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Difference between revisions of "Mock AIME 3 Pre 2005 Problems/Problem 15"
(wrote solution)
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==Solution==
 
==Solution==
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{{solution}}
 
  
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Let <cmath> x := k^2 + k + 1</cmath>
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<cmath>a_k := \cos^{-1}\left(\frac{k^2 + k + 1}{\sqrt{k^4 + 2k^3 + 3k^2 + 2k + 2}} \right )</cmath>
 +
 +
Factoring the radicand, we have
 +
<cmath>a_k = \cos^{-1}\left(\frac{x}{\sqrt{x^2+1}} \right )</cmath>
 +
The fraction looks remarkably apt for a trigonometric substitution; namely, define <math>\theta</math> such that <math> x = \tan{\theta}</math>. Then the RHS becomes
 +
<cmath>a_k = \cos^{-1}\left(\frac{\tan{\theta}}{\sec{\theta}} \right ) = \cos^{-1}{(\sin{\theta})}</cmath>
 +
But <cmath>\cos{(\pi/2-\theta)} = \sin{\theta}</cmath>
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Therefore,
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<cmath> a_k = \pi/2 - \theta = \pi/2 - \tan^{-1}{x} </cmath>
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This gives us
 +
<cmath> \tan{a_k} = \tan{(\pi/2 - \tan^{-1}{x})} </cmath>
 +
<cmath> = \cot{(\tan^{-1}{x})} = \dfrac{1}{\tan{(\tan^{-1}{x})} } </cmath>
 +
<cmath>  = \dfrac{1}{x} = \dfrac{1}{k^2+k+1} </cmath>
 +
So now
 +
<cmath> a_k = \tan^{-1}{\left( \dfrac{1}{k^2+k+1} \right) } </cmath>
 +
<cmath> = \tan^{-1}{ \left( \dfrac{ (k+1) + (-k) }{ 1 - (-k)(k+1) } \right ) } </cmath>
 +
<cmath> = \tan^{-1}{(k+1)} - \tan^{-1}{k} </cmath>
 +
When we sum <math>a_k</math>, this sum now telescopes:
 +
<cmath>\Omega = \sum_{k=1}^{40} a_k </cmath>
 +
<cmath> = \sum_{k=1}^{40} \tan^{-1}{(k+1)} - \tan^{-1}{k} </cmath>
 +
<cmath> = \tan^{-1}{40} - \tan^{-1}{1} </cmath>
 +
Therefore, the required value
 +
<cmath> \tan{\Omega} = \tan{(\tan^{-1}{40} - \tan^{-1}{1})} </cmath>
 +
<cmath> = \dfrac{ 40 - 1}{1 + 40 \cdot 1 } </cmath>
 +
<cmath> = \dfrac{39}{41} </cmath>
 +
giving us the desired answer of <math>\boxed{80}</math>.
 
==See also==
 
==See also==

Revision as of 00:25, 11 March 2009

Problem

Let $\Omega$ denote the value of the sum

$\sum_{k=1}^{40} \cos^{-1}\left(\frac{k^2 + k + 1}{\sqrt{k^4 + 2k^3 + 3k^2 + 2k + 2}}\right)$

The value of $\tan\left(\Omega\right)$ can be expressed as $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Compute $m + n$.

Solution

Let \[x := k^2 + k + 1\] \[a_k := \cos^{-1}\left(\frac{k^2 + k + 1}{\sqrt{k^4 + 2k^3 + 3k^2 + 2k + 2}} \right )\]

Factoring the radicand, we have \[a_k = \cos^{-1}\left(\frac{x}{\sqrt{x^2+1}} \right )\] The fraction looks remarkably apt for a trigonometric substitution; namely, define $\theta$ such that $x = \tan{\theta}$. Then the RHS becomes \[a_k = \cos^{-1}\left(\frac{\tan{\theta}}{\sec{\theta}} \right ) = \cos^{-1}{(\sin{\theta})}\] But \[\cos{(\pi/2-\theta)} = \sin{\theta}\] Therefore, \[a_k = \pi/2 - \theta = \pi/2 - \tan^{-1}{x}\] This gives us \[\tan{a_k} = \tan{(\pi/2 - \tan^{-1}{x})}\] \[= \cot{(\tan^{-1}{x})} = \dfrac{1}{\tan{(\tan^{-1}{x})} }\] \[= \dfrac{1}{x} = \dfrac{1}{k^2+k+1}\] So now \[a_k = \tan^{-1}{\left( \dfrac{1}{k^2+k+1} \right) }\] \[= \tan^{-1}{ \left( \dfrac{ (k+1) + (-k) }{ 1 - (-k)(k+1) } \right ) }\] \[= \tan^{-1}{(k+1)} - \tan^{-1}{k}\] When we sum $a_k$, this sum now telescopes: \[\Omega = \sum_{k=1}^{40} a_k\] \[= \sum_{k=1}^{40} \tan^{-1}{(k+1)} - \tan^{-1}{k}\] \[= \tan^{-1}{40} - \tan^{-1}{1}\] Therefore, the required value \[\tan{\Omega} = \tan{(\tan^{-1}{40} - \tan^{-1}{1})}\] \[= \dfrac{ 40 - 1}{1 + 40 \cdot 1 }\] \[= \dfrac{39}{41}\] giving us the desired answer of $\boxed{80}$.

See also

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