AoPS Academy
be a triangle. Let 
be a circle passing through 
intersecting 
at 
and 
at 
. Let 
be the intersection of 
and . Further, let 
and 
be the intersections of 
and 
with the tangent to 
at 
. Now, let 
be the second intersection of 
and . Then, prove that , , , 
and are concyclic.
be a polynomial with integer coefficients such that 
, and let 
be an integer. Define 
and 
for all integers 
. Show that there are infinitely many positive integers 
such that 
.
table filled with natural numbers, we say it is a divisor table if:
-th row are exactly all the divisors of some natural number 
,
-th column are exactly all the divisors of some natural number 
,
for every 
.
is given. Determine the smallest natural number , divisible by , such that there exists an divisor table, or prove that such does not exist. be an acute triangle with circumcenter 
. Points and are chosen on segments 
and 
such that 
. If is the midpoint of the arc 
and is the midpoint of the arc 
, prove that 
.
, 
, 
be non-negative real numbers, no two of which are zero. Prove that :
that, for all and positive real numbers 
, there is![\[\sum_{i=1}^n \frac{x_i^2}{\sum_{j=1}^i x_j}\geq C \ln n\left(\prod_{i=1}^n x_i\right)^{\frac{1}{n}}\]](http://latex.artofproblemsolving.com/4/8/b/48b9fc4af5a20de7fd9dfc45a74d5aa27e615e46.png)
squares with two game pieces. At the beginning, Alice’s piece is on the first square while Bob’s piece is on the last square. The figure shows the starting position for a strip of 
squares. can Bob ensure a win no matter how Alice plays? can Alice ensure a win no matter how Bob plays?
at and 
meet at . The circumcircle of 
meets sides 
and again at and respectively. Let be the circumcentre of . Show that 
is perpendicular to 
.
is defined as the intersection of the tangents. We can replace the circumcircle of with any circle that passes through and . Indeed, we know that 
, and since 
is cyclic, we have that 
, from which the result follows.
by reflecting 
about its 
angle bisector and taking a homothety at . Now since the circumcentre and orthocentre are isogonal conjugates, has to lie on the altitude of , and therefore we are done. 
be the orthocenter of .
is concyclic 
.
, such that goes to . Under this transformation, line 
goes to line because they are isogonal in . Since 
, we have 
.
lies on the circumcircle of 
, we know that
so
we know that
hence we are done 
hence lies on perpendicular bisector of and lies on perpendicular bisector of moreover also lies on these lines
is the orthocentre of hence 
is perpendicular to 
such that 
coprime, 
and 
.
![\[
\text{Find all functions } f : \mathbb{R} \to \mathbb{R} \text{ such that:} \\
f(a^4 + a^2b^2 + b^4) = f\left((a^2 - f(ab) + b^2)(a^2 + f(ab) + b^2)\right)
\]](http://latex.artofproblemsolving.com/2/a/7/2a73a5e58eab4a994fbd83160fb79a0b00152951.png)
is the 
are concyclic. So 
.
,
since 
are concyclic. Therefore, 
.
(because CFEB is cyclic)
