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.
, point 
is the midpoint of 
, 
the point of intersection of the tangents at points 
and 
of the circumscribed circle of , 
is the midpoint of the segment 
. The segment 
meets the circumcircle at the point 
. Prove that 
.
Prove that: 
is the sum of 
square numbers
Let 
and 
is the sum of square numbers. Prove that: 
is the sum of square numbers
such that 
that satisfy the equation
with incenter 
. The incircle of triangle is tangent to at 
. Let and be points on the side BC such that 
and 
, respectively. Let 
and 
be the incenter of triangles 
and 
, respectively. Prove that 
is the Euler line of triangle 
.
, , 
, 
, where and 
are odd primes and![\[2^ap^b=(p+2)^c-1.\]](http://latex.artofproblemsolving.com/5/8/5/58518a8ebed14836b72af743ff100eae2efba5e2.png)
we have 
.
.
be reals such that 
and 
Prove that
and intersect at the midpoint of arc of the circumcircle of 
.
is the intersection of two perpendicular bisectors of and .
and 
.
and 
are cyclic. transform circle to circle , transform to , 
to . to midpoint 
and 
are similar
and then 
. Calculate 
and 
. It's well known that 
so that the displacement vector 
is 
. Displacement vector 
is 
. Then it's clear that the vectors are multiples of each other.
(
lies on 
, 
such that 
)
, 
, is a rhombus, so lies on angle bisector of 
and result follow
:
Where 
,![\[\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}\geq 4+\frac{8(a-c)^2}{(a+b+c+d)^2}.\]](http://latex.artofproblemsolving.com/7/b/6/7b6e32d32d89ca058a06974e7276f0a5874d90a7.png)
and 
