# Maximum Area Property of Equilateral Triangles

A particular case of the Isoperimetric Theorem tells us that among all triangles with the same perimeter, the equilateral one has the largest area. A related theorem concerning the triangles inscribed into a given circle is also true:

Among all triangles inscribed in a given circle, the equilateral one has the largest area.

|Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny

Among all triangles inscribed in a given circle, the equilateral one has the largest area.

### Proof

The proof depends on the following

### Lemma

Among all triangles inscribed in a given circle, with a given base, the isosceles one has the largest area.

The assertion of the lemma is quite obvious:

Among all inscribed triangles with a given base, the tallest one is isosceles and, therefore, it has the largest area, due to the standard formula

The lemma shows that for a triangle that has two unequal sides, there is another triangle (an isosceles one at that) with the same circumcircle but larger area. The only triangle for which no improvement is possible is equilateral.

The lemma also shows that in order to prove the statement we only need to look among isosceles triangles. Consider the diagram below:

Let α be the base angle of an isosceles triangle ABC. Then half the apex angle at C equals

∠ACB = 180° - 2α,

∠BCE = ∠OCD = 90° - α,

OB = OC, BD = CD,

∠OBD = ∠OCD = 90° - α,

∠OBE = α - (90° - α) = 2α - 90°,90° - 2α,2α - 90°,2α

OE = R sin(2α - 90°) = R -cos,sin,-cos,cos,tan(2α),

CE = R(1 - cos(2α)),

AB = 2 EB = 2R cos(2α - 90°) = 2R sin,sin,cos,tan(2α),

Area( ΔABC) = AB×CE / 2 = R² (1 - cos(2α)) sin(2α).

So to maximize the area of triangle ABC we need to find the maximum of function

f'(β) = cos(β)(1 - cos(β)) + sin(β) sin(β) = -cos²(β) + cos(β) + (1 - cos²(β)).

Letting x = cos(β) we are left with solving the quadratic equation

(The second root x = 1, leads to cos(2α) = 1. Allowing for an expanded angle range that includes 0,

It goes without saying (see the discussion of the general Isoperimetric Theorem) that our statement admits an equivalent formulation:

Among all triangles with the given area, the equilateral one has the smallest circumscribed circle.

- Isoperimetric Theorem and Inequality
- An Isoperimetric theorem
- Isoperimetric theorem and its variants
- Isoperimetric Property of Equilateral Triangles
- Maximum Area Property of Equilateral Triangles
- Isoperimetric Theorem For Quadrilaterals
- Isoperimetric Theorem For Quadrilaterals II
- An Isoperimetric Problem in Quadrilateral

|Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny

64849295 |