122. An identity for an equilateral triangle. Extension.

by Virgil Nicula, Sep 13, 2010, 3:26 AM

Quote:
Let $P$ be a point in the plane of an equilateral triangle $ABC$ with the length $a$ of the side. Show that :

$3\left(PA^{4}+PB^{4}+PC^{4}+a^{4}\right)=\left(PA^{2}+PB^{2}+PC^{2}+a^{2}\right)^{2}\ (*)$ .
Proof. Here is a short proof of this identity, a generalization and some interesting inequalities. Denote

$PA=x$ , $PB=y$ , $PC=z$ and the point $R$ so that $AR=RP=PA=x$ and $AP\cap BR\ne \emptyset$ . Observe that

$\triangle ABP\equiv \triangle ACR$ $\Longrightarrow$ $CR=PB=y$. Denote $m(\widehat {CPR})=\phi$ and the area $\sigma\equiv \sigma (x,y,z)$ of the Pompeiu's triangle $CPR$ .

Thus, $2xz\cos \phi=x^2+z^2-y^2$ and $xz\sin \phi=2\sigma$ obtain $\boxed {\ 16\sigma ^2=2(x^2y^2+y^2z^2+z^2x^2)-(x^4+y^4+z^4)\ }\ \ (1)$ .

$AC^2=PA^2+PC^2-2\cdot PA\cdot PC\cdot \cos (60^{\circ}+\phi )\Longrightarrow$ $a^2=x^2+z^2-\frac 12\cdot 2xz\cos \phi +\sqrt 3\cdot xz\sin \phi\Longrightarrow$

$\boxed {\ 2a^2=x^2+y^2+z^2+4\sigma\sqrt 3\ }\ \ (2)$ . The proposed identity becomes $3(a^4+\sum x^4)=\left( a^2+\sum x^2\right)^2\Longleftrightarrow$

$\boxed {\ a^4+\sum x^4=a^2\sum x^2+\sum x^2y^2\ }\ \ (3)$ . From the relations $(1)$ and $(2)$ we will eliminate the area $\sigma$ . Therefore,

$\left(\sum x^2-2a^2\right)^2=3\cdot 16\sigma ^2=3\left(2\sum x^2y^2-\sum x^4\right)\Longleftrightarrow$

$\sum x^4+2\sum x^2y^2+4a^4-4a^2\sum x^2=6\sum x^2y^2-3\sum x^4$ .

This last relation is equivalently with the proposed identity $(3)$ .

A generalization. Let $ABC$ be a triangle and a inner point $P$. Denote $PA=x$, $PB=y$, $PC=z$

and the circumradius $R$. Then $\boxed {\ 2\sum a^2b^2x^2y^2-\sum a^4x^4=4R^2\cdot \left(abc-\sum ax^2\cos A\right)^2\ }\ .$

Particular cases.

$\blacktriangleright\ x=y=z=R\Longrightarrow \sum a\cos A=\frac{2S}{R}$ .

$\blacktriangleright\ a=b=c\Longrightarrow 2\sum x^2y^2-\sum x^4=\frac 13\cdot \left(2a^2-\sum x^2\right)^2\Longrightarrow$ $a^4+\sum x^4=a^2\sum x^2+\sum x^2y^2,$ i.e. $(*)$ .

Some interesting consequencies. From the Cebasev's inequality results :

$\blacktriangleright\ \ 3\cdot \left(a^4+\sum PA^4\right)=$ $\left(\sum PA^2+3\cdot \frac{a^2}{3}\right)^2\le 6\cdot \left(\frac{a^4}{3}+\sum PA^4\right)\Longrightarrow$

$3a^4+3\cdot \sum PA^4\le 2a^4+6\cdot \sum PA^4\Longrightarrow \boxed {\ a^4\le 3\cdot \sum PA^4\ }\ .$

$\blacktriangleright$ Weitzenbock's inequality $PA^2+PB^2+PC^2\ge 4\sqrt 3\sigma\ \wedge\ (2)$ $\implies$ $\boxed {\ 4\sigma \sqrt 3\le a^2\le PA^2+PB^2+PC^2\ }$ . See here
This post has been edited 5 times. Last edited by Virgil Nicula, Nov 23, 2015, 7:37 AM

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