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 
,
,
of the
,
,
, 
.
such that
Y by farhad.fritl
Given is triangle 
with centroid 
, such that 
. Let 
and 
be the second points of intersection of lines 
and 
with the circumcircle of triangle , respectively. Prove that the line 
is tangent to the Euler circle of triangle .
Proposed by Andrija Živadinović
with centroid 
, such that 
. Let 
and 
be the second points of intersection of lines 
and 
with the circumcircle of triangle , respectively. Prove that the line 
is tangent to the Euler circle of triangle .Proposed by Andrija Živadinović
,
,
be the midpoints of 
,
,
,
be the reflection of 
the point 
,
.
.
,
is the midpoint of 
.![[asy]
import geometry;
size(7cm);
triangle t=triangleabc(3,4,.5*sqrt(50), A=(0,0));
draw(t,red+linewidth(1));
pair A=t.C;
pair B=t.A;
pair C=t.B;
circle w=circle(A,B,C);
pair Ap=.5(B+C);
pair Bp=.5(A+C); pair Cp=.5(A+B);
circle wp=circle(Ap,Bp,Cp);
pair T=intersectionpoint(line(C,Cp),line(B,Bp));
pair S=2Ap-T;
pair X=.5(A+T);
point[] G=intersectionpoints(line(C,Cp),w);
pair G=G[1]; point[] H=intersectionpoints(line(B,Bp),w);
pair H=H[1];
draw(w,black+linewidth(1));draw(wp,deepgreen+linewidth(1));
draw(A--S,StickIntervalMarker(4,1)); draw(C--G,orange); draw(B--H,blue); draw(G--H,dotted); draw(Cp--X,blue); draw(Bp--X,orange);
dot("A",A,dir(60));
dot("B",B,dir(210));
dot("C",C,dir(-35));
dot("A'",Ap,1.3*dir(290)); dot("B'",Bp,1.3*dir(-15)); dot("C'",Cp,1.2dir(189));
dot("T",T,1.3*dir(105));dot("S",S,dir(250)); dot("G",G,dir(160)); dot("H",H,dir(10)); dot("X",X,dir(120));
[/asy]](http://latex.artofproblemsolving.com/9/e/b/9eba82d56f3cbaecce9ade2273e39a3fb52a7f2c.png)
is parallel to 
is parallel to 
which implies that 
,
giving 
at 
reflection of 
over 
respectively, let 
midpoints of 
respectively.
a point such that 
is a parallelogram then in fact 
colinear and also 
so from PoP and centroid lenghts we get 
so from Midbase we get 
and 
colinear so from pascal we get 
colinear but so finish notice that from Reim's we have 
cyclic and thus from radax on 
we get that tangent from 
and 
a point on 
is tangent to 
are colinear but notice 
is tangent to 
as 
and that follows from reflection properties it has, so scaling back we trivially get 
.
