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.
. There after each term is derived from the previous term by substituting 
for and for 
Thus the first few terms are , , 
,
?
is called frameable if it is possible to draw a regular polygon with 
sides all whose vertices lie on these lines, and no line contains more than one vertex of the polygon.
are frameable.
is not frameable.
is frameable.
, determine all polynomials 
with integer coefficients such that 
divides 
for every positive integer .
with the smaller base 
are constructed the squares 
and 
. Prove that the perpendicular bisector of passes through the midpoint of 
.
such that
for all real numbers 
satisfying the equation
has circumcenter 
. The lines 
and 
intersect at 
, while 
and 
intersect at 
and 
and 
intersect at 
. Show that if the triangles and 
are similar(with vertices in that order), than is equilateral.
candidates and judges, where 
is an odd integer. Each candidate is evaluated by each judge as either pass or fail. Suppose that each pair of judges agrees on at most candidates. Prove that ![\[{\frac{k}{m}} \geq {\frac{n-1}{2n}}. \]](http://latex.artofproblemsolving.com/3/8/7/387c8824b28f6422306e3c25b34fccb0a841f4cd.png)
be a triangle. Let 
be the feet of the altitudes from 
respectively. Let 
be the projections of onto line 
. Show that 
.
have the property that 
is a perfect square for infinitely many positive integers 
?
is the common point of a straight edge and a fold line. Construct a perpendicular to this line at point . Do this without drawing tools, but only by bending the paper.
is the common point of a straight edge and a fold line. Construct a perpendicular to this line at point . Do this without drawing tools, but only by bending the paper.
.
?
