![\[x^4-(2\cdot10^{10}+1)x^2-x+10^{20}+10^{10}-1=0\]](http://latex.artofproblemsolving.com/b/9/d/b9d355433e896e8b97f28c2e4235140fd2348e77.png)
correct to four decimal places.
be a triangle with circumcircle 
.
intersect at 
.
intersects the line 
at 
and 
is the midpoint of 
.
and 
intersect on 
be an integer, 
a real number, and 
be positive real numbers such that 
.![\[
\frac{1 + x_1^k}{1 + x_2} + \frac{1 + x_2^k}{1 + x_3} + \cdots + \frac{1 + x_n^k}{1 + x_1} \geq n.
\]](http://latex.artofproblemsolving.com/0/8/d/08de0c74a4e36b50a64d17875d3fd93eeb5b52de.png)
Y by mariam10, son7, A4r43r457yhyt, centslordm, mathematicsy, jhu08, HWenslawski, chessgocube, lian_the_noob12, ItsBesi, NicoN9
Let 
be the circumcircle of 
. Let 
be a point on the side 
. The tangent to at 
intersects the parallel line to 
through at point 
. The segment 
intersects again at 
. Suppose 
, , , are concyclic. Prove that 
, 
, 
are concurrent.
be the circumcircle of 
. Let 
be a point on the side 
. The tangent to at 
intersects the parallel line to 
through at point 
. The segment 
intersects again at 
. Suppose 
, , , are concyclic. Prove that 
, 
, 
are concurrent.
and 
we get 
are concyclic. Done by radical axis on 
,
,
.
so 
is cyclic. Radical Center finishes.
.
so 
is cyclic. Then 
so 
is cyclic. But it's also a trapezoid, so 
finishes.
![\[
\angle EAC = \angle ABC = \angle EDC
\]](http://latex.artofproblemsolving.com/d/9/5/d953e0425549f8033cc17f2ee0124eb6c5555a7a.png)
so quadrilateral 
,
,
,
is cyclic.Hence by Radical Axis Theorm on 
we are done
were not in the place it were.
is cyclic is equivalent to 
since 
then 
.
.
we just need to show that there exists a spiral similiarity centred at 
but this is trivial since 
and 
.
is cyclic, then![\[
\measuredangle{CBA}=\measuredangle{BDE}=\measuredangle{BFE}=\measuredangle{BFC}=\measuredangle{BAC}
\]](http://latex.artofproblemsolving.com/e/0/3/e0307e8b888a020a0490b75fd0b1bb87468d46a8.png)
so 
is 
.
and 
.
be the point at infinity along line 
.
,
.
But 
,
as desired.![\[
\measuredangle{QAE}=\measuredangle{CBA}=\measuredangle{BFC}
\]](http://latex.artofproblemsolving.com/1/6/c/16cd5f1244097917084382fc884d8e46eb2eca41.png)
so 
is cyclic. It follows that![\[
\measuredangle{FQE}=\measuredangle{FAE}=\measuredangle{FBA}
\]](http://latex.artofproblemsolving.com/e/b/f/ebfcc8fe6f0403cba74b6180c56377a363b67389.png)
so 
as desired.![\[
\measuredangle{CAE}=\measuredangle{CBA}=\measuredangle{CDE},
\]](http://latex.artofproblemsolving.com/a/a/8/aa8ad3274be957381d967922901bd6ccf41ff01a.png)
is cyclic. Then the radical center of 
is the desired concurrency point. Note that this approach does not use the fact that 
for some constant 
and side 
.
intersects the circumcircle at 
.![\[
\frac{-\frac{b^2k(1-k)}{a^2k+c^2(1-k)}}{1-k}=-\frac{b^2k}{a^2k+c^2(1-k)}=-\frac{b^2k}{b^2k+c^2(1-k)}=\frac{-\frac{b^2k}{c^2}}{1+\frac{b^2k}{c^2}-k},
\]](http://latex.artofproblemsolving.com/3/0/f/30fa9e596cba33c3837b86f556769c58a1608735.png)
so 
collinear which gives 
as desired.
,
,
.
.
and 
is at 
.
is represented by the complex number![\[
\frac{a^2+a^2f-2af-a-a^2f+af+f}{a^2-f}=\frac{(a-f)(a-1)}{a^2-f}
\]](http://latex.artofproblemsolving.com/1/a/6/1a62694850f452d9cdcfac03fe5fce3b6ea68445.png)
which is pure imaginary by conjugating it. It follows that 
then 
so 
.
then 
so 
.
so 
concur.
are cyclic.
are cyclic and 
so 
is an isosceles trapezoid.
are cyclic.
are collinear.
and we have power 
from angle chasing 
it means 
and 
so we found 
and 
cyclic so 
collinear 
.![\[
\angle BAC = \angle BFC =
180 - \angle BFE
= 180 - \angle EDB = \angle EDC = \angle ABC
\]](http://latex.artofproblemsolving.com/0/3/8/0382c350259bca2a7cf86f474b377ceff2e69acb.png)
shows that 
,
is an isosceles trapezoid and in particular 
.
,
is cyclic. We have 
.![\[
\angle PBC = \angle DAC = \angle DEC = \angle DEF
=\angle FBC,
\]](http://latex.artofproblemsolving.com/e/1/2/e120f110bd581680b4adf9916f58244553ec7df9.png)
so 
are collinear.
![\[ \angle EAC = \angle ABC = \angle EDC, \]](http://latex.artofproblemsolving.com/f/a/9/fa92b6415dd58fa95767409336c29dc1ac050d03.png)
so 
,
,
,
and 
we have 
.
.
.
,
.
.
.
implying 
are collinear.
so 
are concyclic, and then radical axis theorem on the three circles 
implies the result.
are concyclic.
,
is cyclic, and notice that the pairwise radical axis of 
are 
,
.![\[\measuredangle CFX = \measuredangle CFB = \measuredangle EDB = \measuredangle ABC = \measuredangle EAC = \measuredangle EAX\]](http://latex.artofproblemsolving.com/f/2/e/f2e90e9532ed8cc6e0824d00a6282a0513973652.png)
which implies the claim.![\[\measuredangle XEA = \measuredangle XFA = \measuredangle BFA = \measuredangle BCA = \measuredangle BAE = \measuredangle DEA\]](http://latex.artofproblemsolving.com/b/b/d/bbd254f38998156eafad757fb8bc93dba2a9000e.png)
which implies that points 
,
,
are concyclic, respectively. First one is ![\[
\measuredangle CAE =\measuredangle CBA=\measuredangle CDE
\]](http://latex.artofproblemsolving.com/1/7/1/171d8ed7ea2a7f6bcc355d94f601c533c11e00c6.png)
and we are done.
.
are collinear.![\[\angle EAF = \angle ABF = \angle BFD + \angle EDF = \angle ETF,\]](http://latex.artofproblemsolving.com/6/c/5/6c50e97ab05740ac4e384768fd498fd1ef7f5471.png)
so quadrilateral 
i![\[\angle EAC = \angle ABC = \angle EDC,\]](http://latex.artofproblemsolving.com/d/9/e/d9ee3c2819ec530a129a4dd572630e512a91c2ed.png)
so quadrilateral ![\[\angle ACE = \angle ADE = \angle ABF,\]](http://latex.artofproblemsolving.com/d/a/d/dad22a19f15a81be6462ec57772756fec1a725f8.png)
so quadrilateral 
is cyclic.![\[\angle ATE = \angle AFE = \angle ABC = \angle DTC.\]](http://latex.artofproblemsolving.com/9/9/d/99d96a54eb6b4d05499a4c357df519276be83cef.png)
