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Question Number 83767 by mr W last updated on 05/Mar/20

Commented by mr W last updated on 06/Mar/20

$F i n d t h e m i n i m u m o f t h e p e r i m e t e r$ $o f t r i a n g l e P Q R .$
	Answered by mr W last updated on 06/Mar/20

	$M e t h o d 1$ $s a y Q (a, a^{2}), R (b, 0)$ $p e r i m e t e r o f t r i a n g l e p$ $p = \sqrt{{(4 - a)}^{2} + {(3 - a^{2})}^{2}} + \sqrt{{(4 - b)}^{2} + 3^{2}} + \sqrt{{(a - b)}^{2} + a^{4}}$ $\frac{\partial p}{\partial a} = \frac{- (4 - a) - 2 a (3 - a^{2})}{\sqrt{{(4 - a)}^{2} + {(3 - a^{2})}^{2}}} + \frac{(a - b) + 2 a^{3}}{\sqrt{{(a - b)}^{2} + a^{4}}} = 0$ $\Rightarrow \frac{4 + 5 a - 2 a^{3}}{\sqrt{{(4 - a)}^{2} + {(3 - a^{2})}^{2}}} = \frac{a - b + 2 a^{3}}{\sqrt{{(a - b)}^{2} + a^{4}}} . . (i)$ $\frac{\partial p}{\partial b} = \frac{- (4 - b)}{\sqrt{{(4 - b)}^{2} + 3^{2}}} + \frac{- (a - b)}{\sqrt{{(a - b)}^{2} + a^{4}}} = 0$ $\Rightarrow \frac{b - 4}{\sqrt{{(4 - b)}^{2} + 3^{2}}} = \frac{a - b}{\sqrt{{(a - b)}^{2} + a^{4}}} . . (i i)$ $\frac{{(b - 4)}^{2}}{{(4 - b)}^{2} + 3^{2}} = \frac{{(a - b)}^{2}}{{(a - b)}^{2} + a^{4}}$ $\frac{3^{2}}{{(b - 4)}^{2}} = \frac{a^{4}}{{(a - b)}^{2}}$ $\frac{3}{4 - b} = \frac{a^{2}}{b - a}$ $(3 + a^{2}) b = a (4 a + 3)$ $\Rightarrow b = \frac{a (4 a + 3)}{a^{2} + 3}$ $p u t t h i s i n t o (i) :$ $\Rightarrow a \approx 1.3869$ $\Rightarrow b \approx 2.4078$
	Commented by jagoll last updated on 06/Mar/20

	$method 2 ?$
	Answered by mr W last updated on 06/Mar/20

	Commented by mr W last updated on 06/Mar/20

	$M e t h o d 2$ $t h e t r i n g l e w i t h m i n i m u m p e r i m e t e r$ $i s t h e p a t h w h i c h a l i g h t r a y f o l l o w s .$ $i f P^{'} i s t h e m i r r o r i m a g e o f P a b o u t$ $x - a x i s, t h e n P^{'} R Q P r e p r e s e n t s t h e$ $m i n i m u m p e r i m e t e r w h e n a l i g h t$ $r a y f r o m P^{'} f o l l o w s t h i s w a y .$ $s a y Q (a, a^{2})$ $\tan α = \frac{a^{2} + 3}{4 - a}$ $\tan θ = y^{'} = 2 a \Rightarrow \tan 2 θ = \frac{4 a}{1 - 4 a^{2}}$ $\frac{π}{2} - β = π - θ - α$ $2 β - α = 2 θ + α - π$ $\tan (2 β - α) = \frac{3 - a^{2}}{4 - a}$ $\tan (2 θ + α) = \frac{3 - a^{2}}{4 - a}$ $\frac{\frac{4 a}{1 - 4 a^{2}} + \frac{a^{2} + 3}{4 - a}}{1 - \frac{4 a}{1 - 4 a^{2}} \times \frac{a^{2} + 3}{4 - a}} = \frac{3 - a^{2}}{4 - a}$ $\frac{4 a (4 - a) + (a^{2} + 3) (1 - 4 a^{2})}{(1 - 4 a^{2}) (4 - a) - 4 a (a^{2} + 3)} = \frac{3 - a^{2}}{4 - a}$ $\frac{16 a + 3 - 4 a^{4} - 15 a^{2}}{4 - 16 a^{2} - 13 a} = \frac{3 - a^{2}}{4 - a}$ $\Rightarrow 2 a^{4} - 16 a^{3} + a^{2} - 12 a + 50 = 0$ $\Rightarrow a \approx 1.3869$
	Commented by jagoll last updated on 06/Mar/20

	$waw great sir$
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