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.

Call the three sides x-y, x, and x+y.

Using Heron's formula, the area of the triangle is:
√((3x/2)(3x/2-x+y)(3x/2-x)(3x/2-x-y))=
√((3x²/4)(x²/4-y²))=
√(3x^4/16-3x²y²/4)

The area of the equilateral triangle is:
x²√3/4

Putting this together:
x²√3/4=(3/5)*√(3x^4/16-3x²y²/4)
5x²√3/12=√(3x^4/16-3x²y²/4)
75x^4/144=3x^4/16-3x²y²/4
48x^4/144+3x²y²/4=0
x²(x²/3+3y²/4)=0
So, either x=0(obviously not forming a triangle), or x²/3+3y²/4=0

Would anyone like to check my work or finish it?

Posted by Tristan on 2003-11-23 12:22:33