,
,
be nonzero real numbers such that 
and 
.
.
be non-negative real numbers such that ![\[b+c\leqslant a+1,\quad c+a\leqslant b+1,\quad a+b\leqslant c+1.\]](http://latex.artofproblemsolving.com/b/9/e/b9e86e898c0536b7323a03611d5bdbf679caa710.png)
Prove that 
be a function satisfying
for all integers 
and 
.
and 
such that 
for all integers 
such that ![\[f(xy-1)+f(x)f(y)=2xy-1,\]](http://latex.artofproblemsolving.com/8/8/8/888ca39f2b7f8cec6d6426bee28d40eade40a66e.png)
for any real numbers 
and 
![\[
f : \mathbb{R}^+ \to \mathbb{R}
\]](http://latex.artofproblemsolving.com/e/7/b/e7bfc5b236fb6f00ef328f850b1b14632cbf8416.png)
satisfying the condition that for all 
,![\[
f(x + y^2) \geq f(x) + y.
\]](http://latex.artofproblemsolving.com/3/a/a/3aa20083835682dbd81be692eba65cab19e923e5.png)
of positive integers with 
prime, 
,
,
is a multiple of 
.
a point in the space and 
the centroid of a tetrahedron 
.
be a 
on it, line 
makes a fixed angle with the tangent to 
be a rectangular hyperbola centered at the origin, scaled such that it is tangent to the logarithmic spiral at some point.
on the spiral, the polar of 
be a monic polynomial so that there exists another real coefficients 
that satisfy![\[P(x^2-2)=P(x)Q(x)\]](http://latex.artofproblemsolving.com/6/9/7/697faa929e4fda7a6e9b1cd97849bd42ffc14306.png)
Determine all complex roots that are possible from 
Y by Davi-8191, icematrix2, Adventure10, Mango247
Let 
be a convex polygon with 
sides, 
. Any set of 
diagonals of that do not intersect in the interior of the polygon determine a triangulation of into 
triangles. If is regular and there is a triangulation of consisting of only isosceles triangles, find all the possible values of .
be a convex polygon with 
sides, 
. Any set of 
diagonals of that do not intersect in the interior of the polygon determine a triangulation of into 
triangles. If is regular and there is a triangulation of consisting of only isosceles triangles, find all the possible values of .This post has been edited 1 time. Last edited by worthawholebean, May 1, 2008, 8:54 PM
be the vertices of a regular 
to return 
if the polygon 
of a regular 
otherwise. We claim that 
if and only if 
.
.
-
and 
,
cannot be paired with a vertex opposite it to make an isosceles triangle.
,
points in total, or 
,
are not both 0.
,
.
can be done, 
can be done, so everything follows.
where a1=3 and its for n>=1
where integer 
doesn't fit the criteria. Later, I managed to prove that if 
work and are the only solutions, where 
.
where integer 
doesn't fit the criteria. Later, I managed to prove that if 
work and are the only solutions, where 
.
yields 
which violates 
.
where a1=3 and its for n>=1
.
,
is a power-of-2. Then I basically proved 
in the beginning
,
,
work.
for 
being non-negative integers. Before showing that this satisfies the given conditions, we first set up the following claims.
If we can triangulate in such a manner for 
.
We just form a construction based on parity. Label the parities from 
to 
,
and 
,
shorter sides can be triangulated in such a manner if and only if 
,
.
,
and connecting that to 
.
will works for any non-negative 
.
shorter sides, and an isosceles triangle.
.
for nonnegative 
work.
to 
.
,
When 
has a triangulation.
is the edge that uses option 3, so we can't have vertex 
be blocked by 
,
,
.
for some nonnegative 
.
or 
,
,
for some 
.
outer isosceles triangles and an inner 
work. \bigskip
congruent angles and an 
-
are odd, which is precisely equivalent to having 
for nonnegative integers 
.
-
in that order. Define a \textit{small} triangle to be a triangle likewise to 
and define a \textit{large} triangle to be a triangle likewise to 
.
sided polygons with 
for a nonnegative integer 
.
triangles) on the side. Now, if 
and 
,
are in the same triangle. There is no point diametrically opposite an edge on a 
must be adjacent to either 
clearly works, all powers of 
Therefore, we only need to consider odd 
work, and proving that the rest don't.
it is obvious how to generalize.
-
which we will call 
Call a triangle 
very long. Note that every 
edges. Therefore, there is exactly one very long triangle in such a triangulation.
is even, and so is 
and so on. This forces 
so 
As a result, the only 
as claimed at the beginning.
-
an 
-
.
,
.
small sides and one long side. This long side is longer than any diagonal in this new polygon, so it must be used as the base. Hence, if 
,
for a positive integer 
.
works.
.
vertices for some p, we can triangulate it. Then the outer perimeter of the equal polygons concurrent with the side of the n-gon must be triangulated into small isosceles triangles, and we must continue this process 
it is only possible when 
.
.
is 
(
)
-
(
,
for any 
.
is base for some isosceles triangle. Then note it has to be 
.
so set 
as \empty{good} triangle. Note for this we should have 
even, hence 
(for one side). And if we consider it's side, then by same logic we still have to make \empty{good} triangles only. Doing this many time, we will be always making \empty{good} triangle on exiting polygon made up. Hence we only increase 
each time. We finally end up with 
for some 
for nonnegative integers 
.
for some integer 
are funny. Consider one side of this polygon and form an isosceles triangle with the diametrically opposite vertice. We then form these two hemispheres on the side (see the image in the beginning for a visual I don't feel like asying right now). We induct on these hemispheres. Note that the number of vertices in these hemispheres are also one more than a power of two. The base case is obvious, and to generalize to the next hemisphere (i.e. upping the power of two by one) by creating edge triangles on the sides of this new hemisphere. Thus by induction all 
.
vertice hemisphere works as well. This process continues until we reach a triangle, i.e 
.
.
be a sequence of numbers such that 
for 
then at some point 
It is easy to show that only 
then 
work while 
also clearly works.)
for nonnegative integers 
.
in clockwise order. All proceeding indices referring to vertices of the 
.
where 
.
A 
can be triangulated if and only if 
so 
and 
are drawn. This yields a triangle and two 
-
,
dissecting the 
-
we can fully triangulate these polygons as wanted.
,
.
which dissects the 
-
-
where 
are positive integers is the isoceles triangle with 
that (not strictly) contains the center of the 
are bowls, and thus we require 
to be powers of 
and hence is expressible in the form 
-
,
line segments. There must be another triangle such that the base of the new triangle is one of the line segments of the older triangle, meaning that we can split the region with 
in which case the recursive method stops. Therefore, 
,
