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Question Number 78060 by aliesam last updated on 13/Jan/20

Commented by aliesam last updated on 13/Jan/20

the triangle is equilateral

$${the}\:{triangle}\:{is}\:{equilateral} \\ $$

Answered by ajfour last updated on 13/Jan/20

answer is  ((1+(√5))/2) .

$${answer}\:{is}\:\:\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\:. \\ $$

Answered by mr W last updated on 13/Jan/20

r=incircle radius  a=side length of equilateral  ((3ar)/2)=(((√3)a^2 )/4)  a=2(√3)r  GH=2r((√3)/2)=(√3)r  let BH=x  a^2 =x^2 +((√3)r+x)^2 +2x((√3)r+x)(1/2)  x^2 +(√3)rx−3r^2 =0  x=((((√5)−1)(√3)r)/2)  ((GH)/(BH))=((2(√3)r)/(((√5)−1)(√3)r))=(2/((√5)−1))=(((√5)+1)/2) (=Φ)

$${r}={incircle}\:{radius} \\ $$$${a}={side}\:{length}\:{of}\:{equilateral} \\ $$$$\frac{\mathrm{3}{ar}}{\mathrm{2}}=\frac{\sqrt{\mathrm{3}}{a}^{\mathrm{2}} }{\mathrm{4}} \\ $$$${a}=\mathrm{2}\sqrt{\mathrm{3}}{r} \\ $$$${GH}=\mathrm{2}{r}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}=\sqrt{\mathrm{3}}{r} \\ $$$${let}\:{BH}={x} \\ $$$${a}^{\mathrm{2}} ={x}^{\mathrm{2}} +\left(\sqrt{\mathrm{3}}{r}+{x}\right)^{\mathrm{2}} +\mathrm{2}{x}\left(\sqrt{\mathrm{3}}{r}+{x}\right)\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} +\sqrt{\mathrm{3}}{rx}−\mathrm{3}{r}^{\mathrm{2}} =\mathrm{0} \\ $$$${x}=\frac{\left(\sqrt{\mathrm{5}}−\mathrm{1}\right)\sqrt{\mathrm{3}}{r}}{\mathrm{2}} \\ $$$$\frac{{GH}}{{BH}}=\frac{\mathrm{2}\sqrt{\mathrm{3}}{r}}{\left(\sqrt{\mathrm{5}}−\mathrm{1}\right)\sqrt{\mathrm{3}}{r}}=\frac{\mathrm{2}}{\sqrt{\mathrm{5}}−\mathrm{1}}=\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{2}}\:\left(=\Phi\right) \\ $$

Commented by aliesam last updated on 13/Jan/20

god bless you sir

$${god}\:{bless}\:{you}\:{sir} \\ $$

Commented by jagoll last updated on 13/Jan/20

waw the golden ratio

$${waw}\:{the}\:{golden}\:{ratio} \\ $$

Commented by jagoll last updated on 14/Jan/20

sir W how your got equation  ((3ar)/2)=(((√3) a^2 )/4)?

$${sir}\:{W}\:{how}\:{your}\:{got}\:{equation} \\ $$$$\frac{\mathrm{3}{ar}}{\mathrm{2}}=\frac{\sqrt{\mathrm{3}}\:{a}^{\mathrm{2}} }{\mathrm{4}}? \\ $$

Commented by mr W last updated on 14/Jan/20

area of triangle ABC=Δ=((3ar)/2) or Δ=(((√3) a^2 )/4)

$${area}\:{of}\:{triangle}\:{ABC}=\Delta=\frac{\mathrm{3}{ar}}{\mathrm{2}}\:{or}\:\Delta=\frac{\sqrt{\mathrm{3}}\:{a}^{\mathrm{2}} }{\mathrm{4}} \\ $$

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