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Question Number 6384 by Rasheed Soomro last updated on 25/Jun/16

Commented by Rasheed Soomro last updated on 25/Jun/16

$$Questionreposted.$$

Commented by nburiburu last updated on 25/Jun/16

$$thebiseasierthana$$$$sinceABCisequilateraltrianglethen$$$$\sphericalangle HAB=30°andbyconstruction\sphericalangle GHA=30°\left(itcanbedemonstratedbyalternanglesbetweenparallellinesalso\right)$$$$then,sincetheconstructioniscompletelypolarandsimetric$$$$GHI=HIG=IGH=2\times 30°=60°$$$$soitisequilateral.$$

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

$$\mathcal{THANKS}!$$$${Waiting}^{...}forotherpart.$$

Answered by Yozzii last updated on 27/Jun/16

$$\left(a\right)Weknowthat△ABCisequilateral$$$$and\mid AB\mid =x>0.Thus,\mid CB\mid =\mid AC\mid =x.$$$$FisthemidpointofABand\mathrm{\angle }CAF=60°.$$$$Let\alpha =\mathrm{\angle }ACFand\beta =\mathrm{\angle }CFA.$$$$Since\mathrm{\angle }ACF+\mathrm{\angle }CFA+\mathrm{\angle }CAF=180°$$$$\Rightarrow \alpha +\beta +60°=180°\Rightarrow \alpha =120°-\beta .$$$$Notethat0<\alpha ⩽90°and0<\beta ⩽90°.$$$$Bythelawofsines,$$$$\frac{sin\mathrm{\angle }CFA}{\mid AC\mid }=\frac{sin\mathrm{\angle }ACF}{\mid AF\mid }$$$$or\frac{sin\beta }{x}=\frac{sin(120°-\beta )}{0.5x}$$$$sin\beta =2sin(120°-\beta )$$$$sin\beta =2(sin120°cos\beta -sin\beta cos120°)$$$$sin\beta =\sqrt{3}cos\beta +sin\beta$$$$0<\beta ⩽90°\Rightarrow \beta =90°.$$$$Since\beta =90°and\mid AF\mid =\mid FB\mid ,then$$$$FCisaperpendicularbisectorofAB.$$$$Bysymmetryof△ABC,AHandGB$$$$arealsoperpendicularbisectors.$$$$LetObethepointofintersectionofthe$$$$threeperpendicularbisectorsof△ABC.$$$$Then,\mathrm{\angle }HAB=\mathrm{\angle }OAF=\frac{1}{2}\mathrm{\angle }CAF=30°.$$$$Since\mathrm{\angle }AFO=\mathrm{\angle }AFC=90°,△AOFis$$$$right-angledandwecanwrite$$$$cos\mathrm{\angle }OAF=\frac{\mid AF\mid }{\mid AO\mid }\Rightarrow \mid AO\mid =\frac{x/2}{cos30°}=\frac{x}{\sqrt{3}}.$$$$Alsonotethat\mathrm{\angle }AOF=90°-\mathrm{\angle }OAF=60°$$$$and\mathrm{\angle }COH=\mathrm{\angle }AOF=60°since$$$$\mathrm{\angle }AOFand\mathrm{\angle }COHareverticallyopposite$$$$angles.Wecanlikewisededucethat$$$$\mathrm{\angle }GOC=60°byvirtueofthesymmetry$$$$of△ABCaboutthelineCF.$$$$So,\mathrm{\angle }GOH=\mathrm{\angle }GOC+\mathrm{\angle }COH=2\times 60°=120°.$$$$$$$$AisthecentreofthearcHBsothat$$$$\mid AH\mid =\mid AB\mid =x.A,OandHarecollinear$$$$sothat\mid OH\mid =\mid AH\mid -\mid AO\mid =x-\frac{x}{\sqrt{3}}=x(1-\frac{1}{\sqrt{3}}).$$$$Wecansimilarlyfindthat\mid OG\mid =x(1-\frac{1}{\sqrt{3}}).$$$$Bythelawofcosinesin△GOH$$$$\mid GH{\mid }^2=\mid OH{\mid }^2+\mid OG{\mid }^2-2\mid OH\mid \mid OG\mid cos\mathrm{\angle }GOH.$$$$\therefore \mid GH{\mid }^2=2{\left(x(1-\frac{1}{\sqrt{3}})\right)}^2(1-cos120°)$$$$=\frac{2x^2{(\sqrt{3}-1)}^2}{3}\times (1+\frac{1}{2})$$$$Or\mid GH\mid =x(\sqrt{3}-1).$$$$\left(b\right)Rotating△ABCaboutOthrough$$$$120°andthen120°oncemoreshows$$$$thatthelinesHIandGIcoincideexactlywith$$$$theinitialpositionofthelineGH,$$$$havingthesamelengthafterboth$$$$rotations.$$$$Hence,△GHIisequilateral.$$$$Or,bysuchsymmetryof△ABC,$$$$weshouldfindthatboth\mid HI\mid and$$$$\mid GI\mid equalx(\sqrt{3}-1).So,△GHIisequilateral.$$$$$$

Commented by Rasheed Soomro last updated on 28/Jun/16

$$\mathcal{GREAT}!!....Ihaverepostedthe$$$$requestedquestion.$$

Commented by Yozzii last updated on 27/Jun/16

$$Youshouldrepostthequestionconcerning$$$$thetreatmentofthegeneralcase.$$

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