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Question-6384

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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
Question reposted.
$${Question}\:{reposted}. \\ $$
Commented by nburiburu last updated on 25/Jun/16
the b is easier than a since ABC is equilateral triangle then ∢HAB=30° and by construction ∢ GHA=30° (it can be demonstrated by altern angles between parallel lines also) then, since the construction is completely polar and simetric GHI=HIG=IGH=2×30°=60° so it is equilateral.
$${the}\:{b}\:{is}\:{easier}\:{than}\:{a} \\ $$$${since}\:{ABC}\:{is}\:{equilateral}\:{triangle}\:{then} \\ $$$$\sphericalangle{HAB}=\mathrm{30}°\:\:{and}\:{by}\:{construction}\:\sphericalangle\:{GHA}=\mathrm{30}°\:\left({it}\:{can}\:{be}\:{demonstrated}\:{by}\:{altern}\:{angles}\:{between}\:{parallel}\:{lines}\:{also}\right) \\ $$$${then},\:{since}\:{the}\:{construction}\:{is}\:{completely}\:{polar}\:{and}\:{simetric} \\ $$$${GHI}={HIG}={IGH}=\mathrm{2}×\mathrm{30}°=\mathrm{60}°\: \\ $$$${so}\:\:{it}\:{is}\:{equilateral}. \\ $$
Commented by Rasheed Soomro last updated on 27/Jun/16
THANKS! Waiting^(...) for other part.
$$\mathscr{THANKS}!\:\: \\ $$$${Waiting}^{…} \:{for}\:{other}\:{part}. \\ $$
Answered by Yozzii last updated on 27/Jun/16
(a)We know that △ABC is equilateral and ∣AB∣=x>0. Thus, ∣CB∣=∣AC∣=x. F is the midpoint of AB and ∠CAF=60°. Let α=∠ACF and β=∠CFA. Since ∠ACF+∠CFA+∠CAF=180° ⇒α+β+60°=180°⇒α=120°−β. Note that 0<α≤90° and 0<β≤90°. By the law of sines, ((sin∠CFA)/(∣AC∣))=((sin∠ACF)/(∣AF∣)) or ((sinβ)/x)=((sin(120°−β))/(0.5x)) sinβ=2sin(120°−β) sinβ=2(sin120°cosβ−sinβcos120°) sinβ=(√3)cosβ+sinβ 0<β≤90°⇒β=90°. Since β=90° and ∣AF∣=∣FB∣, then FC is a perpendicular bisector of AB. By symmetry of △ABC, AH and GB are also perpendicular bisectors. Let O be the point of intersection of the three perpendicular bisectors of △ABC. Then, ∠HAB=∠OAF=(1/2)∠CAF=30°. Since ∠AFO=∠AFC=90°, △AOF is right−angled and we can write cos∠OAF=((∣AF∣)/(∣AO∣))⇒∣AO∣=((x/2)/(cos30°))=(x/( (√3))). Also note that ∠AOF=90°−∠OAF=60° and ∠COH=∠AOF=60° since ∠AOF and ∠COH are vertically opposite angles. We can likewise deduce that ∠GOC=60° by virtue of the symmetry of △ABC about the line CF. So, ∠GOH=∠GOC+∠COH=2×60°=120°. A is the centre of the arc HB so that ∣AH∣=∣AB∣=x. A,O and H are collinear so that ∣OH∣=∣AH∣−∣AO∣=x−(x/( (√3)))=x(1−(1/( (√3)))). We can similarly find that ∣OG∣=x(1−(1/( (√3)))). By the law of cosines in △GOH ∣GH∣^2 =∣OH∣^2 +∣OG∣^2 −2∣OH∣∣OG∣cos∠GOH. ∴∣GH∣^2 =2(x(1−(1/( (√3)))))^2 (1−cos120°) =((2x^2 ((√3)−1)^2 )/3)×(1+(1/2)) Or ∣GH∣=x((√3)−1). (b)Rotating △ABC about O through 120° and then 120° once more shows that the lines HI and GI coincide exactly with the initial position of the line GH, having the same length after both rotations. Hence, △GHI is equilateral. Or, by such symmetry of △ABC, we should find that both ∣HI∣ and ∣GI∣ equal x((√3)−1). So, △GHI is equilateral.
$$\left({a}\right){We}\:{know}\:{that}\:\bigtriangleup{ABC}\:{is}\:{equilateral} \\ $$$${and}\:\mid{AB}\mid={x}>\mathrm{0}.\:{Thus},\:\mid{CB}\mid=\mid{AC}\mid={x}. \\ $$$${F}\:{is}\:{the}\:{midpoint}\:{of}\:{AB}\:{and}\:\angle{CAF}=\mathrm{60}°. \\ $$$${Let}\:\alpha=\angle{ACF}\:{and}\:\beta=\angle{CFA}.\: \\ $$$${Since}\:\angle{ACF}+\angle{CFA}+\angle{CAF}=\mathrm{180}° \\ $$$$\Rightarrow\alpha+\beta+\mathrm{60}°=\mathrm{180}°\Rightarrow\alpha=\mathrm{120}°−\beta. \\ $$$${Note}\:{that}\:\mathrm{0}<\alpha\leqslant\mathrm{90}°\:{and}\:\mathrm{0}<\beta\leqslant\mathrm{90}°. \\ $$$${By}\:{the}\:{law}\:{of}\:{sines}, \\ $$$$\frac{{sin}\angle{CFA}}{\mid{AC}\mid}=\frac{{sin}\angle{ACF}}{\mid{AF}\mid} \\ $$$${or}\:\:\frac{{sin}\beta}{{x}}=\frac{{sin}\left(\mathrm{120}°−\beta\right)}{\mathrm{0}.\mathrm{5}{x}} \\ $$$${sin}\beta=\mathrm{2}{sin}\left(\mathrm{120}°−\beta\right) \\ $$$${sin}\beta=\mathrm{2}\left({sin}\mathrm{120}°{cos}\beta−{sin}\beta{cos}\mathrm{120}°\right) \\ $$$${sin}\beta=\sqrt{\mathrm{3}}{cos}\beta+{sin}\beta \\ $$$$\mathrm{0}<\beta\leqslant\mathrm{90}°\Rightarrow\beta=\mathrm{90}°. \\ $$$${Since}\:\beta=\mathrm{90}°\:{and}\:\mid{AF}\mid=\mid{FB}\mid,\:{then} \\ $$$${FC}\:{is}\:{a}\:{perpendicular}\:{bisector}\:{of}\:{AB}. \\ $$$${By}\:{symmetry}\:{of}\:\bigtriangleup{ABC},\:{AH}\:{and}\:{GB} \\ $$$${are}\:{also}\:{perpendicular}\:{bisectors}. \\ $$$${Let}\:{O}\:{be}\:{the}\:{point}\:{of}\:{intersection}\:{of}\:{the} \\ $$$${three}\:{perpendicular}\:{bisectors}\:{of}\:\bigtriangleup{ABC}. \\ $$$${Then},\:\angle{HAB}=\angle{OAF}=\frac{\mathrm{1}}{\mathrm{2}}\angle{CAF}=\mathrm{30}°. \\ $$$${Since}\:\angle{AFO}=\angle{AFC}=\mathrm{90}°,\:\bigtriangleup{AOF}\:{is} \\ $$$${right}−{angled}\:{and}\:{we}\:{can}\:{write} \\ $$$${cos}\angle{OAF}=\frac{\mid{AF}\mid}{\mid{AO}\mid}\Rightarrow\mid{AO}\mid=\frac{{x}/\mathrm{2}}{{cos}\mathrm{30}°}=\frac{{x}}{\:\sqrt{\mathrm{3}}}. \\ $$$${Also}\:{note}\:{that}\:\angle{AOF}=\mathrm{90}°−\angle{OAF}=\mathrm{60}° \\ $$$${and}\:\angle{COH}=\angle{AOF}=\mathrm{60}°\:{since}\: \\ $$$$\angle{AOF}\:{and}\:\angle{COH}\:{are}\:{vertically}\:{opposite} \\ $$$${angles}.\:{We}\:{can}\:{likewise}\:{deduce}\:{that}\: \\ $$$$\angle{GOC}=\mathrm{60}°\:{by}\:{virtue}\:{of}\:{the}\:{symmetry} \\ $$$${of}\:\bigtriangleup{ABC}\:{about}\:{the}\:{line}\:{CF}.\: \\ $$$${So},\:\angle{GOH}=\angle{GOC}+\angle{COH}=\mathrm{2}×\mathrm{60}°=\mathrm{120}°. \\ $$$$ \\ $$$${A}\:{is}\:{the}\:{centre}\:{of}\:{the}\:{arc}\:{HB}\:{so}\:{that} \\ $$$$\mid{AH}\mid=\mid{AB}\mid={x}.\:{A},{O}\:{and}\:{H}\:{are}\:{collinear} \\ $$$${so}\:{that}\:\mid{OH}\mid=\mid{AH}\mid−\mid{AO}\mid={x}−\frac{{x}}{\:\sqrt{\mathrm{3}}}={x}\left(\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right). \\ $$$${We}\:{can}\:{similarly}\:{find}\:{that}\:\mid{OG}\mid={x}\left(\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right). \\ $$$${By}\:{the}\:{law}\:{of}\:{cosines}\:{in}\:\bigtriangleup{GOH} \\ $$$$\mid{GH}\mid^{\mathrm{2}} =\mid{OH}\mid^{\mathrm{2}} +\mid{OG}\mid^{\mathrm{2}} −\mathrm{2}\mid{OH}\mid\mid{OG}\mid{cos}\angle{GOH}. \\ $$$$\therefore\mid{GH}\mid^{\mathrm{2}} =\mathrm{2}\left({x}\left(\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)\right)^{\mathrm{2}} \left(\mathrm{1}−{cos}\mathrm{120}°\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{2}{x}^{\mathrm{2}} \left(\sqrt{\mathrm{3}}−\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{3}}×\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$${Or}\:\mid{GH}\mid={x}\left(\sqrt{\mathrm{3}}−\mathrm{1}\right). \\ $$$$\left({b}\right){Rotating}\:\bigtriangleup{ABC}\:{about}\:{O}\:{through} \\ $$$$\mathrm{120}°\:{and}\:{then}\:\mathrm{120}°\:{once}\:{more}\:{shows} \\ $$$${that}\:{the}\:{lines}\:{HI}\:{and}\:{GI}\:{coincide}\:{exactly}\:{with} \\ $$$${the}\:\:{initial}\:{position}\:{of}\:{the}\:{line}\:{GH}, \\ $$$${having}\:{the}\:{same}\:{length}\:{after}\:{both} \\ $$$${rotations}. \\ $$$${Hence},\:\bigtriangleup{GHI}\:{is}\:{equilateral}. \\ $$$${Or},\:{by}\:{such}\:{symmetry}\:{of}\:\bigtriangleup{ABC}, \\ $$$${we}\:{should}\:{find}\:{that}\:{both}\:\mid{HI}\mid\:{and} \\ $$$$\mid{GI}\mid\:{equal}\:{x}\left(\sqrt{\mathrm{3}}−\mathrm{1}\right).\:{So},\:\bigtriangleup{GHI}\:{is}\:{equilateral}. \\ $$$$ \\ $$
Commented by Rasheed Soomro last updated on 28/Jun/16
GREAT !!....I have reposted the requested question.
$$\mathcal{GREAT}\:!!….{I}\:{have}\:{reposted}\:{the} \\ $$$${requested}\:\:{question}. \\ $$
Commented by Yozzii last updated on 27/Jun/16
You should repost the question concerning the treatment of the general case.
$${You}\:{should}\:{repost}\:{the}\:{question}\:{concerning} \\ $$$${the}\:{treatment}\:{of}\:{the}\:{general}\:{case}. \\ $$

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