[/quote]
,
and 
such that 
and 
is a prime number
and 
Find the value of 
such that 
for all 
be an even integer and let 
be real numbers satisfying 
.
be a positive integer. Ana and Banana play a game. Banana thinks of a function 
and a prime number 
.
is nonconstant, 
,
for all integers 
.
.
in order. Ana wins if she can determine the value of 
Let 
be positive reals such that 
.
of positive integers is called central if for every positive integer 
terms of the sequence is equal to 
of positive integers such that for every central sequence 
there are infinitely many positive integers 
.
Prove that
be a set of 
points in the plane, no three of which are collinear. Initially these points are connected with 
and 
sharing a common interior point and replace them by the segments 
and 
if none of them is present at this moment. Prove that it is impossible to perform 
or more such moves.
is the center of the inscribed circle and holds 
.
.
and 
.
and 
.
and 
intersect the circumcircle of 
and 
,
and 
such that 
and 
are parallelograms (with 
,
)
be the point of intersection of lines 
and 
.
,
,
be all positive integers smaller than 
such that 
for all 
is the largest positive integer that divides both 
and 
.
.
Y by The-X-squared-factor, Adventure10, Mango247
Let 
be a non-equilateral triangle with integer sides. Let 
and 
be respectively the mid-points of 
and 
; let 
be the centroid of 
. Suppose, , 
, , are concyclic. Find the least possible perimeter of .
be a non-equilateral triangle with integer sides. Let 
and 
be respectively the mid-points of 
and 
; let 
be the centroid of 
. Suppose, , 
, , are concyclic. Find the least possible perimeter of .This post has been edited 2 times. Last edited by integrated_JRC, Jan 22, 2018, 2:10 AM
Let 
,
;
and 
are tangent to 
.
,
,
,
.
.
is for 
.
leads to 
,
leads to 
,
.
![\[1 = \frac{BG\cdot BE}{BD\cdot BC} = \frac{\frac23 BE^2}{\frac12 BC^2} \implies \frac{BE}{BC}= \frac{\sqrt3}2\]](http://latex.artofproblemsolving.com/b/2/4/b245dbe53e1711a3e15c1b320a1f6428d4af1cf1.png)
Let 
,![\[ a^2 + c^2 = 2(BE^2 + EC^2) = \frac32 a^2 + \frac12 b^2 \iff a^2 + b^2 = 2c^2\]](http://latex.artofproblemsolving.com/a/a/0/aa0b88ce32510cd6116f81175ca96a22921bb15c.png)
![\[a^2+b^2=2c^2 \iff (a-b)^2+(a+b)^2 = (2c)^2 \]](http://latex.artofproblemsolving.com/9/c/8/9c8d74eabdedcd4c106d8405ddb7a848217e90ae.png)
WLOG 
![\[3,4,5 \implies a-b = 3, a+b = 4, 2c = 5 \implies a = \frac72, b=\frac12, c=\frac52\]](http://latex.artofproblemsolving.com/4/5/6/456548ee3ea5bbde350fa7643a88a006bb9a5227.png)
![\[\text{But } a + c = \frac62 < \frac72 = b \]](http://latex.artofproblemsolving.com/a/9/9/a995285fdda944fb4568649df53fa07da75db6ba.png)
Therefore, this can not be a solution![\[(a-b,a+b, 2c) = (5, 12, 13) \implies a = \frac{17}2, b=\frac72, c=\frac{13}2 \]](http://latex.artofproblemsolving.com/3/3/6/336ed59ed6fad4502d6009ef197de50772c625e5.png)
![\[\text{and }b + c = 10 > \frac{17}2 = a\]](http://latex.artofproblemsolving.com/d/d/f/ddfc480f280abd1fcc8293087498475ad09ffd01.png)
Therefore 
Thus, the smallest value of 
.
.
is a valid solution with the smallest value of 
)
is required for least perimeter. The triplet 
was also possible
I read the article about the 
points by anantmudgal09 but i did not see this fact.... can you suggest an article with all the properties of 
in 5 to 10 minutes by a barycentric bash.
.
.
(
we get 
and 
.
.
.
.
,
are concyclic applying ptolemy's theorem get,
.
.
.
.
.
.
.
.
is obtuse.
.
.
.
)
.
with 
.
.
so 
which is not possible since 
are sides of triangles. The next pythagorean triplet is 
,
to get 
and hence, the minimal perimeter is 
.
and 
since 
is not a triangle but a straight line..
.
we have 
and they simply prove our claim.
and we know 
must be pythagorean.
.
,
and 
and 
is less than 
.
so least possible perimeter of 
.
![\[a^2 + b^2 = 2c^2\]](http://latex.artofproblemsolving.com/0/d/c/0dc6b76de74c1c4317f5c441e97d7d282e67b9e8.png)
Now we just have to check for pythagorean triplets and we find the smallest solution to be 
