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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.
$${general}\:{case} \\ $$$${OA}={a} \\ $$$${OB}={b}\:>\:{a} \\ $$$$\angle{AOB}=\theta\:<\frac{\pi}{\mathrm{2}} \\ $$$${n}={even}\:{number}=\mathrm{2}{k} \\ $$$${find}\:{the}\:{minimum}\:{of} \\ $$$${AC}_{\mathrm{1}} +{C}_{\mathrm{1}} {C}_{\mathrm{2}} +…+{C}_{{n}−\mathrm{1}} {C}_{{n}} +{C}_{{n}} {B} \\ $$$${in}\:{terms}\:{of}\:{a},\:{b},\:{n},\:\theta. \\ $$
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)
$${if}\:{A}={C}_{\mathrm{2}} ={C}_{\mathrm{4}} =…={C}_{{n}} \:\:\:\:\:\:\left({n}=\mathrm{2}{k}\right) \\ $$$${and}\:\:\:\:{C}_{\mathrm{1}} ={C}_{\mathrm{3}} =…={C}_{{n}−\mathrm{1}} ={D}\:\:\:\left({OD}={OA}={a}\right) \\ $$$${L}=\overset{} {\sum}{C}_{{n}−\mathrm{1}} {C}_{{n}} =\mathrm{2}.{n}.{a}.{sin}\frac{\theta}{\mathrm{2}}+\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{ab}.{cos}\theta} \\ $$$${is}\:{minimum}\:\:\left({a}<{b}\right)\: \\ $$
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
$${you}\:{mean} \\ $$$${L}_{{min}} =\mathrm{2}{na}\:\mathrm{sin}\:\frac{\theta}{\mathrm{2}}+\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{ab}\:\mathrm{cos}\:\theta} \\ $$$${but}\:{this}\:{is}\:{not}\:{true}.\:{example}: \\ $$$${a}=\mathrm{2},\:{b}=\mathrm{6},\:{n}=\mathrm{2},\:\theta=\mathrm{20}° \\ $$$${L}_{{min}} =\mathrm{2}×\mathrm{2}×\mathrm{2}\:\mathrm{sin}\:\frac{\mathrm{20}°}{\mathrm{2}}+\sqrt{\mathrm{2}^{\mathrm{2}} +\mathrm{6}^{\mathrm{2}} −\mathrm{2}×\mathrm{2}×\mathrm{6}×\mathrm{cos}\:\mathrm{20}°} \\ $$$$\:\:\:=\mathrm{8}\:\mathrm{sin}\:\frac{\mathrm{20}°}{\mathrm{2}}+\sqrt{\mathrm{40}−\mathrm{24}\:\mathrm{cos}\:\mathrm{20}°}\:\approx\mathrm{5}.\mathrm{56} \\ $$$${but}\:{minimum}\:=\:\mathrm{2}\sqrt{\mathrm{7}}\approx\mathrm{5}.\mathrm{29} \\ $$$${see}\:{Q}\mathrm{209637} \\ $$
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
$$\begin{cases}{{if}\:\:\left({n}+\mathrm{1}\right)\theta\leqslant\mathrm{180}\:\:\Rightarrow{L}_{{min}} =\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{ab}.{cos}\left(\left({n}+\mathrm{1}\right)\theta\right)}}\\{{if}\:\:\left({n}+\mathrm{1}\right)\theta\geqslant\mathrm{180}\:\:\Rightarrow{L}_{{min}} ={a}+{b}}\end{cases} \\ $$$$\mathrm{i}\:\mathrm{will}\:\mathrm{try}\:\mathrm{to}\:\mathrm{send}\:\mathrm{photo}\:,\:\mathrm{that}\:\mathrm{is}\:\mathrm{explain}\:\mathrm{my}\:\mathrm{answer} \\ $$
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.
$${the}\:{shortest}\:{path}\:{from}\:{A}\:{to}\:{B}\:{is} \\ $$$${the}\:{path}\:{of}\:{a}\:{light}\:{between}\:{two} \\ $$$${mirrors}. \\ $$$${if}\:{the}\:{light}\:{ray}\:{goes}\:{directly}\:{from} \\ $$$${A}\:{to}\:{B}\:{the}\:{shortest}\:{path}\:{length}\:{is} \\ $$$${L}_{{min}} ={AB}=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{ab}\:\mathrm{cos}\:\theta} \\ $$$${if}\:{the}\:{light}\:{ray}\:{are}\:{reflected} \\ $$$${n}=\mathrm{2}{k}\:{times}\:{by}\:{the}\:{mirrors},\:{the} \\ $$$${shortest}\:{path}\:{length}\:{is} \\ $$$${L}_{{min}} =\left({AC}_{\mathrm{1}} +{C}_{\mathrm{1}} {C}_{\mathrm{2}} +…+{C}_{{n}} {B}\right)_{{min}} \\ $$$$\:\:\:\:\:\:\:\:\:=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{ab}\:\mathrm{cos}\:\left(\mathrm{2}{k}+\mathrm{1}\right)\theta} \\ $$$$\left({see}\:{also}\:{Q}\mathrm{209637}\right) \\ $$$${such}\:{a}\:{light}\:{ray}\:{path}\:{is}\:{only}\:{possible} \\ $$$${when}\:\left(\mathrm{2}{k}+\mathrm{1}\right)\theta\leqslant\mathrm{180}°,\:{i}.{e}.\:{k}\leqslant\frac{\mathrm{90}}{\theta}−\frac{\mathrm{1}}{\mathrm{2}}. \\ $$$${that}\:{means},\: \\ $$$${when}\:{k}\leqslant{k}_{\mathrm{1}} =\lfloor\frac{\mathrm{90}}{\theta}−\frac{\mathrm{1}}{\mathrm{2}}\rfloor, \\ $$$$\left({AC}_{\mathrm{1}} +{C}_{\mathrm{1}} {C}_{\mathrm{2}} +…+{C}_{{n}} {B}\right)_{{min}} \\ $$$$\:\:\:\:=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{ab}\:\mathrm{cos}\:\left(\mathrm{2}{k}+\mathrm{1}\right)\theta} \\ $$$${when}\:{k}>{k}_{\mathrm{1}} =\lfloor\frac{\mathrm{90}}{\theta}−\frac{\mathrm{1}}{\mathrm{2}}\rfloor,\: \\ $$$$\left({AC}_{\mathrm{1}} +{C}_{\mathrm{1}} {C}_{\mathrm{2}} +…+{C}_{{n}} {B}\right)_{{min}} \\ $$$$\:\:\:\:=\mathrm{2}\left({k}−{k}_{\mathrm{1}} \right){a}\:\mathrm{sin}\:\theta+\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{ab}\:\mathrm{cos}\:\left(\mathrm{2}{k}_{\mathrm{1}} +\mathrm{1}\right)\theta} \\ $$$${in}\:{this}\:{case}\:{we}\:{repeat}\:\mathrm{2}\left({k}−{k}_{\mathrm{1}} \right)\:{times} \\ $$$${the}\:{shortest}\:{distance}\:{AD}\:{and}\:{then} \\ $$$${following}\:{the}\:{path}\:{of}\:{the}\:{light}\:{ray}. \\ $$
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
$$ \\ $$$${if}\:\left(\mathrm{2}{k}+\mathrm{1}\right)\theta\geqslant\mathrm{180}\:,\:{my}\:{answer}\:{is}\:: \\ $$$${when}\:{C}_{\mathrm{1}} ={C}_{\mathrm{2}} =…={C}_{{n}} ={O}\:\Rightarrow\:{L}={min}={a}+{b} \\ $$$${for}\:{example}\::\:{a}=\mathrm{2}\:,\:{b}=\mathrm{6}\:,\:\theta=\mathrm{45}\:,\:{n}=\mathrm{4}\:\left({k}=\mathrm{2}\right) \\ $$$${my}\:{ans}:\:{L}_{{min}} =\mathrm{8} \\ $$$${your}\:{ans}\::\:{k}_{\mathrm{1}} =\mathrm{1}\Rightarrow{L}_{{min}} =\mathrm{2}\sqrt{\mathrm{2}}+\sqrt{\mathrm{40}+\mathrm{12}\sqrt{\mathrm{2}}}=\mathrm{10}.\mathrm{4} \\ $$
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.
$${the}\:{points}\:{between}\:{A}\:{and}\:{B}\:{should} \\ $$$${lie}\:{alternately}\:{on}\:{both}\:{lines},\:{therefore} \\ $$$${point}\:{O}\:{should}\:{be}\:{excepted}. \\ $$
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 ...
$$\mathrm{i}\:\mathrm{think}\:\mathrm{we}\:\mathrm{must}\:\mathrm{accept}\:\mathrm{point}\:\mathrm{O}\:\mathrm{is}\:\mathrm{on}\: \\ $$$$\mathrm{both}\:\mathrm{line}\:,\:\mathrm{then}\:\:\forall\mathrm{C}_{{i}} =\mathrm{O}\:\mathrm{alternately}\:\mathrm{on}\:\mathrm{both}\:\mathrm{line} \\ $$$$\mathrm{if}\:\mathrm{O}\:\mathrm{is}\:\mathrm{excepted}\:;\:\mathrm{as}\:\mathrm{you}\:\mathrm{sey}\:\left({C}_{{i}} \neq{O}\right) \\ $$$${C}_{\mathrm{2}{i}+\mathrm{1}} \in{OB}\:,\:{OC}_{\mathrm{2}{i}+\mathrm{1}} ={very}\:{small}=\beta \\ $$$${C}_{\mathrm{2}{i}} \in{OA}\:,\:{OC}_{\mathrm{2}{i}} ={very}\:{small}=\alpha \\ $$$${L}_{{min}} =\left[{a}^{\mathrm{2}} +\beta^{\mathrm{2}} −\mathrm{2}{a}\beta.{cos}\theta\right]^{\mathrm{0}.\mathrm{5}} +\left[{b}^{\mathrm{2}} +\alpha^{\mathrm{2}} −\mathrm{2}{b}\alpha.{cos}\theta\right]^{\mathrm{0}.\mathrm{5}} + \\ $$$$\frac{{n}}{\mathrm{2}}\sqrt{\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} −\mathrm{2}\alpha\beta.{cos}\theta} \\ $$$${for}\:{last}\:{example}\:\left({a}=\mathrm{2}\:,\:{b}=\mathrm{6}\:,\:\theta=\mathrm{45}\:,\:{n}=\mathrm{4}\right) \\ $$$${for}\:\beta=\alpha=\mathrm{1}\:\Rightarrow\:{L}_{{min}} =\mathrm{8}.\mathrm{34}\:<\:\mathrm{10}.\mathrm{4} \\ $$$${for}\:\beta=\alpha=\mathrm{0}.\mathrm{1}\:\Rightarrow\:{L}_{{min}} =\mathrm{8}.\mathrm{01}\:<\mathrm{10}.\mathrm{4} \\ $$$$… \\ $$
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
$${you}\:{are}\:{right}.\:{we}\:{can}\:{select}\:{the} \\ $$$${points}\:{very}\:{close}\:{to}\:{point}\:{O}. \\ $$$${L}_{{min}} \approx{a}−\epsilon+\sqrt{\epsilon^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}\epsilon{b}\:\mathrm{cos}\:\left(\mathrm{2}{k}_{\mathrm{1}} −\mathrm{1}\right)\theta} \\ $$$$\:\:\:\:\:\:\:\:\:\:\rightarrow{a}+{b} \\ $$

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