AoPS Academy
be a function such that for every regular 
-gon 
we have 
. Prove that 
for all reals 
.
be a triangle (
). 
are the altitudes of the triangle. The bisector of 
intersects 
at 
. Let 
be a point such that 
and 
. Prove that 
passes through the midpoint of 
.
with 
and at most one of them being zero, show that![\[\frac{1}{\max(a^2,b)}+\frac{1}{\max(b^2,c)}+\frac{1}{\max(c^2,a)}\geq\frac{27}{4}\]](http://latex.artofproblemsolving.com/9/e/1/9e14a9ef2d1a0a98006bcf716fa33c3d9ea29bd5.png)
be a triangle and let denote the midpoint of the side . Suppose that there exist two points 
and 
interior to the side 
and 
respectively, such that 
where 
is the intersection point of the lines 
and . Show that 
.
be defined such that 
and 
. Since 
we have that 
. Let 
be a point on 
such that 
and 
. This implies 
so we have that 
(
and 
so 
is a parallelogram) and 
. Take 
to be the midpoint of . We have that 
and 
, which implies that 
. From 
we have that 
where 
is the foot of the 
altitude. Now we have that 
so 
which implies that 
.
which sends to 
, to 
and to 
. This would preserve the length ratios, and thus 
would be another set of points that satisfy this condition.
. Then the median cuts the segment 
in the ratio 
, hence is also a median, and since 
, we have 
, and 
, so 
as required.
& 
.
