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Question Number 213606 by CrispyXYZ last updated on 10/Nov/24

$$△ABC.2a+b=2c.\mathrm{Find}\mathrm{the}\mathrm{minimum}\mathrm{of}$$$$\frac{3}{\sin C}+\frac{1}{\tan A}.$$

Answered by Ghisom last updated on 10/Nov/24

$$\mathrm{wlog}b=1\Rightarrow c=a+\frac{1}{2}$$$$a^2+b^2-2ab\cos \gamma =c^2$$$$b^2+c^2-2bc\cos \alpha =a^2$$$$\cos \gamma =\frac{3-4a}{8a}$$$$\cos \alpha =\frac{4a+5}{4(2a+1)}$$$$\frac{3}{\sin \gamma }+\frac{1}{\tan \alpha }=\frac{28a+5}{\sqrt{3(4a-1)(4a+3)}}$$$$\frac{d\left[\frac{28a+5}{\sqrt{3(4a-1)(4a+3)}}\right]}{da}=0$$$$\frac{8(4a-13)}{\sqrt{3{(4a-1)}^3{(4a+3)}^3}}=0$$$$a=\frac{13}{4}$$$$\min (\frac{3}{\sin \gamma }+\frac{1}{\tan \alpha })=4$$

Answered by CrispyXYZ last updated on 11/Nov/24

$$\sin B=2(\sin C-\sin A)$$$$\sin (A+C)=2(\sin C-\sin A)$$$$2\sin \frac{A+C}{2}\cos \frac{A+C}{2}=4\cos \frac{A+C}{2}\sin \frac{C-A}{2}$$$$\sin \frac{A+C}{2}=2\sin \frac{C-A}{2}$$$$\sin \frac{A}{2}\cos \frac{C}{2}+\cos \frac{A}{2}\sin \frac{C}{2}=2\sin \frac{C}{2}\cos \frac{A}{2}-2\cos \frac{C}{2}\sin \frac{A}{2}$$$$3\tan \frac{A}{2}=\tan \frac{C}{2}$$$$\mathrm{Let}x=\tan \frac{A}{2}.$$$$\frac{3}{\sin C}+\frac{1}{\tan A}$$$$=\frac{3(1+{\tan }^2\frac{C}{2})}{2\tan \frac{C}{2}}+\frac{1-{\tan }^2\frac{A}{2}}{2\tan \frac{A}{2}}$$$$=\frac{1+9x^2}{2x}+\frac{1-x^2}{2x}$$$$=\frac{1}{x}+4x$$$$⩾4$$

Commented by mnjuly1970 last updated on 10/Nov/24

$${(\frac{1}{x}+4x)}_{min}=4$$

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