2008 AIME I Problems/Problem 8: Difference between revisions

Revision as of 17:46, 26 July 2012

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.

Solution 3

Adding a series of angles is the same as multiplying the complex numbers whose arguments they are. In general, $\arctan\frac{1}{n}$ , is the argument of $n+i$ . The sum of these angles is then just the argument of the product

$(3+i)(4+i)(5+i)(n+i)$

and expansion give us $(48n-46)+(48+46n)i$ . Since the argument of this complex number is $\frac{\pi}{4}$ , its real and imaginary parts must be equal. Setting them equal and solving gives the answer.

@@ Line 29: / Line 29: @@
 \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},
 </cmath>
 which makes for more direct, less error-prone computations. Substitution gives the desired answer.
+=== Solution 3 ===
+Adding a series of angles is the same as multiplying the complex numbers whose arguments they are. In general, <math>\arctan\frac{1}{n}</math>, is the argument of <math>n+i</math>. The sum of these angles is then just the argument of the product
+<cmath>(3+i)(4+i)(5+i)(n+i)</cmath>
+and expansion give us <math>(48n-46)+(48+46n)i</math>. Since the argument of this complex number is <math>\frac{\pi}{4}</math>, its real and imaginary parts must be equal. Setting them equal and solving gives the answer.
 == See also ==

2008 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
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