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2008 AIME I Problems/Problem 8

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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}.

Contents

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 \... 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 (ProblemsResources)
Preceded by
Problem 7
Followed by
Problem 9
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
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