[/quote]
,
.
and 
.
,
and 
.
be the circumcenter of triangle 
,
the orthocenter of triangle 
the point of intersection of the lines 
and 
.
consecutive even numbers, such that the square of their product divides the sum of their squares.
be a positive integer. Find all real solutions 
to the system:![\[a_1^2 + a_1 - 1 = a_2\]](http://latex.artofproblemsolving.com/a/4/e/a4e26b83c929b82800576bac11e12de0c3aa1cf6.png)
![\[ a_2^2 + a_2 - 1 = a_3\]](http://latex.artofproblemsolving.com/9/9/5/995e8eec93828b82901b0639463f907f5ff95d08.png)
![\[\hspace*{3.3em} \vdots \]](http://latex.artofproblemsolving.com/e/e/1/ee15e1fefe2ad550d3c9b1c51207eaf66fbfa41f.png)
![\[a_{n}^2 + a_n - 1 = a_1\]](http://latex.artofproblemsolving.com/7/c/7/7c727d1f2044c1bf7c80981e1074bb3cde807b8f.png)
and 
are the feet of the altitudes through 
and 
and 
are 
and 
respectively; the circumcenters of the triangles 
and 
are 
and 
respectively. Prove that 
and 
are parallel.
with 
prime and ![\[a^p+b^p=p!.\]](http://latex.artofproblemsolving.com/f/e/b/feb27894b1f6497559eeb094cfe92256bc26688e.png)
such that 
.
such that 
.
.
be a cyclic quadrilateral with circumcenter 
,
is not a diameter of its circumcircle. The lines 
and 
lies between 
and 
is acute and let 
and 
and 
.
,
which satisfy the condition![\[ f(m + n) \geq f(m) + f(f(n)) - 1
\]](http://latex.artofproblemsolving.com/4/b/d/4bddce124cf0da82cb8bb6c6ae2325239354f27c.png)
Find all possible values of 
be a positive integer and 
be a positive integer coprime to 
.
,
,
Find, in terms of 
,
such that 
is divisible by 
.
table are initially white. Alice and Bob play a game. First Alice paints 
Y by sami1618
Let 
be an isosceles triangle with 
. The circle 
, passing through 
and 
, intersects segment 
at 
. The circle 
is tangent to 
at and passes through 
. Let 
and 
be the midpoints of segments and 
, respectively. The line 
intersects and at points 
and 
, respectively, where and are the intersections closer to . Prove that 
.
Proposed by Hooman Fattahi
be an isosceles triangle with 
. The circle 
, passing through 
and 
, intersects segment 
at 
. The circle 
is tangent to 
at and passes through 
. Let 
and 
be the midpoints of segments and 
, respectively. The line 
intersects and at points 
and 
, respectively, where and are the intersections closer to . Prove that 
.Proposed by Hooman Fattahi
such that 
at the same time, between 
.
the midpoints of 
and 
.
passes through the center of 
is isoceles. By symmetry 
is also isoceles. We have that 
is a parallelogram, thus 
.![\[\overline{MP} + \overline{MQ} + \overline{MX} + \overline{MY} = 0\]](http://latex.artofproblemsolving.com/7/d/f/7dfb6cf96aa7c742a4cba735de5ebe91feeb7bb2.png)
But then, by power of a point: 
.
and 
.
.
is obviously ruled out as 
are collinear in that very same order.
,
or 
,
meet 
Now let 
.
and 
so we have 
which implies 
.
so 
so now we have 
as wanted.
and 
,
be the midpoint of 
and 
be the midpoint 
.![[asy]
import geometry;
size(10cm);
pair A=dir(90);
pair B=dir(220);
pair C=dir(320);
pair K=B+.6(A-B);
pair O_1=circumcenter(B,K,C);
circle w_1=circle(K,B,C);
pair M=.5(A+B);
pair N=.5(A+C);
point[] T=intersectionpoints(w_1,line(N,M));
pair P=T[0];
pair Pp=T[1];
line b=line(.5(B+K),.5(B+K)+rotate(90)*(B-K));
line l=line(B,B+rotate(90)*(C-B));
pair O_2=intersectionpoint(b,l);
circle w_2=circle(B,K,O_2+rotate(90)*(K-O_2));
point[] X=intersectionpoints(w_2,line(M,N));
pair Q=X[1];
pair Qp=X[0];
pair M_p=.5(P+Pp);
pair M_q=.5(Q+Qp);
fill(A--B--C--cycle, palered);
draw(A--B--C--cycle);
draw(w_1,deepblue);
draw(w_2,deepgreen);
draw(Pp--Qp);
draw(B--M_q, dashed); draw(O_1--A, dashed);
dot("$A$",A,dir(A));
dot("$B$",B,dir(B));
dot("$C$",C,dir(C));
dot("$K$",K,2*dir(90));
dot("$O_1$",O_1,dir(270));
dot("$M$",M,dir(120));
dot("$N$",N,dir(60));
dot("$P$",P,dir(120));
dot("$P'$",Pp,dir(60));
dot("$O_2$",O_2,dir(180));
dot("$Q$",Q,dir(60));
dot("$Q'$",Qp,dir(120));
dot("$M_p$",M_p,dir(60));
dot("$M_q$",M_q,dir(120));
[/asy]](http://latex.artofproblemsolving.com/c/7/6/c7603b5a491ab07b031d59d44a97ed3fec553942.png)
,
.
so 
.
then the signs would need to be reversed).
