AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
have the same incenter 
.Prove that 
is concyclic iff 
is concurrent
be the nonnegative integers in 
where 
. Define 
. Prove that 
.
such that 
where 
, and the statement is still true when 
? such that 
where 
, and the statement is still true when ?.
be a scalene triangle with circumcircle 
and circumcircle 
. Denote 
and 
as the midpoints of 
and 
, respectively. The altitude from 
to intersects again at 
. The line 
meets a second time at 
, and 
intersects at 
. The perpendicular bisector of intersects 
at 
and 
at 
. Let the circumcircle of 
intersect again at 
, and let the line 
intersect at 
.
is the reflection of over , prove that the intersection 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 
. be a positive integer and 
be positive real numbers such that 
Prove that 
be prime numbers such that their sum isn't divisible by 
. Find the all 
positive integer quadruples satisfy:
, let 
be the midpoint of the median corresponding to vertex 
and 
the intersection point of 
with the side 
. Show that 
.
, let be the midpoint of the median corresponding to vertex and the intersection point of with the side . Show that .
, 
, and 
Then, the midpoint 
of 
has barycentric coordinates 
so the barycentric coordinates of are 
It follows that the equation of the line is 
where 
We also have that the equation of line is 
and so the second entry of the barycentric coordinates of is 
. Looking back at the equation of , this means that 
and so the barycentric coordinates of are 
and so it follows that 
which implies the desired.
![$a,b,c \in [-\frac{1}{2}, \frac{1}{2}]$](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)
.
![\[\sum_{k=1}^n\frac{\phi(k)}k\ln\frac{2^k}{2^k-1}\ge 1-\frac 1{2^n}.\]](http://latex.artofproblemsolving.com/3/d/3/3d32207efff8052214e65aca2505c9e4e7bb581c.png)
![$[ASB] =[BSD]= [CSD] =k$](http://latex.artofproblemsolving.com/1/d/d/1dd25b3381584a3f8126bc2899e79d4985fb806c.png)
,![$[.]$](http://latex.artofproblemsolving.com/0/6/0/06030c213549a6aaae52b1daf08720ec217c42a3.png)
denotes the area of the figure. Now ![$\frac{AQ}{QC} =\frac{[ASB]}{BSC]} = \frac{k}{2k} =\frac{1}{2}$](http://latex.artofproblemsolving.com/a/8/f/a8f0d16c7c7e8bdcc39e0420e48e74305052706b.png)
.![$[ASC] =k$](http://latex.artofproblemsolving.com/3/e/7/3e70ceb7b03e0b5d69f36bdffc90d81e9799aa4d.png)
.![$[ASQ] =\frac{k}{3}$](http://latex.artofproblemsolving.com/5/a/a/5aaa6b4c947eab22b107ad71ec7d09da938da420.png)
.![$\frac{BS}{SQ}=\frac{[ASC]}{[ASQ]
} =3$](http://latex.artofproblemsolving.com/2/b/b/2bbf4aed90aaba403e1f55a51d6514cfa912262b.png)
, 
.