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Question Number 39458 by MJS last updated on 06/Jul/18
find the greatest possible square insribed in a triangle with sides a b c
$$\mathrm{find}\:\mathrm{the}\:\mathrm{greatest}\:\mathrm{possible}\:\mathrm{square}\:\mathrm{insribed}\:\mathrm{in} \\ $$$$\mathrm{a}\:\mathrm{triangle}\:\mathrm{with}\:\mathrm{sides}\:{a}\:{b}\:{c} \\ $$
Commented by MJS last updated on 06/Jul/18
the side s of the square not necessarily on one of the sides of the triangle
$$\mathrm{the}\:\mathrm{side}\:{s}\:\mathrm{of}\:\mathrm{the}\:\mathrm{square}\:\mathrm{not}\:\mathrm{necessarily}\:\mathrm{on} \\ $$$$\mathrm{one}\:\mathrm{of}\:\mathrm{the}\:\mathrm{sides}\:\mathrm{of}\:\mathrm{the}\:\mathrm{triangle} \\ $$
Commented by ajfour last updated on 06/Jul/18
In △PCR ((s(√2))/(sin γ)) = ((b−y)/(sin (θ+(π/4)))) ....(i) In △SRA (s/(sin α)) = (y/(sin (β−θ))) .....(ii) eliminating y amongst (i), (ii) ((s(√2)sin (θ+(π/4)))/(sin γ)) =b− ((s sin (β−θ))/(sin α)) ⇒ s=(b/((((√2)sin (𝛉+(𝛑/4)))/(sin 𝛄))+((sin (𝛃−𝛉))/(sin 𝛂)))) =(b/(((sin θ+cos θ)/(sin γ))+((sin βcos θ−cos βsin θ)/(sin α)))) =((bsin γ sin α)/(sin θ(sin α−cos β sin γ)+cos θ(sin α+sin β sin γ))) s=((bsin γ sin α)/(Asin (θ+φ))) where tan 𝛗 = ((sin α+sin β sin γ)/(sin α−cos β sin γ)) A=(√((sin α−cos β sin γ)^2 +(sin α+sin β sin γ)^2 )) for s to be maximum sin (θ+φ) has to be minimum. For many of the cases(though not all), sin (θ+φ) is minimum if θ=0 Then s_(max) = ((bsin 𝛄 sin 𝛂)/(Asin 𝛗)) = ((bsin γ sin α)/(sin α+sin β sin γ)) and since ((sin α)/a)= ((sin β)/b) = ((sin γ)/c) = (1/(2R)) s_(max) = ((abc)/(2aR+bc)) = (a/(((2aR)/(bc))+1)) . This formula will have to be amended if sin (θ+φ)< sin φ for nonzero θ .
$${In}\:\bigtriangleup{PCR} \\ $$$$\:\:\:\:\:\frac{{s}\sqrt{\mathrm{2}}}{\mathrm{sin}\:\gamma}\:=\:\frac{{b}−{y}}{\mathrm{sin}\:\left(\theta+\frac{\pi}{\mathrm{4}}\right)}\:\:\:\:\:\:\:\:….\left({i}\right) \\ $$$${In}\:\bigtriangleup{SRA} \\ $$$$\:\:\:\:\:\:\frac{{s}}{\mathrm{sin}\:\alpha}\:=\:\frac{{y}}{\mathrm{sin}\:\left(\beta−\theta\right)}\:\:\:\:\:\:\:…..\left({ii}\right) \\ $$$${eliminating}\:{y}\:{amongst}\:\left({i}\right),\:\left({ii}\right) \\ $$$$\:\:\:\:\frac{{s}\sqrt{\mathrm{2}}\mathrm{sin}\:\left(\theta+\frac{\pi}{\mathrm{4}}\right)}{\mathrm{sin}\:\gamma}\:={b}−\:\frac{{s}\:\mathrm{sin}\:\left(\beta−\theta\right)}{\mathrm{sin}\:\alpha} \\ $$$$\Rightarrow\:\:\boldsymbol{{s}}=\frac{\boldsymbol{{b}}}{\frac{\sqrt{\mathrm{2}}\mathrm{sin}\:\left(\boldsymbol{\theta}+\frac{\boldsymbol{\pi}}{\mathrm{4}}\right)}{\mathrm{sin}\:\boldsymbol{\gamma}}+\frac{\mathrm{sin}\:\left(\boldsymbol{\beta}−\boldsymbol{\theta}\right)}{\mathrm{sin}\:\boldsymbol{\alpha}}} \\ $$$$\:\:\:\:=\frac{{b}}{\frac{\mathrm{sin}\:\theta+\mathrm{cos}\:\theta}{\mathrm{sin}\:\gamma}+\frac{\mathrm{sin}\:\beta\mathrm{cos}\:\theta−\mathrm{cos}\:\beta\mathrm{sin}\:\theta}{\mathrm{sin}\:\alpha}} \\ $$$$=\frac{{b}\mathrm{sin}\:\gamma\:\mathrm{sin}\:\alpha}{\mathrm{sin}\:\theta\left(\mathrm{sin}\:\alpha−\mathrm{cos}\:\beta\:\mathrm{sin}\:\gamma\right)+\mathrm{cos}\:\theta\left(\mathrm{sin}\:\alpha+\mathrm{sin}\:\beta\:\mathrm{sin}\:\gamma\right)} \\ $$$$\:\:\:\:\:\:\:\boldsymbol{{s}}=\frac{{b}\mathrm{sin}\:\gamma\:\mathrm{sin}\:\alpha}{{A}\mathrm{sin}\:\left(\theta+\phi\right)} \\ $$$$\boldsymbol{{where}}\: \\ $$$$\:\:\mathrm{tan}\:\boldsymbol{\phi}\:=\:\frac{\mathrm{sin}\:\alpha+\mathrm{sin}\:\beta\:\mathrm{sin}\:\gamma}{\mathrm{sin}\:\alpha−\mathrm{cos}\:\beta\:\mathrm{sin}\:\gamma} \\ $$$$\:\boldsymbol{{A}}=\sqrt{\left(\mathrm{sin}\:\alpha−\mathrm{cos}\:\beta\:\mathrm{sin}\:\gamma\right)^{\mathrm{2}} +\left(\mathrm{sin}\:\alpha+\mathrm{sin}\:\beta\:\mathrm{sin}\:\gamma\right)^{\mathrm{2}} } \\ $$$${for}\:{s}\:{to}\:{be}\:{maximum} \\ $$$$\:\:\:\:\mathrm{sin}\:\left(\theta+\phi\right)\:{has}\:{to}\:{be}\:{minimum}. \\ $$$${For}\:{many}\:{of}\:{the}\:{cases}\left({though}\right. \\ $$$$\left.{not}\:{all}\right), \\ $$$$\:\:\:\:\mathrm{sin}\:\left(\theta+\phi\right)\:{is}\:{minimum}\:{if}\:\theta=\mathrm{0} \\ $$$${Then} \\ $$$$\:\:\:\:\:\:\boldsymbol{{s}}_{{max}} \:=\:\frac{\boldsymbol{{b}}\mathrm{sin}\:\boldsymbol{\gamma}\:\mathrm{sin}\:\boldsymbol{\alpha}}{{A}\mathrm{sin}\:\boldsymbol{\phi}} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\:\frac{{b}\mathrm{sin}\:\gamma\:\mathrm{sin}\:\alpha}{\mathrm{sin}\:\alpha+\mathrm{sin}\:\beta\:\mathrm{sin}\:\gamma} \\ $$$${and}\:{since}\: \\ $$$$\:\:\:\:\:\frac{\mathrm{sin}\:\alpha}{{a}}=\:\frac{\mathrm{sin}\:\beta}{{b}}\:=\:\frac{\mathrm{sin}\:\gamma}{{c}}\:=\:\frac{\mathrm{1}}{\mathrm{2}{R}} \\ $$$$\:\:\:\boldsymbol{{s}}_{{max}} =\:\:\frac{{abc}}{\mathrm{2}{aR}+{bc}}\:=\:\frac{{a}}{\frac{\mathrm{2}{aR}}{{bc}}+\mathrm{1}}\:. \\ $$$$\:{This}\:{formula}\:{will}\:{have}\:{to}\:{be} \\ $$$${amended}\:{if}\:\mathrm{sin}\:\left(\theta+\phi\right)<\:\mathrm{sin}\:\phi \\ $$$${for}\:{nonzero}\:\theta\:. \\ $$
Commented by ajfour last updated on 06/Jul/18
Commented by MJS last updated on 06/Jul/18
thanks. I already thought there would be several different cases...
$$\mathrm{thanks}.\:\mathrm{I}\:\mathrm{already}\:\mathrm{thought}\:\mathrm{there}\:\mathrm{would}\:\mathrm{be} \\ $$$$\mathrm{several}\:\mathrm{different}\:\mathrm{cases}… \\ $$

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