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.
'is a perfect square.
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
, the point 
is the midpoint of the diagonal 
, 
is the midpoint of side 
. Prove that triangle 
is equilateral.
![[asy]draw((0,0)--(10,0)--(15,8.66)--(10,17.32)--(0,17.32)--(-5,8.66)--cycle);
label("A",(0,0),SW);
label("B",(10,0),SE);
label("C",(15.8,8.66),E);
label("D",(10,17.32),NE);
label("E",(0,17.32),NW);
label("F",(-5,8.66),W);
draw((0,0)--(15,8.66));
dot((7.9,4.33));
label("M",(7.9,4.33),NW);
dot((5,17.32));
label("N",(5,17.32),N);
draw((5,17.32)--(7.9,4.33)--(-5,8.66)--cycle);[/asy]](http://latex.artofproblemsolving.com/0/4/3/0430965d229ab82827b9ff53cea876748ea90c97.png)
. Examine 
. Law of cosines gives: 
Next, we examine 
. Note that 
. Notice that 
. Thus 
.
and 
. Then, 
Since we know the sum of the angles is less than 
, the angles sum to 
. This means that 
, so by SAS, 
is equilateral.
be the center of the hexagon. Clearly, 
and 
are equilateral. Furthermore, is the midpoint of 
, so we are done by the gliding principal.
Prove that
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 
.
Let 
.
