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denote the set of positive rational numbers. Find, with proof, all functions 
such that, for all positive rational numbers 
and 
we have ![\[f(x)=f(x+y)+f(x+x^2f(y)).\]](http://latex.artofproblemsolving.com/9/e/a/9ea71d2ee277dc3c689ba2d791b451e29353edb3.png)
+1 w
be a triangle with circumcenter 
. Let 
be a point on the perpendicular bisector of 
(see figure) and 
, 
be the intersections of the perpendicular bisectors of 
and 
, respectively, with 
and 
. Prove that the midpoints of 
and 
and the point are collinear.
is given. Let 
be set of all 
segments of real line of type ![$[i, j]$](http://latex.artofproblemsolving.com/a/6/4/a644edd67caa59eaded765c417b6d9822f782ec2.png)
, where 
and 
are integers, 
. A subset 
is said to be valuable if the intersection of any two segments from 
is either empty, or is a segment of nonzero length belonging to . Find the number of valuable subsets of set .
are there such that 
and 
?
be real numbers such that 
Prove that
Let be real numbers such that 
Prove that
and 
, then we can deduce 
\le 2 $ ?
satisfying 
. Prove that:![\[
0 \le (x^2 + 1)(y^2 + 1) + 4(x - 1)(y - 1) \le 164.
\]](http://latex.artofproblemsolving.com/4/2/d/42d0b88388462f16dd99d928b94024e689ba7c2e.png)
be a sequence defined by 
. Then the product of four consecutive terms in can be written as the product of two terms in . Find 
if 
.
and 
be nonintersecting chords of a circle and let 
be a point on . Construct (with straightedge and compass) a point on the circle such that is the midpoint of the part of segment lying inside triangle 
.
and be nonintersecting chords of a circle and let be a point on . Construct (with straightedge and compass) a point on the circle such that is the midpoint of the part of segment lying inside triangle .
appear in this order. Let the line through perpendicular to meet the perpendicular bisector of at . The circle passing through 
centered at be 
. 
and polar of 
wrt be 
. Line meet arc (not containing ) at 
. Tangent at meet at 
. Circle centered at with radius 
meet 
again at 
. 
meet again at . 
and 
.
