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 ![\[\left(g(m)+n\right)\left(g(n)+m\right)\]](http://latex.artofproblemsolving.com/5/a/c/5ac9fce315e8a524a3a01b345cfe76bc76056d68.png)
is a perfect square for all 
be an acute triangle inscribed in a circle of center 
. If the altitudes 
intersect at 
and the circumcenter of 
is 
, prove that 
is a parallelogram.
is a perfect square number.
, 
and 
are the midpoints of sides 
and 
respectively. Point 
lies inside the triangle. Let 
. Prove that 
.
is good if there exists a permutation 
of 
such that 
is perfect square for all 
. Determine all the good numbers in the set 
.
of positive integers such that 
and 
are both powers of 
.
the set of all positive real numbers.
which have the property:![\[f(x)f(y)=2f(x+yf(x))\]](http://latex.artofproblemsolving.com/4/f/f/4ffbde35ff082585a188c1a4368d39338d1b07a4.png)
and 
.
be a function that satisfies the following equation for all real numbers 
:![\[
f(x^2 + x + 3) + 2f(x^2 - 3x + 5) = 6x^2 - 10x + 17.
\]](http://latex.artofproblemsolving.com/f/6/2/f625b922656061221157fdb5a3b958cfa7d57108.png)
Find the value of 
.
and 
such that does not divide 
for all positive integers .
consecutive even numbers, such that the square of their product is divisible by the sum of their squares.
consecutive even numbers, such that the square of their product divides the sum of their squares.
is a triangle, let 
is orthocenter of 
, let 
is midpoint of 
. Let 
is a line perpendicular with 
at point . Let meet 
at 
respectively. Prove that 
.
be, the orthogonal projections of 
respectively, on the line segment 
and we denote the points 
from 
and 
and then, the points 
and 
are the reflexions of respectively, with respect to 
with 
as their orthocenters respectively and so, their distances from their homologous vertices, are equal.
and the proof is completed.
is a chord of a circle 
and is a point on 
Two distintc chords 
and 
pass through 
and 
cut at 
respectively. Then we have![\[ \frac{1}{PA}-\frac{1}{PX}=\frac{1}{PB}-\frac{1}{PY} \]](http://latex.artofproblemsolving.com/0/f/6/0f6e88f2cc6b36865336290a1ea0bfab7c9359f6.png)
the diametrically oposite point of 
in the circumcircle of 
. 
are collinear since by angle chasing 
is a parallelogram. Also, 
and 
are both cyclic (opposite angles equal 
, sum 
. Since 
this implies is angle bisector of 
then 
.
the diametrically oposite point of in the circumcircle of . are collinear since by angle chasing is a parallelogram. Also, and are both cyclic (opposite angles equal 
, sum . Since 
this implies is angle bisector of then 
.
and 
.![$\sqrt[3]{ab-1}+\sqrt[3]{bc-1}+\sqrt[3]{ca-1} \le \sqrt[3]{(a+b+c)^2}$](http://latex.artofproblemsolving.com/9/6/3/9633b5849794285e44519c745c1f616eceb77711.png)
.
Let 
.
cut 
at 
and 
cut 
at 
.
so the circle from Butterfly Theorem is of center 
.
and 
.
perpendicular to (d), we have that 
.
and 
,
.
