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ABC-with-sides-a-b-c-Z-and-A-3-B-If-its-circumference-is-minimum-then-find-a-b-c-

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Question Number 21973 by Joel577 last updated on 08/Oct/17
ΔABC with sides a, b, c ∈ Z and ∠A = 3∠B If its circumference is minimum, then find a, b, c
$$\Delta{ABC}\:\mathrm{with}\:\mathrm{sides}\:{a},\:{b},\:{c}\:\in\:\mathbb{Z}\:\mathrm{and}\:\angle{A}\:=\:\mathrm{3}\angle{B} \\ $$$$\mathrm{If}\:\mathrm{its}\:\mathrm{circumference}\:\mathrm{is}\:\mathrm{minimum},\:\mathrm{then} \\ $$$$\mathrm{find}\:{a},\:{b},\:{c} \\ $$
Commented by Rasheed.Sindhi last updated on 08/Oct/17
Let ∠B=θ⇒∠A=3θ , ∠C=π−4θ Let a,b & c are respective opposite sides of ∠A,∠B & ∠C. sine law:(a/(sin(3θ)))=(b/(sin(θ)))=(c/(sin(π−4θ)))=k(say) a=k sin(3θ), b=k sin(θ), c=k sin(π−4θ) a,b,c∈Z⇒(a/b),(b/c),(a/c) ...∈Q ((sin(θ))/(sin(3θ))),((sin(θ))/(sin(π−4θ))),((sin(3θ))/(sin(π−4θ))) etc are rational numbers. ⋮ a=(√(b^2 +c^2 −2bc cos(3θ) )) ⋮
$$\mathrm{Let}\:\angle\mathrm{B}=\theta\Rightarrow\angle\mathrm{A}=\mathrm{3}\theta\:,\:\angle\mathrm{C}=\pi−\mathrm{4}\theta \\ $$$$\mathrm{Let}\:{a},{b}\:\&\:{c}\:\mathrm{are}\:\mathrm{respective}\:\mathrm{opposite} \\ $$$$\mathrm{sides}\:\mathrm{of}\:\angle\mathrm{A},\angle\mathrm{B}\:\&\:\angle\mathrm{C}. \\ $$$$\mathrm{sine}\:\mathrm{law}:\frac{{a}}{\mathrm{sin}\left(\mathrm{3}\theta\right)}=\frac{{b}}{\mathrm{sin}\left(\theta\right)}=\frac{{c}}{\mathrm{sin}\left(\pi−\mathrm{4}\theta\right)}=\mathrm{k}\left(\mathrm{say}\right) \\ $$$$\:\:\:\:{a}=\mathrm{k}\:\mathrm{sin}\left(\mathrm{3}\theta\right),\:{b}=\mathrm{k}\:\mathrm{sin}\left(\theta\right), \\ $$$$\:\:\:\:{c}=\mathrm{k}\:\mathrm{sin}\left(\pi−\mathrm{4}\theta\right) \\ $$$${a},{b},{c}\in\mathbb{Z}\Rightarrow\frac{{a}}{{b}},\frac{{b}}{{c}},\frac{{a}}{{c}}\:…\in\mathbb{Q} \\ $$$$\frac{\mathrm{sin}\left(\theta\right)}{\mathrm{sin}\left(\mathrm{3}\theta\right)},\frac{\mathrm{sin}\left(\theta\right)}{\mathrm{sin}\left(\pi−\mathrm{4}\theta\right)},\frac{\mathrm{sin}\left(\mathrm{3}\theta\right)}{\mathrm{sin}\left(\pi−\mathrm{4}\theta\right)}\:\mathrm{etc} \\ $$$$\mathrm{are}\:\mathrm{rational}\:\mathrm{numbers}. \\ $$$$\vdots \\ $$$$\:\:\:{a}=\sqrt{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −\mathrm{2}{bc}\:\mathrm{cos}\left(\mathrm{3}\theta\right)\:} \\ $$$$\:\:\vdots \\ $$
Answered by ajfour last updated on 09/Oct/17
((sin 3θ)/a)=((sin θ)/b)=((sin (π−4θ))/c) (∠B=θ) p=a+b+c =((bsin 3θ)/(sin θ))+b+((bsin 4θ)/(sin θ)) p is minimum , ⇒ 3−4sin^2 θ+1+4cos θcos 2θ is min. ⇒ −8sin θcos θ−4sin θcos 2θ −8cos θsin 2θ=0 ⇒ 2cos θ+cos 2θ+4cos^2 θ=0 ⇒ 6cos^2 θ+2cos θ−1=0 cos θ=((−2±2(√7))/(12)) cos θ=((−1±(√7))/6) let us choose cos θ=(((√7)−1)/6) sin θ=(√(1−(((8−2(√7)))/(36)))) =((√(28+2(√7)))/6) ........
$$\frac{\mathrm{sin}\:\mathrm{3}\theta}{{a}}=\frac{\mathrm{sin}\:\theta}{{b}}=\frac{\mathrm{sin}\:\left(\pi−\mathrm{4}\theta\right)}{{c}}\:\:\:\left(\angle{B}=\theta\right) \\ $$$${p}={a}+{b}+{c} \\ $$$$\:=\frac{{b}\mathrm{sin}\:\mathrm{3}\theta}{\mathrm{sin}\:\theta}+{b}+\frac{{b}\mathrm{sin}\:\mathrm{4}\theta}{\mathrm{sin}\:\theta} \\ $$$${p}\:{is}\:{minimum}\:,\:\Rightarrow\:\: \\ $$$$\mathrm{3}−\mathrm{4sin}\:^{\mathrm{2}} \theta+\mathrm{1}+\mathrm{4cos}\:\theta\mathrm{cos}\:\mathrm{2}\theta\:\:{is}\:{min}. \\ $$$$\Rightarrow\:\:−\mathrm{8sin}\:\theta\mathrm{cos}\:\theta−\mathrm{4sin}\:\theta\mathrm{cos}\:\mathrm{2}\theta \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{8cos}\:\theta\mathrm{sin}\:\mathrm{2}\theta=\mathrm{0} \\ $$$$\Rightarrow\:\:\:\mathrm{2cos}\:\theta+\mathrm{cos}\:\mathrm{2}\theta+\mathrm{4cos}\:^{\mathrm{2}} \theta=\mathrm{0} \\ $$$$\Rightarrow\:\:\:\mathrm{6cos}\:^{\mathrm{2}} \theta+\mathrm{2cos}\:\theta−\mathrm{1}=\mathrm{0} \\ $$$$\:\:\:\:\:\mathrm{cos}\:\theta=\frac{−\mathrm{2}\pm\mathrm{2}\sqrt{\mathrm{7}}}{\mathrm{12}}\: \\ $$$$\:\:\:\:\:\mathrm{cos}\:\theta=\frac{−\mathrm{1}\pm\sqrt{\mathrm{7}}}{\mathrm{6}} \\ $$$${let}\:{us}\:{choose}\:\mathrm{cos}\:\theta=\frac{\sqrt{\mathrm{7}}−\mathrm{1}}{\mathrm{6}} \\ $$$$\mathrm{sin}\:\theta=\sqrt{\mathrm{1}−\frac{\left(\mathrm{8}−\mathrm{2}\sqrt{\mathrm{7}}\right)}{\mathrm{36}}}\: \\ $$$$\:\:\:\:\:\:\:\:=\frac{\sqrt{\mathrm{28}+\mathrm{2}\sqrt{\mathrm{7}}}}{\mathrm{6}}\: \\ $$$$…….. \\ $$

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