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

## 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, $23n + 24 = 24n - 23 \Longrightarrow n = \boxed{47}$.

### 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: Complex Numbers

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. So we set them equal and expand the product to get $48n - 46 = 48 + 46n.$ Therefore, $n$ equals $\boxed{47}$.

## See also

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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