Trigonometry (3)

Geometry Level 3

Triangle $ABC$ is such that $\tan \dfrac{A}{2}$ , $\tan \dfrac{B}{2}$ , $\tan \dfrac{C}{2}$ are in harmonic progression , then find the minimum value of $\cot \dfrac{B}{2}$ to 3 decimal places.

The answer is 1.732.

2 solutions

Chew-Seong Cheong
Apr 19, 2017

When $\tan \dfrac A2$ , $\tan \dfrac B2$ and $\tan \dfrac C2$ are in harmonic progression, then

$\begin{aligned} \frac 1{\tan \frac A2} + \frac 1{\tan \frac C2} & = \frac 2{\tan \color{#3D99F6} \frac B2} & \small \color{#3D99F6} \text{Note that for }\triangle ABC: \ \frac B2 = \frac \pi 2 - \left(\frac A2 + \frac C2\right) \\ & = \frac 2{\tan \color{#3D99F6}\left(\frac \pi 2 - \left(\frac A2 + \frac C2\right)\right)} \\ & = \frac 2{\cot \left(\frac A2 + \frac C2\right)} \\ \implies \frac 1{\tan \frac A2} + \frac 1{\tan \frac C2} & = 2 \tan \left(\frac A2 + \frac C2\right) \\ \frac {\tan \frac A2 + \tan \frac C2}{\tan \frac A2 \tan \frac C2 } & = \frac {2\left(\tan \frac A2 + \tan \frac C2 \right)}{1-\tan \frac A2 \tan \frac C2} \\ 1-\tan \frac A2 \tan \frac C2 & = 2 \tan \frac A2 \tan \frac C2 \\ \implies \tan \frac A2 \tan \frac C2 & = \frac 13 \end{aligned}$

Using AM-GM inequality, we have:

$\begin{aligned} \frac 1{\tan \frac A2} + \frac 1{\tan \frac C2} \ge \frac 2{\sqrt{\tan \frac A2 \tan \frac C2}} = 2\sqrt 3 \\ \frac 2{\tan \frac B2} \ge 2\sqrt 3 \\ \implies \cot \frac B2 \ge \sqrt 3 \approx \boxed{1.732} \end{aligned}$

Thank you sir for such a nice solution (+1). :)

Rahil Sehgal - 4 years, 1 month ago

But the problem may be wrong. Because $\tan \frac A2 = \tan \frac B2 = \tan \frac C2 = \frac 1{\sqrt 3}$

Chew-Seong Cheong - 4 years, 1 month ago

Sir , that means that all the angles are equal to 60 degrees than what is the issue?

Ankit Kumar Jain - 4 years, 1 month ago

@Ankit Kumar Jain – Are they in harmonic progression then? The common difference is 0.

Chew-Seong Cheong - 4 years, 1 month ago

@Chew-Seong Cheong – But sir , we can still consider it to be an HP though the common difference is 0. Because it satisfies the conditions of being an HP..Point out the mistakes if I am wrong. Thanks

Ankit Kumar Jain - 4 years, 1 month ago

@Ankit Kumar Jain – I suppose so. I was trying to get one with difference not equal to zero but couldn't.

Chew-Seong Cheong - 4 years, 1 month ago

Md Zuhair
Apr 18, 2017

Solution

$\implies$ $\dfrac{2}{\tan \dfrac{B}{2}} = \dfrac{1}{\tan \dfrac{A}{2}} + \dfrac{1}{\tan \dfrac{C}{2}}$

$\implies$ $\dfrac{2}{\tan \dfrac{180-A-C}{2}} = \dfrac{1}{\tan \dfrac{A}{2}} + \dfrac{1}{\tan \dfrac{C}{2}}$

$\implies$ $\dfrac{2}{\tan (90 -\dfrac{A+C}{2})} = \dfrac{1}{\tan \dfrac{A}{2}} + \dfrac{1}{\tan \dfrac{C}{2}}$

$\implies$ $2 \tan \dfrac{A+C}{2} = \dfrac{1}{\tan \dfrac{A}{2}} + \dfrac{1}{\tan \dfrac{C}{2}}$

$\implies$ $2 \tan \dfrac{A+C}{2} = \dfrac{\tan \dfrac{A}{2} + \tan \dfrac{C}{2}}{\tan\dfrac{A}{2} \times \tan\dfrac{C}{2}}$

$\implies$ $2 \dfrac{ \tan\dfrac{A}{2} + \tan \dfrac{C}{2}}{1- \tan \dfrac{A}{2} \times \tan \dfrac{C}{2}} = \dfrac{\tan \dfrac{A}{2} + \tan \dfrac{C}{2}}{\tan\dfrac{A}{2} \times \tan\dfrac{C}{2}}$

$\implies$ $\dfrac{2}{1- \tan \dfrac{A}{2} \times \tan \dfrac{C}{2}} = \dfrac{1}{\tan\dfrac{A}{2} \times \tan\dfrac{C}{2}}$

Crossmultiplying and simplyfying we get we get

$\implies$ $\tan \dfrac{A}{2} \times \tan\dfrac{C}{2} = \dfrac{1}{3}$

$\implies$ $\dfrac{A}{2} = \dfrac{C}{2} = 30$

$\implies \boxed{A=B=C=60}$

$\implies \cot\dfrac{B}{2} = \cot 30 = \boxed{\sqrt{3} = 1.732}$

Trigonometry (3)

The answer is 1.732.

2 solutions

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