AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
+1 w
be an acute triangle with 
and let 
be its orthocenter. Let 
be a point on the perpendicular bisector of 
such that 
and and 
are on different sides of 
, 
a point on the perpendicular bisector of 
such that 
and 
a point on the perpendicular bisector of 
such that 
and 
lie on the opposite side of 
w.r.t 
. Prove that 
and are collinear.
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? with incenter 
,
= 
= incenter of 
= incenter of 
= intersection of 
and 
= intersection of 
and 
= midpoint of 
concycle
with circumcenter . 
and 
intersect the altitude from to at points and respectively. 
is the circumcenter of triangle 
and 
is the reflection of over . 
is the second intersection of circumcircles of triangles 
and 
. Show that 
are collinear.
, let 
be all positive integers smaller than that are coprime to . Find all 
such that 
for all 
is the largest positive integer that divides both and . Integers and are coprime if 
.
and arbitrary radius. Let 
respectively.
and 
.
is a parallelogram and 
So, 
.
and 
.
and 
.
.
and 
which is true since 
.
This gives 
Hence, we get
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)
such that:
Prove that:
such that 
and 
.
is a cyclic quadrilateral. Now we have 
are collinear and so 
and we are done.
be the circumcenters of 
.
.
are concyclic (by simple angle chasing)
at 
are conllinear. so 
so 
thus 
so 
of 
.
be the point such that 
and 
be the intersection of 
Clearly, 
so from 
we get 
hence 
with oriented angles .