Problem to shake your mind

Geometry Level 5

If $a$ is a real constant and $A,B$ and $C$ are variable angles. They are related to the constraint $\sqrt{a^{2}-13}\tan{A} + a\tan{B} + \sqrt {a^{2}+13}\tan{C} = 6a$ . Then find minimum positive value of $\tan^{2}{A} + \tan^{2}{B} + \tan^{2}{C}$ .

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The answer is 12.00.

3 solutions

Righved K
Dec 24, 2015

I love cauchy Schwartz inequality! Did by this way ! One of the best questions which was there in vectors set for jee!

How did you do it using cauchy schwartz please explain your solution

neelesh vij - 5 years, 5 months ago

This vector dot product inequality that u used in your solution is basically an inference of cauchy Schwartz....the reference is present in 12th ncert mathematics book under the section of vectors in one of the examples :D

Righved K - 5 years, 5 months ago

Can you post your solution? It's not clear how Cauchy Schwartz is applied. Thanks.

Pi Han Goh - 5 years, 5 months ago

@Pi Han Goh – By statement of Cauchy-Schwarz inequality, $(\sqrt{a^{2}-13}\tan{A} + a\tan{B} + \sqrt {a^{2}+13}\tan{C} = 6a)^{2} \le (a^{2}-13 + a^{2} + a^{2}+13)(\sum_{cyc}tan^{2}A)$ , thus leading to final answer.

Harsh Shrivastava - 4 years, 12 months ago

@Pi Han Goh – Actually the inequality- Vector A . Vector B <= |vector A| ×|Vector B| this inequality is itself a cauchy Schwartz inequality....seee this hope this helps..... https://en.m.wikipedia.org/wiki/Cauchy%E2%80%93Schwarz_inequality

Righved K - 5 years, 5 months ago

See how $\color{#D61F06}{\text{Cauchy Swarz Inequality}}$ is proved using $\color{#3D99F6}{\text{Linear Algebra}}$ and then compare your solution to the proof . You would find that your solution and the proof are essentially the same.

Aditya Sky - 5 years ago

Neelesh Vij
Dec 24, 2015

Consider vectors $\overrightarrow{X} =$ $\sqrt{a^{2} - 13} \hat{i} + a\hat{j} + \sqrt{a^{2} + 3}\hat{k}$ and $\overrightarrow{Y} =$ $\tan{A}^{2}\hat{i} + \tan{B}^{2}\hat{j} + \tan{C}^{2}\hat{k}$ taking their dot product we get $\overrightarrow{X} . \overrightarrow{Y} = |\overrightarrow{X}||\overrightarrow{Y}|\cos{\theta}$ where $\theta$ is the angle between the vectors now evaluating the dot product we get $\sqrt{a^{2} - 13 + a^{2} + a^{2} - 13} \times \sqrt{\tan{A}^{2} + \tan{B}^{2} + \tan{C}^{2}} \times \cos{\theta} = 6a$
$\longleftrightarrow$ $\sqrt{3} \times a \times \sqrt{\tan{A}^{2} + \tan{B}^{2} + \tan{C}^{2}} = 6a \sec{\theta}$ squaring both sides
$\tan{A}^{2} + \tan{B}^{2} + \tan{C}^{2} = 12 \sec{\theta}$
Now we know that $\sec{\theta} \geq 1$ so for minimum value we get $\sec{\theta} = 1$ so our minimum value is $\boxed{12}$
Also it is tan square A not tan A square

Did the same way ! :)

Prakhar Bindal - 5 years, 5 months ago

Ashutosh Sharma
Jan 26, 2018

use cauchy schwatz inequality,becomes lot easier

Problem to shake your mind

The answer is 12.00.

3 solutions

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