A Two-Triangle Inequality

D. Pedoe
Problem E 1562,
The American Mathematical Monthly
Vol. 70, No. 9 (Nov., 1963), p. 1012

A two-triangle inequality by Dan Pedoe

Proof

If one of the triangles is equilateral, the inequality becomes one attributed to Weitzenboeck:

$a^2+b^2+c^2\ge 4\sqrt{3}K,$

withe the equality if and only if the triangle is equilateral.

Two prove the two-triangle inequality, erect on $BC,\;$ and on t he same side of $BC\;$ as the vertex $A,\;$ a triangle $A''BC\;$ similar to triangle $A'B'C'.$

Pedoe: a two-triangle inequality

By the Law of Cosines in $\Delta ACA'',$

$\displaystyle (AA'')^2=b^2+\left(b'\frac{a}{a'}\right)^2-2bb'\frac{a}{a'}\cos (C'-C),$

or else

$\displaystyle\begin{align} a'^2(AA'')^2&=a'^2b^2+a^2b'^2-2aa'bb'\cos (C'-C)\\ &=a'^2b^2+a^2b'^2-2aa'bb'\cos C\cos C'-2aa'bb'\sin C\sin C'\ge 0. \end{align}$

Therefore, $2aa'bb'\cos C\cos C' \ge 8KK',\;$ or

(*)

$\displaystyle a'^2b^2+a^2b'^2-\frac{1}{2}(b^2+a^2-c^2)(b'^2+a'^2-c'^2)\ge 8KK',$

or

$a^2(-a'^2+b'^2+c'^2)+b^2(a'^2-b'^2+c'^2)+c^2(a'^2+b'^2-c'^2)\ge 16KK'.$

For other proofs see D. Pedoe, "An inequality for two triangles," Proc. Cambridge Philos. Soc., vol. 38, part 4, p.397.

Note that (*) is symmetric in $ABC\;$ and $A'B'C',\;$ making $a,b,c\;$ and $a',b',c'\;$ interchangeable.

 

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

Copyright © 1996-2017 Alexander Bogomolny

 61628317

Search by google: