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-lt-x-lt-1-as-shown-what-is-the-length-of-the-shortest-path-begins-fromX-touching-all-s

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

$$let{A}_1{A}_2\dots A_{n}aregularpolygonofnsides$$$$withlengthleach,letXapointsuch$$$$that{A}_{1}X=x\cdot lwhere0<x<1\left(asshown\right)$$$$whatisthelengthoftheshortestpath$$$$beginsfromXtouchingallsidesonce$$$$andendsatX$$$$\left({there}^{\prime }saproblemshape{can}^{\prime }tbeloaded\right)$$

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

$$2\{nx+(l-2x)\lfloor \frac{n}{2}\rfloor \}\cos \left(\frac{180°}{n}\right)$$

Commented by MAB last updated on 10/Jul/20

$$anydemonstrationsir?$$

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

$$answercorrect?$$

Commented by MAB last updated on 10/Jul/20

$$seemstobecorrectsir$$

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

$${let}^{\prime }ssaythepathstartsatpoint{B}_1(=X)$$$$andtouchestheothersidesatpoints$$$$B_2,B_3,\dots ,B_n.thepathisalsoanside$$$$polygon.$$$$theshortestpathisthatonewhich$$$$alightrayfollowsfrom{B}_{1}to{B}_1$$$$whenreflectedbyallsidesofthe$$$$regularpolygonasmirrors.$$$$ifniseven,wecaneasilyseewhich$$$$paththelightraywillfollow,{that}^{\prime }s$$$$whatthefollowingdiagramshows.$$

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

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

$$let{A}_1{B}_1=x$$$$\theta =180°-\frac{360°}{n}$$$$b_1=b_3=b_5=\dots =b_{n-1}=2x\sin \frac{\theta }{2}=2x\cos \frac{180°}{n}$$$$b_2=b_4=b_6=\dots =b_n=2(l-x)\sin \frac{\theta }{2}=2(l-x)\cos \frac{180°}{n}$$$$totallengthoftheshortestpathis$$$$therefore$$$$L_{min}=\frac{n}{2}\times 2(x+l-x)\cos \frac{180°}{n}$$$$\Rightarrow L_{min}=nl\cos \frac{180°}{n}$$

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

$$ifnisodd,thepathofthelightray$$$$followsfrom{B}_{1}to{B}_{1}isnotsoeasily$$$$todetermine.weneedsomenew$$$$consideration.$$

Commented by MAB last updated on 10/Jul/20

$$perfectsir$$

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

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

$$thisisapossiblepath,butitisnot$$$$theshortest,becauseit{doesn}^{\prime }t$$$$followalight{ray}^{\prime }spath.$$$$b^2=x^2+{(l-x)}^2+2x(l-x)\cos \theta$$$$b^2=l^2-2x(l-x)(1+\cos \frac{360°}{n})$$$$L=nb=n\sqrt{l^2-2x(l-x)(1+\cos \frac{360°}{n})}$$$$\dots \dots$$

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