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Question Number 111926 by ajfour last updated on 05/Sep/20

Commented by ajfour last updated on 05/Sep/20

If △ABC is equilateral with side s, find radii of the three circles in terms of s.

$${If}\:\bigtriangleup{ABC}\:{is}\:{equilateral}\:{with}\:{side}\:\boldsymbol{{s}}, \\ $$$${find}\:{radii}\:{of}\:{the}\:{three}\:{circles}\:{in} \\ $$$${terms}\:{of}\:{s}. \\ $$

Answered by mr W last updated on 05/Sep/20

Commented by ajfour last updated on 05/Sep/20

from (ii): (β+γ)^2 =β^2 +(1−γ)^2 −β(1−γ) ...(ii) 2βγ=−2γ+1−β+βγ 𝛃=((1−2𝛄)/(1+𝛄)) ....(I) & from γ(1−8β)=4β −1 4𝛃=((1+𝛄)/(1+2𝛄)) ...(II) ⇒ ((4(1−2γ))/(1+γ))=((1+γ)/(1+2γ)) ⇒ 4−16γ^2 =1+2γ+γ^2 ⇒ 17γ^2 +2γ−3=0 ⇒ 𝛄 = ((2(√(13))−1)/(17)) β=((1−2γ)/(1+γ)) = ((1−(((4(√(13))−2))/(17)))/(1+(((2(√(13))−1))/(17)))) =((19−4(√(13)))/(16+2(√(13))))=(((19−4(√(13)))(16−2(√(13))))/(204)) =((304+104−102(√(13)))/(204)) =((4−(√(13)))/2) 𝛃 =((4−(√(13)))/2) 𝛂=(1/2)−(((4−(√(13))))/2) =(((√(13))−3)/2) 𝛄 = ((2(√(13))−1)/(17)) Thank you very much, mrW Sir! great solution!

$${from}\:\left({ii}\right): \\ $$$$\left(\beta+\gamma\right)^{\mathrm{2}} =\beta^{\mathrm{2}} +\left(\mathrm{1}−\gamma\right)^{\mathrm{2}} −\beta\left(\mathrm{1}−\gamma\right)\:\:\:...\left({ii}\right) \\ $$$$\mathrm{2}\beta\gamma=−\mathrm{2}\gamma+\mathrm{1}−\beta+\beta\gamma \\ $$$$\boldsymbol{\beta}=\frac{\mathrm{1}−\mathrm{2}\boldsymbol{\gamma}}{\mathrm{1}+\boldsymbol{\gamma}}\:\:\:\:\:....\left({I}\right) \\ $$$$\&\:\:\:{from}\:\:\:\gamma\left(\mathrm{1}−\mathrm{8}\beta\right)=\mathrm{4}\beta\:−\mathrm{1} \\ $$$$\mathrm{4}\boldsymbol{\beta}=\frac{\mathrm{1}+\boldsymbol{\gamma}}{\mathrm{1}+\mathrm{2}\boldsymbol{\gamma}}\:\:\:\:...\left({II}\right) \\ $$$$\Rightarrow\:\:\:\frac{\mathrm{4}\left(\mathrm{1}−\mathrm{2}\gamma\right)}{\mathrm{1}+\gamma}=\frac{\mathrm{1}+\gamma}{\mathrm{1}+\mathrm{2}\gamma} \\ $$$$\Rightarrow\:\:\mathrm{4}−\mathrm{16}\gamma^{\mathrm{2}} =\mathrm{1}+\mathrm{2}\gamma+\gamma^{\mathrm{2}} \\ $$$$\Rightarrow\:\:\:\mathrm{17}\gamma^{\mathrm{2}} +\mathrm{2}\gamma−\mathrm{3}=\mathrm{0} \\ $$$$\Rightarrow\:\:\boldsymbol{\gamma}\:=\:\frac{\mathrm{2}\sqrt{\mathrm{13}}−\mathrm{1}}{\mathrm{17}} \\ $$$$\:\:\:\:\:\:\beta=\frac{\mathrm{1}−\mathrm{2}\gamma}{\mathrm{1}+\gamma}\:=\:\frac{\mathrm{1}−\frac{\left(\mathrm{4}\sqrt{\mathrm{13}}−\mathrm{2}\right)}{\mathrm{17}}}{\mathrm{1}+\frac{\left(\mathrm{2}\sqrt{\mathrm{13}}−\mathrm{1}\right)}{\mathrm{17}}} \\ $$$$\:\:\:\:=\frac{\mathrm{19}−\mathrm{4}\sqrt{\mathrm{13}}}{\mathrm{16}+\mathrm{2}\sqrt{\mathrm{13}}}=\frac{\left(\mathrm{19}−\mathrm{4}\sqrt{\mathrm{13}}\right)\left(\mathrm{16}−\mathrm{2}\sqrt{\mathrm{13}}\right)}{\mathrm{204}} \\ $$$$\:\:\:=\frac{\mathrm{304}+\mathrm{104}−\mathrm{102}\sqrt{\mathrm{13}}}{\mathrm{204}}\:=\frac{\mathrm{4}−\sqrt{\mathrm{13}}}{\mathrm{2}} \\ $$$$\:\:\:\boldsymbol{\beta}\:=\frac{\mathrm{4}−\sqrt{\mathrm{13}}}{\mathrm{2}} \\ $$$$\:\:\:\boldsymbol{\alpha}=\frac{\mathrm{1}}{\mathrm{2}}−\frac{\left(\mathrm{4}−\sqrt{\mathrm{13}}\right)}{\mathrm{2}}\:=\frac{\sqrt{\mathrm{13}}−\mathrm{3}}{\mathrm{2}} \\ $$$$\:\:\:\boldsymbol{\gamma}\:=\:\frac{\mathrm{2}\sqrt{\mathrm{13}}−\mathrm{1}}{\mathrm{17}} \\ $$$$\mathcal{T}{hank}\:{you}\:{very}\:{much},\:{mrW}\:{Sir}! \\ $$$${great}\:{solution}! \\ $$$$ \\ $$

Commented by mr W last updated on 06/Sep/20

wow! i didn′t expect that we can get the exact solution. thanks sir!

$${wow}!\:{i}\:{didn}'{t}\:{expect}\:{that}\:{we}\:{can}\:{get} \\ $$$${the}\:{exact}\:{solution}.\:{thanks}\:{sir}! \\ $$

Commented by mr W last updated on 05/Sep/20

a+b=(s/2) α+β=(1/2) ...(i) BD=s−c FD=b+c FD^2 =BF^2 +BD^2 −2×BF×BD×cos ∠B (b+c)^2 =b^2 +(s−c)^2 −b(s−c) (β+γ)^2 =β^2 +(1−γ)^2 −β(1−γ) ...(ii) ED=a+c (a+c)^2 =(a+2b)^2 +(s−c)^2 −(a+2b)(s−c) ((s/2)−b+c)^2 =((s/2)+b)^2 +(s−c)^2 −((s/2)+b)(s−c) ((1/2)−β+γ)^2 =((1/2)+β)^2 +(1−γ)^2 −((1/2)+β)(1−γ) ...(iii) (iii)−(ii): γ(1−8β)=4β −1 ⇒γ=((4β−1)/(1−8β)) ⇒β≈0.19722=(b/s) ⇒γ≈0.36541=(c/s) ⇒α≈0.30378=(a/s)

$${a}+{b}=\frac{{s}}{\mathrm{2}} \\ $$$$\alpha+\beta=\frac{\mathrm{1}}{\mathrm{2}}\:\:\:...\left({i}\right) \\ $$$$ \\ $$$${BD}={s}−{c} \\ $$$${FD}={b}+{c} \\ $$$${FD}^{\mathrm{2}} ={BF}^{\mathrm{2}} +{BD}^{\mathrm{2}} −\mathrm{2}×{BF}×{BD}×\mathrm{cos}\:\angle{B} \\ $$$$\left({b}+{c}\right)^{\mathrm{2}} ={b}^{\mathrm{2}} +\left({s}−{c}\right)^{\mathrm{2}} −{b}\left({s}−{c}\right) \\ $$$$\left(\beta+\gamma\right)^{\mathrm{2}} =\beta^{\mathrm{2}} +\left(\mathrm{1}−\gamma\right)^{\mathrm{2}} −\beta\left(\mathrm{1}−\gamma\right)\:\:\:...\left({ii}\right) \\ $$$$ \\ $$$${ED}={a}+{c} \\ $$$$\left({a}+{c}\right)^{\mathrm{2}} =\left({a}+\mathrm{2}{b}\right)^{\mathrm{2}} +\left({s}−{c}\right)^{\mathrm{2}} −\left({a}+\mathrm{2}{b}\right)\left({s}−{c}\right) \\ $$$$\left(\frac{{s}}{\mathrm{2}}−{b}+{c}\right)^{\mathrm{2}} =\left(\frac{{s}}{\mathrm{2}}+{b}\right)^{\mathrm{2}} +\left({s}−{c}\right)^{\mathrm{2}} −\left(\frac{{s}}{\mathrm{2}}+{b}\right)\left({s}−{c}\right) \\ $$$$\left(\frac{\mathrm{1}}{\mathrm{2}}−\beta+\gamma\right)^{\mathrm{2}} =\left(\frac{\mathrm{1}}{\mathrm{2}}+\beta\right)^{\mathrm{2}} +\left(\mathrm{1}−\gamma\right)^{\mathrm{2}} −\left(\frac{\mathrm{1}}{\mathrm{2}}+\beta\right)\left(\mathrm{1}−\gamma\right)\:\:\:...\left({iii}\right) \\ $$$$ \\ $$$$\left({iii}\right)−\left({ii}\right): \\ $$$$\gamma\left(\mathrm{1}−\mathrm{8}\beta\right)=\mathrm{4}\beta\:−\mathrm{1} \\ $$$$\Rightarrow\gamma=\frac{\mathrm{4}\beta−\mathrm{1}}{\mathrm{1}−\mathrm{8}\beta} \\ $$$$ \\ $$$$\Rightarrow\beta\approx\mathrm{0}.\mathrm{19722}=\frac{{b}}{{s}} \\ $$$$\Rightarrow\gamma\approx\mathrm{0}.\mathrm{36541}=\frac{{c}}{{s}} \\ $$$$\Rightarrow\alpha\approx\mathrm{0}.\mathrm{30378}=\frac{{a}}{{s}} \\ $$

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