Question Number 78060 by aliesam last updated on 13/Jan/20
Commented by aliesam last updated on 13/Jan/20
$${the}\:{triangle}\:{is}\:{equilateral} \\ $$
Answered by ajfour last updated on 13/Jan/20
$${answer}\:{is}\:\:\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\:. \\ $$
Answered by mr W last updated on 13/Jan/20
$${r}={incircle}\:{radius} \\ $$$${a}={side}\:{length}\:{of}\:{equilateral} \\ $$$$\frac{\mathrm{3}{ar}}{\mathrm{2}}=\frac{\sqrt{\mathrm{3}}{a}^{\mathrm{2}} }{\mathrm{4}} \\ $$$${a}=\mathrm{2}\sqrt{\mathrm{3}}{r} \\ $$$${GH}=\mathrm{2}{r}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}=\sqrt{\mathrm{3}}{r} \\ $$$${let}\:{BH}={x} \\ $$$${a}^{\mathrm{2}} ={x}^{\mathrm{2}} +\left(\sqrt{\mathrm{3}}{r}+{x}\right)^{\mathrm{2}} +\mathrm{2}{x}\left(\sqrt{\mathrm{3}}{r}+{x}\right)\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} +\sqrt{\mathrm{3}}{rx}−\mathrm{3}{r}^{\mathrm{2}} =\mathrm{0} \\ $$$${x}=\frac{\left(\sqrt{\mathrm{5}}−\mathrm{1}\right)\sqrt{\mathrm{3}}{r}}{\mathrm{2}} \\ $$$$\frac{{GH}}{{BH}}=\frac{\mathrm{2}\sqrt{\mathrm{3}}{r}}{\left(\sqrt{\mathrm{5}}−\mathrm{1}\right)\sqrt{\mathrm{3}}{r}}=\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}−\mathrm{1}}=\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{2}}\:\left(=\Phi\right) \\ $$
Commented by aliesam last updated on 13/Jan/20
$${god}\:{bless}\:{you}\:{sir} \\ $$
Commented by jagoll last updated on 13/Jan/20
$${waw}\:{the}\:{golden}\:{ratio} \\ $$
Commented by jagoll last updated on 14/Jan/20
$${sir}\:{W}\:{how}\:{your}\:{got}\:{equation} \\ $$$$\frac{\mathrm{3}{ar}}{\mathrm{2}}=\frac{\sqrt{\mathrm{3}}\:{a}^{\mathrm{2}} }{\mathrm{4}}? \\ $$
Commented by mr W last updated on 14/Jan/20
$${area}\:{of}\:{triangle}\:{ABC}=\Delta=\frac{\mathrm{3}{ar}}{\mathrm{2}}\:{or}\:\Delta=\frac{\sqrt{\mathrm{3}}\:{a}^{\mathrm{2}} }{\mathrm{4}} \\ $$