[/quote]
,
be a triangle, and let 
and 
be points on segment 
symmetric with respect to the midpoint of 
denote the circle passing through 
at 
.
denote the circle passing through 
at 
.
)
.
and 
,
[list=a]
and 
,
satisfying this equation.
denotes the greatest common divisor of positive integers 
and 
.
that are each continuous and last on a certain time interval. Each event has a weight 
However, one can only choose to attend activities that do not overlap with each other. The goal is to maximize the sum of weights of all activities attended. Prove or disprove that the following algorithm allows for an optimal selection:
consider 
the sum of 
over all 
such that 
and 
spheres centered at 
,
and 
and 
,
points 
outside of 
as the number of sets 
such that the midpoint of 
lies entirely inside of 
for 
Thus the first few terms are 
,
is called frameable if it is possible to draw a regular polygon with 
are frameable.
is not frameable.
is frameable.
,
with integer coefficients such that 
divides 
for every positive integer 
with the smaller base 
and 
.
.
such that
for all real numbers 
satisfying the equation
?
has circumcenter 
.
and 
intersect at 
,
and 
intersect at 
and 
and 
intersect at 
.
are similar(with vertices in that order), than 
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 ![\[{\frac{k}{m}} \geq {\frac{n-1}{2n}}. \]](http://latex.artofproblemsolving.com/3/8/7/387c8824b28f6422306e3c25b34fccb0a841f4cd.png)
be the feet of the altitudes from 
respectively. Let 
be the projections of 
.
.
is a perfect square for infinitely many positive integers 
?
Y by Adventure10, Mango247
Let 
be an isosceles trapezium with 
and 
The parallel to 
through 
meets the perpendicular to through 
in point 
The line through 
drawn which is parallel to 
meets the perpendicular to through in point 
Prove that points 
and 
lie on a common circle.
be an isosceles trapezium with 
and 
The parallel to 
through 
meets the perpendicular to through 
in point 
The line through 
drawn which is parallel to 
meets the perpendicular to through in point 
Prove that points 
and 
lie on a common circle.This post has been edited 1 time. Last edited by orl, Jan 3, 2009, 12:01 PM
.
be the intersection point of 
and 
and 
the intersection point of 
and 
.
is inscribed (1). We have 
and 
hence 
=> 
which hence 
is inscribed which combined with (1) gives solution.
