ABC-2a-b-2c-Find-the-minimum-of-3-sin-C-1-tan-A-

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

△ A B C . 2 a + b = 2 c . Find the minimum of

\frac{3}{\sin C} + \frac{1}{\tan A} .

Answered by Ghisom last updated on 10/Nov/24

wlog b = 1 \Rightarrow c = a + \frac{1}{2}

a^{2} + b^{2} - 2 a b \cos γ = c^{2}

b^{2} + c^{2} - 2 b c \cos α = a^{2}

\cos γ = \frac{3 - 4 a}{8 a}

\cos α = \frac{4 a + 5}{4 (2 a + 1)}

\frac{3}{\sin γ} + \frac{1}{\tan α} = \frac{28 a + 5}{\sqrt{3 (4 a - 1) (4 a + 3)}}

\frac{d [\frac{28 a + 5}{\sqrt{3 (4 a - 1) (4 a + 3)}}]}{d a} = 0

\frac{8 (4 a - 13)}{\sqrt{3 {(4 a - 1)}^{3} {(4 a + 3)}^{3}}} = 0

a = \frac{13}{4}

\min (\frac{3}{\sin γ} + \frac{1}{\tan α}) = 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}

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 + 9 x^{2}}{2 x} + \frac{1 - x^{2}}{2 x}

= \frac{1}{x} + 4 x

⩾ 4

Commented by mnjuly1970 last updated on 10/Nov/24

{(\frac{1}{x} + 4 x)}_{m i n} = 4