarity) followed by reflection over the angle bisector of 
(they coincide because of spiral similarity).
and the other similarities
,
is cyclic. Next is to consider that the spiral similarity centered at 
to 
and 
to 
will maps 
.
is cyclic. Clearly, 
is the perpendicular bisector of 
and our claim is proved.
is the symmedian of 
,
.
,
so 
.
and we get 
.
,
and 
have common angle bisector and this implies 
is sent to 
and 
.
by claim 1, the angle of 
be the center of 
and 
be the intersection of 
are points such that 
then 
is the spiral center sending 
to 
sending 
so let 
and 
Since 
is the 
it's well known that 
is cyclic; however, 
also lies on 
so 
have to be collinear again due to Lemma 1. Now it suffices to show that 
are collinear, but this is trivial:
from 
so we're done. 
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is diameter of circle 
.
such that 
and 
so 
and it get us 
, and the claim is proved .
so 
, so 
are collinear 
,
and 
,
,
and make their midpoints X, Y
,
.
and 
the angle bisector of 
(which is also the angle bissector of 
)
followed by a reflection through 
then 
and 
.
the second intersection of 
.
.
such that 
is harmonic quadrilateral. Then 
.
on 
,
which give us that 
collinear. Let 
)
cyclic, then 
so 
.
s
,
Since 
is cyclic, 
Therefore, 
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meet 
and 
so 
so we need to prove 
or 
is cyclic. Note that 
and 
so 
and 
are similar so 
are collinear so 
and 
and 
are similar so 
so 
