,
.
be 
g
on the circle different than 
is the greatest number on at least one of the two arcs 
.
such that regardless of the values of 
are written on a board. 
and 
plays the following game: They take turns choosing a number from the board and deleting them. 
and 
be prime numbers. Prove that if 
is a perfect square, then 
is also a perfect square.
is a fifth degree polynomial. 
,
,
,
.
is differentiable and 
for all 
.
,
.
a prime number, with 
and 
,
vector space.
is a group?
and the distribution function is 
then
and find the expression of 
![\[
\int_0^1 \int_{x^2}^x (x^2 + y^2)^{-1/2} \, dy \, dx
\]](http://latex.artofproblemsolving.com/e/a/9/ea9f3a093b1d4c0fa39e5b6d42afa2f71dace1e0.png)
has no roots in the disc of unity.
Y by
Let 
be fixed points on this order on a line. Let 
be a variable circle through 
and 
and suppose it meets the perpendicular bisector of 
at the points 
and 
. Let 
and 
be the other points of intersection of 
and 
with . Prove that 
passes through a fixed point independent of .
be fixed points on this order on a line. Let 
be a variable circle through 
and 
and suppose it meets the perpendicular bisector of 
at the points 
and 
. Let 
and 
be the other points of intersection of 
and 
with . Prove that 
passes through a fixed point independent of .
be the midpoint of 
be the intersection of 
,
,
are reciprocal pairs. In other words, there exists an inversion on the line 
passing through the points 
that maps the following points to each other: ![\[ A \leftrightarrow B, \quad C\leftrightarrow D, \quad N \leftrightarrow M. \]](http://latex.artofproblemsolving.com/5/f/e/5fee2568d0598577a0a88872ffc96a2267dfc55a.png)
Note that defining an inversion on a line only requires two pairs of points. So 
a
is constant so 
,
is fixed. Consider two circles 
and 
.
is cyclic, and a quick angle chase gives that 
is also cyclic.
,
and 
is fixed.
and 
are cyclic. Let 
.
,
as a variable, this gives us a linear equation for 
