Question Number 224108 by MirHasibulHossain last updated on 21/Aug/25
Answered by mr W last updated on 21/Aug/25
$${i}\:{think}\:{if}\:{the}\:{three}\:{big}\:{blue}\:{circles}\: \\ $$$${have}\:{different}\:{radii}\:{a}_{\mathrm{1}} ,\:{a}_{\mathrm{2}} ,\:{a}_{\mathrm{3}} ,\: \\ $$$${generally}\:{we}\:{have} \\ $$$${bc}=\frac{\mathrm{1}}{\frac{\mathrm{2}}{{a}_{\mathrm{1}} {a}_{\mathrm{2}} }+\frac{\mathrm{2}}{{a}_{\mathrm{2}} {a}_{\mathrm{3}} }+\frac{\mathrm{2}}{{a}_{\mathrm{3}} {a}_{\mathrm{1}} }−\frac{\mathrm{1}}{{a}_{\mathrm{1}} ^{\mathrm{2}} }−\frac{\mathrm{1}}{{a}_{\mathrm{2}} ^{\mathrm{2}} }−\frac{\mathrm{1}}{{a}_{\mathrm{3}} ^{\mathrm{2}} }} \\ $$
Commented by fantastic last updated on 21/Aug/25
$${I}\:{got} \\ $$$$\mathrm{sin}^{−\mathrm{1}} \left(\frac{{a}_{\mathrm{1}} }{{a}_{\mathrm{1}} +{c}}\right)+\mathrm{sin}^{−\mathrm{1}} \left(\frac{{a}_{\mathrm{2}} }{{a}_{\mathrm{2}} +{c}}\right)+\mathrm{sin}^{−\mathrm{1}} \left(\frac{{a}_{\mathrm{3}} }{{a}_{\mathrm{3}} +{c}}\right)=\mathrm{180} \\ $$
Answered by mr W last updated on 21/Aug/25
$${a}=\frac{\sqrt{\mathrm{3}}\left({a}+{c}\right)}{\mathrm{2}}\:\Rightarrow{c}=\left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}−\mathrm{1}\right){a} \\ $$$${b}=\mathrm{2}{a}+{c}=\left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}+\mathrm{1}\right){a} \\ $$$$\mathrm{3}{bc}=\mathrm{3}\left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}+\mathrm{1}\right)\left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}−\mathrm{1}\right){a}^{\mathrm{2}} =\mathrm{3}\left(\frac{\mathrm{4}}{\mathrm{3}}−\mathrm{1}\right){a}^{\mathrm{2}} ={a}^{\mathrm{2}} \:\checkmark \\ $$
Commented by fantastic last updated on 21/Aug/25
$${Is}\:{your}\:{problem}\:{fixed}\:{now}? \\ $$
Commented by mr W last updated on 21/Aug/25
$${no}.\:{it}'{s}\:{still}\:{there}. \\ $$
Answered by fantastic last updated on 21/Aug/25
$$\left(\mathrm{2}{a}\right)^{\mathrm{2}} =\left({c}+{a}\right)^{\mathrm{2}} +\left({c}+{a}\right)^{\mathrm{2}} −\mathrm{2}\left({c}+{a}\right)^{\mathrm{2}} \mathrm{cos}\:\mathrm{120}^{\mathrm{0}} \\ $$$$\Rightarrow\mathrm{3}\left({c}+{a}\right)^{\mathrm{2}} \\ $$$$\mathrm{2}{a}=\sqrt{\mathrm{3}}\left({c}+{a}\right) \\ $$$${a}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\left({c}+{a}\right)\Rightarrow{c}=\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}{a}−{a}=\left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}−\mathrm{1}\right){a} \\ $$$$\mathrm{2}{a}+{c}={b} \\ $$$$\mathrm{2}{a}+{a}\left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}−\mathrm{1}\right)={b} \\ $$$${a}\left(\mathrm{1}+\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\right)={b} \\ $$$${bc}={a}\left(\mathrm{1}+\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\right)×{a}\left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}−\mathrm{1}\right)={a}^{\mathrm{2}} \left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}+\mathrm{1}\right)\left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}−\mathrm{1}\right) \\ $$$$={a}^{\mathrm{2}} \left(\frac{\mathrm{4}}{\mathrm{3}}−\mathrm{1}\right)={a}^{\mathrm{2}} \frac{\mathrm{1}}{\mathrm{3}} \\ $$$${a}^{\mathrm{2}} =\mathrm{3}{bc} \\ $$