[/quote]
,
have the same incenter 
.
is concyclic iff 
is concurrent
be the nonnegative integers in 
where 
.
.
.
such that 
where 
,
?
where 
,
be a scalene triangle with circumcircle 
and circumcircle 
.
and 
as the midpoints of 
and 
,
to 
.
meets 
,
intersects 
.
at 
and 
at 
.
intersect 
,
intersect 
.
is the reflection of 
and 
lies on the circumcircle of 
.
be real numbers such that 
and 
and determine the equality cases.
be positive real integer such that 
Prove that
such that 
and 
satisfying :![\[n^{2p}=m^2+n^2+p+1\]](http://latex.artofproblemsolving.com/7/4/e/74ee592cb8b3f3a3996ea34f40b78fde9d3672fc.png)
and 
are integers and 
is a prime number.
Approach: I'm trying to factor 
to get 
.
such that 
divides 
.
![$a,b,c \in [-\frac{1}{2}, \frac{1}{2}]$](http://latex.artofproblemsolving.com/5/0/b/50bc691baa4e4bbef868e4799929c8c93d1b6f1a.png)
.
Let ![$a,b,c \in [-2,2]$](http://latex.artofproblemsolving.com/8/0/6/8066d4353091772c401e3e34f5499a8fa2bd9991.png)
.
be positive real numbers such that 
Prove that 
be prime numbers such that their sum isn't divisible by 
.
positive integer quadruples satisfy:
![\[\sum_{k=1}^n\frac{\phi(k)}k\ln\frac{2^k}{2^k-1}\ge 1-\frac 1{2^n}.\]](http://latex.artofproblemsolving.com/3/d/3/3d32207efff8052214e65aca2505c9e4e7bb581c.png)
Y by Adventure10, Mango247
Let there be a scalene triangle 
, and denote 
by the midpoint of 
. The perpendicular bisector of meets the circumcircle of at point 
, on the same side with 
with respect to . Let the incenters of 
and 
be 
, respectively. Let 
, 
, 
. Find 
.
, and denote 
by the midpoint of 
. The perpendicular bisector of meets the circumcircle of at point 
, on the same side with 
with respect to . Let the incenters of 
and 
be 
, respectively. Let 
, 
, 
. Find 
.
and 
triangles according to 
and 
be the reflection of 
.
.
and
.
.
![[asy]
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[/asy]](http://latex.artofproblemsolving.com/7/a/4/7a43daf68cbd0ca2102db8063c842fb9a6abcff5.png)
be the point outside the triangle 
such that 
.
be the reflection of 
and 
.
and 
Therefore, 
and 
.
