is interesting if 2018 divides 
(the number of positive divisors of 
such that there exists an infinite arithmetic progression with common difference 
,
.
that satisfies:
be an odd prime number and a sequence 
satisfy 
and 
Prove that always exists positive integer 
.
is "not congruence"
and 
Prove that
Solution:
and 
then 
and 
.
![$$(\sqrt{x}+1)[2-(x-6)\sqrt{x-3}]=x+8$$](http://latex.artofproblemsolving.com/b/1/d/b1d283f2e7b8e574b6c0ff72236c52a39d3da9e8.png)
Prove that
Prove that
and 
intersect at points 
and 
.
and 
intersects 
,
intersects 
.
is the circumcenter of 
,
.
Y by itslumi, GeoKing, Rounak_iitr, ItsBesi
We are given an acute triangle 
with 
. Let 
be a point of 
such that 
is tangent to the circumcircle of . Let 
and 
be the circumcenters of triangles 
and 
, respectively, and let 
be the midpoints 
. Prove that the line tangent to the circumcircle of 
through is also tangent to the circumcircle of .
Proposed by Patrik Bak, Slovakia
with 
. Let 
be a point of 
such that 
is tangent to the circumcircle of . Let 
and 
be the circumcenters of triangles 
and 
, respectively, and let 
be the midpoints 
. Prove that the line tangent to the circumcircle of 
through is also tangent to the circumcircle of .Proposed by Patrik Bak, Slovakia
This post has been edited 1 time. Last edited by Tafi_ak, Dec 23, 2022, 10:13 AM
rest is easy chase 
be the tangent from 
.
is tangent to 
iff 
bisects 
.
is isosceles, so we need 
,
is sufficient. We have that 
,
.
of 
,
,
and 
are isogonal in 
)
be the other tangent line through 
,
.
for the obvious reason. Similarly 
.
is a parallelogram. So 
which bisects 
.
.
and 
be the point of 
distinct from 
is tangent to 
be reflection of 
be reflection of 
)
.
be circumcenters of 
,
Proof: Note 
,
.
It follows
which implies 
,
![[asy]
size(250);
pair B=dir(-165),C=dir(-15),A=dir(120),O=(0,0),N=1/2*(B+C),S=2*foot(A,B,C)-A,K=IP(N--S,unitcircle),T=2*foot(A,N,O)-A,D=2*A*K/(A+K);
pair E=circumcenter(A,B,D),F=circumcenter(A,D,C),M=1/2*(E+F);
fill(A--D--M--A--cycle,cyan+white+white);
fill(A--S--N--cycle,green+white+white);
draw(A--D--K,brown);
draw(E--F,magenta);
draw(D--C,red);
draw(S--T);
draw(A--M--D,purple);
draw(S--A--N,purple);
draw(A--B--S--C--A,orange+linewidth(0.8));
draw(A--E--D--F--A,fuchsia+linewidth(0.8));
draw(unitcircle,royalblue);
draw(circumcircle(A,O,K),grey);
dot("$A$",A,dir(90));
dot("$B$",B,dir(-110));
dot("$C$",C,dir(C));
dot("$O$",O,dir(60));
dot("$N$",N,dir(-60));
dot("$S$",S,dir(S));
dot("$T$",T,dir(T));
dot("$K$",K,dir(-70));
dot("$D$",D,dir(D));
dot("$E$",E,dir(E));
dot("$F$",F,dir(F));
dot("$M$",M,dir(-140));
[/asy]](http://latex.artofproblemsolving.com/9/d/7/9d7f7c0a68e2cc33c573077b45a33a898384c044.png)
be the spiral similarity centered at 
and 
.
.
.
In particular, this gives
are collinear.
is symmedian (in 
)
.
with 
,
.
lie on 
,
lie on circle with diameter 
It follows 
.
is perpendicular bisector of 
and 
is perpendicular bisector of 
and 
we have that 
is cyclic, where 
is a parallelogram
and 
.
.
but 
at 
.
.
,
but 
thus 
is tangent to the circumcircle of 
Since 
Similarly 
Now we wanna prove that 
is tangent to 
.
too. Then, we make the following key claim.
Then,
.
so, 
.
and 
be the midpoints of 
and 
are collinear. Thus,
as well. Establishing the claim.
and 
.
and 
implying 
.
So, 
too. So points 
are concylic with 
which implies 
which gives our desired tangency.
.
are collinear.
armonic quadrilateral, so 
,
,
are collinear.
,
are collinear.
are concyclic.
,
,
so 
,
is tangent to 
,
,
.
.![\[2\measuredangle EDB = \measuredangle DEB = 2\measuredangle DAB = 2\measuredangle ACB = 90+ \measuredangle ACB\]](http://latex.artofproblemsolving.com/5/4/1/54188efc667ef6f3b3d327c05790aacfcd426aa5.png)
Thus, 
.![\[2\measuredangle BDF = \measuredangle CFD = 2\measuredangle CAD - 180\]](http://latex.artofproblemsolving.com/7/5/1/7514bceea2bbed2bb0bf584cc0bd679ec50eed38.png)
Then, we have that 
.![\[\measuredangle EDF = \measuredangle EDB + \measuredangle BDF = 90 + \measuredangle ACB + \measuredangle CAD - 90 = \measuredangle CAB\]](http://latex.artofproblemsolving.com/1/c/e/1ce477155bbb68f134a836765bd3b817c4cf36c5.png)
Thus,![\[\measuredangle FOE = \measuredangle EDF\]](http://latex.artofproblemsolving.com/f/d/9/fd945770bb77604e6deb05ebdd4f09cfe6c5c4e5.png)
Further note that,![\[\measuredangle OED = 360 - ( \measuredangle BDE +90 + \measuredangle ABD) = 180 + \measuredangle BCA + CBA = 2\measuredangle CBA + \measuredangle BAC\]](http://latex.artofproblemsolving.com/2/3/1/23147fbfff8e2b47b69cfa1277b542697133e1eb.png)
with a similar angle chase, we obtain that 
.![\[\measuredangle OEF = \measuredangle DFO\]](http://latex.artofproblemsolving.com/8/0/8/80804b307a10ebded250233c502f1cd80fb16dad.png)
as well. This implies the required result.
is a cyclic quadrilateral.
.
!
.![\[\measuredangle MDK = \measuredangle ADK + \measuredangle KDM = \measuredangle ADM = \measuredangle MAD\]](http://latex.artofproblemsolving.com/2/2/9/2292c50e6cfa266f493d9f01e36a3498908b252d.png)
This means that the line tangent to 
in 
invert we have that from easy angle chase that 
is cyclic, redefine 
which implies the tangency.
,
to 
is the same as 
by the midpoint, and by radax we have 
colinear so by thales we have 
as desired thus we are done 
.
and 
.
.
.
and 
,
and 
[using EM 
]
,
and we are done. 
.
is tangent to 
.
are collinear.
is a parallelogram. This implies that 
.
such that 
.
where the last equality comes from symmetry.
so 
.
bisects 
so 
as desired.
is a parallelogram
and 
thus by Salmon Theorem we get that 
hence we get 
we get that 
hence the quadrilateral 
is a parallelogram 
are collinear and 
and 
we get that 
lies on the perpendicular bisector of 
lies on the perpendicular bisector of 
so 
and 
lies on the perpendicular bisector of 
lies on the perpendicular bisector of 
hence 
is the side bisector of 
(because they are tangent to 
