such that ![\[f(2x+y) + f(x+f(2y)) = f(x)f(y) - xy\]](http://latex.artofproblemsolving.com/0/6/2/0623b7c328af6d088cfc5dd288afe3e17a94ff29.png)
for all reals 
and 
.
and 
be two sequences of integers such that 
and 
for all 
.
divides 
for an infinity of integers 
such that 
for all 
are coprime. Consider the sequence 
given by 
.
such that 
is even and 
is an integer. Prove that the numerator in its simplest form of 
is an odd integer greater than 
.
be an integer and let 
be 
distinct subsets of 
.
such that the n subsets 
are also disjoint.
s
.
,
from all multiples of 
-
.
which are simultaneously 
-
-
-
.
satisfying 
and![\[ a^2b + a^2c + b^2a + b^2c + c^2a + c^2b = 6{,}000{,}000.\]](http://latex.artofproblemsolving.com/0/4/4/0446861fb2e63782803c4f182a529b79e213fdfd.png)
satisfies 
,
.
and 
intersect at point 
,
.
Y by Adventure10, megarnie, Mango247
A set 
of points from the space will be called completely symmetric if it has at least three elements and fulfills the condition that for every two distinct points 
and 
from , the perpendicular bisector plane of the segment 
is a plane of symmetry for . Prove that if a completely symmetric set is finite, then it consists of the vertices of either a regular polygon, or a regular tetrahedron or a regular octahedron.
of points from the space will be called completely symmetric if it has at least three elements and fulfills the condition that for every two distinct points 
and 
from , the perpendicular bisector plane of the segment 
is a plane of symmetry for . Prove that if a completely symmetric set is finite, then it consists of the vertices of either a regular polygon, or a regular tetrahedron or a regular octahedron.Attachments:
This post has been edited 1 time. Last edited by orl, Nov 14, 2004, 10:18 PM
of the points 
of 
.
,
,
.
the mass center of 
the perpendicular bisector of 
is an axis of symmetry for 
,
;
is an axis of symmetry for 
lies on it, hence 
.
over the perpendicular bisector of BC to get another point 
lying on the circle because the perpendicular bisector of 
passes through 
on AC, which is a contradiction, so AB=BC. In either case, AB=BC, and so we can take any other 3 points, A,B,D (D on the opposite side of C) and arrive at the same conclusion. lnductively, all the sides are equal and it is circumscriptible, so it is a regular polygon. 
be the centroid of 
and 
in that order around 
,
and 
consecutive, and
.![[asy]
draw(unitcircle);
pair A=dir(162),B=dir(234),C=dir(354),D=dir(42);
pair Z=(5/4)*sqrt(A*D), Cp=A*D/C;
draw(-Z--Z,blue+dashed+(1/2)*bp);
draw(C--Cp,blue+(1/2)*bp,Arrow,Margins);
dot("$A$",A,A); dot("$B$",B,B); dot("$C$",C,C); dot("$D$",D,D);
dot("$C'$",Cp,Cp,blue);
[/asy]](http://latex.artofproblemsolving.com/6/7/f/67fb85536a31cd9f7fc49ff9f62614a94eb94c69.png)
gives a point between 
and let the neighbors of 
on the convex hull be 
in that order for some 
Thus, 
so the faces containing these angles have the same number of sides and are congruent regular polygons. It follows that the convex hull of 
be the centroid of the points in 
;
,
for all 
,
in 
,
,
.
by looking at the perpendicular bisector of 
that 
,
not on the plane, then there is a point 
with side length 
.
on the plane containing 
yet still ``above'' the plane with 
.
,
,
intersects segment 
;
be the centroid of 
centered at 
.
.
inscribed in 
.
has 
.
in 
or 
.
.
is longer than arc 
,
be a subset of 
,
,
be adjacent points appearing in that order, taking cyclics if needed. By the perpendicular bisector of 
,
.
,
which means 
,
.
sends neighbors of 
to neighbors of 
,
is constant, meaning 
