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Question Number 193464 by Mingma last updated on 14/Jun/23

Commented by Frix last updated on 14/Jun/23

For any triangle r_n =((cδ)/(2((a+b+c)c+(n−1)δ))) [δ=(√((a+b+c)(−a+b+c)(a−b+c)(a+b−c))) For a rectangular triangle with c=(√(a^2 +b^2 )) r_n =((ab(√(a^2 +b^2 )))/(a^2 +2(n−1)ab+b^2 +(a+b)(√(a^2 +b^2 )))) With a=8∧b=15 ⇒ c=17 r_n =((51)/(6n+11))

$$\mathrm{For}\:\mathrm{any}\:\mathrm{triangle} \\ $$$${r}_{{n}} =\frac{{c}\delta}{\mathrm{2}\left(\left({a}+{b}+{c}\right){c}+\left({n}−\mathrm{1}\right)\delta\right)} \\ $$$$\:\:\:\:\:\left[\delta=\sqrt{\left({a}+{b}+{c}\right)\left(−{a}+{b}+{c}\right)\left({a}−{b}+{c}\right)\left({a}+{b}−{c}\right)}\right. \\ $$$$\mathrm{For}\:\mathrm{a}\:\mathrm{rectangular}\:\mathrm{triangle}\:\mathrm{with}\:{c}=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$${r}_{{n}} =\frac{{ab}\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}{{a}^{\mathrm{2}} +\mathrm{2}\left({n}−\mathrm{1}\right){ab}+{b}^{\mathrm{2}} +\left({a}+{b}\right)\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }} \\ $$$$\mathrm{With}\:{a}=\mathrm{8}\wedge{b}=\mathrm{15}\:\Rightarrow\:{c}=\mathrm{17} \\ $$$${r}_{{n}} =\frac{\mathrm{51}}{\mathrm{6}{n}+\mathrm{11}} \\ $$

Commented by Mingma last updated on 15/Jun/23

Perfect ��

Commented by a.lgnaoui last updated on 15/Jun/23

in terme a b c?

$$\mathrm{in}\:\mathrm{terme}\:\mathrm{a}\:\mathrm{b}\:\mathrm{c}? \\ $$

Answered by a.lgnaoui last updated on 15/Jun/23

∡DCM=((∡C)/2) ∡EBN=((∡B)/2) BC=BMcos ((∡C)/2)+CNcos ((∡B)/2)+12r posons ∡C=θ ∡B=ϕ=90−θ ⇒cos (𝛉/2)=(√((a+c)/(2c))) cos (𝛟/2)=(√((b+c)/(2c))) ⇒BC=BM(√(((a+c)/(2c)) )) +CN(√((b+c)/(2c))) +12r(1) sin (𝛉/2)=(r/(BM)) BM=(r/(sin (𝛉/2)))=r(√((2c)/(c−a))) CN=r(√((2c)/(c−b))) (1)⇔ BC=r(√((2c)/(c−a))) (√((a+c)/(2c))) +r(√((2c)/(c−b))) (√((b+c)/(2c))) +12r BC =r(12+(√((a+c)/(c−a))) +(√(((b+c)/(c−b)) )) ) ABC triangld rectangle BC^2 =a^2 +b^2 =c^2 c=(√(a^2 +b^2 )) (√(a^2 +b^2 )) =r(12+(√((a+c)/(c−a))) +(√((b+c)/(c−b))) ) soit r=(c/(12+(√((a+c)/(c−a))) +(√((b+c)/(c−b))))) { ((a=8 b= 15 c=17)),(( r=((17)/(12+(√((25)/9)) +(√((32)/2))))=((17)/(16+(5/3))))) :} soit r =((51)/(53)) cas general pour n cercles de rayon r r =(c/(2(n−1)+(√((a+c)/(c−a))) +(√((b+c)/(c−b)))))

$$\measuredangle\mathrm{DCM}=\frac{\measuredangle\mathrm{C}}{\mathrm{2}}\:\:\:\:\:\:\:\measuredangle\mathrm{EBN}=\frac{\measuredangle\mathrm{B}}{\mathrm{2}} \\ $$$$\mathrm{BC}=\boldsymbol{\mathrm{B}}\mathrm{Mcos}\:\frac{\measuredangle\mathrm{C}}{\mathrm{2}}+\boldsymbol{\mathrm{C}}\mathrm{Ncos}\:\frac{\measuredangle\mathrm{B}}{\mathrm{2}}+\mathrm{12}\boldsymbol{\mathrm{r}} \\ $$$$\mathrm{posons}\:\:\measuredangle\mathrm{C}=\theta\:\:\:\:\:\:\measuredangle\mathrm{B}=\varphi=\mathrm{90}−\theta \\ $$$$\Rightarrow\mathrm{cos}\:\frac{\boldsymbol{\theta}}{\mathrm{2}}=\sqrt{\frac{\mathrm{a}+\mathrm{c}}{\mathrm{2c}}}\:\:\:\:\:\:\mathrm{cos}\:\frac{\boldsymbol{\varphi}}{\mathrm{2}}=\sqrt{\frac{\mathrm{b}+\mathrm{c}}{\mathrm{2c}}} \\ $$$$ \\ $$$$\Rightarrow\boldsymbol{\mathrm{BC}}=\boldsymbol{\mathrm{BM}}\sqrt{\frac{\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{c}}}{\mathrm{2}\boldsymbol{\mathrm{c}}}\:}\:+\boldsymbol{\mathrm{CN}}\sqrt{\frac{\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{c}}}{\mathrm{2}\boldsymbol{\mathrm{c}}}}\:+\mathrm{12}\boldsymbol{\mathrm{r}}\left(\mathrm{1}\right) \\ $$$$ \\ $$$$\:\:\mathrm{sin}\:\frac{\boldsymbol{\theta}}{\mathrm{2}}=\frac{\mathrm{r}}{\boldsymbol{\mathrm{BM}}}\:\:\:\:\boldsymbol{\mathrm{BM}}=\frac{\boldsymbol{\mathrm{r}}}{\mathrm{sin}\:\frac{\boldsymbol{\theta}}{\mathrm{2}}}=\boldsymbol{\mathrm{r}}\sqrt{\frac{\mathrm{2}\boldsymbol{\mathrm{c}}}{\boldsymbol{\mathrm{c}}−\boldsymbol{\mathrm{a}}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{CN}}=\boldsymbol{\mathrm{r}}\sqrt{\frac{\mathrm{2}\boldsymbol{\mathrm{c}}}{\boldsymbol{\mathrm{c}}−\boldsymbol{\mathrm{b}}}}\: \\ $$$$\left(\mathrm{1}\right)\Leftrightarrow\:\:\boldsymbol{\mathrm{BC}}=\boldsymbol{\mathrm{r}}\sqrt{\frac{\mathrm{2}\boldsymbol{\mathrm{c}}}{\boldsymbol{\mathrm{c}}−\boldsymbol{\mathrm{a}}}}\:\sqrt{\frac{\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{c}}}{\mathrm{2}\boldsymbol{\mathrm{c}}}}\:+\boldsymbol{\mathrm{r}}\sqrt{\frac{\mathrm{2}\boldsymbol{\mathrm{c}}}{\boldsymbol{\mathrm{c}}−\boldsymbol{\mathrm{b}}}}\:\sqrt{\frac{\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{c}}}{\mathrm{2}\boldsymbol{\mathrm{c}}}}\:+\mathrm{12}\boldsymbol{\mathrm{r}} \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{BC}}\:=\boldsymbol{\mathrm{r}}\left(\mathrm{12}+\sqrt{\frac{\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{c}}}{\boldsymbol{\mathrm{c}}−\boldsymbol{\mathrm{a}}}}\:+\sqrt{\frac{\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{c}}}{\boldsymbol{\mathrm{c}}−\boldsymbol{\mathrm{b}}}\:}\:\:\right) \\ $$$$\:\:\:\:\:\boldsymbol{\mathrm{ABC}}\:\:\boldsymbol{\mathrm{triangld}}\:\boldsymbol{\mathrm{rectangle}}\:\: \\ $$$$\:\:\:\:\:\boldsymbol{\mathrm{BC}}^{\mathrm{2}} =\boldsymbol{\mathrm{a}}^{\mathrm{2}} +\boldsymbol{\mathrm{b}}^{\mathrm{2}} =\boldsymbol{\mathrm{c}}^{\mathrm{2}} \:\:\:\:\:\:\:\boldsymbol{\mathrm{c}}=\sqrt{\boldsymbol{\mathrm{a}}^{\mathrm{2}} +\boldsymbol{\mathrm{b}}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\sqrt{\boldsymbol{\mathrm{a}}^{\mathrm{2}} +\boldsymbol{\mathrm{b}}^{\mathrm{2}} \:}\:=\boldsymbol{\mathrm{r}}\left(\mathrm{12}+\sqrt{\frac{\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{c}}}{\boldsymbol{\mathrm{c}}−\boldsymbol{\mathrm{a}}}}\:+\sqrt{\frac{\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{c}}}{\boldsymbol{\mathrm{c}}−\boldsymbol{\mathrm{b}}}}\:\right) \\ $$$$\:\:\:\:\boldsymbol{\mathrm{soit}}\: \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{r}}=\frac{\boldsymbol{\mathrm{c}}}{\mathrm{12}+\sqrt{\frac{\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{c}}}{\boldsymbol{\mathrm{c}}−\boldsymbol{\mathrm{a}}}}\:+\sqrt{\frac{\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{c}}}{\boldsymbol{\mathrm{c}}−\boldsymbol{\mathrm{b}}}}} \\ $$$$\:\:\:\:\begin{cases}{\boldsymbol{\mathrm{a}}=\mathrm{8}\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{b}}=\:\:\mathrm{15}\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{c}}=\mathrm{17}}\\{\:\:\:\:\:\boldsymbol{\mathrm{r}}=\frac{\mathrm{17}}{\mathrm{12}+\sqrt{\frac{\mathrm{25}}{\mathrm{9}}}\:+\sqrt{\frac{\mathrm{32}}{\mathrm{2}}}}=\frac{\mathrm{17}}{\mathrm{16}+\frac{\mathrm{5}}{\mathrm{3}}}}\end{cases} \\ $$$$\:\:\boldsymbol{\mathrm{soit}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{r}}\:\:=\frac{\mathrm{51}}{\mathrm{53}} \\ $$$$\boldsymbol{\mathrm{cas}}\:\boldsymbol{\mathrm{general}}\:\:\:\boldsymbol{\mathrm{pour}}\:\boldsymbol{\mathrm{n}}\:\boldsymbol{\mathrm{cercles}}\:\boldsymbol{\mathrm{de}}\:\boldsymbol{\mathrm{rayon}}\:\boldsymbol{\mathrm{r}} \\ $$$$ \\ $$$$\:\:\:\:\:\boldsymbol{\mathrm{r}}\:\:\:\:=\frac{\boldsymbol{\mathrm{c}}}{\mathrm{2}\left(\boldsymbol{\mathrm{n}}−\mathrm{1}\right)+\sqrt{\frac{\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{c}}}{\boldsymbol{\mathrm{c}}−\boldsymbol{\mathrm{a}}}}\:+\sqrt{\frac{\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{c}}}{\boldsymbol{\mathrm{c}}−\boldsymbol{\mathrm{b}}}}} \\ $$$$ \\ $$$$ \\ $$

Commented by a.lgnaoui last updated on 15/Jun/23

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