[/quote]
,
of positive integers, such that for every positive integer 
there exists an integer 
with ![\[\sum^{2011}_{j=1} j x^n_j = a^{n+1} + 1\]](http://latex.artofproblemsolving.com/e/8/6/e866c59c4603d1fe6a92f12cd2de89c0f34c6845.png)
be a sequence of positive integers satisfying the following property: for all positive integers 
,
and for all distinct integers 
,![\[
a_{m_1} + a_{m_2} + \cdots + a_{m_k} \leqslant a_{n_1} + a_{n_2} + \cdots + a_{n_\ell}.
\]](http://latex.artofproblemsolving.com/d/1/6/d1632ff7e5e55319cdbf56f45dca9ed84343aa47.png)
Prove that there exist two integers 
and 
such that 
for all 
.
,
,
be the centers of the circles circumscribing triangles ADE, BDF, CEF. Prove that the orthocenter of triangle 
for which the following statement holds: there exists at least one triple 
of integers such that
and all triples 
are integers.
be a triangle, 
its circumcircle and 
be a point on the segment 
.
the second intersection point of the circumcircles of triangles 
and 
.
and the tangent to 
intersect on the circumcircle of triangle 
.
.![\[\frac{1}{(2a+b+c)^2}+\frac{1}{(a+2b+c)^2}+\frac{1}{(a+b+2c)^2}\leq \frac{3}{16}.\]](http://latex.artofproblemsolving.com/d/1/8/d18152c5635ee2f2059a860bf2fa362fe525ebb3.png)
such that, for all indices 
,
.
and for any index ![\[
a_n a_{n+1} \cdots a_{n+p-1} - a_{n+p}
\]](http://latex.artofproblemsolving.com/5/4/a/54ac17063104439114f3c7b07de8fd27450ea1eb.png)
is a multiple of 
the intersection point of lines 
and 
.
be the circumcircle of triangle 
.
be the second intersection point of 
.
intersects 
.
intersects line 
at 
,
intersects 
the second intersection point of the circumcircle of triangle 
with 
into two regions, two parallel planes divide it into three regions, and two intersecting planes divide space into four regions. Consider the six planes which the faces of the cube 
lie on, and the four planes that the tetrahedron 
lie on. How many regions do these ten planes split the space into?
such that 
.
on the board. Charlotte has two operations available: the GCD operation and the LCM operation.
.
.
and, for each of them, determine the minimum number of GCD operations Charlotte must use.
be an integer. We consider a square grid of size 
divided into 
unit squares. The grid is called balanced if:
,
or 
.
.
such that any balanced grid always contains an 
square whose absolute sum of the 
cells is less than or equal to 
.
Y by Adventure10
Let 
be a triangle inscribed in circle 
with center 
. Let 
be the orthocenter of triangle and let 
be the midpoint of 
. Tangent of at 
intersects the perpendicular bisector of 
at 
. Tangent of at 
intersects the perpendicular bisector of 
at 
. Prove that 
and 
are perpendicular.
by Michael Sarantis, Greece
be a triangle inscribed in circle 
with center 
. Let 
be the orthocenter of triangle and let 
be the midpoint of 
. Tangent of at 
intersects the perpendicular bisector of 
at 
. Tangent of at 
intersects the perpendicular bisector of 
at 
. Prove that 
and 
are perpendicular.by Michael Sarantis, Greece
This post has been edited 1 time. Last edited by parmenides51, May 5, 2019, 6:32 PM
Reason: author added
Reason: author added
be the feet of the altitudes from 
respectively and denote by 
be the center of 
.
and 
.
collinear. Next, note that 
,
and 
.
.
and 
and 
.
.
.
.
be a point such that 
is a parallelogram. Let 
be points on 
,
.
are collinear. Also, 
as 
form a parallelogram. So, 
.
are the polars of 
respectively. Since 
,
.
.
Two equations give 
.
Finally
Clearly 
![[asy]
size(250);
pair P=dir(105),Q=dir(-120),R=dir(-60),O=incenter(P,Q,R),A=foot(O,Q,R),B=foot(O,R,P),C=foot(O,P,Q),K=circumcenter( 1/2*(B+C),1/2*(C+A),1/2*(A+B)),L=extension(P,B,Q,O),M=extension(P,C,R,O),IP=2*dir(-90)-O,T=circumcenter(P,Q,R),IR=extension(Q,IP,R,O),IQ=extension(R,IP,P,IR);
draw(circumcircle(A,B,C),blue);
dot("$P$",P,dir(P));
dot("$Q$",Q,dir(Q));
dot("$R$",R,dir(R));
dot("$A$",A,dir(A));
dot("$B$",B,dir(B));
dot("$C$",C,dir(C));
dot("$K$",K,dir(230));
dot("$M$",M,dir(180));
dot("$L$",L,dir(90));
dot("$I_P$",IP,dir(IP));
draw(P--Q--R--P,red);
draw(M--L,brown);
draw(IP--T^^A--K,green);
draw(Q--L^^R--M,purple);
[/asy]](http://latex.artofproblemsolving.com/e/f/0/ef0e400710fdb6d62066c0aef8ef3537b5c9398a.png)
be the tangential triangle of 
are angle bisectors. Let 
be the excenters of 
.
and nine-point center of 
is perpendicular to 
.
are homothetic, thus 
.
are at 
with 
,
and 
.
be the midpoint of 
at 
.
and 
.
we have 
.
and rearrange to get 
Moreover, since 
we have
Combining 
and 
we get that
Due to symmetry we also obtain
Now we have to prove that 
,
.
.
Now 
.
Therefore 
and thus 
,
be the reflection of 
the reflection of 
and 
.
.
be the intersection of 
and 
.
is a parallelogram. So 
.
,
but this is obvious because 
is a parallelogram so we are done.
