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Question Number 182192 by mr W last updated on 05/Dec/22
Commented by mr W last updated on 05/Dec/22
find the smallest area of inscribed right−angle triangle in a given triangle with sides a,b,c. (a≥b≥c)
$${find}\:{the}\:{smallest}\:{area}\:{of}\:{inscribed} \\ $$$${right}−{angle}\:{triangle}\:{in}\:{a}\:{given} \\ $$$${triangle}\:{with}\:{sides}\:{a},{b},{c}.\:\left({a}\geqslant{b}\geqslant{c}\right) \\ $$
Answered by mr W last updated on 06/Dec/22
Commented by mr W last updated on 06/Dec/22
BD=((p sin (β+θ))/(sin β)) DC=((q sin (γ+(π/2)−θ))/(sin γ))=((q cos (γ−θ))/(sin γ)) ((p sin (β+θ))/(sin β))+((q cos (γ−θ))/(sin γ))=a ((p(sin β cos θ+cos β sin θ))/(sin β))+((q(cos γ cos θ+sin γ sin θ))/(sin γ))=a (p/a)(cos θ+((sin θ)/(tan β)))+(q/a)(((cos θ)/(tan γ))+sin θ)=1 ξ(cos θ+((sin θ)/(tan β)))+η(((cos θ)/(tan γ))+sin θ)=1 ξu+ηv=1 S=((pq)/2)=((ξηa^2 )/2)=(((ξu)(ηv)a^2 )/(2uv))=((((√((ξu)(ηv))))^2 a^2 )/(2uv)) ≤(((ξu+ηv)^2 a^2 )/(8uv))=(a^2 /(8uv)) maximum (a^2 /(8uv)) at ξu=ηv=(1/2) S=(a^2 /(8uv))=(a^2 /(8(cos θ+((sin θ)/(tan β)))(((cos θ)/(tan γ))+sin θ))) Ψ=(cos θ+((sin θ)/(tan β)))(((cos θ)/(tan γ))+sin θ) (dΨ/dθ)=(−sin θ+((cos θ)/(tan β)))(((cos θ)/(tan γ))+sin θ)+(cos θ+((sin θ)/(tan β)))(−((sin θ)/(tan γ))+cos θ)=0 −((sin θ cos θ)/(tan γ))+((cos^2 θ)/(tan β tan γ))−sin^2 θ+((sin θ cos θ)/(tan β))−((sin θ cos θ)/(tan γ))−((sin^2 θ)/(tan β tan γ))+cos^2 θ+((sin θ cos θ)/(tan β))=0 tan 2θ(((tan β−tan γ)/(1+tan β tan γ)))=1 tan 2θ tan (β−γ)=1 2θ+β−γ=(π/2) ⇒θ=(π/4)−((β−γ)/2)
$${BD}=\frac{{p}\:\mathrm{sin}\:\left(\beta+\theta\right)}{\mathrm{sin}\:\beta} \\ $$$${DC}=\frac{{q}\:\mathrm{sin}\:\left(\gamma+\frac{\pi}{\mathrm{2}}−\theta\right)}{\mathrm{sin}\:\gamma}=\frac{{q}\:\mathrm{cos}\:\left(\gamma−\theta\right)}{\mathrm{sin}\:\gamma} \\ $$$$\frac{{p}\:\mathrm{sin}\:\left(\beta+\theta\right)}{\mathrm{sin}\:\beta}+\frac{{q}\:\mathrm{cos}\:\left(\gamma−\theta\right)}{\mathrm{sin}\:\gamma}={a} \\ $$$$\frac{{p}\left(\mathrm{sin}\:\beta\:\mathrm{cos}\:\theta+\mathrm{cos}\:\beta\:\mathrm{sin}\:\theta\right)}{\mathrm{sin}\:\beta}+\frac{{q}\left(\mathrm{cos}\:\gamma\:\mathrm{cos}\:\theta+\mathrm{sin}\:\gamma\:\mathrm{sin}\:\theta\right)}{\mathrm{sin}\:\gamma}={a} \\ $$$$\frac{{p}}{{a}}\left(\mathrm{cos}\:\theta+\frac{\mathrm{sin}\:\theta}{\mathrm{tan}\:\beta}\right)+\frac{{q}}{{a}}\left(\frac{\mathrm{cos}\:\theta}{\mathrm{tan}\:\gamma}+\mathrm{sin}\:\theta\right)=\mathrm{1} \\ $$$$\xi\left(\mathrm{cos}\:\theta+\frac{\mathrm{sin}\:\theta}{\mathrm{tan}\:\beta}\right)+\eta\left(\frac{\mathrm{cos}\:\theta}{\mathrm{tan}\:\gamma}+\mathrm{sin}\:\theta\right)=\mathrm{1} \\ $$$$\xi{u}+\eta{v}=\mathrm{1} \\ $$$${S}=\frac{{pq}}{\mathrm{2}}=\frac{\xi\eta{a}^{\mathrm{2}} }{\mathrm{2}}=\frac{\left(\xi{u}\right)\left(\eta{v}\right){a}^{\mathrm{2}} }{\mathrm{2}{uv}}=\frac{\left(\sqrt{\left(\xi{u}\right)\left(\eta{v}\right)}\right)^{\mathrm{2}} {a}^{\mathrm{2}} }{\mathrm{2}{uv}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\leqslant\frac{\left(\xi{u}+\eta{v}\right)^{\mathrm{2}} {a}^{\mathrm{2}} }{\mathrm{8}{uv}}=\frac{{a}^{\mathrm{2}} }{\mathrm{8}{uv}} \\ $$$${maximum}\:\frac{{a}^{\mathrm{2}} }{\mathrm{8}{uv}}\:{at}\:\xi{u}=\eta{v}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${S}=\frac{{a}^{\mathrm{2}} }{\mathrm{8}{uv}}=\frac{{a}^{\mathrm{2}} }{\mathrm{8}\left(\mathrm{cos}\:\theta+\frac{\mathrm{sin}\:\theta}{\mathrm{tan}\:\beta}\right)\left(\frac{\mathrm{cos}\:\theta}{\mathrm{tan}\:\gamma}+\mathrm{sin}\:\theta\right)} \\ $$$$\Psi=\left(\mathrm{cos}\:\theta+\frac{\mathrm{sin}\:\theta}{\mathrm{tan}\:\beta}\right)\left(\frac{\mathrm{cos}\:\theta}{\mathrm{tan}\:\gamma}+\mathrm{sin}\:\theta\right) \\ $$$$\frac{{d}\Psi}{{d}\theta}=\left(−\mathrm{sin}\:\theta+\frac{\mathrm{cos}\:\theta}{\mathrm{tan}\:\beta}\right)\left(\frac{\mathrm{cos}\:\theta}{\mathrm{tan}\:\gamma}+\mathrm{sin}\:\theta\right)+\left(\mathrm{cos}\:\theta+\frac{\mathrm{sin}\:\theta}{\mathrm{tan}\:\beta}\right)\left(−\frac{\mathrm{sin}\:\theta}{\mathrm{tan}\:\gamma}+\mathrm{cos}\:\theta\right)=\mathrm{0} \\ $$$$−\frac{\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta}{\mathrm{tan}\:\gamma}+\frac{\mathrm{cos}^{\mathrm{2}} \:\theta}{\mathrm{tan}\:\beta\:\mathrm{tan}\:\gamma}−\mathrm{sin}^{\mathrm{2}} \:\theta+\frac{\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta}{\mathrm{tan}\:\beta}−\frac{\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta}{\mathrm{tan}\:\gamma}−\frac{\mathrm{sin}^{\mathrm{2}} \:\theta}{\mathrm{tan}\:\beta\:\mathrm{tan}\:\gamma}+\mathrm{cos}^{\mathrm{2}} \:\theta+\frac{\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta}{\mathrm{tan}\:\beta}=\mathrm{0} \\ $$$$\mathrm{tan}\:\mathrm{2}\theta\left(\frac{\mathrm{tan}\:\beta−\mathrm{tan}\:\gamma}{\mathrm{1}+\mathrm{tan}\:\beta\:\mathrm{tan}\:\gamma}\right)=\mathrm{1} \\ $$$$\mathrm{tan}\:\mathrm{2}\theta\:\mathrm{tan}\:\left(\beta−\gamma\right)=\mathrm{1} \\ $$$$\mathrm{2}\theta+\beta−\gamma=\frac{\pi}{\mathrm{2}} \\ $$$$\Rightarrow\theta=\frac{\pi}{\mathrm{4}}−\frac{\beta−\gamma}{\mathrm{2}} \\ $$
Commented by mr W last updated on 06/Dec/22
Commented by mr W last updated on 07/Dec/22
ΔDEF is the right angled triangle with smallest area. AD=((√(bc(b+c+a)(b+c−a)))/(b+c)) DE=DF=((sin (α/2)(√(bc(b+c+a)(b+c−a))))/((b+c)sin ((π/4)+(α/2)))) =((sin (α/2)(√(2bc(b+c+a)(b+c−a))))/((b+c)(sin (α/2)+cos (α/2)))) =(((√(1−cos α))(√(bc(b+c+a)(b+c−a))))/((b+c)(√(1+sin α)))) S_(min) =((bc(b+c+a)(b+c−a)(1−cos α))/(2(b+c)^2 (1+sin α))) ⇒S_(min) =(((((2Δ)/(b+c)))^2 )/(1+((2Δ)/(bc)))) with Δ=area of ΔABC. Δ=((√((a+b+c)(−a+b+c)(a−b+c)(a+b−c)))/4)
$$\Delta{DEF}\:{is}\:{the}\:{right}\:{angled}\:{triangle} \\ $$$${with}\:{smallest}\:{area}. \\ $$$${AD}=\frac{\sqrt{{bc}\left({b}+{c}+{a}\right)\left({b}+{c}−{a}\right)}}{{b}+{c}} \\ $$$${DE}={DF}=\frac{\mathrm{sin}\:\frac{\alpha}{\mathrm{2}}\sqrt{{bc}\left({b}+{c}+{a}\right)\left({b}+{c}−{a}\right)}}{\left({b}+{c}\right)\mathrm{sin}\:\left(\frac{\pi}{\mathrm{4}}+\frac{\alpha}{\mathrm{2}}\right)} \\ $$$$\:\:\:=\frac{\mathrm{sin}\:\frac{\alpha}{\mathrm{2}}\sqrt{\mathrm{2}{bc}\left({b}+{c}+{a}\right)\left({b}+{c}−{a}\right)}}{\left({b}+{c}\right)\left(\mathrm{sin}\:\frac{\alpha}{\mathrm{2}}+\mathrm{cos}\:\frac{\alpha}{\mathrm{2}}\right)} \\ $$$$\:\:\:=\frac{\sqrt{\mathrm{1}−\mathrm{cos}\:\alpha}\sqrt{{bc}\left({b}+{c}+{a}\right)\left({b}+{c}−{a}\right)}}{\left({b}+{c}\right)\sqrt{\mathrm{1}+\mathrm{sin}\:\alpha}} \\ $$$${S}_{{min}} =\frac{{bc}\left({b}+{c}+{a}\right)\left({b}+{c}−{a}\right)\left(\mathrm{1}−\mathrm{cos}\:\alpha\right)}{\mathrm{2}\left({b}+{c}\right)^{\mathrm{2}} \left(\mathrm{1}+\mathrm{sin}\:\alpha\right)} \\ $$$$\Rightarrow{S}_{{min}} =\frac{\left(\frac{\mathrm{2}\Delta}{{b}+{c}}\right)^{\mathrm{2}} }{\mathrm{1}+\frac{\mathrm{2}\Delta}{{bc}}} \\ $$$${with}\:\Delta={area}\:{of}\:\Delta{ABC}. \\ $$$$\Delta=\frac{\sqrt{\left({a}+{b}+{c}\right)\left(−{a}+{b}+{c}\right)\left({a}−{b}+{c}\right)\left({a}+{b}−{c}\right)}}{\mathrm{4}} \\ $$

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