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.
satisfying 
. Prove that:![\[
0 \le (x^2 + 1)(y^2 + 1) + 4(x - 1)(y - 1) \le 164.
\]](http://latex.artofproblemsolving.com/4/2/d/42d0b88388462f16dd99d928b94024e689ba7c2e.png)
and 
, we define 
as follows:
where 
are all the positive divisors of . For example,
.
Find all positive integers such that 
.
Find all positive integers such that 
.
be a convex quadrilateral. Suppose that the circles with diameters 
and 
intersect at points 
and 
. Let 
and 
. Prove that the points 
, 
, and are concyclic. and are not the diagnols)
attaining all positive values and positive numbers 
are given. Numbers 
satisfy
for every 
. Prove that there exists a positive integer satisfying 
.
?
and colors are given. We will say that a set of 
points on a plane is 
, if it contains exactly 2 points of each color and if lines connecting every two points of the same color are pairwise distinct. Find, in terms of the least integer 
such that: in every set of 
points of a plane, no three of which are collinear, consisting of points of every color there exists a subset.
, let 
be the set of all positive integers not exceeding that are relatively prime to . Consider the polynomial![\[
P_n(x) = \sum_{k \in S_n} x^{k - 1}
\]](http://latex.artofproblemsolving.com/3/a/c/3ac60cb70baf4dfcf297e9af87480a8ee34ba70c.png)
and a polynomial 
with integer coefficients such that![\[
P_n(x) = (x^{r_n} + 1) Q_n(x).
\]](http://latex.artofproblemsolving.com/7/0/e/70e19a9d3e0a7e97abc7aa08d3f23126c14ea7d0.png)
such that 
is irreducible in ![$\mathbb{Z}[x]$](http://latex.artofproblemsolving.com/4/b/7/4b7d768fffbb7a1a6163af392ed348f2d49d8783.png)
.
meet at point 
. In triangles 
medians 
and angle bisectors 
are drawn respectively. Prove that angles 
and 
are equal.
We have by similar triangles that 
and 
and therefore 
gives 
and applying the same theorem to 
gives 
Then, by properties of parallelograms, 
so we can equate the above ratios. So 
It follows that 
Similarly, 
By similar triangles, 
and 
So 
