Arccot and Arctans!

Geometry Level 4

cot ⁡ − 1 [ a b + 1 a − b ] + cot ⁡ − 1 [ b c + 1 b − c ] + cot ⁡ − 1 [ c a + 1 c − a ] \displaystyle\large \cot^{-1}\left[\dfrac{ab+1}{a-b}\right]+\cot^{-1} \left[\dfrac{bc+1}{b-c}\right]+\cot^{-1}\left[\dfrac{ca+1}{c-a}\right]

If a , b , c a,~b,~c are distinct non zero numbers having same sign, then the expression above returns two distinct values, m m and n n , find the value of m + n m+n .

Note: This problem uses the convention 0 < cot ⁡ − 1 ( x ) < π 0 < \cot^{-1}(x) < \pi for the range of the inverse cotangent function.


The answer is 9.42477796.

Kishore S. Shenoy by Kishore S. Shenoy , 22, India

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1 solution

Kishore S. Shenoy
Oct 15, 2015

We know that, tan ⁡ − 1 ( 1 x ) = { − π + cot ⁡ − 1 x x < 0 cot ⁡ − 1 x x > 0 \tan^{-1} \left(\dfrac{1}{x}\right) = \begin{cases}-\pi+\cot^{-1}x&x<0\\\cot^{-1}x &x>0\end{cases}

So, if a > b > c a>b>c cot ⁡ − 1 [ a b + 1 a − b ] + cot ⁡ − 1 [ b c + 1 b − c ] + cot ⁡ − 1 [ c a + 1 c − a ] = tan ⁡ − 1 [ a b + 1 a − b ] + tan ⁡ − 1 [ b c + 1 b − c ] + tan ⁡ − 1 [ c a + 1 c − a ] + π = tan ⁡ − 1 a − tan ⁡ − 1 b + tan ⁡ − 1 b − tan ⁡ − 1 c + tan ⁡ − 1 c − tan ⁡ − 1 a + π = π \displaystyle\cot^{-1}\left[\dfrac{ab+1}{a-b}\right]+\cot^{-1} \left[\dfrac{bc+1}{b-c}\right]+\cot^{-1}\left[\dfrac{ca+1}{c-a}\right] \\~~~~~~~~~~~~~~~= \tan^{-1}\left[\dfrac{ab+1}{a-b}\right]+\tan^{-1} \left[\dfrac{bc+1}{b-c}\right]+\tan^{-1}\left[\dfrac{ca+1}{c-a}\right]+\pi\\~~~~~~~~~~~~~~~=\tan^{-1}a-\tan^{-1}b+\tan^{-1}b-\tan^{-1}c+\tan^{-1}c-\tan^{-1}a+\pi\\\Large=\boxed{\pi}

Or else, if you say c > b > a c>b>a cot ⁡ − 1 [ a b + 1 a − b ] + cot ⁡ − 1 [ b c + 1 b − c ] + cot ⁡ − 1 [ c a + 1 c − a ] = π + tan ⁡ − 1 [ a b + 1 a − b ] + π + tan ⁡ − 1 [ b c + 1 b − c ] + tan ⁡ − 1 [ c a + 1 c − a ] = tan ⁡ − 1 a − tan ⁡ − 1 b + tan ⁡ − 1 b − tan ⁡ − 1 c + tan ⁡ − 1 c − tan ⁡ − 1 a + π = 2 π \displaystyle\cot^{-1}\left[\dfrac{ab+1}{a-b}\right]+\cot^{-1} \left[\dfrac{bc+1}{b-c}\right]+\cot^{-1}\left[\dfrac{ca+1}{c-a}\right] \\~~~~~~~~~~~~~~~= \pi+\tan^{-1}\left[\dfrac{ab+1}{a-b}\right]+\pi+\tan^{-1} \left[\dfrac{bc+1}{b-c} \right]+\tan^{-1}\left[\dfrac{ca+1}{c-a}\right]\\~~~~~~~~~~~~~~~=\tan^{-1}a-\tan^{-1}b+\tan^{-1}b-\tan^{-1}c+\tan^{-1}c-\tan^{-1}a+\pi\\\Large=\boxed{2\pi}

Hence m + n = 3 π \Huge \boxed{m+n = 3\pi}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

Good job keeping track of the signage.

Why are the only 2 cases that we care about a > b > c a > b > c and c > b > a c > b > a ?

@Calvin Lin sir, in the expression given, the values occur in cyclic form so that they can be taken in only two ways: Clockwise and Anticlockwise i.e. a > b > c a>b>c and c > b > a c>b>a

Kishore S. Shenoy - 5 years, 8 months ago

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Right, that has to be stated in the solution. Note that it is not true when we have 4 or more variables.

Calvin Lin Staff - 5 years, 8 months ago

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Yeah it's not... Thanks!

Kishore S. Shenoy - 5 years, 8 months ago

Cool problem! Btw you have a typo where you forgot to invert the argument.

James Wilson - 3 years, 7 months ago

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