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Geometry Level 5

Let θ 1 \theta _{1} , θ 2 \theta_{2} and θ 3 \theta_{3} be three distinct solutions to the equation tan ⁡ θ = 17 100 \tan{\theta} = \dfrac{17}{100} such that θ i ∈ ( 0 , 3 π ) \theta _{i} \in \left( 0 , 3\pi \right) , for all i = 1 , 2 , 3 i={1,2,3} . Find the value of

tan ⁡ θ 1 3 × tan ⁡ θ 2 3 + tan ⁡ θ 2 3 × tan ⁡ θ 3 3 + tan ⁡ θ 3 3 × tan ⁡ θ 1 3 \tan{\dfrac{\theta_{1}}{3}} \times \tan{\dfrac{\theta_{2}}{3}} + \tan{\dfrac{\theta_{2}}{3}} \times \tan{\dfrac{\theta_{3}}{3}} + \tan{\dfrac{\theta_{3}}{3}} \times \tan{\dfrac{\theta_{1}}{3}}

This is not my original problem.


The answer is -3.

Surya Prakash by Surya Prakash , 22, India

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2 solutions

Surya Prakash
Oct 13, 2015

Since, tan ⁡ 3 θ = 3 tan ⁡ θ − tan ⁡ 3 θ 1 − 3 tan ⁡ 2 θ \tan{3\theta} = \dfrac{3\tan{\theta} - \tan^3 \theta}{1- 3 \tan ^2 \theta} . So,

tan ⁡ θ = 3 tan ⁡ θ 3 − tan ⁡ 3 θ 3 1 − 3 tan ⁡ 2 θ 3 \tan{\theta}= \dfrac{3\tan{\dfrac{\theta}{3}} - \tan^3 \dfrac{\theta}{3}}{1- 3 \tan ^2 \dfrac{\theta}{3}}

Let tan ⁡ θ 3 = x \tan{\dfrac{\theta}{3}} = x and it is given that tan ⁡ θ = 17 100 \tan{\theta} = \dfrac{17}{100} . So, the above equation turns into

100 x 3 − 51 x 2 − 300 x + 17 = 0 100 x^3 - 51 x^2 - 300 x +17 = 0

But tan ⁡ θ 1 3 \tan{\dfrac{\theta_{1}}{3}} , tan ⁡ θ 2 3 \tan{\dfrac{\theta_{2}}{3}} and tan ⁡ θ 3 3 \tan{\dfrac{\theta_{3}}{3}} are the roots of this equation. So, Vieta is going to help us here. Since they are the roots we get

tan ⁡ θ 1 3 × tan ⁡ θ 2 3 + tan ⁡ θ 2 3 × tan ⁡ θ 3 3 + tan ⁡ θ 3 3 × tan ⁡ θ 1 3 = − 300 100 = − 3 \tan{\dfrac{\theta_{1}}{3}} \times \tan{\dfrac{\theta_{2}}{3}} + \tan{\dfrac{\theta_{2}}{3}} \times \tan{\dfrac{\theta_{3}}{3}} + \tan{\dfrac{\theta_{3}}{3}} \times \tan{\dfrac{\theta_{1}}{3}} = \dfrac{-300}{100} = \boxed{-3}

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Moderator note:

Using this approach, we can see that the answer is independent of the exact value of tan ⁡ θ \tan \theta .

Now, how can we (directly / more easily) show that \sum \tan \frac \tan { \theta_i}{3} \times \frac \tan { \theta_j}{3}= - 3 ?

\sum \tan \frac \tan { \theta_i}{3} \times \frac \tan { \theta_j}{3}= - 3 can you pleased make this expression clear?
Do you mean. ∑ tan ⁡ θ i 3 × tan ⁡ θ j 3 = − 3 \sum \tan \frac { \theta_i}{3} \times \tan \frac { \theta_j}{3}= - 3 ?

Niranjan Khanderia - 5 years, 8 months ago

Same solution .

Aakash Khandelwal - 5 years, 8 months ago

tan ⁡ θ = 3 tan ⁡ θ 3 − tan ⁡ 3 θ 3 1 − 3 tan ⁡ 2 θ 3 \tan{\theta}= \dfrac{3\tan{\dfrac{\theta}{3}} - \tan^3 \dfrac{\theta}{3}}{1- 3 \tan ^2 \dfrac{\theta}{3}} For convenience let us say X = tan ⁡ θ 3 a n d Y = tan ⁡ θ . ∴ X 3 − 3 ∗ Y ∗ X 2 − 3 ∗ X + Y = 0. By Vieta, the required result = − 3 1 = − 3. \text{For convenience let us say }X= \tan{\dfrac{\theta}{3}} ~~and~~Y=\tan\theta.\\ \therefore~X^3 - 3*Y*X^2 - 3*X + Y=0.\\ \text{By Vieta, the required result }=\dfrac {-3} 1=- 3.

3 Helpful 0 Interesting 0 Brilliant 0 Confused

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