139. A caracterization of a right angle in a triangle.

by Virgil Nicula, Oct 3, 2010, 5:41 PM

Let $ABC$ be a triangle with $a\ge b$ and $a\ge c$ . Prove that the equivalence

$\boxed{\left(\sqrt {a+b}+\sqrt {a-b}\right)\left(\sqrt {a+c}+\sqrt {a-c}\right)=\sqrt 2\cdot (a+b+c)\ \iff\ AB\perp AC}$ .

Proof. Denote the "defect" $D$ , where $4\cdot D=b^2+c^2-a^2$ and $\left\{\begin{array}{c} X=(a+b)(a+c)\ ;\ Y=(a-b)(a-c)\\\\ Z=(a-b)(a+c)\ ;\ T=(a+b)(a-c)\\\\ E=\left(\sqrt {a+b}+\sqrt {a-b}\right)\left(\sqrt {a+c}+\sqrt {a-c}\right)\end{array}\right\|$ . Observe that

$E=\sqrt X+\sqrt Y+\sqrt Z+\sqrt T$ . I"ll prove generaly that $\mathrm{sign}\ \left(2s\sqrt 2-E\right)=\mathrm{sign}\ D$ , i.e. $\left\{\begin{array}{ccccc} E<2s\sqrt 2 & \iff & D>0 & \iff & A>90^{\circ}\\\\ E=2s\sqrt 2 & \iff & D=0 & \iff & A=90^{\circ}\\\\ E>2s\sqrt 2 & \iff & D<0 & \iff & A>90^{\circ}\end{array}\right\|$ .

Prove easily that the relations $\boxed{2s^2-X=2(s-a)^2-Y=2(s-b)^2-Z=2(s-c)^2-T=2\cdot D}$ . Therefore,

$\mathrm{sign}\left(s\sqrt 2-\sqrt X\right)=$ $\mathrm{sign}\left[(s-a)\sqrt 2-\sqrt Y\right]=$ $\mathrm{sign}\left[(s-b)\sqrt 2-\sqrt Z\right]=$ $\mathrm{sign}\left[(s-c)\sqrt 2-\sqrt T\right]=\mathrm{sign}\ D$ . Since

$\left(s\sqrt 2-\sqrt X\right)+\left[(s-a)\sqrt 2-\sqrt Y\right]+\left[(s-b)\sqrt 2-\sqrt Z\right]+$ $\left[(s-c)\sqrt 2-\sqrt T\right]=2s\sqrt 2-E$ obtain $\mathrm{sign}\left(2s\sqrt 2-E\right)=\mathrm{sign}\ D$ .
This post has been edited 13 times. Last edited by Virgil Nicula, Dec 1, 2015, 11:09 AM

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