and 
.![\[ab+bc+ac\leqslant 27\]](http://latex.artofproblemsolving.com/c/b/5/cb5e2a1efd462abd2354f4136f5fa64408f3138c.png)
of degree 
has exactly 
real roots, counted with multiplicity. We know that the coefficient of 
is 
,
.
take?
and 
.
.
.
satisfy,
Where 
}
Y by Adventure10, Mango247
In the triangle 
, the respective mid points of the sides 
, 
and 
are 
, 
and 
. Let 
be the point where the internal bisector of the angle 
intersects the side , and 
the point where the internal bisector of the angle 
intersects the side . Also, let 
be the intersection point of 
and 
, 
the intersection point of and 
, and 
the intersection point of and 
. Prove that 
.
, the respective mid points of the sides 
, 
and 
are 
, 
and 
. Let 
be the point where the internal bisector of the angle 
intersects the side , and 
the point where the internal bisector of the angle 
intersects the side . Also, let 
be the intersection point of 
and 
, 
the intersection point of and 
, and 
the intersection point of and 
. Prove that 
.
.
Multiplying 
,
and 
we get 
be a point on 
.
as the midpoint of 
,
,
.
and also 
.
.
be the infinity point on 
(here 
)
and 
and 
then The intersection of 
and 
![[asy]
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[/asy]](http://latex.artofproblemsolving.com/1/3/8/1383265b95fd451ff8e421eca1a3406e3bc89a57.png)
so we have ![\[A:= (1, 0, 0), \ B:= (0, 1, 0), \ C:= (0, 0, 1), \ D:= (0, \tfrac{1}{2}, \tfrac{1}{2}), \ E:= (\tfrac{1}{2}, \tfrac{1}{2}, 0), \ F:=(\tfrac{1}{2}, 0, \tfrac{1}{2}).\]](http://latex.artofproblemsolving.com/2/9/d/29d39572fba4ffc9ad51cd98417ae6a2018c1509.png)
From Stewart's Theorem we can easily compute 
.
.
and 
.
.
and 
.
is 
so we have ![\[PR^2=\frac{k^2}{4}=AD^2,\]](http://latex.artofproblemsolving.com/1/c/1/1c10e2d070b269f62def68857e570823ee53911a.png)
as desired.
