Question-213884

Question Number 213884 by mr W last updated on 20/Nov/24

Commented by mr W last updated on 20/Nov/24

$${find}\:{maximum}\:{radius}\:{of}\:{circle}\:{C} \\ $$

Commented by Frix last updated on 21/Nov/24

I think it′s at r_A =(3/5) r_B =(3/8) r_C =(1/4)

$$\mathrm{I}\:\mathrm{think}\:\mathrm{it}'\mathrm{s}\:\mathrm{at} \\ $$$${r}_{{A}} =\frac{\mathrm{3}}{\mathrm{5}} \\ $$$${r}_{{B}} =\frac{\mathrm{3}}{\mathrm{8}} \\ $$$${r}_{{C}} =\frac{\mathrm{1}}{\mathrm{4}} \\ $$

Commented by mr W last updated on 21/Nov/24

$${any}\:{easy}\:{way}\:{to}\:{prove}\:{this}? \\ $$

Answered by Ghisom last updated on 21/Nov/24

r_A =p∧0<p<1 A= (((1−p)),(0) ) r_B =((4p(1−p))/((1+p)^2 )) [ _(of circles)^(use tangents) ] B= ((((1−3p)/(1+p))),(r_B ) ) now use Descartes to get r_C =((4p(1−p))/(9p^2 −14p+9)) ⇒ max r_C =(3/5) C= (((((3−p)(3−5p))/(9p^2 −14p+9))),((3r_C )) ) we can find other interesting values like properties of the triangle ABC etc

$${r}_{{A}} ={p}\wedge\mathrm{0}<{p}<\mathrm{1} \\ $$$${A}=\begin{pmatrix}{\mathrm{1}−{p}}\\{\mathrm{0}}\end{pmatrix} \\ $$$${r}_{{B}} =\frac{\mathrm{4}{p}\left(\mathrm{1}−{p}\right)}{\left(\mathrm{1}+{p}\right)^{\mathrm{2}} }\:\:\:\:\:\left[\:_{\mathrm{of}\:\mathrm{circles}} ^{\mathrm{use}\:\mathrm{tangents}} \right] \\ $$$${B}=\begin{pmatrix}{\frac{\mathrm{1}−\mathrm{3}{p}}{\mathrm{1}+{p}}}\\{{r}_{{B}} }\end{pmatrix} \\ $$$$\mathrm{now}\:\mathrm{use}\:\mathrm{Descartes}\:\mathrm{to}\:\mathrm{get} \\ $$$${r}_{{C}} =\frac{\mathrm{4}{p}\left(\mathrm{1}−{p}\right)}{\mathrm{9}{p}^{\mathrm{2}} −\mathrm{14}{p}+\mathrm{9}}\:\Rightarrow\:\mathrm{max}\:{r}_{{C}} \:=\frac{\mathrm{3}}{\mathrm{5}} \\ $$$${C}=\begin{pmatrix}{\frac{\left(\mathrm{3}−{p}\right)\left(\mathrm{3}−\mathrm{5}{p}\right)}{\mathrm{9}{p}^{\mathrm{2}} −\mathrm{14}{p}+\mathrm{9}}}\\{\mathrm{3}{r}_{{C}} }\end{pmatrix} \\ $$$$ \\ $$$$\mathrm{we}\:\mathrm{can}\:\mathrm{find}\:\mathrm{other}\:\mathrm{interesting}\:\mathrm{values}\:\mathrm{like} \\ $$$$\mathrm{properties}\:\mathrm{of}\:\mathrm{the}\:\mathrm{triangle}\:{ABC}\:\mathrm{etc} \\ $$

Answered by mr W last updated on 21/Nov/24

((1/a)+(2/b)−(1/R))^2 =2((1/a^2 )+(2/b^2 )+(1/R^2 )) (4/b)((1/a)−(1/R))=((1/a)+(1/R))^2 ⇒(1/b)=((((1/a)+(1/R))^2 )/(4((1/a)−(1/R)))) let α=(R/a), β=(R/b), γ=(R/c) ⇒β=(((α+1)^2 )/(4(α−1))) ((1/a)+(1/b)+(1/c)−(1/R))^2 =2((1/a^2 )+(1/b^2 )+(1/c^2 )+(1/R^2 )) (R^2 /c^2 )−2((R/a)+(R/b)−1)(R/c)+1+((R/a)−(R/b))^2 +2((R/a)+(R/b)) γ^2 −2(α+β−1)γ+1+(α−β)^2 +2(α+β)=0 γ=α+β−1−2(√(αβ−α−β)) γ=((9α^2 −14α+9)/(4(α−1))) =1+((9(α−1))/4)+(1/(α−1))≥1+2(√(9/4))=4 ⇒γ_(min) =4 ⇒c_(max) =(R/4) at a=((3R)/5)

$$\left(\frac{\mathrm{1}}{{a}}+\frac{\mathrm{2}}{{b}}−\frac{\mathrm{1}}{{R}}\right)^{\mathrm{2}} =\mathrm{2}\left(\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{2}}{{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{{R}^{\mathrm{2}} }\right) \\ $$$$\frac{\mathrm{4}}{{b}}\left(\frac{\mathrm{1}}{{a}}−\frac{\mathrm{1}}{{R}}\right)=\left(\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{R}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{b}}=\frac{\left(\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{R}}\right)^{\mathrm{2}} }{\mathrm{4}\left(\frac{\mathrm{1}}{{a}}−\frac{\mathrm{1}}{{R}}\right)} \\ $$$${let}\:\alpha=\frac{{R}}{{a}},\:\beta=\frac{{R}}{{b}},\:\gamma=\frac{{R}}{{c}} \\ $$$$\Rightarrow\beta=\frac{\left(\alpha+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}\left(\alpha−\mathrm{1}\right)} \\ $$$$\left(\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}−\frac{\mathrm{1}}{{R}}\right)^{\mathrm{2}} =\mathrm{2}\left(\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{{c}^{\mathrm{2}} }+\frac{\mathrm{1}}{{R}^{\mathrm{2}} }\right) \\ $$$$\frac{{R}^{\mathrm{2}} }{{c}^{\mathrm{2}} }−\mathrm{2}\left(\frac{{R}}{{a}}+\frac{{R}}{{b}}−\mathrm{1}\right)\frac{{R}}{{c}}+\mathrm{1}+\left(\frac{{R}}{{a}}−\frac{{R}}{{b}}\right)^{\mathrm{2}} +\mathrm{2}\left(\frac{{R}}{{a}}+\frac{{R}}{{b}}\right) \\ $$$$\gamma^{\mathrm{2}} −\mathrm{2}\left(\alpha+\beta−\mathrm{1}\right)\gamma+\mathrm{1}+\left(\alpha−\beta\right)^{\mathrm{2}} +\mathrm{2}\left(\alpha+\beta\right)=\mathrm{0} \\ $$$$\gamma=\alpha+\beta−\mathrm{1}−\mathrm{2}\sqrt{\alpha\beta−\alpha−\beta} \\ $$$$\gamma=\frac{\mathrm{9}\alpha^{\mathrm{2}} −\mathrm{14}\alpha+\mathrm{9}}{\mathrm{4}\left(\alpha−\mathrm{1}\right)} \\ $$$$\:\:\:=\mathrm{1}+\frac{\mathrm{9}\left(\alpha−\mathrm{1}\right)}{\mathrm{4}}+\frac{\mathrm{1}}{\alpha−\mathrm{1}}\geqslant\mathrm{1}+\mathrm{2}\sqrt{\frac{\mathrm{9}}{\mathrm{4}}}=\mathrm{4} \\ $$$$\Rightarrow\gamma_{{min}} =\mathrm{4}\:\Rightarrow{c}_{{max}} =\frac{{R}}{\mathrm{4}}\:{at}\:{a}=\frac{\mathrm{3}{R}}{\mathrm{5}} \\ $$

Commented by mr W last updated on 21/Nov/24

Commented by mr W last updated on 21/Nov/24

Commented by mr W last updated on 21/Nov/24

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