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Question Number 62850 by ajfour last updated on 25/Jun/19

Commented by ajfour last updated on 25/Jun/19

$F i n d p o s s i b l e v a l u e s o f s i d e s o f$ $e q u i l a t e r a l △ P Q R i n t e r m s o f$ $a a n d b . (f o r e x a m p l e l e t a = 4, b = 3) .$
	Answered by mr W last updated on 26/Jun/19

	$Q (u, 0)$ $R (v, 0)$ $t = s^{2} = {(u + a)}^{2} + b^{2} = {(v + b)}^{2} + a^{2} = u^{2} + v^{2}$ $\Rightarrow u = \sqrt{t - b^{2}} - a$ $\Rightarrow v = \sqrt{t - a^{2}} - b$ $\Rightarrow t - b^{2} + a^{2} - 2 a \sqrt{t - b^{2}} + t - a^{2} + b^{2} - 2 b \sqrt{t - a^{2}} = t$ $\Rightarrow t - 2 a \sqrt{t - b^{2}} = 2 b \sqrt{t - a^{2}}$ $\Rightarrow t^{2} - 4 a t \sqrt{t - b^{2}} + 4 a^{2} (t - b^{2}) = 4 b^{2} (t - a^{2})$ $\Rightarrow t + 4 (a^{2} - b^{2}) = 4 a \sqrt{t - b^{2}}$ $\Rightarrow t^{2} + 8 (a^{2} - b^{2}) t + 16 {(a^{2} - b^{2})}^{2} = 16 a^{2} (t - b^{2})$ $\Rightarrow t^{2} - 8 (a^{2} + b^{2}) t + 16 [{(a^{2} - b^{2})}^{2} + a^{2} b^{2}] = 0$ $\Rightarrow t = 4 (a^{2} + b^{2}) \pm \sqrt{16 {(a^{2} + b^{2})}^{2} - 16 {(a^{2} - b^{2})}^{2} - 16 a^{2} b^{2}}$ $\Rightarrow t = 4 (a^{2} + b^{2} \pm \sqrt{3} a b)$ $\Rightarrow s = 2 \sqrt{a^{2} + b^{2} \pm \sqrt{3} a b}$
	Commented by ajfour last updated on 26/Jun/19

	$T h a n k s S i r, y o u m a d e i t v e r y e a s y .$ $h o w m a n y v a l u e s f o r s i f,$ $u, v \in R ?$
	Commented by mr W last updated on 26/Jun/19

	$i t h i n k t h e r e a r e a l w a y s 2 v a l u e s f o r s$ $i f a b \neq 0 .$
	Commented by ajfour last updated on 26/Jun/19
	yes sir, i am inclined to believe it.
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