Question Number 53792 by behi83417@gmail.com last updated on 25/Jan/19
Commented by behi83417@gmail.com last updated on 25/Jan/19
$${A}\overset{\blacktriangle} {{B}C},{equilateral}\:{with}:{AB}=\mathrm{1}. \\ $$$${circle}\:{with}\:{radius}:\:\boldsymbol{{r}}=\frac{\mathrm{1}}{\mathrm{2}\pi},\:{tangents}\:{to} \\ $$$${AC}\:{and}\:{AB}\:{at}:\:{D},{E}. \\ $$$$\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{min}}\:\boldsymbol{\mathrm{distance}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{circle}}\:\boldsymbol{\mathrm{from}}\:\boldsymbol{\mathrm{vertices}} \\ $$$$\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{A}}\overset{\blacktriangle} {\boldsymbol{\mathrm{B}C}}. \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 26/Jan/19
Commented by tanmay.chaudhury50@gmail.com last updated on 26/Jan/19
$${A}\left(\mathrm{0},\mathrm{0}\right)\:\:{B}\left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)\:\:{C}\left(\frac{−\mathrm{1}}{\mathrm{2}},\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right) \\ $$$${centre}\:{of}\:{circle}\:{on}\:{y}\:{axis}\left(\mathrm{0},{b}\right) \\ $$$${sin}\mathrm{30}^{{o}} =\frac{{r}}{{b}}\:\:{b}=\mathrm{2}{r}=\mathrm{2}×\frac{\mathrm{1}}{\mathrm{2}\pi}=\frac{\mathrm{1}}{\pi} \\ $$$${so}\:{minimum}\:{distance}={b}=\mathrm{2}{r}=\frac{\mathrm{1}}{\pi} \\ $$