The sides of a triangle are in arithmetic progression and its area is 3/5th the area of an equilateral triangle with the same perimeter.

Find the ratio of the sides of the triangle.

(In reply to Starters by Tristan)

Oh, I just realized I made an obvious mistake. I put 3/5 on the wrong side of the equation!

Back to "Putting this together:"
(3/5)*x²√3/4=√(3x^4/16-3x²y²/4)
3x²√3/20=√;(3x^4/16-3x²y²/4)
27x^4/400=3x^4/16-3x²y²/4
0=48x^4/400-3x²y²/4
0=3x²(x²/25-y²/4)
0=3x²(x/5+y/2)(x/5-y/2)
So x=0, or x/5±y/2=0
x/5=±y/2
x=±5y/2

I doesn't make a difference whether y is positive or negative, so just call it positive. The ratio is:
5y/2+y:5y/2:5y/2-y
7y:5y:3y
7:5:3

The ratio is 7:5:3

Posted by Tristan on 2003-11-23 12:36:11