[/quote]
,
and 
Prove that
Let 
and 
Prove that
which satisfy 
![$ \frac{a+b+c}{3} \ge \sqrt[10]{\frac{a^3+b^3+c^3}{3}}$](http://latex.artofproblemsolving.com/c/d/a/cdae05ad34239366b0dd3921243522c711b6ae58.png)
method, I can assum 
![$ \frac{2x+\frac{1}{x^2}}{3} \ge \sqrt[10]{ \frac{2x^3+ \frac{1}{x^6}}{3}} $](http://latex.artofproblemsolving.com/c/d/4/cd481672bee1e87d8c97ac623436e0ec75cfa73a.png)
are given such that 
.
in natural numbers m and n.
,
denote the smallest possible value of 
where 
are sets such that 
and 
whenever 
.
and 
intersect at 
.
at 
respectively. Let 
be the symmetric points of 
intersects circles 
respectively which are distinct from 
and 
are equal.
,
.
at 
.
cuts segment 
at 
respectively. If 
and 
,
equals 
.
are points on a circle 
,
,
.
at 
.
is on 
such that 
.
intersects 
.
is the reflection of 
wrt 
.
lies on the same circle.
in its exterio
so that 
,
,
.
and
and 
.
be the points on 
and 
,
and 
.
and 
is on the circumference of triangle 
has a right angle at 
.
such that 
and 
lies between 
.
is chosen so that 
and 
.
is chosen so that 
and 
is the bisector of 
.
be the point such that 
is a parallelogram. Prove that 
and 
are concurrent.
such that the number of 
which satisfy the following condition is minimal:
has periodic 
and 
are tangent to the circumcircle of 
be the reflection of 
be the reflection of 
is the circumcenter of 
,
is perpendicular to 
be a positive integer. Lexi has a dictionary 
consisting of some 
grid so that each column contains a string from 
such that if 
Y by Adventure10, Mango247
Let 
be an acuteangled triangle with 
inscribed in a circle. The perpendicular bisector of the side 
cuts arc 
containing 
in 
Let 
be a point on the segment 
such that 
Prove that the circumcircle of the triangle 
cuts 
in its midpoint.
be an acuteangled triangle with 
inscribed in a circle. The perpendicular bisector of the side 
cuts arc 
containing 
in 
Let 
be a point on the segment 
such that 
Prove that the circumcircle of the triangle 
cuts 
in its midpoint.
and 
since 
and 
with 
and 
since 
are concyclic 
through midpoint of 
the midpoint of 
and 
.
cyclic
,
is cyclic, Since 
to 
by observing that 
is a cyclic pentagon and we are done.
