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Question Number 223962 by behi834171 last updated on 11/Aug/25
Commented by behi834171 last updated on 11/Aug/25
Area of circles: { ((yellow=1)),((blue=(√2))),((green=(√3))) :} (any unit)^2 Area of gray(big) circle=? circles and lines are tangent to each other at contact points.
$$\boldsymbol{{Area}}\:\boldsymbol{{of}}\:\boldsymbol{{circles}}: \\ $$$$\begin{cases}{\boldsymbol{{yellow}}=\mathrm{1}}\\{\boldsymbol{{blue}}=\sqrt{\mathrm{2}}}\\{\boldsymbol{{green}}=\sqrt{\mathrm{3}}}\end{cases}\:\:\:\:\:\:\left(\boldsymbol{{any}}\:\boldsymbol{{unit}}\right)^{\mathrm{2}} \\ $$$$\boldsymbol{{Area}}\:\boldsymbol{{of}}\:\boldsymbol{{gray}}\left(\boldsymbol{{big}}\right)\:\:\boldsymbol{{circle}}=? \\ $$$$ \\ $$$$\boldsymbol{{circles}}\:\boldsymbol{{and}}\:\boldsymbol{{lines}}\:\boldsymbol{{are}}\:\boldsymbol{{tangent}}\:\boldsymbol{{to}} \\ $$$$\boldsymbol{{each}}\:\boldsymbol{{other}}\:\boldsymbol{{at}}\:\boldsymbol{{contact}}\:\boldsymbol{{points}}. \\ $$
Answered by behi834171 last updated on 12/Aug/25
Radius of circles: yellow=r_1 ,blue=r_2 ,green=r_3 ,gray=R ⇒ { ((cos(A/2)=((2(√(r_1 .R)))/(R+r_1 )),sin(A/2)=((R−r_1 )/(R+r_1 )))),((cos(B/2)=((2(√(r_2 .R)))/(R+r_2 )),sin(B/2)=((R−r_2 )/(R+r_2 )))),((cos(C/2)=((2(√(r_3 .R)))/(R+r_3 )),sin(C/2)=((R−r_3 )/(R+r_3 )))) :} (A/2)+(B/2)=90−(C/2)⇒sin((A/2)+(B/2))=cos(C/2)⇒ ⇒((R−r_1 )/(R+r_1 )).((2(√(r_2 .R)))/(R+r_2 ))+((R−r_2 )/(R+r_2 )).((2(√(r_1 .R)))/(R+r_1 ))=((2(√(r_3 .R)))/(R+r_3 ))⇒ (((√r_1 )(R−r_2 )+(√r_2 )(R−r_1 ))/((R+r_1 ).(R+r_2 )))=((√r_3 )/(R+r_3 ))⇒ ((((√r_1 )+(√r_2 ))(R−(√(r_1 .r_2 ))))/((R+r_1 )(R+r_2 )))=((√r_3 )/(R+r_3 )) according to mr.proph: mr.W′s commnet and refer to Q77425,this is leads to: R=(√(r_1 r_2 ))+(√(r_3 r_2 ))+(√(r_1 r_3 ))
$${Radius}\:{of}\:{circles}: \\ $$$$\boldsymbol{{yellow}}=\boldsymbol{{r}}_{\mathrm{1}} ,\boldsymbol{{blue}}=\boldsymbol{{r}}_{\mathrm{2}} ,\boldsymbol{{green}}=\boldsymbol{{r}}_{\mathrm{3}} ,\boldsymbol{{gray}}=\boldsymbol{{R}} \\ $$$$\Rightarrow\begin{cases}{\boldsymbol{{cos}}\frac{\boldsymbol{{A}}}{\mathrm{2}}=\frac{\mathrm{2}\sqrt{\boldsymbol{{r}}_{\mathrm{1}} .\boldsymbol{{R}}}}{\boldsymbol{{R}}+\boldsymbol{{r}}_{\mathrm{1}} },\boldsymbol{{sin}}\frac{\boldsymbol{{A}}}{\mathrm{2}}=\frac{\boldsymbol{{R}}−\boldsymbol{{r}}_{\mathrm{1}} }{\boldsymbol{{R}}+\boldsymbol{{r}}_{\mathrm{1}} }}\\{\boldsymbol{{cos}}\frac{\boldsymbol{{B}}}{\mathrm{2}}=\frac{\mathrm{2}\sqrt{\boldsymbol{{r}}_{\mathrm{2}} .\boldsymbol{{R}}}}{\boldsymbol{{R}}+\boldsymbol{{r}}_{\mathrm{2}} },\boldsymbol{{sin}}\frac{\boldsymbol{{B}}}{\mathrm{2}}=\frac{\boldsymbol{{R}}−\boldsymbol{{r}}_{\mathrm{2}} }{\boldsymbol{{R}}+\boldsymbol{{r}}_{\mathrm{2}} }}\\{\boldsymbol{{cos}}\frac{\boldsymbol{{C}}}{\mathrm{2}}=\frac{\mathrm{2}\sqrt{\boldsymbol{{r}}_{\mathrm{3}} .\boldsymbol{{R}}}}{\boldsymbol{{R}}+\boldsymbol{{r}}_{\mathrm{3}} },\boldsymbol{{sin}}\frac{\boldsymbol{{C}}}{\mathrm{2}}=\frac{\boldsymbol{{R}}−\boldsymbol{{r}}_{\mathrm{3}} }{\boldsymbol{{R}}+\boldsymbol{{r}}_{\mathrm{3}} }}\end{cases} \\ $$$$\frac{\boldsymbol{{A}}}{\mathrm{2}}+\frac{\boldsymbol{{B}}}{\mathrm{2}}=\mathrm{90}−\frac{\boldsymbol{{C}}}{\mathrm{2}}\Rightarrow\boldsymbol{{sin}}\left(\frac{\boldsymbol{{A}}}{\mathrm{2}}+\frac{\boldsymbol{{B}}}{\mathrm{2}}\right)=\boldsymbol{{cos}}\frac{\boldsymbol{{C}}}{\mathrm{2}}\Rightarrow \\ $$$$\Rightarrow\frac{\boldsymbol{{R}}−\boldsymbol{{r}}_{\mathrm{1}} }{\boldsymbol{{R}}+\boldsymbol{{r}}_{\mathrm{1}} }.\frac{\mathrm{2}\sqrt{\boldsymbol{{r}}_{\mathrm{2}} .\boldsymbol{{R}}}}{\boldsymbol{{R}}+\boldsymbol{{r}}_{\mathrm{2}} }+\frac{\boldsymbol{{R}}−\boldsymbol{{r}}_{\mathrm{2}} }{\boldsymbol{{R}}+\boldsymbol{{r}}_{\mathrm{2}} }.\frac{\mathrm{2}\sqrt{\boldsymbol{{r}}_{\mathrm{1}} .\boldsymbol{{R}}}}{\boldsymbol{{R}}+\boldsymbol{{r}}_{\mathrm{1}} }=\frac{\mathrm{2}\sqrt{\boldsymbol{{r}}_{\mathrm{3}} .\boldsymbol{{R}}}}{\boldsymbol{{R}}+\boldsymbol{{r}}_{\mathrm{3}} }\Rightarrow \\ $$$$\frac{\sqrt{\boldsymbol{{r}}_{\mathrm{1}} }\left(\boldsymbol{{R}}−\boldsymbol{{r}}_{\mathrm{2}} \right)+\sqrt{\boldsymbol{{r}}_{\mathrm{2}} }\left(\boldsymbol{{R}}−\boldsymbol{{r}}_{\mathrm{1}} \right)}{\left(\boldsymbol{{R}}+\boldsymbol{{r}}_{\mathrm{1}} \right).\left(\boldsymbol{{R}}+\boldsymbol{{r}}_{\mathrm{2}} \right)}=\frac{\sqrt{\boldsymbol{{r}}_{\mathrm{3}} }}{\boldsymbol{{R}}+\boldsymbol{{r}}_{\mathrm{3}} }\Rightarrow \\ $$$$\frac{\left(\sqrt{\boldsymbol{{r}}_{\mathrm{1}} }+\sqrt{\boldsymbol{{r}}_{\mathrm{2}} }\right)\left(\boldsymbol{{R}}−\sqrt{\boldsymbol{{r}}_{\mathrm{1}} .\boldsymbol{{r}}_{\mathrm{2}} }\right)}{\left(\boldsymbol{{R}}+\boldsymbol{{r}}_{\mathrm{1}} \right)\left(\boldsymbol{{R}}+\boldsymbol{{r}}_{\mathrm{2}} \right)}=\frac{\sqrt{\boldsymbol{{r}}_{\mathrm{3}} }}{\boldsymbol{{R}}+\boldsymbol{{r}}_{\mathrm{3}} } \\ $$$${according}\:\:{to}\:{mr}.{proph}:\:{mr}.{W}'{s}\:{commnet} \\ $$$${and}\:{refer}\:{to}\:{Q}\mathrm{77425},{this}\:{is}\:{leads}\:{to}: \\ $$$$ \\ $$$$\boldsymbol{{R}}=\sqrt{\boldsymbol{{r}}_{\mathrm{1}} \boldsymbol{{r}}_{\mathrm{2}} }+\sqrt{\boldsymbol{{r}}_{\mathrm{3}} \boldsymbol{{r}}_{\mathrm{2}} }+\sqrt{\boldsymbol{{r}}_{\mathrm{1}} \boldsymbol{{r}}_{\mathrm{3}} } \\ $$$$ \\ $$
Answered by mr W last updated on 11/Aug/25
we have R=(√(r_1 r_2 ))+(√(r_2 r_3 ))+(√(r_3 r_1 )) =(1/( (√π)))(((A_1 A_2 ))^(1/4) +((A_2 A_3 ))^(1/4) +((A_3 A_1 ))^(1/4) ) =(1/( (√π)))(((1×(√2)))^(1/4) +(((√2)×(√3)))^(1/4) +(((√3)×1))^(1/4) ) =(1/( (√π)))(2^(1/8) +6^(1/8) +3^(1/8) ) πR^2 =(2^(1/8) +6^(1/8) +3^(1/8) )^2 ≈12.171
$${we}\:{have} \\ $$$${R}=\sqrt{{r}_{\mathrm{1}} {r}_{\mathrm{2}} }+\sqrt{{r}_{\mathrm{2}} {r}_{\mathrm{3}} }+\sqrt{{r}_{\mathrm{3}} {r}_{\mathrm{1}} } \\ $$$$\:\:\:=\frac{\mathrm{1}}{\:\sqrt{\pi}}\left(\sqrt[{\mathrm{4}}]{{A}_{\mathrm{1}} {A}_{\mathrm{2}} }+\sqrt[{\mathrm{4}}]{{A}_{\mathrm{2}} {A}_{\mathrm{3}} }+\sqrt[{\mathrm{4}}]{{A}_{\mathrm{3}} {A}_{\mathrm{1}} }\right) \\ $$$$\:\:\:=\frac{\mathrm{1}}{\:\sqrt{\pi}}\left(\sqrt[{\mathrm{4}}]{\mathrm{1}×\sqrt{\mathrm{2}}}+\sqrt[{\mathrm{4}}]{\sqrt{\mathrm{2}}×\sqrt{\mathrm{3}}}+\sqrt[{\mathrm{4}}]{\sqrt{\mathrm{3}}×\mathrm{1}}\right) \\ $$$$\:\:\:=\frac{\mathrm{1}}{\:\sqrt{\pi}}\left(\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{8}}} +\mathrm{6}^{\frac{\mathrm{1}}{\mathrm{8}}} +\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{8}}} \right) \\ $$$$\pi{R}^{\mathrm{2}} =\left(\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{8}}} +\mathrm{6}^{\frac{\mathrm{1}}{\mathrm{8}}} +\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{8}}} \right)^{\mathrm{2}} \approx\mathrm{12}.\mathrm{171} \\ $$
Commented by behi834171 last updated on 11/Aug/25
thank you so much,dear master. how can get your formula from my last equation?(R, in term of r_i ′s) i cant handle this ...
$$\boldsymbol{{thank}}\:\boldsymbol{{you}}\:\boldsymbol{{so}}\:\boldsymbol{{much}},\boldsymbol{{dear}}\:\boldsymbol{{master}}. \\ $$$$ \\ $$$$\boldsymbol{{how}}\:\boldsymbol{{can}}\:\boldsymbol{{get}}\:\boldsymbol{{your}}\:\boldsymbol{{formula}}\:\boldsymbol{{from}} \\ $$$$\boldsymbol{{my}}\:\boldsymbol{{last}}\:\boldsymbol{{equation}}?\left(\boldsymbol{{R}},\:\boldsymbol{{in}}\:\boldsymbol{{term}}\:\boldsymbol{{of}}\:\boldsymbol{{r}}_{\boldsymbol{{i}}} '\boldsymbol{{s}}\right) \\ $$$$\boldsymbol{{i}}\:\boldsymbol{{cant}}\:\boldsymbol{{handle}}\:\boldsymbol{{this}}\:… \\ $$
Commented by mr W last updated on 12/Aug/25
your equation should lead to the formula i applied.
$${your}\:{equation}\:{should}\:{lead}\:{to}\:{the} \\ $$$${formula}\:{i}\:{applied}. \\ $$
Commented by mr W last updated on 12/Aug/25
see also Q77425
$${see}\:{also}\:{Q}\mathrm{77425} \\ $$

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