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
, then 
, hence 
, or 
.
, with on 
, 
, 
the midpoint of 
, take the symmetrical of about and get 
, but is equivalent to 
, or 
, but 
, hence 
, or 
; from the way of construction, it's obvious that the problem is iff type.
:
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)
has an obtuse angle at 
.
meets 
at 
,
.
if and only if 
.
then let 
on ray 
so 
and 
so 
be the circle through 
with center 
so 
at 
so 
is an inscribed quadrilateral
