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Geometry Level 4

The 9 elements of the set S = { 5 , 25 , 45 , … , 165 } S= \{5,25,45, \ldots, 165 \} follows an arithmetic progression.

Evaluate ∑ k ∈ S tan ⁡ ( k ∘ ) \displaystyle \sum_{k \in S} \tan(k^\circ) .


The answer is 9.

Pi Han Goh by Pi Han Goh , 30, Malaysia

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1 solution

David Vreken
Jan 5, 2019

The sum, after rearranging, using the identity − tan ⁡ ( 180 ° − x ) = tan ⁡ x -\tan(180° - x) = \tan x , converting to radians, and expressing in sigma notation is

= tan ⁡ 5 ° + tan ⁡ 25 ° + tan ⁡ 45 ° + tan ⁡ 65 ° + tan ⁡ 85 ° + tan ⁡ 105 ° + tan ⁡ 125 ° + tan ⁡ 145 ° + tan ⁡ 165 ° = \tan 5° + \tan 25° + \tan 45° + \tan 65° + \tan 85° + \tan 105° + \tan 125° + \tan 145° + \tan 165°

= tan ⁡ 5 ° + tan ⁡ 165 ° + tan ⁡ 25 ° + tan ⁡ 145 ° + tan ⁡ 45 ° + tan ⁡ 125 ° + tan ⁡ 65 ° + tan ⁡ 105 ° + tan ⁡ 85 ° = \tan 5° + \tan 165° + \tan 25° + \tan 145° + \tan 45° + \tan 125° + \tan 65° + \tan 105° + \tan 85°

= tan ⁡ 5 ° − tan ⁡ 15 ° + tan ⁡ 25 ° − tan ⁡ 35 ° + tan ⁡ 45 ° − tan ⁡ 55 ° + tan ⁡ 65 ° − tan ⁡ 75 ° + tan ⁡ 85 ° = \tan 5° - \tan 15° + \tan 25° - \tan 35° + \tan 45° - \tan 55° + \tan 65° - \tan 75° + \tan 85°

= tan ⁡ π 36 − tan ⁡ 3 π 36 + tan ⁡ 5 π 36 − tan ⁡ 7 π 36 + tan ⁡ 9 π 36 − tan ⁡ 11 π 36 + tan ⁡ 13 π 36 − tan ⁡ 15 π 36 + tan ⁡ 17 π 36 = \tan \frac{\pi}{36} - \tan \frac{3\pi}{36} + \tan \frac{5\pi}{36} - \tan \frac{7\pi}{36} + \tan \frac{9\pi}{36} - \tan \frac{11\pi}{36} + \tan \frac{13\pi}{36} - \tan \frac{15\pi}{36} + \tan \frac{17\pi}{36}

= ∑ 0 8 ( − 1 ) k tan ⁡ ( 2 k + 1 ) π 36 = \displaystyle \sum_{0}^{8} (-1)^k \tan \frac{(2k + 1)\pi}{36}

From this website we find that ∑ 0 n − 1 ( − 1 ) k tan ⁡ ( 2 k + 1 ) π 4 n = ( − 1 ) n − 1 n \displaystyle \sum_{0}^{n-1} (-1)^k \tan \frac{(2k + 1)\pi}{4n} = (-1)^{n - 1}n for all natural numbers n n .

In this case, n = 9 n = 9 , so the sum is ( − 1 ) 9 − 1 9 = 9 (-1)^{9 - 1}9 = \boxed{9} .

3 Helpful 0 Interesting 0 Brilliant 2 Confused

The most crucial part is left unexplained.

Atomsky Jahid - 2 years, 5 months ago

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Hint: Notice that the tangent of 9 times the value of any of the elements in S S is 1. For example, tan ⁡ ( 9 × 10 5 ∘ ) = 1 \tan(9\times 105^\circ) = 1 .

Hint 2: Express tan ⁡ ( 3 x ) \tan(3x) in terms of tan ⁡ ( x ) \tan(x) . Then express tan ⁡ ( 9 x ) \tan(9x) in terms of tan ⁡ ( x ) \tan(x) .

Pi Han Goh - 2 years, 5 months ago

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Interesting! So tan ⁡ 3 x = 3 tan ⁡ x − tan ⁡ 3 x 1 − 3 tan ⁡ 2 x \tan 3x = \frac{3 \tan x - \tan^3 x}{1 - 3\tan^2 x} and tan ⁡ 9 x = tan ⁡ 9 x − 36 tan ⁡ 7 x + 126 tan ⁡ 5 x − 84 tan ⁡ 3 x + 9 tan ⁡ x 9 tan ⁡ 8 x − 84 tan ⁡ 6 x + 126 tan ⁡ 4 x − 36 tan ⁡ 2 x + 1 \tan 9x = \frac{\tan^9 x - 36\tan^7 x + 126\tan^5 x - 84\tan^3 x + 9\tan x}{9\tan^8 x - 84\tan^6 x + 126 \tan^4 x - 36 \tan^2 x + 1} . For tan ⁡ 9 x = 1 \tan 9x = 1 , this solves to 1 = ( tan ⁡ x − 1 ) 9 1 = (\tan x - 1)^9 , so the 9 9 given tangents are the 9 9 roots of 1 1 offset by 1 1 each time which adds up to 9 9 .

David Vreken - 2 years, 5 months ago

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