Let ABC be any triangle. Let D be a point on side AB and let E be a point on side AC. Draw lines CD and BE and call their intersection F. Triangle ABC is then divided into three smaller triangles BDF, CEF, BCF and a quadrilateral ADFE.

Let the area of BDF equal r, the area of CEF equal q, and the area of BCF equal p. Express the area of the whole triangle ABC in terms of p, q, and r.

the approach I used involved some rather nasty equations so I resorted to using Mathematica to solve them.  I hope someone is able to find a more elegant solution than my own, but till then here is mine.

I started by placing A at (0,0), B at (x,0) and C at (y,z) then D is at point (b,zb/y) (on AB), and E is at (a,0) (on AC).  So using these I found the coordinates for F to be

( (axy+bxy-abx-aby)/(ab-xy) , (bxz-abz)/(ab-xy) )

using these points I then calculated the areas for BDF, CEF, and BCF in terms of x,y,z,a,b.  I then set them equal to r,q,p respectively and used Mathematica to solve for x,y,z.

I then had x,y,z in terms of a,p,q,r (b term was eliminated in the solution)

now the area of ABC is simply xz/2 and fortunately in my solution above x=f(p,q,r)*a and z=q/a so when you take xz the a term is eliminated and we get the following equation for the area of ABC in terms of p,q,r

area=

[(2q*r^(3/2)-2qr^(1/2))*sqrt(9qr-8pr-8pq+8p^2)+6q^2r^2-10pq^2r+4p^2q^2]/[4qr^2-(5q^2-4pq)r+4pq^2-q^(3/2)*r^(1/2)*sqrt(9qr-8pr-8pq+8p^2)]

now I have a feeling that this is able to be simplified further but I am unable to see how at the moment

Posted by Daniel on 2008-11-29 05:58:58