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: 2 𝑏ℎ, 2 𝑎𝑏 sin 𝐶 and Heron’s Formula (which
uses all three sides): √𝑠(𝑠 − 𝑎)(𝑠 − 𝑏)(𝑠 − 𝑐) where 𝑠 is half the perimeter of the triangle, i.e.
1
(𝑎
2
+ 𝑏 + 𝑐).
Our intuition might be to use Heron’s formula, partly because of the nice algebraic symmetry in
𝑎, 𝑏, 𝑐, 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, 𝑛 (𝑎1 + 𝑎2 + ⋯ + 𝑎𝑛 )) is at least the Geometric Mean of the numbers
( 𝑛√𝑎1 × 𝑎2 × … × 𝑎𝑛 ). That is:
𝑎1 + 𝑎2 + ⋯ + 𝑎𝑛 𝑛
≥ √𝑎1 𝑎2 … 𝑎𝑛
𝑛
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 𝑠 − 𝑎, 𝑠 − 𝑏 and 𝑠 − 𝑐
while the perimeter involves the sum. It should also be noted that ‘equality is achieved’, 𝐴𝑀 = 𝐺𝑀
(i.e. the maximum the Geometric Mean can be) when 𝑎1 = 𝑎2 = ⋯ = 𝑎𝑛 .
1
1
1
Given the symmetry of 𝑠 − 𝑎, 𝑠 − 𝑏, 𝑠 − 𝑐 combined with 𝑠 = 2 𝑎 + 2 𝑏 + 2 𝑐 (recalling that 𝑠 is half
the perimeter of the triangle), it might seem sensible to use 𝑠 − 𝑎, 𝑠 − 𝑏, 𝑠 − 𝑐 as the terms in our
AM-GM inequality.
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The Proof
Using AM-GM:
(𝑠 − 𝑎) + (𝑠 − 𝑏) + (𝑠 − 𝑐) 3
≥ √(𝑠 − 𝑎)(𝑠 − 𝑏)(𝑠 − 𝑐)
3
1
1
1
1
1
1
1
1
1
2 𝑎 + 2 𝑏 + 2 𝑐 − 𝑎 + 2 𝑎 + 2 𝑏 + 2 𝑐 − 𝑏 + 2 𝑎 + 2 𝑏 + 𝑐 − 𝑐 ≥ 3√(𝑠 − 𝑎)(𝑠 − 𝑏)(𝑠 − 𝑐)
3
𝑎+𝑏+𝑐 3
≥ √(𝑠 − 𝑎)(𝑠 − 𝑏)(𝑠 − 𝑐)
6
𝑠 3
≥ √(𝑠 − 𝑎)(𝑠 − 𝑏)(𝑠 − 𝑐)
3
𝑠3
≥ (𝑠 − 𝑎)(𝑠 − 𝑏)(𝑠 − 𝑐)
27
𝑠4
≥ 𝑠(𝑠 − 𝑎)(𝑠 − 𝑏)(𝑠 − 𝑐)
27
2
𝑠
≥ √𝑠(𝑠 − 𝑎)(𝑠 − 𝑏)(𝑠 − 𝑐)
3√3
𝑠2
This is great, because we’ve shown that the area of the triangle cannot exceed 3 3. Even better,
√
given that equality is achieved in the first line when 𝑠 − 𝑎 = 𝑠 − 𝑏 = 𝑠 − 𝑐 (recall the equality
condition of the AM-GM equality described earlier), i.e. 𝑎 = 𝑏 = 𝑐, we’ve shown the area is
maximised when the triangle is equilateral. □
1
If we make all the sides of a triangle 𝑎, then we can see that the area is 2 × 𝑎 × 𝑎 × sin 60° =
1
1
1
3
𝑠2
√3
But since 𝑠 = 2 𝑎 + 2 𝑎 + 2 𝑎 = 2 𝑎, then 3
=
√3 2
𝑎 .
4
√3 2
𝑎 .
4
This confirms we do indeed achieve this upper
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
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