[/quote]
,
has a right angle at 
.
be the point on line 
such that 
and 
lies between 
.
is chosen so that 
and 
is the bisector of 
.
is chosen so that 
and 
is the bisector of 
.
be the midpoint of 
be the point such that 
is a parallelogram. Prove that 
and 
are concurrent.
and 
Prove that
Let 
Prove that
.
the number of changes of signs in the
Then prove that
.
epic if for any four different positive integers 
,
such that 
can be partitioned into 
be positive real numbers such that 
.
![\[\frac {x^{2}}{\left(x - 1\right)^{2}} + \frac {y^{2}}{\left(y - 1\right)^{2}} + \frac {z^{2}}{\left(z - 1\right)^{2}} \geq 1\]](http://latex.artofproblemsolving.com/f/a/6/fa6c46c12e7e6bf0b04f6f9a910df6c2b452fe6a.png)
for all real numbers 
,
,
,
,
.
be a prime and suppose 
is prime. Prove that 
is prim
are positive real numbers such that 
prove that
girls on one side of a tennis table, and a queue of 
to 
grid. We color the squares with three colors: blue, red, and yellow. Two squares are defined as opposite if they share a vertex but not an edge. A valid coloring is a coloring such that no two squares that are red and blue are opposite each other. Count the number of valid colorings.
,
and 
.
be the intersection of 
and 
and 
and 
.
is a cyclic quadrilateral prove that 
,
and 
,
Y by
For given positive integers 
and 
, let us consider the equation
denotes the greatest common divisor of positive integers 
and 
.)
and 
, let us consider the equation
- For
and 
, find the least positive integer 
satisfying this equation. - Prove that for any positive integers and , there exist infinitely many positive integers satisfying this equation.
denotes the greatest common divisor of positive integers 
and 
.)
.
.
,
.
.
satisfies the equation for any positive integer 
,
.
.
is either 
or 
and 
,
and hence 
.
and 
,
.
for some positive integer ![\[
a + \gcd(b,x) = a + b = \gcd(a,x) + b.
\]](http://latex.artofproblemsolving.com/b/6/b/b6bafa07f1d0d916f7d9574612736cf4f743c793.png)
Since there are infinitely many choices of 
W
.
so since the divisors of 
we must have 
,
.
so 
.
,
works, so 
is the smallest 
J
for any 
