Question Number 6432 by Rasheed Soomro last updated on 27/Jun/16
Commented by Rasheed Soomro last updated on 27/Jun/16
$${Question}\:{reposted} \\ $$$$\mid{GH}\mid=?\:,\:\mid{HI}\mid=?\:,\mid{IG}\mid=? \\ $$
Commented by Yozzii last updated on 01/Jul/16
$${Vector}\:{Analysis}\:{Method} \\ $$$$−−−−−−−−−−−−−−−−−−−−−−−− \\ $$$${Let}\:\langle{AB}\rangle=\boldsymbol{{a}}\:{and}\:\langle{AC}\rangle=\boldsymbol{{b}}.\:\left(\boldsymbol{{a}}\neq−\boldsymbol{{b}}\right) \\ $$$$\langle{AF}\rangle=\boldsymbol{{a}}+\mathrm{0}.\mathrm{5}\left(\boldsymbol{{b}}−\boldsymbol{{a}}\right)=\mathrm{0}.\mathrm{5}\left(\boldsymbol{{a}}+\boldsymbol{{b}}\right). \\ $$$${Let}\:\langle{AH}\rangle=\mu\langle{AF}\rangle,\:\mu>\mathrm{0}.\:\mid{AH}\mid=\mid{AB}\mid=\mid\boldsymbol{{a}}\mid. \\ $$$$\mid{AH}\mid=\mu\mid{AF}\mid\Rightarrow\mu=\frac{\mid{AH}\mid}{\mid{AF}\mid}=\frac{\mathrm{2}\mid\boldsymbol{{a}}\mid}{\mid\boldsymbol{{a}}+\boldsymbol{{b}}\mid}. \\ $$$$\therefore\:\langle{AH}\rangle=\frac{\mathrm{2}\mid\boldsymbol{{a}}\mid}{\mid\boldsymbol{{a}}+\boldsymbol{{b}}\mid}\mathrm{0}.\mathrm{5}\left(\boldsymbol{{a}}+\boldsymbol{{b}}\right) \\ $$$$\langle{AH}\rangle=\frac{\mid\boldsymbol{{a}}\mid}{\mid\boldsymbol{{a}}+\boldsymbol{{b}}\mid}\left(\boldsymbol{{a}}+\boldsymbol{{b}}\right). \\ $$$$\langle{BD}\rangle=−\boldsymbol{{a}}+\mathrm{0}.\mathrm{5}\boldsymbol{{b}}. \\ $$$${Let}\:\langle{BG}\rangle=\lambda\langle{BD}\rangle,\:\lambda>\mathrm{0}. \\ $$$$\Rightarrow\lambda=\frac{\mid{BG}\mid}{\mid{BD}\mid}=\frac{\mid{CB}\mid}{\mathrm{0}.\mathrm{5}\mid\boldsymbol{{b}}−\mathrm{2}\boldsymbol{{a}}\mid}=\mathrm{2}\frac{\mid\boldsymbol{{a}}−\boldsymbol{{b}}\mid}{\mid\boldsymbol{{b}}−\mathrm{2}\boldsymbol{{a}}\mid}. \\ $$$$\therefore\langle{BG}\rangle=\frac{\mid\boldsymbol{{a}}−\boldsymbol{{b}}\mid}{\mid\boldsymbol{{b}}−\mathrm{2}\boldsymbol{{a}}\mid}\left(\boldsymbol{{b}}−\mathrm{2}\boldsymbol{{a}}\right). \\ $$$$\langle{AG}\rangle=\langle{AB}\rangle+\langle{BG}\rangle \\ $$$$\langle{AG}\rangle=\boldsymbol{{a}}+\frac{\mid\boldsymbol{{a}}−\boldsymbol{{b}}\mid}{\mid\boldsymbol{{b}}−\mathrm{2}\boldsymbol{{a}}\mid}\left(\boldsymbol{{b}}−\mathrm{2}\boldsymbol{{a}}\right) \\ $$$$\langle{AG}\rangle=\left(\mathrm{1}−\frac{\mathrm{2}\mid\boldsymbol{{a}}−\boldsymbol{{b}}\mid}{\mid\boldsymbol{{b}}−\mathrm{2}\boldsymbol{{a}}\mid}\right)\boldsymbol{{a}}+\frac{\mid\boldsymbol{{a}}−\boldsymbol{{b}}\mid}{\mid\boldsymbol{{b}}−\mathrm{2}\boldsymbol{{a}}\mid}\boldsymbol{{b}}. \\ $$$$\langle{AH}\rangle=\frac{\mid\boldsymbol{{a}}\mid}{\mid\boldsymbol{{a}}+\boldsymbol{{b}}\mid}\boldsymbol{{a}}+\frac{\mid\boldsymbol{{a}}\mid}{\mid\boldsymbol{{a}}+\boldsymbol{{b}}\mid}\boldsymbol{{b}} \\ $$$$\langle{GH}\rangle=\langle{AH}\rangle−\langle{AG}\rangle \\ $$$$\langle{GH}\rangle=\left(−\mathrm{1}+\frac{\mathrm{2}\mid\boldsymbol{{a}}−\boldsymbol{{b}}\mid}{\mid\boldsymbol{{b}}−\mathrm{2}\boldsymbol{{a}}\mid}+\frac{\mid\boldsymbol{{a}}\mid}{\mid\boldsymbol{{a}}+\boldsymbol{{b}}\mid}\right)\boldsymbol{{a}}+\left(\frac{\mid\boldsymbol{{a}}\mid}{\mid\boldsymbol{{a}}+\boldsymbol{{b}}\mid}−\frac{\mid\boldsymbol{{a}}−\boldsymbol{{b}}\mid}{\mid\boldsymbol{{b}}−\mathrm{2}\boldsymbol{{a}}\mid}\right)\boldsymbol{{b}} \\ $$$${Let}\:{n}=−\mathrm{1}+\frac{\mathrm{2}\mid\boldsymbol{{a}}−\boldsymbol{{b}}\mid}{\mid\boldsymbol{{b}}−\mathrm{2}\boldsymbol{{a}}\mid}+\frac{\mid\boldsymbol{{a}}\mid}{\mid\boldsymbol{{a}}+\boldsymbol{{b}}\mid}\:{and}\:{m}=\frac{\mid\boldsymbol{{a}}\mid}{\mid\boldsymbol{{a}}+\boldsymbol{{b}}\mid}−\frac{\mid\boldsymbol{{a}}−\boldsymbol{{b}}\mid}{\mid\boldsymbol{{b}}−\mathrm{2}\boldsymbol{{a}}\mid}. \\ $$$$\therefore\:\langle{GH}\rangle={n}\boldsymbol{{a}}+{m}\boldsymbol{{b}} \\ $$$$\Rightarrow\mid{GH}\mid^{\mathrm{2}} =\langle{GH}\rangle.\langle{GH}\rangle \\ $$$$\mid{GH}\mid^{\mathrm{2}} ={n}^{\mathrm{2}} \boldsymbol{{a}}.\boldsymbol{{a}}+\mathrm{2}{nm}\boldsymbol{{a}}.\boldsymbol{{b}}+{m}^{\mathrm{2}} \boldsymbol{{b}}.\boldsymbol{{b}} \\ $$$$\mid{GH}\mid=\sqrt{{n}^{\mathrm{2}} \mid\boldsymbol{{a}}\mid^{\mathrm{2}} +{m}^{\mathrm{2}} \mid\boldsymbol{{b}}\mid^{\mathrm{2}} +\mathrm{2}{mn}\boldsymbol{{a}}.\boldsymbol{{b}}} \\ $$$${Let}\:\theta=\angle{CAB}. \\ $$$$\therefore\:\boldsymbol{{a}}.\boldsymbol{{b}}=\mid\boldsymbol{{a}}\mid\mid\boldsymbol{{b}}\mid{cos}\theta\: \\ $$$$\therefore\mid{GH}\mid=\sqrt{{n}^{\mathrm{2}} \mid\boldsymbol{{a}}\mid^{\mathrm{2}} +{m}^{\mathrm{2}} \mid\boldsymbol{{b}}\mid^{\mathrm{2}} +{mn}\left(\mathrm{2}\mid\boldsymbol{{a}}\mid\mid\boldsymbol{{b}}\mid{cos}\theta\right)} \\ $$$${By}\:{law}\:{of}\:{cosines},\: \\ $$$$\mathrm{2}\mid\boldsymbol{{a}}\mid\mid\boldsymbol{{b}}\mid{cos}\theta=−\mid\boldsymbol{{a}}−\boldsymbol{{b}}\mid^{\mathrm{2}} +\mid\boldsymbol{{a}}\mid^{\mathrm{2}} +\mid\boldsymbol{{b}}\mid^{\mathrm{2}} \\ $$$$\therefore\mid{GH}\mid=\sqrt{{n}\left({n}+{m}\right)\mid\boldsymbol{{a}}\mid^{\mathrm{2}} +{m}\left({m}+{n}\right)\mid\boldsymbol{{b}}\mid^{\mathrm{2}} −{mn}\mid\boldsymbol{{a}}−\boldsymbol{{b}}\mid^{\mathrm{2}} } \\ $$$$\mid{GH}\mid=\sqrt{\left({n}+{m}\right)\left({n}\mid\boldsymbol{{a}}\mid^{\mathrm{2}} +{m}\mid\boldsymbol{{b}}\mid^{\mathrm{2}} \right)−{mn}\mid\boldsymbol{{a}}−\boldsymbol{{b}}\mid^{\mathrm{2}} } \\ $$$${It}\:{can}\:{be}\:{shown}\:{that}\: \\ $$$$\mid\boldsymbol{{a}}+\boldsymbol{{b}}\mid^{\mathrm{2}} =\mathrm{2}\left(\mid\boldsymbol{{a}}\mid^{\mathrm{2}} +\mid\boldsymbol{{b}}\mid^{\mathrm{2}} \right)−\mid\boldsymbol{{a}}−\boldsymbol{{b}}\mid^{\mathrm{2}} \\ $$$$\Rightarrow\mid\boldsymbol{{a}}+\boldsymbol{{b}}\mid=\sqrt{\mathrm{2}\left(\mid\boldsymbol{{a}}\mid^{\mathrm{2}} +\mid\boldsymbol{{b}}\mid^{\mathrm{2}} \right)−\mid\boldsymbol{{a}}−\boldsymbol{{b}}\mid^{\mathrm{2}} }>\mathrm{0} \\ $$$$\mid\boldsymbol{{b}}−\mathrm{2}\boldsymbol{{a}}\mid^{\mathrm{2}} =\mid\boldsymbol{{b}}\mid^{\mathrm{2}} +\mathrm{4}\mid\boldsymbol{{a}}\mid^{\mathrm{2}} −\mathrm{2}×\mathrm{2}\mid\boldsymbol{{b}}\mid\mid\boldsymbol{{a}}\mid{cos}\theta \\ $$$$\mid\boldsymbol{{b}}−\mathrm{2}\boldsymbol{{a}}\mid^{\mathrm{2}} =−\mid\boldsymbol{{b}}\mid^{\mathrm{2}} +\mathrm{2}\mid\boldsymbol{{a}}\mid^{\mathrm{2}} +\mathrm{2}\mid\boldsymbol{{a}}−\boldsymbol{{b}}\mid^{\mathrm{2}} \\ $$$$\mid\boldsymbol{{b}}−\mathrm{2}\boldsymbol{{a}}\mid=\sqrt{\mathrm{2}\left(\mid\boldsymbol{{a}}\mid^{\mathrm{2}} +\mid\boldsymbol{{a}}−\boldsymbol{{b}}\mid^{\mathrm{2}} \right)−\mid\boldsymbol{{b}}\mid^{\mathrm{2}} } \\ $$$${Let}\:\mid\boldsymbol{{a}}\mid={c}>\mathrm{0},\:\mid\boldsymbol{{b}}\mid={b}>\mathrm{0}\:{and}\:\mid\boldsymbol{{a}}−\boldsymbol{{b}}\mid={a}>\mathrm{0}. \\ $$$$−−−−−−−−−−−−−−−−−−−−−−−−−−− \\ $$$${For}\:\bigtriangleup{GHI}\:{constructed}\:{from}\:\bigtriangleup{ABC}, \\ $$$${in}\:{such}\:{a}\:{way}\:{as}\:{above}, \\ $$$$\therefore\mid{GH}\mid=\sqrt{\left({n}+{m}\right)\left({nc}^{\mathrm{2}} +{mb}^{\mathrm{2}} \right)−{mna}^{\mathrm{2}} } \\ $$$$\boldsymbol{{where}} \\ $$$${n}=−\mathrm{1}+\frac{\mathrm{2}{a}}{\:\sqrt{\mathrm{2}\left({c}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)−{b}^{\mathrm{2}} }}+\frac{{c}}{\:\sqrt{\mathrm{2}\left({c}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)−{a}^{\mathrm{2}} }} \\ $$$${m}=\frac{{c}}{\:\sqrt{\mathrm{2}\left({c}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)−{a}^{\mathrm{2}} }}−\frac{{a}}{\:\sqrt{\mathrm{2}\left({c}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)−{b}^{\mathrm{2}} }} \\ $$$$−−−−−−−−−−−−−−−−−−−−−−−−− \\ $$$${A}\:{similar}\:{analysis}\:{can}\:{be}\:{used}\:{to} \\ $$$${find}\:{the}\:{remaining}\:{lengths}.\:{Try}\:{it}! \\ $$$$\left({Cyclic}\:{permutation}\:{of}\:{abc}\:{I}\:{guess}\:{is}\:{possible}.\right) \\ $$
Commented by Rasheed Soomro last updated on 01/Jul/16
$$\mathcal{TH}{ank}\mathcal{S}! \\ $$$${Your}\:{approach}\:{is}\:{always}\:{reliable} \\ $$$${but}\:{I}\:{need}\:{to}\:{revise}\:{my}\:{knowledge} \\ $$$${about}\:\mathrm{vector}\:\mathrm{analysis}. \\ $$