how to prove that an equilateral triangle has the greatest area of any other triangle with the same perimeter

Question

Just need help with a geometry proof.

Answer 1

Place an equilateral triangle with one vertex at the Origin and other vertices of its base at A(-a,-k) and B(a,-k).

Both a and k are positive numbers.  I'm taking an Analytic Geometry approach (DesCartes married Euclidean geometry and Cartesian coordinates to form this field of mathematics).

 

Consider a translation of the vertex angle at O, translated to P(x,0).  This new triangle APB has the same area as equilateral triangle  AOB (and same base, AB), but the other two sides have total length AP + BP, contrasted with AOB's other two sides having total length AO + BO.

 

Using the distance formula, AO + BO = 2√[a² + b²];  let's call this E.

Using the distance formula, AP + BP = √[(x+a)² + k²] + √[(x-a)² + k²]; let's call this F.

 

To compare E vs. F (E is to be the lesser quantity), square both expressions.  On paper it's easy, but I don't have the patience to type it here.  You will find that E² must be < F², algebraically, because the latter has an extra term which is a radical (non-zero) and thus makes F greater than E.