Given a rectangle ABCD with |AB|=a and |BC|=b. There are two distinct equilateral triangles AEF and AGH with points E and G on line BC and points F and H on line CD. What is the product of the areas of the two triangles in terms of a and b?

(In reply to re: Answer by Charlie)

I said "don't ask" because I consider the method I used as being very complicated and not very rigorous.  However, since it is you asking, Charlie, I will at least give an outline.

For fixed a and b, a typical triangle for the problem has vertices (0,0), (a,y), and (x,b) such that

a² + y² = x² + b² = (x-a)² + (b-y)² .

From the first and last we get

2by = x² -2ax + b²

and from the second and last we get

2bx = y² -2by +a² .

Graphing these two equations with graphing calculator software on my PC, I convinced myself that the two parabolas have two intersection points, each of which corresponds to one of the two triangles that the problem describes.

I was then able to derive a 4th degree equation for x by substituting y from the 1st into the second.  This equation has only two roots, and they are the same as the x-coordinates of the intersections of the two parabolas.  The 4th degree equation has a factor of x² + b² which I divided out to leave a quadratic whose roots are the x-coordinates of interest. This quadratic is

x² - 4ax +4a² - 3b² .

The rest is just algebra to get the product of the squares of the roots and the sum of the squares of the roots so that the product of the squares of the side lengths of the triangles can be found in terms of a and b.

Edited on August 22, 2005, 6:20 pm

Posted by Richard on 2005-08-22 18:18:08