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Geometry Level 4

Find the sum of all θ \theta between 0 0^{\circ} and 36 0 360^{\circ} inclusive for which

tan 2 θ = 3 tan θ . \tan 2\theta = 3\tan\theta.

Details and Assumptions

It is given that 0 θ 36 0 0^{\circ} \leq \theta \leq 360^{\circ} .


The answer is 1260.

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1 solution

Jared Low
Feb 17, 2015

Rewrite the equation as:

2 tan θ 1 tan 2 θ = 3 tan θ \frac{2\tan\theta}{1-\tan^2\theta}=3\tan\theta

Note that the left side is undefined for θ = 4 5 , 13 5 , 22 5 , 31 5 \theta=45^\circ,135^\circ,225^\circ,315^\circ while the right side is undefined for θ = 9 0 , 27 0 \theta=90^\circ,270^\circ . In particular, we can thus multiply both sides of the equation by 1 tan 2 θ 1-\tan^2\theta , since for all such other values of 0 θ 36 0 0^\circ\leq\theta \leq360^\circ , 1 tan 2 θ 0 1-\tan^2\theta \neq 0 . We then have:

2 tan θ = 3 tan θ ( 1 tan 2 θ ) tan θ ( 3 tan 2 θ 1 ) = 0 2\tan\theta=3\tan\theta(1-\tan^2\theta) \Rightarrow \tan\theta(3\tan^2\theta-1)=0

For tan θ = 0 \tan\theta=0 , we have θ = 0 , 18 0 , 36 0 \theta=0^\circ,180^\circ,360^\circ

For tan θ = 1 3 \tan\theta=\sqrt{\frac{1}{3}} , we have θ = 3 0 , 21 0 \theta=30^\circ,210^\circ

For tan θ = 1 3 \tan\theta=-\sqrt{\frac{1}{3}} , we have θ = 15 0 , 33 0 \theta=150^\circ,330^\circ

The sum of all these values is then thus 126 0 \boxed{1260^\circ}

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