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.
of people, we define 
to be the least number of tables needed such that each person from the group sits in one of them with no two friends sitting on the same table, and 
to be the least number of tables needed such that each person from the group sits in one of them with no two enemies sitting on the same table.
such that for each group of 
people we have![\[ f(n) \leqslant f(a(G)) + f(b(G)) \quad \text{and} \quad f(2) = 1.\]](http://latex.artofproblemsolving.com/0/3/d/03dfc700f9284152b39e32031c92c33283d65da6.png)
Find all possible values of 
.
is the set of all positive integers.
such that,
for all non-negative integers 
and 
.
is the set of all non-negative integers and 
is the set of all real numbers. cells is a connected figure consisting of equal-sized cells[1].
cells. It is said to be primitive it its cells cannot be partitioned into two or more dinosaurs. Find with proof the maximum number of cells in a primitive dinosaur. cells, one with 
cells is obtained by adjoining a new cell by making it adjacent to one or more existing cells.
such that 
and 
are coprime if and only if 
?
are positive reals. Consider the following polynomial for each 
in 
:
where indices are taken modulo 
, i.e., 
for any 
in 
. Show that it is impossible that each of these polynomials has all its roots real.. Initially, the numbers 
are written on a board. In each move, two numbers are selected such that their difference is also present on the board. This difference is then erased from the board. (For example, if the numbers 
and 
are on the board, then 
can be erased as 
, or 
as 
, or 
as 
.) is it possible to end with only one number remaining on the board?
and 
such that 
and 
is a prime number
, let 
be the midpoint of side 
. Let the line perpendicular to 
at point 
intersect 
at 
. If 
is tangent to 
at 
, find 
.
of side length 
lies in the union of five equilateral triangles of side length 
, show that there exist four equilateral triangles of side length whose union contains .
is a polynomial of degree 
such that
.
, 
. For Menelaus on CMB with trasversal NE we obtaid 
and for Ceva on ACM and point F we obtain 
and so 
. For Menelaus on AEB and trasversal GF we have 
and so 
. Let 
so 
and 
so 
so AFD and ACE are similar and 
, then DCEF is cyclic with DC = CE so C stay on the angle bisector of 
, 
and 
.
, 
, 
. By solving a system,
and 
and we are done.
symmetrical of about 
is the centroid of 
collinear, so 
, but 
by symmetry; also by symmetry 
, so 
and 
is cyclic; with 
we are done.
be real numbers such that 
.
be an isosceles right triangle and 
and 
are taken on the legs 
and 
.
and 
intersect at 
.
bisects the 
.
