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Question Number 6432 by Rasheed Soomro last updated on 27/Jun/16

Commented by Rasheed Soomro last updated on 27/Jun/16

$$Questionreposted$$$$\mid GH\mid =?,\mid HI\mid =?,\mid IG\mid =?$$

Commented by Yozzii last updated on 01/Jul/16

$$VectorAnalysisMethod$$$$------------------------$$$$Let⟨AB⟩=\mathit{a}and⟨AC⟩=\mathit{b}.(\mathit{a}\ne -\mathit{b})$$$$⟨AF⟩=\mathit{a}+0.5(\mathit{b}-\mathit{a})=0.5(\mathit{a}+\mathit{b}).$$$$Let⟨AH⟩=\mu ⟨AF⟩,\mu >0.\mid AH\mid =\mid AB\mid =\mid \mathit{a}\mid .$$$$\mid AH\mid =\mu \mid AF\mid \Rightarrow \mu =\frac{\mid AH\mid }{\mid AF\mid }=\frac{2\mid \mathit{a}\mid }{\mid \mathit{a}+\mathit{b}\mid }.$$$$\therefore ⟨AH⟩=\frac{2\mid \mathit{a}\mid }{\mid \mathit{a}+\mathit{b}\mid }0.5(\mathit{a}+\mathit{b})$$$$⟨AH⟩=\frac{\mid \mathit{a}\mid }{\mid \mathit{a}+\mathit{b}\mid }(\mathit{a}+\mathit{b}).$$$$⟨BD⟩=-\mathit{a}+0.5\mathit{b}.$$$$Let⟨BG⟩=\lambda ⟨BD⟩,\lambda >0.$$$$\Rightarrow \lambda =\frac{\mid BG\mid }{\mid BD\mid }=\frac{\mid CB\mid }{0.5\mid \mathit{b}-2\mathit{a}\mid }=2\frac{\mid \mathit{a}-\mathit{b}\mid }{\mid \mathit{b}-2\mathit{a}\mid }.$$$$\therefore ⟨BG⟩=\frac{\mid \mathit{a}-\mathit{b}\mid }{\mid \mathit{b}-2\mathit{a}\mid }(\mathit{b}-2\mathit{a}).$$$$⟨AG⟩=⟨AB⟩+⟨BG⟩$$$$⟨AG⟩=\mathit{a}+\frac{\mid \mathit{a}-\mathit{b}\mid }{\mid \mathit{b}-2\mathit{a}\mid }(\mathit{b}-2\mathit{a})$$$$⟨AG⟩=(1-\frac{2\mid \mathit{a}-\mathit{b}\mid }{\mid \mathit{b}-2\mathit{a}\mid })\mathit{a}+\frac{\mid \mathit{a}-\mathit{b}\mid }{\mid \mathit{b}-2\mathit{a}\mid }\mathit{b}.$$$$⟨AH⟩=\frac{\mid \mathit{a}\mid }{\mid \mathit{a}+\mathit{b}\mid }\mathit{a}+\frac{\mid \mathit{a}\mid }{\mid \mathit{a}+\mathit{b}\mid }\mathit{b}$$$$⟨GH⟩=⟨AH⟩-⟨AG⟩$$$$⟨GH⟩=(-1+\frac{2\mid \mathit{a}-\mathit{b}\mid }{\mid \mathit{b}-2\mathit{a}\mid }+\frac{\mid \mathit{a}\mid }{\mid \mathit{a}+\mathit{b}\mid })\mathit{a}+(\frac{\mid \mathit{a}\mid }{\mid \mathit{a}+\mathit{b}\mid }-\frac{\mid \mathit{a}-\mathit{b}\mid }{\mid \mathit{b}-2\mathit{a}\mid })\mathit{b}$$$$Letn=-1+\frac{2\mid \mathit{a}-\mathit{b}\mid }{\mid \mathit{b}-2\mathit{a}\mid }+\frac{\mid \mathit{a}\mid }{\mid \mathit{a}+\mathit{b}\mid }andm=\frac{\mid \mathit{a}\mid }{\mid \mathit{a}+\mathit{b}\mid }-\frac{\mid \mathit{a}-\mathit{b}\mid }{\mid \mathit{b}-2\mathit{a}\mid }.$$$$\therefore ⟨GH⟩=n\mathit{a}+m\mathit{b}$$$$\Rightarrow \mid GH{\mid }^2=⟨GH⟩.⟨GH⟩$$$$\mid GH{\mid }^2=n^2\mathit{a}.\mathit{a}+2nm\mathit{a}.\mathit{b}+m^2\mathit{b}.\mathit{b}$$$$\mid GH\mid =\sqrt{n^2\mid \mathit{a}{\mid }^2+m^2\mid \mathit{b}{\mid }^2+2mn\mathit{a}.\mathit{b}}$$$$Let\theta =\mathrm{\angle }CAB.$$$$\therefore \mathit{a}.\mathit{b}=\mid \mathit{a}\mid \mid \mathit{b}\mid cos\theta$$$$\therefore \mid GH\mid =\sqrt{n^2\mid \mathit{a}{\mid }^2+m^2\mid \mathit{b}{\mid }^2+mn(2\mid \mathit{a}\mid \mid \mathit{b}\mid cos\theta )}$$$$Bylawofcosines,$$$$2\mid \mathit{a}\mid \mid \mathit{b}\mid cos\theta =-\mid \mathit{a}-\mathit{b}{\mid }^2+\mid \mathit{a}{\mid }^2+\mid \mathit{b}{\mid }^2$$$$\therefore \mid GH\mid =\sqrt{n(n+m)\mid \mathit{a}{\mid }^2+m(m+n)\mid \mathit{b}{\mid }^2-mn\mid \mathit{a}-\mathit{b}{\mid }^2}$$$$\mid GH\mid =\sqrt{(n+m)(n\mid \mathit{a}{\mid }^2+m\mid \mathit{b}{\mid }^2)-mn\mid \mathit{a}-\mathit{b}{\mid }^2}$$$$Itcanbeshownthat$$$$\mid \mathit{a}+\mathit{b}{\mid }^2=2(\mid \mathit{a}{\mid }^2+\mid \mathit{b}{\mid }^2)-\mid \mathit{a}-\mathit{b}{\mid }^2$$$$\Rightarrow \mid \mathit{a}+\mathit{b}\mid =\sqrt{2(\mid \mathit{a}{\mid }^2+\mid \mathit{b}{\mid }^2)-\mid \mathit{a}-\mathit{b}{\mid }^2}>0$$$$\mid \mathit{b}-2\mathit{a}{\mid }^2=\mid \mathit{b}{\mid }^2+4\mid \mathit{a}{\mid }^2-2\times 2\mid \mathit{b}\mid \mid \mathit{a}\mid cos\theta$$$$\mid \mathit{b}-2\mathit{a}{\mid }^2=-\mid \mathit{b}{\mid }^2+2\mid \mathit{a}{\mid }^2+2\mid \mathit{a}-\mathit{b}{\mid }^2$$$$\mid \mathit{b}-2\mathit{a}\mid =\sqrt{2(\mid \mathit{a}{\mid }^2+\mid \mathit{a}-\mathit{b}{\mid }^2)-\mid \mathit{b}{\mid }^2}$$$$Let\mid \mathit{a}\mid =c>0,\mid \mathit{b}\mid =b>0and\mid \mathit{a}-\mathit{b}\mid =a>0.$$$$---------------------------$$$$For△GHIconstructedfrom△ABC,$$$$insuchawayasabove,$$$$\therefore \mid GH\mid =\sqrt{(n+m)({nc}^2+{mb}^2)-{mna}^2}$$$$\mathit{w}\mathit{h}\mathit{e}\mathit{r}\mathit{e}$$$$n=-1+\frac{2a}{\sqrt{2(c^2+a^2)-b^2}}+\frac{c}{\sqrt{2(c^2+b^2)-a^2}}$$$$m=\frac{c}{\sqrt{2(c^2+b^2)-a^2}}-\frac{a}{\sqrt{2(c^2+a^2)-b^2}}$$$$-------------------------$$$$Asimilaranalysiscanbeusedto$$$$findtheremaininglengths.Tryit!$$$$(CyclicpermutationofabcIguessispossible.)$$

Commented by Rasheed Soomro last updated on 01/Jul/16

$$\mathcal{TH}ank\mathcal{S}!$$$$Yourapproachisalwaysreliable$$$$butIneedtorevisemyknowledge$$$$about\mathrm{vector}\mathrm{analysis}.$$

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