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Question-209637




Question Number 209637 by mr W last updated on 17/Jul/24
Commented by mr W last updated on 17/Jul/24
find minimum of AB+BC+CD=?
$${find}\:{minimum}\:{of}\:{AB}+{BC}+{CD}=? \\ $$
Commented by Frix last updated on 17/Jul/24
I get 2(√7)
$$\mathrm{I}\:\mathrm{get}\:\mathrm{2}\sqrt{\mathrm{7}} \\ $$
Commented by mr W last updated on 17/Jul/24
that′s correct sir!  your method?
$${that}'{s}\:{correct}\:{sir}! \\ $$$${your}\:{method}? \\ $$
Commented by Frix last updated on 17/Jul/24
I will post it later.
$$\mathrm{I}\:\mathrm{will}\:\mathrm{post}\:\mathrm{it}\:\mathrm{later}. \\ $$
Commented by Frix last updated on 19/Jul/24
I “fold” it  A= ((2),(0) )  B∈y=xtan 20°  C∈y=xtan 40°  D∈y=xtan 60° ∧ ∣OD∣=6 ⇒ D= ((3),((3(√3))) )  The shortest path is  ∣AD∣=2(√7)
$$\mathrm{I}\:“\mathrm{fold}''\:\mathrm{it} \\ $$$${A}=\begin{pmatrix}{\mathrm{2}}\\{\mathrm{0}}\end{pmatrix} \\ $$$${B}\in{y}={x}\mathrm{tan}\:\mathrm{20}° \\ $$$${C}\in{y}={x}\mathrm{tan}\:\mathrm{40}° \\ $$$${D}\in{y}={x}\mathrm{tan}\:\mathrm{60}°\:\wedge\:\mid{OD}\mid=\mathrm{6}\:\Rightarrow\:{D}=\begin{pmatrix}{\mathrm{3}}\\{\mathrm{3}\sqrt{\mathrm{3}}}\end{pmatrix} \\ $$$$\mathrm{The}\:\mathrm{shortest}\:\mathrm{path}\:\mathrm{is} \\ $$$$\mid{AD}\mid=\mathrm{2}\sqrt{\mathrm{7}} \\ $$
Commented by mr W last updated on 19/Jul/24
thanks sir!
$${thanks}\:{sir}! \\ $$
Commented by Frix last updated on 19/Jul/24
But this method doesn′t work for n points  with (n−1)α≥180°
$$\mathrm{But}\:\mathrm{this}\:\mathrm{method}\:\mathrm{doesn}'\mathrm{t}\:\mathrm{work}\:\mathrm{for}\:{n}\:\mathrm{points} \\ $$$$\mathrm{with}\:\left({n}−\mathrm{1}\right)\alpha\geqslant\mathrm{180}° \\ $$
Commented by mr W last updated on 19/Jul/24
for (n−1)α>180° there exists no  minimum at all.
$${for}\:\left({n}−\mathrm{1}\right)\alpha>\mathrm{180}°\:{there}\:{exists}\:{no} \\ $$$${minimum}\:{at}\:{all}. \\ $$
Answered by mr W last updated on 19/Jul/24
Commented by mr W last updated on 19/Jul/24
let′s treat OA, OD as mirrors.  (AB+BC+CD)_(min) =A′D′  =(√(2^2 +6^2 −2×2×6 cos (3×20°)))  =2(√7)
$${let}'{s}\:{treat}\:{OA},\:{OD}\:{as}\:{mirrors}. \\ $$$$\left({AB}+{BC}+{CD}\right)_{{min}} ={A}'{D}' \\ $$$$=\sqrt{\mathrm{2}^{\mathrm{2}} +\mathrm{6}^{\mathrm{2}} −\mathrm{2}×\mathrm{2}×\mathrm{6}\:\mathrm{cos}\:\left(\mathrm{3}×\mathrm{20}°\right)} \\ $$$$=\mathrm{2}\sqrt{\mathrm{7}} \\ $$

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