Let $ABC$ be a triangle with side lengths $a,b,c$, such that $a$ is the longest side. Prove that $\angle BAC = 90^\circ$ if and only if \[ (\sqrt { a+b } + \sqrt { a-b} )(\sqrt {a+c } + \sqrt { a-c } ) = (a+b+c) \sqrt 2. \]
Problem
Source: Romanian Junior BkMO TST 2004, problem 6
Tags: inequalities
spider_boy
17.02.2005 15:22
THIS CAN BE EASILY TREATED WITH INEQUALITIES!!! THERE ARE TWO CASES. A <sup>2</sup> <B <sup>2</sup> +C <sup>2</sup> A <sup>2</sup> >B <sup>2</sup> +C <sup>2</sup>
spider_boy
17.02.2005 15:25
spider_boy wrote: THIS CAN BE EASILY TREATED WITH INEQUALITIES!!! THERE ARE TWO CASES. A <sup>2</sup> <B <sup>2</sup> +C <sup>2</sup> A <sup>2</sup> >B <sup>2</sup> +C <sup>2</sup> SQUARING BOTH SIDES THEN APPLYING THE INEQUALITIES INTHE CASES GIVES THE ANSWER!!!
Magnara
18.02.2005 07:30
AHHHHHH CAPITAL LETTERS
shobber
18.02.2005 12:38
$(\sqrt{a+b}+\sqrt{a-b})^2(\sqrt{a+c}+\sqrt{a-c})^2$
$=(a+b+a-b+2\sqrt{(a+b)(a-b)})(a+c+a-c+2\sqrt{(a+c)(a-c)})$
$=(2a+2\sqrt{a^2-c^2})(2a+2\sqrt{a^2-b^2})$
$=(2a+2b)(2a+2c)$
$=2(a+b)(2a+2c)$
$=2(2a^2+2ac+2ab+2bc)$
$=2(a^2+a^2+2ac+2ab+2bc)$
$=2(a^2+b^2+c^2+2ac+2ab+2bc)$
$=2(a+b+c)^2$
Hence:
\[\displaystyle (\sqrt { a+b } + \sqrt { a-b} )(\sqrt {a+c } + \sqrt { a-c } ) = (a+b+c) \sqrt 2. \]
huashiliao2020
04.04.2023 03:43
Nice @above! I did the same, any hints for the converse?
Danielzh
04.04.2023 03:47
huashiliao2020 wrote: Nice @above! I did the same, any hints for the converse? I doubt they are actually active lmaooo