[/quote]
,
be positive integers and 
be prime. We know that 
and 
.
is a perfect square.
such that 
but 
does not divide 
.
and ![\[ 3n^2+3n+7=m^3 \]](http://latex.artofproblemsolving.com/6/f/8/6f84f81f857094b3ea432284a7881697167b2418.png)
be a point on the Neuberg cubic. Show that as 
with the property that 
is divisible by 
.
be a tangent line to 
)
be midpoint of 
.
be coprime positive integers such that:![\[ \dfrac{p}{q}=1-\frac12+\frac13-\frac14 \cdots -\dfrac{1}{1334}+\dfrac{1}{1335} \]](http://latex.artofproblemsolving.com/b/d/3/bd374388fe6b84fbcd66e23a7956f2fbeebe4c28.png)
and 
satisfying 
and 
.
is chosen. Prove that there exists a prime number that divides infinitely many terms of the sequence 
,![\[
b_k = a^{k^k} \cdot 2^{2^k - k} + 1.
\]](http://latex.artofproblemsolving.com/1/7/5/1751a60482d729a36c71b77ac9c978e724f40da0.png)
(the number of positive divisors of 
such that there exists an infinite arithmetic progression with common difference 
,
.
that satisfies:
satisfy 
and 
Prove that always exists positive integer 
.
is "not congruence"
Y by Rounak_iitr, buddyram, ItsBesi
Acute-angled triangle 
with circumcircle 
is given. Let 
be the midpoint of 
, 
be the foot of altitude from 
to 
, and 
be the intersection point of 
and 
. Point 
lies on the arc of (the one that does not contain ) such that 
. Prove that 
.
with circumcircle 
is given. Let 
be the midpoint of 
, 
be the foot of altitude from 
to 
, and 
be the intersection point of 
and 
. Point 
lies on the arc of (the one that does not contain ) such that 
. Prove that 
.
.
is an isosceles trapezoid.
.
be the symmetry of 
is a rectangle. Hence, 
are collinear.
.
.
.
lies on 
.
.
.
,
and 
be the antipode of 
wrt 
are collinear.![[asy]
size(200);
pair A = dir(120),B=dir(-155),C=dir(-25),O=(0,0);
draw(unitcircle);
dot("$A$",A,dir(A));
dot("$B$",B,dir(B));
dot("$C$",C,dir(C));
pair Ap = dir(60), E = foot(A,B,C), D = 1/2*(A+C), F = extension(A,B,D,E),H = IP(E--2*E-Ap,unitcircle);
dot("$A'$",Ap,dir(Ap));
dot("$H$",H,dir(-90));
dot("$E$",E,dir(130));
dot("$D$",D,dir(D));
dot("$F$",F,dir(F));
pair L = 2*E - A - B - C, B0 = -B;
dot("$L$",L,dir(20));
dot("$B_0$",B0,dir(B0));
draw(B--C^^A--Ap,red);
draw(C--A--F,royalblue);
draw(B--B0^^L--Ap,purple);
draw(B--Ap^^F--D,green);
draw(A--L^^H--Ap,brown);
draw(F--B0,dashed);
[/asy]](http://latex.artofproblemsolving.com/2/9/0/2905eb7808cea9753c97680d22949c23909f1f6e.png)
is the reflection of 
is a diameter of 
.
gives that our problem is equivalent to 
.
which gives 
.
,
.
Let's set 
.
This implies that 
,
is cyclic. Now just note that:
Thus 
as desired.
be such that 
.
are collinear.
.
.
.
is the antipode of 
.
.
.
and 
as needed
as desired. 
be the second intersection of the circumcircles of 
and 
.
and 
,
.
.
.
Thus from condition of tangency we have 
is tangent to the line 
at 
Thus 
as desired.
,
,
is cyclic.So, done.
meet 
.
Note that 
.
is tangent to 
so 
.
.
then 
.
.
is an isosceles trapezoid with 
.
.
.
.
.
.
an inversion centered at 
followed by a reflection wrt to the 
for each point 
.
be the circumcenter of 
be the midpoint of 
,
,
and 
such that 
.
,
,
are collinear. Hence, 
.
,
so we get 
are concyclic. Thus, 
as desired.
.
and 
intersect at 
intersect 
.
.
.
.
,
,
by 
cyclic. then 
that means 
.
,
.
that 
and then 
.
and we are done.
