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.
be a given positive integer. Is is true that for every -colouring of the natural numbers there exists a monochromatic solution of the equation 
?
has a list 
of their friends, where 
is the number of distinct friends has. Additionally, any two people are connected by a series of friendships. Each person also has a water balloon. The following game is played until someone ends up with more than one water balloon: on round , each person throws the current water balloon they have to their friend 
such that divides 
. Show that if the game never ends, then everyone has the same number of friends.
be a segment with midpoint 
, and consider the circle tangent to on . Let 
, 
be two points on the circle such that 
is a convex quadrilateral. The circumcircles of 
and 
intersect at points and 
. Let 
, 
be points on 
such that 
, and 
, 
being the intersection points for 
, 
and 
, 
respectively. Prove that lines 
, 
, 
are concurrent.
are the sides of a triangle, prove that![\[\frac{\sqrt{b+c-a}}{\sqrt{b}+\sqrt{c}-\sqrt{a}}+\frac{\sqrt{c+a-b}}{\sqrt{c}+\sqrt{a}-\sqrt{b}}+\frac{\sqrt{a+b-c}}{\sqrt{a}+\sqrt{b}-\sqrt{c}}\leq 3 \]](http://latex.artofproblemsolving.com/f/d/0/fd08baa6b9711d60b2552affcbf26614e66672d1.png)
does there exist a completely multiplicative, complex-valued arithmetic function 
such that![\[
\sum_{n<x}f(n)=\alpha x+O(1)\,\,?
\]](http://latex.artofproblemsolving.com/0/6/7/067fe16c586d94f9f27536046b0590801db58d4a.png)
be the set of real numbers. We denote by 
the set of all functions 
such that
for every 
Find all rational numbers 
such that for every function 
, there exists some 
satisfying 
.
be a point inside a triangle 
such that![\[ \angle APB - \angle ACB = \angle APC - \angle ABC.
\]](http://latex.artofproblemsolving.com/a/6/e/a6e3dc27d0457682fa22af4f918f58d9cd8403bc.png)
, 
be the incenters of triangles 
, 
, respectively. Show that the lines 
, 
, 
meet at a point.
be a triangle and 
, 
, and 
be the midpoints of 
, 
, and . Let 
be the circle through 
, , and , and define 
and 
similarly. Assume 
is the intersection of 
and 
different from 
for any 
. If 
is the centroid of , prove that , 
, 
, and 
are concyclic.
, let 
be the second smallest positive integer such that 
is a perfect square. Prove that 
is a perfect square.
such that for any 
,
Proposed by gghx and EmilXM
be numbers of factors 
of the number 
(that is, 
and 
). Find the least 
such that 
.
grid, one can find 
points such that no four of them are vertices of a square with sides parallel to lines of the grid. Imagine yourself as Erdos (!) and guess what is the best exponent instead of 
!
grid, one can find points such that no four of them are vertices of a square with sides parallel to lines of the grid. Imagine yourself as Erdos (!) and guess what is the best exponent instead of !
or 
points per row, but asymptotically this is irrelevant; I'll assume there are points per row.
points from the grid uniformly and at random, where 
is a positive constant to be determined later.![\[\frac{\frac{(n - 1)n(2n - 1)}{6}}{\frac{(n^2)(n^2 - 1)(n^2 - 2)(n^2 - 3)}{24}} \le Cn^{-5}\]](http://latex.artofproblemsolving.com/6/d/f/6df6f010753baa49ea21e442264eb27576338675.png)
for some positive constant 
points, the expected number of squares with sides parallel to the grid is ![\[Cn^{-5}\binom{cn^{\frac{5}{3}}}{4} \le \frac{c^4Cn^{\frac{5}{3}}}{24} \]](http://latex.artofproblemsolving.com/e/c/1/ec1c90ecc98e5bd78395dd10da8b0f580f6da558.png)
In particular, there exists a configuration in which there are at most this many squares. Removing all vertices of every such square, we are left with ![\[cn^{\frac{5}{3}} - \frac{c^4Cn^{\frac{5}{3}}}{6} = (c - \frac{c^4C}{6})n^{\frac{5}{3}}\]](http://latex.artofproblemsolving.com/9/5/6/956e9b69f4fe41d6401e1761fc668df50d100001.png)
points. For some positive sufficiently small, the constant multiplier is positive, and we are done.
is the best lower bound.
be a real number to be decided later. Choose each of the points randomly and independently with probability . After doing so, remove one vertex of each square present.
as originally we would expect there to be 
points and each square(of which there are 
would be present with probability 
) so LOE gives the stated value. Now its not so hard to see that if we choose 
, then the expected value gives us our construction.
