Isoperimetric Property of Equilateral Triangles II

According to the Isoperimetric Theorem, circle has the maximum among all shapes with a given perimeter. Circle is probably the most regular of all plane shapes. Regularity also plays an important role in restricted families of plane figures. For example, it is known that

Among all triangles of given perimeter, the equilateral one has the largest area.

Proof

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Copyright © 1996-2018 Alexander Bogomolny

Among all triangles of given perimeter, the equilateral one has the largest area.

Proof

The proof below is due to Grégoire Nicollier; it arose out of a discussion that followed the upload of the Isoperimetric Theorem For Quadrilaterals II page.

Suppose that triangle $ABC$ is not equilateral with, say, different side lengths $a=CB$ and $b=CA.$ Consider the ellipse with foci $A$ and $B$ through $C.$ One gets then a triangle of equal perimeter and greater area by replacing $C$ by the point of the ellipse that is on the axis other than $AB.$ We can thus restrict the candidates to isosceles triangles with $a = b$ and $c = 2d.$ By taking a perimeter of $2,$ one has then $a = 1 - d$ and the squared triangle area is

$[ABC]^{2}=d^{2}(a^{2}-d^{2}) = d^{2} - 2d^{3}.$

Noticing that $d\ge 0$ and that

$\displaystyle d^{2} - 2d^{3} = \frac{1}{27} - \bigg(d-\frac{1}{3}\bigg)^{2}\bigg(2d+\frac{1}{3}\bigg),$

which is the Taylor expansion about $\displaystyle\frac{1}{3}$ (but can be easily verified directly), one has at once (without using derivatives) $\displaystyle d^{2} - 2d^{3} \le \frac{1}{27.}$ $\displaystyle\frac{1}{27}$ is attained exactly for $\displaystyle d=\frac{1}{3},$ that is, for $a=b=c=\frac{2}{3}.$

(An alternative proof can be found elsewhere.)


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

Copyright © 1996-2018 Alexander Bogomolny

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