# Question-209637

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