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

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)

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

2{nx+(l−2x)⌊(n/2)⌋}cos (((180°)/n))

Commented byMAB last updated on 10/Jul/20

any demonstration sir?

Commented bymr W last updated on 10/Jul/20

answer correct?

Commented byMAB last updated on 10/Jul/20

seems to be correct sir

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

let′s say the path starts at point B_1 (=X)  and touches the other sides at points  B_2 ,B_3 ,...,B_n . the path is also a n side  polygon.  the shortest path is that one which  a light ray follows from B_1  to B_1   when reflected by all sides of the  regular polygon as mirrors.  if n is even, we can easily see which  path the light ray will follow, that′s  what the following diagram shows.

Commented bymr W last updated on 10/Jul/20

Commented bymr W last updated on 10/Jul/20

let A_1 B_1 =x  θ=180°−((360°)/n)  b_1 =b_3 =b_5 =...=b_(n−1) =2x sin (θ/2)=2x cos ((180°)/n)  b_2 =b_4 =b_6 =...=b_n =2(l−x) sin (θ/2)=2(l−x) cos ((180°)/n)  total length of the shortest path is  therefore  L_(min) =(n/2)×2(x+l−x) cos ((180°)/n)  ⇒L_(min) =nl cos ((180°)/n)

Commented bymr W last updated on 10/Jul/20

if n is odd, the path of the light ray  follows from B_1  to B_1  is not so easily  to determine. we need some new  consideration.

Commented byMAB last updated on 10/Jul/20

perfect sir

Commented bymr W last updated on 10/Jul/20

Commented bymr W last updated on 10/Jul/20

this is a possible path, but it is not  the shortest, because it doesn′t  follow a light ray′s path.  b^2 =x^2 +(l−x)^2 +2x(l−x)cos θ  b^2 =l^2 −2x(l−x)(1+cos ((360°)/n))  L=nb=n(√(l^2 −2x(l−x)(1+cos ((360°)/n))))  ......