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Question Number 209712 by mr W last updated on 19/Jul/24

Commented by mr W last updated on 19/Jul/24

general case  OA=a  OB=b > a  ∠AOB=θ <(π/2)  n=even number=2k  find the minimum of  AC_1 +C_1 C_2 +...+C_(n−1) C_n +C_n B  in terms of a, b, n, θ.

generalcaseOA=aOB=b>aAOB=θ<π2n=evennumber=2kfindtheminimumofAC1+C1C2+...+Cn1Cn+CnBintermsofa,b,n,θ.

Commented by mahdipoor last updated on 19/Jul/24

if A=C_2 =C_4 =...=C_n       (n=2k)  and    C_1 =C_3 =...=C_(n−1) =D   (OD=OA=a)  L=Σ^ C_(n−1) C_n =2.n.a.sin(θ/2)+(√(a^2 +b^2 −2ab.cosθ))  is minimum  (a<b)

ifA=C2=C4=...=Cn(n=2k)andC1=C3=...=Cn1=D(OD=OA=a)L=Cn1Cn=2.n.a.sinθ2+a2+b22ab.cosθisminimum(a<b)

Commented by mr W last updated on 19/Jul/24

you mean  L_(min) =2na sin (θ/2)+(√(a^2 +b^2 −2ab cos θ))  but this is not true. example:  a=2, b=6, n=2, θ=20°  L_(min) =2×2×2 sin ((20°)/2)+(√(2^2 +6^2 −2×2×6×cos 20°))     =8 sin ((20°)/2)+(√(40−24 cos 20°)) ≈5.56  but minimum = 2(√7)≈5.29  see Q209637

youmeanLmin=2nasinθ2+a2+b22abcosθbutthisisnottrue.example:a=2,b=6,n=2,θ=20°Lmin=2×2×2sin20°2+22+622×2×6×cos20°=8sin20°2+4024cos20°5.56butminimum=275.29seeQ209637

Answered by mahdipoor last updated on 20/Jul/24

 { ((if  (n+1)θ≤180  ⇒L_(min) =(√(a^2 +b^2 −2ab.cos((n+1)θ))))),((if  (n+1)θ≥180  ⇒L_(min) =a+b)) :}  i will try to send photo , that is explain my answer

{if(n+1)θ180Lmin=a2+b22ab.cos((n+1)θ)if(n+1)θ180Lmin=a+biwilltrytosendphoto,thatisexplainmyanswer

Answered by mr W last updated on 21/Jul/24

Commented by mr W last updated on 21/Jul/24

the shortest path from A to B is  the path of a light between two  mirrors.  if the light ray goes directly from  A to B the shortest path length is  L_(min) =AB=(√(a^2 +b^2 −2ab cos θ))  if the light ray are reflected  n=2k times by the mirrors, the  shortest path length is  L_(min) =(AC_1 +C_1 C_2 +...+C_n B)_(min)            =(√(a^2 +b^2 −2ab cos (2k+1)θ))  (see also Q209637)  such a light ray path is only possible  when (2k+1)θ≤180°, i.e. k≤((90)/θ)−(1/2).  that means,   when k≤k_1 =⌊((90)/θ)−(1/2)⌋,  (AC_1 +C_1 C_2 +...+C_n B)_(min)       =(√(a^2 +b^2 −2ab cos (2k+1)θ))  when k>k_1 =⌊((90)/θ)−(1/2)⌋,   (AC_1 +C_1 C_2 +...+C_n B)_(min)       =2(k−k_1 )a sin θ+(√(a^2 +b^2 −2ab cos (2k_1 +1)θ))  in this case we repeat 2(k−k_1 ) times  the shortest distance AD and then  following the path of the light ray.

theshortestpathfromAtoBisthepathofalightbetweentwomirrors.ifthelightraygoesdirectlyfromAtoBtheshortestpathlengthisLmin=AB=a2+b22abcosθifthelightrayarereflectedn=2ktimesbythemirrors,theshortestpathlengthisLmin=(AC1+C1C2+...+CnB)min=a2+b22abcos(2k+1)θ(seealsoQ209637)suchalightraypathisonlypossiblewhen(2k+1)θ180°,i.e.k90θ12.thatmeans,whenkk1=90θ12,(AC1+C1C2+...+CnB)min=a2+b22abcos(2k+1)θwhenk>k1=90θ12,(AC1+C1C2+...+CnB)min=2(kk1)asinθ+a2+b22abcos(2k1+1)θinthiscasewerepeat2(kk1)timestheshortestdistanceADandthenfollowingthepathofthelightray.

Commented by mr W last updated on 21/Jul/24

Commented by mahdipoor last updated on 21/Jul/24

  if (2k+1)θ≥180 , my answer is :  when C_1 =C_2 =...=C_n =O ⇒ L=min=a+b  for example : a=2 , b=6 , θ=45 , n=4 (k=2)  my ans: L_(min) =8  your ans : k_1 =1⇒L_(min) =2(√2)+(√(40+12(√2)))=10.4

if(2k+1)θ180,myansweris:whenC1=C2=...=Cn=OL=min=a+bforexample:a=2,b=6,θ=45,n=4(k=2)myans:Lmin=8yourans:k1=1Lmin=22+40+122=10.4

Commented by mr W last updated on 21/Jul/24

the points between A and B should  lie alternately on both lines, therefore  point O should be excepted.

thepointsbetweenAandBshouldliealternatelyonbothlines,thereforepointOshouldbeexcepted.

Commented by mahdipoor last updated on 21/Jul/24

i think we must accept point O is on   both line , then  ∀C_i =O alternately on both line  if O is excepted ; as you sey (C_i ≠O)  C_(2i+1) ∈OB , OC_(2i+1) =very small=β  C_(2i) ∈OA , OC_(2i) =very small=α  L_(min) =[a^2 +β^2 −2aβ.cosθ]^(0.5) +[b^2 +α^2 −2bα.cosθ]^(0.5) +  (n/2)(√(α^2 +β^2 −2αβ.cosθ))  for last example (a=2 , b=6 , θ=45 , n=4)  for β=α=1 ⇒ L_(min) =8.34 < 10.4  for β=α=0.1 ⇒ L_(min) =8.01 <10.4  ...

ithinkwemustacceptpointOisonbothline,thenCi=OalternatelyonbothlineifOisexcepted;asyousey(CiO)C2i+1OB,OC2i+1=verysmall=βC2iOA,OC2i=verysmall=αLmin=[a2+β22aβ.cosθ]0.5+[b2+α22bα.cosθ]0.5+n2α2+β22αβ.cosθforlastexample(a=2,b=6,θ=45,n=4)forβ=α=1Lmin=8.34<10.4forβ=α=0.1Lmin=8.01<10.4...

Commented by mr W last updated on 21/Jul/24

you are right. we can select the  points very close to point O.  L_(min) ≈a−ε+(√(ε^2 +b^2 −2εb cos (2k_1 −1)θ))            →a+b

youareright.wecanselectthepointsveryclosetopointO.Lminaϵ+ϵ2+b22ϵbcos(2k11)θa+b

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