Difference between revisions of "Mock AIME 3 Pre 2005 Problems/Problem 15"
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==Solution== | ==Solution== | ||
| − | |||
| − | ==See | + | Let <cmath> x := k^2 + k + 1</cmath> |
| + | <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> | ||
| + | Therefore, | ||
| + | <cmath> a_k = \pi/2 - \theta = \pi/2 - \tan^{-1}{x} </cmath> | ||
| + | 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}{41} - \tan^{-1}{1} </cmath> | ||
| + | Therefore, the required value | ||
| + | <cmath> \tan{\Omega} = \tan{(\tan^{-1}{41} - \tan^{-1}{1})} </cmath> | ||
| + | <cmath> = \dfrac{ 41 - 1}{1 + 41 \cdot 1 } </cmath> | ||
| + | <cmath> = \dfrac{40}{42} </cmath> | ||
| + | giving us the desired answer of <math>20+21=\boxed{41}</math>. | ||
| + | |||
| + | ==See Also== | ||
| + | {{Mock AIME box|year=Pre 2005|n=3|num-b=14|after=Last Question}} | ||
Latest revision as of 09:38, 4 April 2012
Problem
Let 
denote the value of the sum
The value of 
can be expressed as 
, where 
and 
are relatively prime positive integers. Compute 
.
Solution
Let ![\[x := k^2 + k + 1\]](http://latex.artofproblemsolving.com/b/3/a/b3a5dbe6e4c1cabeb2676cf9ec5e0774195eff8b.png)
![\[a_k := \cos^{-1}\left(\frac{k^2 + k + 1}{\sqrt{k^4 + 2k^3 + 3k^2 + 2k + 2}} \right )\]](http://latex.artofproblemsolving.com/e/c/d/ecd8f9efe5263252b8fb69b68fb71da4f6d761ce.png)
Factoring the radicand, we have
![\[a_k = \cos^{-1}\left(\frac{x}{\sqrt{x^2+1}} \right )\]](http://latex.artofproblemsolving.com/4/1/2/41277565bf5d8c22ae06d1fb9899acf39839ee36.png)
The fraction looks remarkably apt for a trigonometric substitution; namely, define 
such that 
. Then the RHS becomes
![\[a_k = \cos^{-1}\left(\frac{\tan{\theta}}{\sec{\theta}} \right ) = \cos^{-1}{(\sin{\theta})}\]](http://latex.artofproblemsolving.com/f/7/1/f71054771c37d3384eb57ae8866b5620ba8ea0e7.png)
But ![\[\cos{(\pi/2-\theta)} = \sin{\theta}\]](http://latex.artofproblemsolving.com/6/0/c/60cceebd51d69ec31f8a394517746b54437b1b83.png)
Therefore,
![\[a_k = \pi/2 - \theta = \pi/2 - \tan^{-1}{x}\]](http://latex.artofproblemsolving.com/0/d/7/0d7cb58b222e40c73946da71a6704977023d16e3.png)
This gives us
![\[\tan{a_k} = \tan{(\pi/2 - \tan^{-1}{x})}\]](http://latex.artofproblemsolving.com/c/a/3/ca305dbdac1e9c3d40a9b0514f103be0e1c33c64.png)
![\[= \cot{(\tan^{-1}{x})} = \dfrac{1}{\tan{(\tan^{-1}{x})} }\]](http://latex.artofproblemsolving.com/1/7/8/178ad8ace71c8fef0cfdd15c8e53a6b715213df2.png)
![\[= \dfrac{1}{x} = \dfrac{1}{k^2+k+1}\]](http://latex.artofproblemsolving.com/a/e/2/ae2fd08b387d5a1af081c9f542a79379bf4d5e20.png)
So now
![\[a_k = \tan^{-1}{\left( \dfrac{1}{k^2+k+1} \right) }\]](http://latex.artofproblemsolving.com/a/d/6/ad604c93b5bd301217636d312b6898e366461f41.png)
![\[= \tan^{-1}{ \left( \dfrac{ (k+1) + (-k) }{ 1 - (-k)(k+1) } \right ) }\]](http://latex.artofproblemsolving.com/0/6/8/06825b2f63f44bba692c6ab669db2a2a74267222.png)
![\[= \tan^{-1}{(k+1)} - \tan^{-1}{k}\]](http://latex.artofproblemsolving.com/2/2/d/22d8ae36de25efba9a2aed023d8127daec200c62.png)
When we sum 
, this sum now telescopes:
![\[\Omega = \sum_{k=1}^{40} a_k\]](http://latex.artofproblemsolving.com/f/d/e/fdea3ec619d95c9321af1e14994ea599e2679e4f.png)
![\[= \sum_{k=1}^{40} \tan^{-1}{(k+1)} - \tan^{-1}{k}\]](http://latex.artofproblemsolving.com/1/c/b/1cb518e647afda88c64bf03c116b601c26c8ceea.png)
![\[= \tan^{-1}{41} - \tan^{-1}{1}\]](http://latex.artofproblemsolving.com/c/2/8/c287502c69d1f76e6197038ddbb6e5b3579f85db.png)
Therefore, the required value
![\[\tan{\Omega} = \tan{(\tan^{-1}{41} - \tan^{-1}{1})}\]](http://latex.artofproblemsolving.com/1/c/a/1cab3ce310698162e8875eaaa00d72461ac43507.png)
![\[= \dfrac{ 41 - 1}{1 + 41 \cdot 1 }\]](http://latex.artofproblemsolving.com/0/8/9/089b8fd452689369346d6966a68955579b542991.png)
![\[= \dfrac{40}{42}\]](http://latex.artofproblemsolving.com/6/8/9/689d559936cb6ed410d7286e2ae3df8eb09ba158.png)
giving us the desired answer of 
.
See Also
| Mock AIME 3 Pre 2005 (Problems, Source) | ||
| Preceded by Problem 14 |
Followed by Last Question | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
