Question Number 28911 by ajfour last updated on 01/Feb/18
Commented by ajfour last updated on 01/Feb/18
$${Find}\:{area}\:{of}\:{coloured}\:{triangle} \\ $$$${and}\:{radius}\:{of}\:{circle}\:{in}\:{terms}\:{of} \\ $$$$\boldsymbol{{a}},\:\boldsymbol{\theta},\:{and}\:\boldsymbol{\phi}\:.\:{Assume}\:\theta\:,\:\phi\:\leqslant\:\frac{\pi}{\mathrm{4}}\:. \\ $$$${Estimate}\:{maximum}\:{value}\:{of} \\ $$$${radius}\:{in}\:{terms}\:{of}\:\boldsymbol{{a}}. \\ $$
Answered by mrW2 last updated on 01/Feb/18
Commented by mrW2 last updated on 01/Feb/18
![BF=(a/(cos θ)) DF=a(1−tan θ) DE=a(1−tan φ) tan ϕ=((DE)/(DF))=((1−tan φ)/(1−tan θ)) ⇒ϕ=tan^(−1) (((1−tan φ)/(1−tan θ))) α=90−(θ+φ) β=180−(90−θ)−ϕ=90−(ϕ−θ) (r/(tan (α/2)))+(r/(tan (β/2)))=(a/(cos θ)) (1/(tan (α/2)))=((1+cos α)/(sin α))=((1+sin (θ+φ))/(cos (θ+φ))) (1/(tan (β/2)))=((1+cos β)/(sin β))=((1+sin (ϕ−θ))/(cos (ϕ−θ))) ⇒r[((1+sin (φ+θ))/(cos (φ+θ)))+((1+sin (ϕ−θ))/(cos (ϕ−θ)))]=(a/(cos θ)) ⇒r=((a cos (φ+θ) cos (ϕ−θ))/(cos θ [cos (φ+θ)+cos (ϕ−θ)+sin (φ+θ) cos (ϕ−θ)+cos (φ+θ) sin (ϕ−θ)])) ⇒r=((a cos (φ+θ) cos (ϕ−θ))/(cos θ [cos (φ+θ)+cos (ϕ−θ)+sin (φ+ϕ)])) A_Δ =(1/2)×(a/(cos θ))×(a/(cos φ))×sin α=(a^2 /2)×((cos (θ+φ))/(cos θ cos φ)) ⇒A_Δ =(a^2 /2)×((cos (θ+φ))/(cos θ cos φ))=(a^2 /2)(1−tan θ tan φ) For max. of radius r: φ=θ ϕ=(π/4) ⇒r=((a cos (2θ) cos ((π/4)−θ))/(cos θ [cos (2θ)+cos ((π/4)−θ)+sin ((π/4)+θ)])) ⇒r=((a cos (2θ) cos ((π/4)−θ))/(cos θ [cos (2θ)+2cos ((π/4)−θ)])) ⇒r=((a cos (2θ) (cos θ+sin θ)×(1/(√2)))/(cos θ [cos (2θ)+(√2)(cos θ+sin θ)])) ⇒r=((a cos (2θ) (cos θ+sin θ))/(cos θ [(√2) cos (2θ)+2(cos θ+sin θ)])) ⇒r=((a cos (2θ))/(cos θ [(√2) (cos θ−sin θ)+2])) f(θ)=((cos (2θ))/(cos θ [(√2) (cos θ−sin θ)+2])) f(θ)=((cos^2 θ−sin^2 θ)/(cos^2 θ [(√2) (1−tan θ)+(2/(cos θ))])) f(θ)=((1−tan^2 θ)/( (√2)(1−tan θ)+2(√(1+tan^2 θ)))) f(θ)=((1−x^2 )/( (√2)(1−x)+2(√(1+x^2 )))) with x=tan θ f(θ)_(max) ≈0.3032 at θ≈9.3° ⇒r_(max) ≈0.3032 a](Q28918.png)
$${BF}=\frac{{a}}{\mathrm{cos}\:\theta} \\ $$$${DF}={a}\left(\mathrm{1}−\mathrm{tan}\:\theta\right) \\ $$$${DE}={a}\left(\mathrm{1}−\mathrm{tan}\:\phi\right) \\ $$$$\mathrm{tan}\:\varphi=\frac{{DE}}{{DF}}=\frac{\mathrm{1}−\mathrm{tan}\:\phi}{\mathrm{1}−\mathrm{tan}\:\theta} \\ $$$$\Rightarrow\varphi=\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}−\mathrm{tan}\:\phi}{\mathrm{1}−\mathrm{tan}\:\theta}\right) \\ $$$$\alpha=\mathrm{90}−\left(\theta+\phi\right) \\ $$$$\beta=\mathrm{180}−\left(\mathrm{90}−\theta\right)−\varphi=\mathrm{90}−\left(\varphi−\theta\right) \\ $$$$\frac{{r}}{\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}}+\frac{{r}}{\mathrm{tan}\:\frac{\beta}{\mathrm{2}}}=\frac{{a}}{\mathrm{cos}\:\theta} \\ $$$$\frac{\mathrm{1}}{\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}}=\frac{\mathrm{1}+\mathrm{cos}\:\alpha}{\mathrm{sin}\:\alpha}=\frac{\mathrm{1}+\mathrm{sin}\:\left(\theta+\phi\right)}{\mathrm{cos}\:\left(\theta+\phi\right)} \\ $$$$\frac{\mathrm{1}}{\mathrm{tan}\:\frac{\beta}{\mathrm{2}}}=\frac{\mathrm{1}+\mathrm{cos}\:\beta}{\mathrm{sin}\:\beta}=\frac{\mathrm{1}+\mathrm{sin}\:\left(\varphi−\theta\right)}{\mathrm{cos}\:\left(\varphi−\theta\right)} \\ $$$$\Rightarrow{r}\left[\frac{\mathrm{1}+\mathrm{sin}\:\left(\phi+\theta\right)}{\mathrm{cos}\:\left(\phi+\theta\right)}+\frac{\mathrm{1}+\mathrm{sin}\:\left(\varphi−\theta\right)}{\mathrm{cos}\:\left(\varphi−\theta\right)}\right]=\frac{{a}}{\mathrm{cos}\:\theta} \\ $$$$\Rightarrow{r}=\frac{{a}\:\mathrm{cos}\:\left(\phi+\theta\right)\:\mathrm{cos}\:\left(\varphi−\theta\right)}{\mathrm{cos}\:\theta\:\left[\mathrm{cos}\:\left(\phi+\theta\right)+\mathrm{cos}\:\left(\varphi−\theta\right)+\mathrm{sin}\:\left(\phi+\theta\right)\:\mathrm{cos}\:\left(\varphi−\theta\right)+\mathrm{cos}\:\left(\phi+\theta\right)\:\mathrm{sin}\:\left(\varphi−\theta\right)\right]} \\ $$$$\Rightarrow{r}=\frac{{a}\:\mathrm{cos}\:\left(\phi+\theta\right)\:\mathrm{cos}\:\left(\varphi−\theta\right)}{\mathrm{cos}\:\theta\:\left[\mathrm{cos}\:\left(\phi+\theta\right)+\mathrm{cos}\:\left(\varphi−\theta\right)+\mathrm{sin}\:\left(\phi+\varphi\right)\right]} \\ $$$$ \\ $$$${A}_{\Delta} =\frac{\mathrm{1}}{\mathrm{2}}×\frac{{a}}{\mathrm{cos}\:\theta}×\frac{{a}}{\mathrm{cos}\:\phi}×\mathrm{sin}\:\alpha=\frac{{a}^{\mathrm{2}} }{\mathrm{2}}×\frac{\mathrm{cos}\:\left(\theta+\phi\right)}{\mathrm{cos}\:\theta\:\mathrm{cos}\:\phi} \\ $$$$\Rightarrow{A}_{\Delta} =\frac{{a}^{\mathrm{2}} }{\mathrm{2}}×\frac{\mathrm{cos}\:\left(\theta+\phi\right)}{\mathrm{cos}\:\theta\:\mathrm{cos}\:\phi}=\frac{{a}^{\mathrm{2}} }{\mathrm{2}}\left(\mathrm{1}−\mathrm{tan}\:\theta\:\mathrm{tan}\:\phi\right) \\ $$$$ \\ $$$${For}\:{max}.\:{of}\:{radius}\:{r}:\: \\ $$$$\phi=\theta \\ $$$$\varphi=\frac{\pi}{\mathrm{4}} \\ $$$$\Rightarrow{r}=\frac{{a}\:\mathrm{cos}\:\left(\mathrm{2}\theta\right)\:\mathrm{cos}\:\left(\frac{\pi}{\mathrm{4}}−\theta\right)}{\mathrm{cos}\:\theta\:\left[\mathrm{cos}\:\left(\mathrm{2}\theta\right)+\mathrm{cos}\:\left(\frac{\pi}{\mathrm{4}}−\theta\right)+\mathrm{sin}\:\left(\frac{\pi}{\mathrm{4}}+\theta\right)\right]} \\ $$$$\Rightarrow{r}=\frac{{a}\:\mathrm{cos}\:\left(\mathrm{2}\theta\right)\:\mathrm{cos}\:\left(\frac{\pi}{\mathrm{4}}−\theta\right)}{\mathrm{cos}\:\theta\:\left[\mathrm{cos}\:\left(\mathrm{2}\theta\right)+\mathrm{2cos}\:\left(\frac{\pi}{\mathrm{4}}−\theta\right)\right]} \\ $$$$\Rightarrow{r}=\frac{{a}\:\mathrm{cos}\:\left(\mathrm{2}\theta\right)\:\left(\mathrm{cos}\:\theta+\mathrm{sin}\:\theta\right)×\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}}{\mathrm{cos}\:\theta\:\left[\mathrm{cos}\:\left(\mathrm{2}\theta\right)+\sqrt{\mathrm{2}}\left(\mathrm{cos}\:\theta+\mathrm{sin}\:\theta\right)\right]} \\ $$$$\Rightarrow{r}=\frac{{a}\:\mathrm{cos}\:\left(\mathrm{2}\theta\right)\:\left(\mathrm{cos}\:\theta+\mathrm{sin}\:\theta\right)}{\mathrm{cos}\:\theta\:\left[\sqrt{\mathrm{2}}\:\mathrm{cos}\:\left(\mathrm{2}\theta\right)+\mathrm{2}\left(\mathrm{cos}\:\theta+\mathrm{sin}\:\theta\right)\right]} \\ $$$$\Rightarrow{r}=\frac{{a}\:\mathrm{cos}\:\left(\mathrm{2}\theta\right)}{\mathrm{cos}\:\theta\:\left[\sqrt{\mathrm{2}}\:\left(\mathrm{cos}\:\theta−\mathrm{sin}\:\theta\right)+\mathrm{2}\right]} \\ $$$${f}\left(\theta\right)=\frac{\mathrm{cos}\:\left(\mathrm{2}\theta\right)}{\mathrm{cos}\:\theta\:\left[\sqrt{\mathrm{2}}\:\left(\mathrm{cos}\:\theta−\mathrm{sin}\:\theta\right)+\mathrm{2}\right]} \\ $$$${f}\left(\theta\right)=\frac{\mathrm{cos}^{\mathrm{2}} \:\theta−\mathrm{sin}^{\mathrm{2}} \:\theta}{\mathrm{cos}^{\mathrm{2}} \:\theta\:\left[\sqrt{\mathrm{2}}\:\left(\mathrm{1}−\mathrm{tan}\:\theta\right)+\frac{\mathrm{2}}{\mathrm{cos}\:\theta}\right]} \\ $$$${f}\left(\theta\right)=\frac{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \:\theta}{\:\sqrt{\mathrm{2}}\left(\mathrm{1}−\mathrm{tan}\:\theta\right)+\mathrm{2}\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\theta}} \\ $$$${f}\left(\theta\right)=\frac{\mathrm{1}−{x}^{\mathrm{2}} }{\:\sqrt{\mathrm{2}}\left(\mathrm{1}−{x}\right)+\mathrm{2}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\:{with}\:{x}=\mathrm{tan}\:\theta \\ $$$$ \\ $$$${f}\left(\theta\right)_{{max}} \approx\mathrm{0}.\mathrm{3032}\:{at}\:\theta\approx\mathrm{9}.\mathrm{3}° \\ $$$$\Rightarrow{r}_{{max}} \approx\mathrm{0}.\mathrm{3032}\:{a} \\ $$
Commented by ajfour last updated on 01/Feb/18
![Area(△BEF)=(1/2)(BE)(BF)sin α =((a^2 cos (𝛉+𝛗))/(2cos 𝛉cos 𝛗)) =(r/2)(BE+BF+EF) ⇒ r((a/(cos 𝛗))+(a/(cos 𝛉))+EF)=((a^2 cos (𝛉+𝛗))/(cos 𝛉cos 𝛗)) EF=a(√((1−tan 𝛉)^2 +(1−tan 𝛗)^2 )) r=a[((cos (𝛉+𝛗))/(cos 𝛉+cos 𝛗+(√2)((√(cos^2 𝛗 sin^2 (𝛉−𝛑/4)+cos^2 𝛉 sin^2 (𝛗−𝛑/4))) )))] expression for r, if 𝛉=𝛗 , r=a[((cos 2𝛉)/(2cos 𝛉+2cos 𝛉 ∣sin (𝛉−𝛑/4)∣))] =((acos 2𝛉)/(2cos 𝛉[1+sin ((𝛑/4)−𝛉)])) =((a(1−x^2 ))/(2((√(1+x^2 ))+(1/(√2))−(x/(√2))))) (if we let x=tan 𝛉) r(x)=(a/(√2))(((1−x^2 )/((√(2(1+x^2 )))+1−x))) (dr/dx)=0 ⇒ x= ? (please help) (dr/dx)=0 leads to the equation (x−1)^4 (1+x^2 )=2x^2 (3+x^2 )^2 find its roots that belong to (0, 1).](Q28925.png)
$${Area}\left(\bigtriangleup{BEF}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left({BE}\right)\left({BF}\right)\mathrm{sin}\:\alpha \\ $$$$\:=\frac{\boldsymbol{{a}}^{\mathrm{2}} \mathrm{cos}\:\left(\boldsymbol{\theta}+\boldsymbol{\phi}\right)}{\mathrm{2cos}\:\boldsymbol{\theta}\mathrm{cos}\:\boldsymbol{\phi}}\:=\frac{\boldsymbol{{r}}}{\mathrm{2}}\left(\boldsymbol{{BE}}+\boldsymbol{{BF}}+\boldsymbol{{EF}}\right) \\ $$$$\Rightarrow\:\:\boldsymbol{{r}}\left(\frac{\boldsymbol{{a}}}{\mathrm{cos}\:\boldsymbol{\phi}}+\frac{\boldsymbol{{a}}}{\mathrm{cos}\:\boldsymbol{\theta}}+\boldsymbol{{EF}}\right)=\frac{\boldsymbol{{a}}^{\mathrm{2}} \mathrm{cos}\:\left(\boldsymbol{\theta}+\boldsymbol{\phi}\right)}{\mathrm{cos}\:\boldsymbol{\theta}\mathrm{cos}\:\boldsymbol{\phi}} \\ $$$$\boldsymbol{{EF}}=\boldsymbol{{a}}\sqrt{\left(\mathrm{1}−\mathrm{tan}\:\boldsymbol{\theta}\right)^{\mathrm{2}} +\left(\mathrm{1}−\mathrm{tan}\:\boldsymbol{\phi}\right)^{\mathrm{2}} } \\ $$$$\:\:\boldsymbol{{r}}=\boldsymbol{{a}}\left[\frac{\mathrm{cos}\:\left(\boldsymbol{\theta}+\boldsymbol{\phi}\right)}{\mathrm{cos}\:\boldsymbol{\theta}+\mathrm{cos}\:\boldsymbol{\phi}+\sqrt{\mathrm{2}}\left(\sqrt{\mathrm{cos}\:^{\mathrm{2}} \boldsymbol{\phi}\:\mathrm{sin}\:^{\mathrm{2}} \left(\boldsymbol{\theta}−\boldsymbol{\pi}/\mathrm{4}\right)+\mathrm{cos}\:^{\mathrm{2}} \boldsymbol{\theta}\:\mathrm{sin}\:^{\mathrm{2}} \left(\boldsymbol{\phi}−\boldsymbol{\pi}/\mathrm{4}\right)}\:\right)}\right] \\ $$$${expression}\:{for}\:\:\boldsymbol{{r}},\:{if}\:\boldsymbol{\theta}=\boldsymbol{\phi}\:, \\ $$$$\boldsymbol{{r}}=\boldsymbol{{a}}\left[\frac{\mathrm{cos}\:\mathrm{2}\boldsymbol{\theta}}{\mathrm{2cos}\:\boldsymbol{\theta}+\mathrm{2cos}\:\boldsymbol{\theta}\:\mid\mathrm{sin}\:\left(\boldsymbol{\theta}−\boldsymbol{\pi}/\mathrm{4}\right)\mid}\right] \\ $$$$=\frac{\boldsymbol{{a}}\mathrm{cos}\:\mathrm{2}\boldsymbol{\theta}}{\mathrm{2cos}\:\boldsymbol{\theta}\left[\mathrm{1}+\mathrm{sin}\:\left(\frac{\boldsymbol{\pi}}{\mathrm{4}}−\boldsymbol{\theta}\right)\right]} \\ $$$$=\frac{\boldsymbol{{a}}\left(\mathrm{1}−\boldsymbol{{x}}^{\mathrm{2}} \right)}{\mathrm{2}\left(\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}−\frac{\boldsymbol{{x}}}{\sqrt{\mathrm{2}}}\right)}\:\:\:\:\left({if}\:{we}\:{let}\:\boldsymbol{{x}}=\mathrm{tan}\:\boldsymbol{\theta}\right)\: \\ $$$$\boldsymbol{{r}}\left({x}\right)=\frac{\boldsymbol{{a}}}{\sqrt{\mathrm{2}}}\left(\frac{\mathrm{1}−\boldsymbol{{x}}^{\mathrm{2}} }{\sqrt{\mathrm{2}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}+\mathrm{1}−\boldsymbol{{x}}}\right) \\ $$$$\:\:\:\:\frac{{dr}}{{dx}}=\mathrm{0}\:\:\:\:\Rightarrow\:\:\:\:\:{x}=\:?\:\:\left({please}\:{help}\right) \\ $$$$\:\frac{\boldsymbol{{dr}}}{\boldsymbol{{dx}}}=\mathrm{0}\:\:\:{leads}\:{to}\:{the}\:{equation} \\ $$$$\left({x}−\mathrm{1}\right)^{\mathrm{4}} \left(\mathrm{1}+{x}^{\mathrm{2}} \right)=\mathrm{2}{x}^{\mathrm{2}} \left(\mathrm{3}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$${find}\:{its}\:{roots}\:{that}\:{belong}\:{to}\:\left(\mathrm{0},\:\mathrm{1}\right). \\ $$
Commented by ajfour last updated on 01/Feb/18
Thank you mrW2 Sir !
Commented by mrW2 last updated on 01/Feb/18
Thank you sir! I could form my formula to the same expression as yours. I think it is not possible to solve x analytically. I found my solution through graph. x=0.165
Commented by ajfour last updated on 01/Feb/18
Which graph app sir?
Commented by mrW2 last updated on 01/Feb/18
One can use e.g. Geogebra.
Commented by mrW2 last updated on 01/Feb/18
Commented by ajfour last updated on 01/Feb/18
$${thanks}\:{Sir}\:. \\ $$