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Question Number 52061 by ajfour last updated on 02/Jan/19

Commented by ajfour last updated on 02/Jan/19

$F i n d m i n i m u m o f (A P + P B) i n$ $t e r m s o f α, a, b .$
Commented by MJS last updated on 14/Jan/19

$I get \sqrt{a^{2} + b^{2} - 2 a b \cos 2 α}$
Commented by mr W last updated on 15/Jan/19

$p l e a s e r e c h e c k y o u r s o l u t i o n s i r .$ $t h i s r e s u l t i s n o t c o r r e c t .$ $e x a m p l e : b = 0$ $m i n . p a t h i s o b v i o u s l y 2 a \sin α .$ $a c c o r d i n g t o y o u r r e s u l t i t i s a .$ $a c c o r d i n g t o m y r e s u l t i t i s$ $\sqrt{a^{2} + a^{2} - 2 a^{2} \cos 2 α} = a \sqrt{2 (1 - \cos 2 α)}$ $= a \sqrt{2 \times 2 \sin^{2} α} = 2 a \sin α . t h a t m e a n s$ $m y r e s u l t i s c o r r e c t .$
	Answered by mr W last updated on 02/Jan/19

	Commented by mr W last updated on 02/Jan/19

	$A P + P B = A^{'} P + P B$ ${(A P + P B)}_{m i n} = A^{'} B$ $= a^{2} + {(a + b)}^{2} - 2 a (a + b) \cos (2 α)$
	Commented by ajfour last updated on 15/Jan/19

	$y o u m e a n o f c o u r s e {(A^{'} B)}^{2} = . .$
	Commented by mr W last updated on 15/Jan/19

	$y e s s i r .$
	Answered by MJS last updated on 15/Jan/19

	$A = (\begin{matrix} a \\ 0 \end{matrix}) B = (\begin{matrix} a + b = c \\ 0 \end{matrix}) P = (\begin{matrix} p \\ p t \end{matrix})$ $l (p) =∣ A P ∣ + ∣ B P ∣= \sqrt{(t^{2} + 1) p^{2} - 2 a p + a^{2}} + \sqrt{(t^{2} + 1) p^{2} - 2 c p + c^{2}}$ $l^{'} (p) = 0 \Rightarrow p = 0 \lor p = \frac{2 a c}{(a + c) (t^{2} + 1)}$ $maximum is at l (0) = a + c = 2 a + b$ $minimum is at l (\frac{2 a c}{(a + c) (t^{2} + 1)}) = \sqrt{\frac{{(a + c)}^{2} t^{2} + {(a - c)}^{2}}{t^{2} + 1}} = \sqrt{\frac{{(2 a + b)}^{2} t^{2} + b^{2}}{t^{2} + 1}}$ $t = \tan α$ $\sqrt{\frac{{(2 a + b)}^{2} t^{2} + b^{2}}{t^{2} + 1}} = \sqrt{b^{2} + 4 a (a + b) \sin^{2} α}$
	Commented by MJS last updated on 15/Jan/19

	$not sure what happened before . this is correct$ $and should be the same as your result$
	Commented by ajfour last updated on 15/Jan/19

	$T h a n k s f o r t h e a l t e r n a t i v e w a y S i r .$
	Commented by mr W last updated on 15/Jan/19

	${i t}^{'} s p e r f e c t s i r!$ $t h i s i s a r e a l m a t h e m a t i c a l s o l u t i o n$ $w h i l e m y s o l u t i o n i s a ‘ ‘ {p h y s i c a l}^{″} o n e$ $b a s e d o n ‘ ‘ {k n o w l e d g e}^{″} t h a t t h e s t r a i g h t$ $i s t h e s h o r t e s t p a t h b e t w e e n t w o p o i n t s .$
	Commented by mr W last updated on 15/Jan/19

	$b u t a n y w a y, a m a x i m u m {d o e s n}^{'} t e x i t .$
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