An identity in a right triangle.

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An identity in a right triangle.

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Source: From CRUX.

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#1 h Oct 4, 2006, 2:26 AM • 2 Y

Y by Adventure10, Mango247

Let $ABC$ be a triangle for which $a>b$ and $a>c.$ Prove that :

$\boxed{\ A=90^{\circ}\Longleftrightarrow (\sqrt{a+b}+\sqrt{a-b})\cdot (\sqrt{a+c}+\sqrt{a-c})=2p\sqrt 2\ }\ ,$

where ${2p=a+b+c\ }\ .$

This post has been edited 3 times. Last edited by Virgil Nicula, Oct 4, 2006, 4:23 AM

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#2 h Oct 4, 2006, 2:55 AM • 1 Y

Y by Adventure10

Not sure what you mean by [aopsnowrap][math=inline]$p$[/math],[/aopsnowrap] buy here is what I got:

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#3 h Oct 4, 2006, 4:27 AM • 1 Y

Y by Adventure10

And reciprocally, i.e. $\Longleftarrow\ ?$
Only for Me@home.This problem is mine. And my solution has one trick, but which works with the chain of equivalencies !

Proof. Denote $2p=a+b+c\ ,\ 2\Delta=b^{2}+c^{2}-a^{2}$ and $E=(\sqrt{a+b}+\sqrt{a-b})(\sqrt{a+c}+\sqrt{a-c})\ .$
Prove easily that the following interesting chain of identities (here is the trick):
$\boxed{\ 2p^{2}-(a+b)(a+c)=2(p-a)^{2}-(a-b)(a-c)=2(p-b)^{2}-(a-b)(a+c)=2(p-c)^{2}-(a+b)(a-c)=\Delta\ }\ \ (*)\ .$
The relations $(*)$ solve immediately the our proposed problem !
$1.\blacktriangleright \Delta<0\Longrightarrow \sqrt{(a+b)(a+c)}> p\sqrt 2\ ,\ \sqrt{(a-b)(a-c)}>(p-a)\sqrt 2$ a.s.o $\Longrightarrow E> 2p\sqrt 2\ .$
$2.\blacktriangleright \Delta>0\Longrightarrow \sqrt{(a+b)(a+c)}<p\sqrt 2\ ,\ \sqrt{(a-b)(a-c)}<(p-a)\sqrt 2$ a.s.o. $\Longrightarrow E<2p\sqrt 2\ .$
$3.\blacktriangleright \Delta=0\Longrightarrow \sqrt{(a+b)(a+c)}=p\sqrt 2\ ,\ \sqrt{(a-b)(a-c)}=(p-a)\sqrt 2$ a.s.o. $\Longrightarrow E=2p\sqrt 2\ .$

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