[/quote]
,
of positive integers satisfying the following two conditions:
,
is in 
,
and 
;
such that 
is not divisible by 
.
?
then
where 
Also, 
and 
is largest real root of the equation
then 
Equality holds iff 
and any cyclic permutations.
be a triangle, and let 
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 
If there are multiple such events, delete the event with the minimal weight.
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)
Y by Adventure10
The points 
,
,
are middle points of the arcs 
of the circumscribed circle of 
, respectively. The points 
are the reflections in the middle points of 
of the center 
of the inscribed circle in the triangle. Prove that 
, and 
are concurrent.
,
,
are middle points of the arcs 
of the circumscribed circle of 
, respectively. The points 
are the reflections in the middle points of 
of the center 
of the inscribed circle in the triangle. Prove that 
, and 
are concurrent.
as reference triangle.Then trivially we get 
.
,
,
are the reflections in the middle points of 
,
are concurrent.
be the medial triangle.Let 
be midpoints of 
respectively.Clearly 
via mid-point theorm.Hence after a Homothety at 
.
line through 
meets 
again at 
,
and 
are isogonal wrt 
or 
then 
will concurrent at a point on 
be the midpoint of 
symmetry with 
so done
