Point in a triangle 2

[onta...@hotmail.com]

ABC is a triangle with a = BC, b = AC and c = AB.
It is the base of a pyramid which has D as its apex.
Let d = AD, e = BD and f = CD.
A line passes downward perpendicular to the base from D to a point P at the base.
Find expressions for the magnitudes of the polar coordinates PA, PB and PC in terms a,b,c,d,e and f.

[Narasimham]

On Apr 21, 10:22 am, "ontad...@hotmail.com" <ontad...@hotmail.com>
wrote:

Suggested approach is to use three relations:

Height = 3*V/FaceArea---- (1)

where the Cayley-Menger determinant determines Volume V :

288 V^2 = {{0,1,1,1,1},{1,0,C,B,D},{1,C,0,A,E},{1,B,A,0,F},{1,D,E,F,
0}};
(A = a^2, B = b^2, ..., F = f^2); ---- (2)

and FaceArea^2 = s (s-a) (s-b) (s-c);2 s = a+b+c.----(3) so that we
get tetrahedron Height. And now

PA = Sqrt[d^2 - Height^2] ----(4) , two similar required relations for
polar radii PB and PC follow.

Narasimham

[onta...@hotmail.com]

An interesting solution
Sketch the base triangle ABC and attach three triangles to it - in plan view. ABD',BCD'' and ACD'''. Lines can now be drawn perpendicular to AB,BC and CA through the vertices D and to intersect at point P on the base. One of these lines crosses AB at G and another across BC at F.
Let FB = u and GB = v
u = (a^2+e^2-d^2)/(2*a)
v = (c^2+e^2-f^2)/(2*c)
w = cos(ABC) = (a^2+c^2-b^2)/(2*a*c)
PB = SQRT(v^2+u^2-2*v*u*w)/SQRT(1-w^2)
h = SQRT(e^2-PB^2)
PA = SQRT(f^2-h^2) and
PC = SQRT(d^2-h^2)