Triangle area problem

Leon

Let A', B', C' be three points in the interior of a triangle ABC such that:

(1) A, B' and A' are collinear, with B' between A and A'.
(2) B, C' and B' are collinear, with C' between B and B'.
(3) C, A' and C' are collinear, with A' between C and C'.
(4) the four triangles AA'C, CC'B, BB'A, A'B'C' all have equal areas

If AB=c, BC=a, CA=b show that
$AA' = \frac{1}{4}\sqrt {\frac{{\left( {5 + \sqrt 5 } \right)}}{2}\left( {2a^2 + b^2 - c^2 } \right) + \sqrt 5 a^2 + c^2 }$

**Posted:** Wed Jan 26, 2005 2:15 pm

fleeting_guest · Yang-Mills Theory Offline Joined: 14 Dec 2004 Posts: 903

The linear transformation of this configuration onto an equilateral triangle should produce a symmetrical (invariant under 120 degree rotation) configuration. From this you can calculate the barycentric coordinates of A', B', C'. Then $|AA'|^2$ is a simple calculation with vectors and the Cosine Law in triangles.

**Posted:** Wed Jan 26, 2005 4:01 pm

Leon

**Posted:** Thu Jan 27, 2005 10:41 am

Myth

I think, fleeting_guest didn't perform calulations. It was just a plan.

**Posted:** Thu Jan 27, 2005 11:02 am

_________________
Myth is out of here

fleeting_guest · Yang-Mills Theory Offline Joined: 14 Dec 2004 Posts: 903

**Posted:** Thu Jan 27, 2005 5:46 pm