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Difference between revisions of "2008 AIME I Problems/Problem 8"

(solutions by CatalystofNostalgia, and Boy Soprano II respectively)
 
m (correct typo)
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Find the positive integer <math>n</math> such that  
 
Find the positive integer <math>n</math> such that  
  
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<math>\arctan\frac {1}{3} + \arctan\frac {1}{4} + \arctan\frac {1}{5} + \arctan\frac {1}{n} = \frac {\pi}{4}</math>.
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<cmath>\arctan\frac {1}{3} + \arctan\frac {1}{4} + \arctan\frac {1}{5} + \arctan\frac {1}{n} = \frac {\pi}{4}.</cmath>
  
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__TOC__
 
== Solution ==
 
== Solution ==
 
=== Solution 1 ===
 
=== Solution 1 ===
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Applying this to the first two terms, we get <math>\arctan{\dfrac{1}{3}} + \arctan{\dfrac{1}{4}} = \arctan{\dfrac{7}{11}}</math>.
 
Applying this to the first two terms, we get <math>\arctan{\dfrac{1}{3}} + \arctan{\dfrac{1}{4}} = \arctan{\dfrac{7}{11}}</math>.
  
−
Now, <math>\arctan{\dfrac{7}{11}} + \arctan{\dfrac{1}{5}} = \dfrac{23}{27} = \arctan{\dfrac{23}{24}}</math>.
+
Now, <math>\arctan{\dfrac{7}{11}} + \arctan{\dfrac{1}{5}} = \arctan{\dfrac{23}{24}}</math>.
  
 
We now have <math>\arctan{\dfrac{23}{24}} + \arctan{\dfrac{1}{n}} = \dfrac{\pi}{4} = \arctan{1}</math>. Thus, <math>\dfrac{\dfrac{23}{24} + \dfrac{1}{n}}{1 - \dfrac{23}{24n}} = 1</math>; and simplifying, <math>23n + 24 = 24n - 23 \Longrightarrow n = \boxed{047}</math>.
 
We now have <math>\arctan{\dfrac{23}{24}} + \arctan{\dfrac{1}{n}} = \dfrac{\pi}{4} = \arctan{1}</math>. Thus, <math>\dfrac{\dfrac{23}{24} + \dfrac{1}{n}}{1 - \dfrac{23}{24n}} = 1</math>; and simplifying, <math>23n + 24 = 24n - 23 \Longrightarrow n = \boxed{047}</math>.

Revision as of 12:50, 23 March 2008

Problem

Find the positive integer $n$ such that

\[\arctan\frac {1}{3} + \arctan\frac {1}{4} + \arctan\frac {1}{5} + \arctan\frac {1}{n} = \frac {\pi}{4}.\]

Solution

Solution 1

Since we are dealing with acute angles, $\tan(\arctan{a}) = a$.

Note that $\tan(\arctan{a} + \arctan{b}) = \dfrac{a + b}{1 - ab}$, by tangent addition. Thus, $\arctan{a} + \arctan{b} = \arctan{\dfrac{a + b}{1 - ab}}$.

Applying this to the first two terms, we get $\arctan{\dfrac{1}{3}} + \arctan{\dfrac{1}{4}} = \arctan{\dfrac{7}{11}}$.

Now, $\arctan{\dfrac{7}{11}} + \arctan{\dfrac{1}{5}} = \arctan{\dfrac{23}{24}}$.

We now have $\arctan{\dfrac{23}{24}} + \arctan{\dfrac{1}{n}} = \dfrac{\pi}{4} = \arctan{1}$. Thus, $\dfrac{\dfrac{23}{24} + \dfrac{1}{n}}{1 - \dfrac{23}{24n}} = 1$; and simplifying, $23n + 24 = 24n - 23 \Longrightarrow n = \boxed{047}$.

Solution 2 (generalization)

From the expansion of $e^{iA}e^{iB}e^{iC}e^{iD}$, we can see that \[\cos(A + B + C + D) = \cos A \cos B \cos C \cos D - \tfrac{1}{4} \sum_{\rm sym} \sin A \sin B \cos C \cos D + \sin A \sin B \sin C \sin D,\] and \[\sin(A + B + C + D) = \sum_{\rm cyc}\sin A \cos B \cos C \cos D - \sum_{\rm cyc} \sin A \sin B \sin C \cos D .\] If we divide both of these by $\cos A \cos B \cos C \cos D$, then we have \[\tan(A + B + C + D) = \frac {1 - \sum \tan A \tan B + \tan A \tan B \tan C \tan D}{\sum \tan A - \sum \tan A \tan B \tan C},\] which makes for more direct, less error-prone computations. Substitution gives the desired answer.

See also

2008 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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