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.
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 
. such that 
is a perfect square.
be a triangle with incenter . Let be the point of intersection between the incircle and the side , the points and are in the rays 
and 
, respectively, such that 
and 
. Prove that 
.
be real numbers such that 
. Find the value of 
be real numbers such that 
and 
. Find the value of 
with -excenter 
, the foot of the perpendicular from to is . Let the midpoint of segment 
be , 
lies on arc (not containing ) satisfying 
, 
intersects the circumcircle of at 
. If 
and intersect at , the perpendicular bisector of 
intersects 
at 
respectively, prove that 
are concurrent.
, there exists some positive integer 
that makes the following equation have no integer root 
.
are nonnegative because 
. We aim to show that there must exist a less than 
that cannot be equal to the left-hand side of the given equation. We count the number of possible values of each so that 
., assume for the sake of contradiction that 
. When we multiply this by itself 
times we see that 
, which we don't want. to be less than 
. There are possible values for in the range 
. When we take the product of for all such that 
, we get ![\[ \prod\limits_{i=1}^n 2^{2^{n-i}} = 2^{\sum\limits_{i=1}^n 2^{n-i}} = 2^{2^n - 1}\]](http://latex.artofproblemsolving.com/e/4/d/e4d0e5a2f1977c4e12bec6fa560fbe9166f30418.png)
This value is an upper bound for the amount of tuples 
such that the LHS of the given equation is less than . However, this value 
is less than the 
possible values of less than . Thus, by pigeonhole there must exist a less than that cannot be equal to the LHS of the given equation.
![\[
\text{Find all functions } f : \mathbb{R} \to \mathbb{R} \text{ such that:} \\
f(a^4 + a^2b^2 + b^4) = f\left((a^2 - f(ab) + b^2)(a^2 + f(ab) + b^2)\right)
\]](http://latex.artofproblemsolving.com/2/a/7/2a73a5e58eab4a994fbd83160fb79a0b00152951.png)
is always false. If it turns out this equation can be true for some large