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Question Number 214372 by ajfour last updated on 06/Dec/24

Commented by ajfour last updated on 06/Dec/24

Find maximum r. Outer figure is a rhombus.Circle is inscribed in an isosceles triangle as shown above.

$${Find}\:{maximum}\:{r}.\:{Outer}\:{figure}\:{is}\:{a} \\ $$$${rhombus}.{Circle}\:{is}\:{inscribed}\:{in}\:{an}\: \\ $$$${isosceles}\:{triangle}\:{as}\:{shown}\:{above}. \\ $$

Answered by a.lgnaoui last updated on 07/Dec/24

on peut verifier facilement que le rayon r est maximum lorsqaue y=x=45° alors AB=(a/2) tan 45=(r/(a/2))=((2r)/a) alors r=(a/2)

$$\mathrm{on}\:\mathrm{peut}\:\mathrm{verifier}\:\:\mathrm{facilement}\:\mathrm{que}\:\mathrm{le}\:\mathrm{rayon}\:\boldsymbol{\mathrm{r}}\:\boldsymbol{\mathrm{est}}\: \\ $$$$\boldsymbol{\mathrm{maximum}}\:\mathrm{lorsqaue}\:\:\boldsymbol{\mathrm{y}}=\boldsymbol{\mathrm{x}}=\mathrm{45}° \\ $$$$\mathrm{alors}\:\:\mathrm{AB}=\frac{\mathrm{a}}{\mathrm{2}}\:\:\:\mathrm{tan}\:\mathrm{45}=\frac{\boldsymbol{\mathrm{r}}}{\boldsymbol{\mathrm{a}}/\mathrm{2}}=\frac{\mathrm{2}\boldsymbol{\mathrm{r}}}{\boldsymbol{\mathrm{a}}} \\ $$$$\boldsymbol{\mathrm{alors}}\:\:\:\:\:\:\:\boldsymbol{\mathrm{r}}=\frac{\boldsymbol{\mathrm{a}}}{\mathrm{2}} \\ $$$$\:\: \\ $$

Commented by a.lgnaoui last updated on 07/Dec/24

Commented by mr W last updated on 07/Dec/24

non−sense again. the circle should be inscribed in the isosceles triangle, not in the rhombus!

$${non}−{sense}\:{again}.\:{the}\:{circle}\:{should} \\ $$$${be}\:{inscribed}\:{in}\:{the}\:{isosceles}\: \\ $$$${triangle},\:{not}\:{in}\:{the}\:{rhombus}! \\ $$

Answered by mr W last updated on 07/Dec/24

Commented by mr W last updated on 07/Dec/24

0<θ≤α<(π/2) OB=((r+(r/(sin θ)))/(cos θ))=(((1+sin θ)r)/(sin θ cos θ)) ((OB)/(sin (π−2α)))=(a/(sin (π−2α+α−θ))) (((1+sin θ)r)/(sin θ cos θ sin 2α))=(a/(sin (θ+α)))=(a/(sin θ cos α+cos θ sin α)) ⇒(r/(2a))=((sin θ sin α)/((1+sin θ)(tan θ+ tan α))) Φ=ln ((r/(2a)))=ln (sin θ)+ln (sin α)−ln (1+sin θ)−ln (tan θ+tan α) r_(max) ⇔ Φ_(max) (∂Φ/∂α)=(1/(tan α))−(1/((tan θ+tan α)cos^2 α))=0 tan θ=((1/(cos^2 α))−1)tan α=tan^3 α ⇒tan α=((tan θ))^(1/3) (∂Φ/∂θ)=(1/(tan θ))−((cos θ)/(1+sin θ))−(1/((tan θ+tan α)cos^2 θ))=0 ((1/(tan θ))−((cos θ)/(1+sin θ)))cos^2 θ=(1/(tan θ+tan α)) ⇒cos^2 θ((1/(tan θ))−((cos θ)/(1+sin θ)))=(1/(tan θ+((tan θ))^(1/3) )) ⇒θ≈0.5984 (34.286°) ⇒α≈0.7218 (41.356°) ≠45° ⇒(r_(max) /a)≈0.304840051439 ✓ example with a=4 see below

$$\mathrm{0}<\theta\leqslant\alpha<\frac{\pi}{\mathrm{2}} \\ $$$${OB}=\frac{{r}+\frac{{r}}{\mathrm{sin}\:\theta}}{\mathrm{cos}\:\theta}=\frac{\left(\mathrm{1}+\mathrm{sin}\:\theta\right){r}}{\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta} \\ $$$$\frac{{OB}}{\mathrm{sin}\:\left(\pi−\mathrm{2}\alpha\right)}=\frac{{a}}{\mathrm{sin}\:\left(\pi−\mathrm{2}\alpha+\alpha−\theta\right)} \\ $$$$\frac{\left(\mathrm{1}+\mathrm{sin}\:\theta\right){r}}{\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta\:\mathrm{sin}\:\mathrm{2}\alpha}=\frac{{a}}{\mathrm{sin}\:\left(\theta+\alpha\right)}=\frac{{a}}{\mathrm{sin}\:\theta\:\mathrm{cos}\:\alpha+\mathrm{cos}\:\theta\:\mathrm{sin}\:\alpha} \\ $$$$\Rightarrow\frac{{r}}{\mathrm{2}{a}}=\frac{\mathrm{sin}\:\theta\:\mathrm{sin}\:\alpha}{\left(\mathrm{1}+\mathrm{sin}\:\theta\right)\left(\mathrm{tan}\:\theta+\:\mathrm{tan}\:\alpha\right)} \\ $$$$\Phi=\mathrm{ln}\:\left(\frac{{r}}{\mathrm{2}{a}}\right)=\mathrm{ln}\:\left(\mathrm{sin}\:\theta\right)+\mathrm{ln}\:\left(\mathrm{sin}\:\alpha\right)−\mathrm{ln}\:\left(\mathrm{1}+\mathrm{sin}\:\theta\right)−\mathrm{ln}\:\left(\mathrm{tan}\:\theta+\mathrm{tan}\:\alpha\right) \\ $$$${r}_{{max}} \:\Leftrightarrow\:\Phi_{{max}} \\ $$$$\frac{\partial\Phi}{\partial\alpha}=\frac{\mathrm{1}}{\mathrm{tan}\:\alpha}−\frac{\mathrm{1}}{\left(\mathrm{tan}\:\theta+\mathrm{tan}\:\alpha\right)\mathrm{cos}^{\mathrm{2}} \:\alpha}=\mathrm{0} \\ $$$$\mathrm{tan}\:\theta=\left(\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \:\alpha}−\mathrm{1}\right)\mathrm{tan}\:\alpha=\mathrm{tan}^{\mathrm{3}} \:\alpha \\ $$$$\Rightarrow\mathrm{tan}\:\alpha=\sqrt[{\mathrm{3}}]{\mathrm{tan}\:\theta} \\ $$$$\frac{\partial\Phi}{\partial\theta}=\frac{\mathrm{1}}{\mathrm{tan}\:\theta}−\frac{\mathrm{cos}\:\theta}{\mathrm{1}+\mathrm{sin}\:\theta}−\frac{\mathrm{1}}{\left(\mathrm{tan}\:\theta+\mathrm{tan}\:\alpha\right)\mathrm{cos}^{\mathrm{2}} \:\theta}=\mathrm{0} \\ $$$$\left(\frac{\mathrm{1}}{\mathrm{tan}\:\theta}−\frac{\mathrm{cos}\:\theta}{\mathrm{1}+\mathrm{sin}\:\theta}\right)\mathrm{cos}^{\mathrm{2}} \:\theta=\frac{\mathrm{1}}{\mathrm{tan}\:\theta+\mathrm{tan}\:\alpha} \\ $$$$\Rightarrow\mathrm{cos}^{\mathrm{2}} \:\theta\left(\frac{\mathrm{1}}{\mathrm{tan}\:\theta}−\frac{\mathrm{cos}\:\theta}{\mathrm{1}+\mathrm{sin}\:\theta}\right)=\frac{\mathrm{1}}{\mathrm{tan}\:\theta+\sqrt[{\mathrm{3}}]{\mathrm{tan}\:\theta}} \\ $$$$\Rightarrow\theta\approx\mathrm{0}.\mathrm{5984}\:\left(\mathrm{34}.\mathrm{286}°\right) \\ $$$$\Rightarrow\alpha\approx\mathrm{0}.\mathrm{7218}\:\left(\mathrm{41}.\mathrm{356}°\right)\:\neq\mathrm{45}° \\ $$$$\Rightarrow\frac{{r}_{{max}} }{{a}}\approx\mathrm{0}.\mathrm{304840051439}\:\checkmark \\ $$$${example}\:{with}\:{a}=\mathrm{4}\:{see}\:{below} \\ $$

Commented by mr W last updated on 07/Dec/24

Commented by ajfour last updated on 08/Dec/24

Thanks for your solution. I am trying still sometimes..

$${Thanks}\:{for}\:{your}\:{solution}.\:{I}\:{am} \\ $$$${trying}\:{still}\:{sometimes}.. \\ $$

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