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Geometry Level 3

If tan ⁡ θ = n tan ⁡ ϕ \tan \theta = n\tan\phi , then the maximum value of tan ⁡ 2 ( θ − ϕ ) { \tan }^{ 2 }(\theta -\phi ) is equal to:

( n + 1 ) 2 4 n \frac { { (n+1) }^{ 2 } }{ 4n } ( n − 1 ) 2 2 n \frac { { (n-1) }^{ 2 } }{ 2n } ( n + 1 ) 2 2 n \frac { { (n+1) }^{ 2 } }{ 2n } ( n − 1 ) 2 4 n \frac { { (n-1) }^{ 2 } }{ 4n }

Yash Jain by Yash Jain , 21, India

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1 solution

Chew-Seong Cheong
Oct 27, 2018

tan ⁡ 2 ( θ − ϕ ) = ( tan ⁡ θ − tan ⁡ ϕ 1 + tan ⁡ θ tan ⁡ ϕ ) 2 Given that tan ⁡ θ = n tan ⁡ ϕ = ( n − 1 ) 2 tan ⁡ 2 ϕ ( 1 + n tan ⁡ 2 ϕ ) 2 Let x = tan ⁡ 2 ϕ = ( n − 1 ) 2 x ( 1 + n x ) 2 = ( n − 1 ) 2 n ( 1 + n x − 1 ( 1 + n x ) 2 ) = ( n − 1 ) 2 n ( 1 1 + n x − 1 ( 1 + n x ) 2 ) = ( n − 1 ) 2 n ( 1 4 − ( 1 1 + n x − 1 2 ) 2 ) Note that ( 1 1 + n x − 1 2 ) 2 ≥ 0 ≤ ( n − 1 ) 2 4 n Equality occurs when x = 1 n \begin{aligned} \tan^2 (\theta - \phi) & = \left(\frac {\tan \theta - \tan \phi}{1+\tan \theta \tan \phi} \right)^2 & \small \color{#3D99F6} \text{Given that }\tan \theta = n \tan \phi \\ & = \frac {(n-1)^2\tan^2 \phi}{(1+n\tan^2 \phi)^2} & \small \color{#3D99F6} \text{Let }x = \tan^2 \phi \\ & = \frac {(n-1)^2x}{(1+nx)^2} \\ & = \frac {(n-1)^2}n \left(\frac {1+nx - 1}{(1+nx)^2}\right) \\ & = \frac {(n-1)^2}n \left(\frac 1{1+nx}- \frac 1{(1+nx)^2}\right) \\ & = \frac {(n-1)^2}n \left(\frac 14 - \left(\frac 1{1+nx}- \frac 12\right)^2 \right) & \small \color{#3D99F6} \text{Note that }\left(\frac 1{1+nx}- \frac 12\right)^2 \ge 0 \\ & \le \boxed {\dfrac {(n-1)^2}{4n}} & \small \color{#3D99F6} \text{Equality occurs when }x = \frac 1n \end{aligned}

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