let A_1 A_2 ...A_n a regular polygon of n sides with length l each, let X a point such that A_1 X=x∙l where 0<x<1 (as shown) what is the length of the shortest path begins fromX touching all sides once and ends at X (there′s a problem shape can′t be loaded)


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Question Number 102576 by MAB last updated on 10/Jul/20

$l e t A_{1} A_{2} . . . A_{n} a r e g u l a r p o l y g o n o f n s i d e s$ $w i t h l e n g t h l e a c h, l e t X a p o i n t s u c h$ $t h a t A_{1} X = x \cdot l w h e r e 0 < x < 1 (a s s h o w n)$ $w h a t i s t h e l e n g t h o f t h e s h o r t e s t p a t h$ $b e g i n s f r o m X t o u c h i n g a l l s i d e s o n c e$ $a n d e n d s a t X$ $({t h e r e}^{'} s a p r o b l e m s h a p e {c a n}^{'} t b e l o a d e d)$
	Answered by mr W last updated on 10/Jul/20

	$2 {n x + (l - 2 x) ⌊ \frac{n}{2} ⌋} \cos (\frac{180 °}{n})$
	Commented by MAB last updated on 10/Jul/20

	$a n y d e m o n s t r a t i o n s i r ?$
	Commented by mr W last updated on 10/Jul/20

	$a n s w e r c o r r e c t ?$
	Commented by MAB last updated on 10/Jul/20

	$s e e m s t o b e c o r r e c t s i r$
	Answered by mr W last updated on 10/Jul/20

	${l e t}^{'} s s a y t h e p a t h s t a r t s a t p o i n t B_{1} (= X)$ $a n d t o u c h e s t h e o t h e r s i d e s a t p o i n t s$ $B_{2}, B_{3}, . . ., B_{n} . t h e p a t h i s a l s o a n s i d e$ $p o l y g o n .$ $t h e s h o r t e s t p a t h i s t h a t o n e w h i c h$ $a l i g h t r a y f o l l o w s f r o m B_{1} t o B_{1}$ $w h e n r e f l e c t e d b y a l l s i d e s o f t h e$ $r e g u l a r p o l y g o n a s m i r r o r s .$ $i f n i s e v e n, w e c a n e a s i l y s e e w h i c h$ $p a t h t h e l i g h t r a y w i l l f o l l o w, {t h a t}^{'} s$ $w h a t t h e f o l l o w i n g d i a g r a m s h o w s .$
	Commented by mr W last updated on 10/Jul/20

	Commented by mr W last updated on 10/Jul/20

	$l e t A_{1} B_{1} = x$ $θ = 180 ° - \frac{360 °}{n}$ $b_{1} = b_{3} = b_{5} = . . . = b_{n - 1} = 2 x \sin \frac{θ}{2} = 2 x \cos \frac{180 °}{n}$ $b_{2} = b_{4} = b_{6} = . . . = b_{n} = 2 (l - x) \sin \frac{θ}{2} = 2 (l - x) \cos \frac{180 °}{n}$ $t o t a l l e n g t h o f t h e s h o r t e s t p a t h i s$ $t h e r e f o r e$ $L_{m i n} = \frac{n}{2} \times 2 (x + l - x) \cos \frac{180 °}{n}$ $\Rightarrow L_{m i n} = n l \cos \frac{180 °}{n}$
	Commented by mr W last updated on 10/Jul/20

	$i f n i s o d d, t h e p a t h o f t h e l i g h t r a y$ $f o l l o w s f r o m B_{1} t o B_{1} i s n o t s o e a s i l y$ $t o d e t e r m i n e . w e n e e d s o m e n e w$ $c o n s i d e r a t i o n .$
	Commented by MAB last updated on 10/Jul/20

	$p e r f e c t s i r$
	Commented by mr W last updated on 10/Jul/20

	Commented by mr W last updated on 10/Jul/20

	$t h i s i s a p o s s i b l e p a t h, b u t i t i s n o t$ $t h e s h o r t e s t, b e c a u s e i t {d o e s n}^{'} t$ $f o l l o w a l i g h t {r a y}^{'} s p a t h .$ $b^{2} = x^{2} + {(l - x)}^{2} + 2 x (l - x) \cos θ$ $b^{2} = l^{2} - 2 x (l - x) (1 + \cos \frac{360 °}{n})$ $L = n b = n \sqrt{l^{2} - 2 x (l - x) (1 + \cos \frac{360 °}{n})}$ $. . . . . .$
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