Among all triangles of given perimeter, the equilateral one has the largest area.
Among all triangles of given perimeter, the equilateral one has the largest area.
Proof
The proof is based on Heron's formula
Lemma
S² = s(s - a)(s - b)(s - c),
where a, b, c are the sides of a triangle, S its area, and s = (a + b + c)/2, the semiperimeter.
Since s is constant, the question is to maximize S²/s = (s - a)(s - b)(s - c) for a + b + c = 2s. To this end we shall employ the arithmetic mean - geometric mean inequality which, for three terms u, v, w asserts that
(u + v + w) / 3 ≥ (uvw)1/3,
with equality only if u = v = w. For the three terms (s - a), (s - b), (s - c), that add up to s, we have
(s / 3)³ ≥ (s - a)(s - b)(s - c) = S² / s,
so that S2 ≤ s4 / 27, with equality only when s - a = s - b = s - c, i.e., for a = b = c.
To sum up, the area of a triangle with perimeter 2s never exceeds, s²/33/2; as is easy to verify, it exactly equals s²/33/2 for an equilateral triangle with side 2s/3,s/3,1/3,2s/3,3s/2.
It must be understood (see the discussion of the general Isoperimetric Theorem) that our statement admits an equivalent formulation:
Among all triangles with given area, the equilateral one has the least perimeter.
Isoperimetric Theorem and Inequality
Isoperimetric Theorem for Rectangles
An Isoperimetric theorem
Isoperimetric theorem and its variants
Isoperimetric Property of Equilateral Triangles
Maximum Area Property of Equilateral Triangles
|Contact|
|Front page|
|Contents|
|Algebra|
|Geometry|
|Store|
Copyright © 1996-2012 Alexander Bogomolny
