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.
, where a > 0 and 
is differentiable such that 
Then which of the following is correct?
always exists but not necessarily zero. always exists and is equal to zero. may not exist. exists if 
is twice differentiable.
such that for all 
, ![\[xf(y+x)+(y+x)f(y)=f(x^2+y^2)+2f(xy)\]](http://latex.artofproblemsolving.com/5/b/1/5b101ead6195155ccbd99b757016e8f73dac4411.png)
Proposed by Aritra Mondal
+1 w
be positive real numbers such that 
. Prove that
with rational coordinates lie on a plane. It turned out that the distance between every pair of points is an integer. Prove that there exist points 
with integer coordinates such that 
, 
, 
checkered board is cut into several squares with integer side length. The size of a square is it's side length.
of integers, a pair 
with 
is called interesting if there exists a pair 
of integers with 
such that 
For each 
, find the largest possible number of interesting pairs in a sequence of length 
.
such that, for all 
,
cards on which the numbers 
, 
, 
, are written respectively. Alice and Bob play the following game: in each turn, Alice gives two cards to Bob, who must keep one card and discard the other. The game proceeds for four turns in total; in the first two turns, Bob cannot keep both of the cards with the larger numbers, and in the last two turns, Bob also cannot keep both of the cards with the larger numbers. Let 
be the sum of the numbers written on the cards that Bob keeps. Find the greatest positive integer 
for which Bob can guarantee that is at least .
be positive real number such that 
Find the minimum value of 
. The colour of a vertex 
is denoted by 
. Suppose some colour, say red, occurs only once as the colour of some vertex, say 
. Then the vertices from 
to 
must be coloured alternately with the colours 
and 
. Thus the triangulation 
works.
and 
, say. Then by our assumption, 
, 
etc, and the whole colouring is determined by only red and blue. But this would lead to a contradiction, as the colour green occurred nowhere. Thus some vertex has neighbours of different colours. The triangle 
that they form has all three colours different. Separate this triangle, and the remaining polygon is 
-sided, and each colour occurs at least once in it ( this is why we need to make cases; we need each colour to occur at least once). Now this is induction.
, so assume it holds for and we need to prove it for 
, so we get that the next vertices need to be coloured 
and so on, and so all vertices have to be coloured only , which is impossible.
and let them be coloured with colours 
. Then, we can draw diagonal 
and then use the induction hypothesis on all the vertices except 
. This works (almost always) because 
are indeed different colours, as required by the problem. was the only vertex of colour , because then the remaining polygon does not have any vertex of colour and so the induction hypothesis doesnt work.
gon with colours alternating every vertex. Call the remaining vertices 
. Then, just draw the diagonals 
and 
, which can easily be checked to work.
prove inequality 
Prove that
Where 
