[/quote]
,
be a triangle with 
let 
be the circumcircle of 
and let 
be its radius. Point 
is chosen on 
such taht 
and point 
is the foot of the perpendicular from 
.
mets 
is chosen on line 
such that 
and 
lie on a line in that order. Finally, let 
and 
.
,
has two real roots, so that the one is the square of the other.
,
be all positive integers smaller than 
such that 
for all 
is the largest positive integer that divides both 
.
.
such that: ![\[ x = \sqrt{ x - \frac{1}{x} } + \sqrt{ 1 - \frac{1}{x} } \]](http://latex.artofproblemsolving.com/e/c/f/ecf7fb3c2805290f2cb07b7fdd7c73e1c7a55b99.png)
be a convex pentagon such that 
.
is the circumcenter of triangle 
.
be the circumcenter of triangle 
.
passes through the midpoint of segment 
.
such that for all 
the following holds:![\[f(x)+f(y)=f(xy)+f(f(x)+f(y))\]](http://latex.artofproblemsolving.com/a/c/a/aca098172ea7ce5663e9dd0dd10c2f1495630230.png)
.
and 
and 
are the incenter and the 
be the midpoint of arc 
on the circumcircle, and 
intersects the circumcircle again at 
. For
on the circumcircle of triangle 
intersect 
. Prove
are concyclic.
,
lie on a line in this order and satisfy 
.
and 
,
is acute, and let 
be its orthocentre. Points 
and 
,
and 
and 
and 
and 
be a transversal of 
be intersections of 
with 
be a tangent line of 
is cyclic
be integers such that there are consecutive integers 
satisfy 
.
is a perfect square.
a
Prove that
of these coins arranged in a line from left to right. He repeatedly performs the following operation: if there are exactly 
coins showing 
t
the process starting with the configuration 
would be 
,
be the number of operations before Harry stops. For example, 
and 
.
possible initial configurations 
Y by mariam10, son7, A4r43r457yhyt, centslordm, mathematicsy, jhu08, HWenslawski, chessgocube, lian_the_noob12, ItsBesi
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 
.
.![\[
\frac{-\frac{b^2k(1-k)}{a^2k+c^2(1-k)}}{1-k}=-\frac{b^2k}{a^2k+c^2(1-k)}=-\frac{b^2k}{b^2k+c^2(1-k)}=\frac{-\frac{b^2k}{c^2}}{1+\frac{b^2k}{c^2}-k},
\]](http://latex.artofproblemsolving.com/3/0/f/30fa9e596cba33c3837b86f556769c58a1608735.png)
so 
collinear which gives 
as desired.
,
.
.
and 
.
is 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 
.
concur on the radical axes of 
,
is tangent to 
is concyclic. 
are pairwise radical axes, hence they are concurrent at the radical center of 
.
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 
