Proving that a triangle with fixed perimeter has maximal area when it is equilateral The Isoperimetric Theorem states that for a shape with fixed perimeter (“iso” meaning “same”), a circle has the greatest area. We might naturally want to constrain this to triangles: for triangles with a fixed perimeter, which gives the maximal area? The answer would intuitively seem to be an equilateral triangle, but how would we prove it? How intuitively would we know where to begin the proof? We have three different possible formulae for the area of a 1 1 triangle to work with: 푏ℎ, 푎푏 sin 퐶 and Heron’s Formula (which 2 2 uses all three sides): √푠(푠 − 푎)(푠 − 푏)(푠 − 푐) where 푠 is half the perimeter of the triangle, i.e. 1 (푎 + 푏 + 푐). 2 Our intuition might be to use Heron’s formula, partly because of the nice algebraic symmetry in 푎, 푏, 푐, but also because of its use of the perimeter of the triangle. With such problems of maximising/minimising area, it’s also useful to form an inequality, where we state that the area of any triangle is at most some fixed limit, where we hope to find this limit. For those with some knowledge of common inequalities, some have simple conditions where equality is achieved, i.e. giving us this upper limit. AM-GM Inequality If you read the section on Inequalities in my algebra slides (http://www.drfrostmaths.com/resource.php?id=11270), you’ll find a handy inequality known as the AM-GM Inequality. This states that the Arithmetic Mean of some numbers (i.e. the mean you’re 1 used to, (푎 + 푎 + ⋯ + 푎 )) is at least the Geometric Mean of the numbers 푛 1 2 푛 푛 ( √푎1 × 푎2 × … × 푎푛). That is: 푎1 + 푎2 + ⋯ + 푎푛 ≥ 푛√푎 푎 … 푎 푛 1 2 푛 This inequality is particularly useful where we have a sum of expressions and wish to find some inequality involving the product of those expressions, or vice versa. It seems particularly applicable to the problem in hand, given that Heron’s formula involves the product of 푠 − 푎, 푠 − 푏 and 푠 − 푐 while the perimeter involves the sum. It should also be noted that ‘equality is achieved’, 퐴푀 = 퐺푀 (i.e. the maximum the Geometric Mean can be) when 푎1 = 푎2 = ⋯ = 푎푛. 1 1 1 Given the symmetry of 푠 − 푎, 푠 − 푏, 푠 − 푐 combined with 푠 = 푎 + 푏 + 푐 (recalling that 푠 is half 2 2 2 the perimeter of the triangle), it might seem sensible to use 푠 − 푎, 푠 − 푏, 푠 − 푐 as the terms in our AM-GM inequality. www.drfrostmaths.com The Proof Using AM-GM: (푠 − 푎) + (푠 − 푏) + (푠 − 푐) ≥ 3√(푠 − 푎)(푠 − 푏)(푠 − 푐) 3 1 1 1 1 1 1 1 1 1 푎 + 푏 + 푐 − 푎 + 푎 + 푏 + 푐 − 푏 + 푎 + 푏 + − 푐 2 2 2 2 2 2 2 2 푐 ≥ 3√(푠 − 푎)(푠 − 푏)(푠 − 푐) 3 푎 + 푏 + 푐 ≥ 3√(푠 − 푎)(푠 − 푏)(푠 − 푐) 6 푠 ≥ 3√(푠 − 푎)(푠 − 푏)(푠 − 푐) 3 푠3 ≥ (푠 − 푎)(푠 − 푏)(푠 − 푐) 27 푠4 ≥ 푠(푠 − 푎)(푠 − 푏)(푠 − 푐) 27 푠2 ≥ √푠(푠 − 푎)(푠 − 푏)(푠 − 푐) 3√3 푠2 This is great, because we’ve shown that the area of the triangle cannot exceed . Even better, 3√3 given that equality is achieved in the first line when 푠 − 푎 = 푠 − 푏 = 푠 − 푐 (recall the equality condition of the AM-GM equality described earlier), i.e. 푎 = 푏 = 푐, we’ve shown the area is maximised when the triangle is equilateral. □ 1 3 If we make all the sides of a triangle 푎, then we can see that the area is × 푎 × 푎 × sin 60° = √ 푎2. 2 4 1 1 1 3 푠2 3 But since 푠 = 푎 + 푎 + 푎 = 푎, then = √ 푎2. This confirms we do indeed achieve this upper 2 2 2 2 3√3 4 bound on the area when we have an equilateral triangle, although the equality condition of AM-GM was sufficient to complete our proof. Further Reading https://en.wikipedia.org/wiki/Isoperimetric_inequality www.drfrostmaths.com .