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Question Number 21973 by Joel577 last updated on 08/Oct/17

$$\mathrm{\Delta }ABC\mathrm{with}\mathrm{sides}a,b,c\in \mathbb{Z}\mathrm{and}\mathrm{\angle }A=3\mathrm{\angle }B$$$$\mathrm{If}\mathrm{its}\mathrm{circumference}\mathrm{is}\mathrm{minimum},\mathrm{then}$$$$\mathrm{find}a,b,c$$

Commented by Rasheed.Sindhi last updated on 08/Oct/17

$$\mathrm{Let}\mathrm{\angle }\mathrm{B}=\theta \Rightarrow \mathrm{\angle }\mathrm{A}=3\theta ,\mathrm{\angle }\mathrm{C}=\pi -4\theta$$$$\mathrm{Let}a,b\mathrm{\&}c\mathrm{are}\mathrm{respective}\mathrm{opposite}$$$$\mathrm{sides}\mathrm{of}\mathrm{\angle }\mathrm{A},\mathrm{\angle }\mathrm{B}\mathrm{\&}\mathrm{\angle }\mathrm{C}.$$$$\mathrm{sine}\mathrm{law}:\frac{a}{\sin \left(3\theta \right)}=\frac{b}{\sin \left(\theta \right)}=\frac{c}{\sin (\pi -4\theta )}=\mathrm{k}\left(\mathrm{say}\right)$$$$a=\mathrm{k}\sin \left(3\theta \right),b=\mathrm{k}\sin \left(\theta \right),$$$$c=\mathrm{k}\sin (\pi -4\theta )$$$$a,b,c\in \mathbb{Z}\Rightarrow \frac{a}{b},\frac{b}{c},\frac{a}{c}...\in \mathbb{Q}$$$$\frac{\sin \left(\theta \right)}{\sin \left(3\theta \right)},\frac{\sin \left(\theta \right)}{\sin (\pi -4\theta )},\frac{\sin \left(3\theta \right)}{\sin (\pi -4\theta )}\mathrm{etc}$$$$\mathrm{are}\mathrm{rational}\mathrm{numbers}.$$$$⋮$$$$a=\sqrt{b^2+c^2-2bc\cos \left(3\theta \right)}$$$$⋮$$

Answered by ajfour last updated on 09/Oct/17

$$\frac{\sin 3\theta }{a}=\frac{\sin \theta }{b}=\frac{\sin (\pi -4\theta )}{c}(\mathrm{\angle }B=\theta )$$$$p=a+b+c$$$$=\frac{b\sin 3\theta }{\sin \theta }+b+\frac{b\sin 4\theta }{\sin \theta }$$$$pisminimum,\Rightarrow$$$$3-4\sin {}^2\theta +1+4\cos \theta \cos 2\theta ismin.$$$$\Rightarrow -8\sin \theta \cos \theta -4\sin \theta \cos 2\theta$$$$-8\cos \theta \sin 2\theta =0$$$$\Rightarrow 2\cos \theta +\cos 2\theta +4\cos {}^2\theta =0$$$$\Rightarrow 6\cos {}^2\theta +2\cos \theta -1=0$$$$\cos \theta =\frac{-2\pm 2\sqrt{7}}{12}$$$$\cos \theta =\frac{-1\pm \sqrt{7}}{6}$$$$letuschoose\cos \theta =\frac{\sqrt{7}-1}{6}$$$$\sin \theta =\sqrt{1-\frac{(8-2\sqrt{7})}{36}}$$$$=\frac{\sqrt{28+2\sqrt{7}}}{6}$$$$........$$

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