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.
such that for all 
,![\[f(x+yf(x))+y = xy + f(x+y).\]](http://latex.artofproblemsolving.com/c/f/3/cf3d20a041c27244e90876119b4568b7a3e13c03.png)
+1 w
such that for any integer 
there exists an integer 
for which 
is divisible by .
are distinct prime numbers which satisfy
for natural number 
. Find all values of .
, 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 
.
+1 w, , 
, 
, where and 
are odd primes and![\[2^ap^b=(p+2)^c-1.\]](http://latex.artofproblemsolving.com/5/8/5/58518a8ebed14836b72af743ff100eae2efba5e2.png)
lies onto Carnot circles, i.e. symmetrical of the circle 
w.r.t. its sides. The circles 
and 
being equal, there is a translation mapping them, and this translation is perpendicular onto 
, i.e. 
maps . Since 
, the circumcenter will go to the midpoint of the arc not containing , so 
and 
.
why?
,
.
is cyclic? reflected in 
. These three circles coincide at 
. It's easy to see why as 
where 
extended meets 's circumcircle at 
.
and 
are congruent, the translation - preserving size - from to is perpendicular to the radical axis and so is translated onto the line 
.
is cyclic, regardless here is one argument: be the point such that 
is a rhombus. Then taking 
as the origin, 
. So 
is a parallelogram, so 
so 
is isosceles, as is 
. So 
so is the centre of . But also is a rhombus so 
so lies on 
. is cyclic means 
so 
.
being the unit circle on the complex plane, then 
. But 
.
of the acute triangle 
is equidistant from the circumcenter and the orthocenter 
Determine all possible values for the measure of angle 
(
donates the exradius opposite to vertex )
which implies 
, where 
is the foot of the altitude from , is 
, we have that 
. Setting this equal to 
, we have 
, so 
. 
is literally an exercise but fine be the origin. Then, 
is equivalent to 
, or
where 
is implied from 
. Expanding, substituting 
, and dividing out 
yields 
, or 
(from the acute triangle condition). Consequently, 
.
:
be reals such that 
and 
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)
where 
is the midpoint of 
then 
(if its 
then 
xD) then 
and 
because 
meet 
in 
and 
.
(
the triangles 
and 
are congruent
.
(since 
)
is equilatera
or 
.
or 
.
.