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

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== Problem ==
 
== Problem ==
 
Find the positive integer <math>n</math> such that  
 
Find the positive integer <math>n</math> such that  
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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>
 
<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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== Solution 1 ==
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== Solution ==
 
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=== Solution 1 ===
 
 
Since we are dealing with acute angles, <math>\tan(\arctan{a}) = a</math>.
 
Since we are dealing with acute angles, <math>\tan(\arctan{a}) = a</math>.
  
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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{47}</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{47}</math>.
  
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=== Solution 2 (generalization) ===  
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== Solution 2 (generalization) ==
 
From the expansion of <math>e^{iA}e^{iB}e^{iC}e^{iD}</math>, we can see that
 
From the expansion of <math>e^{iA}e^{iB}e^{iC}e^{iD}</math>, we can see that
 
<cmath>
 
<cmath>
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which makes for more direct, less error-prone computations. Substitution gives the desired answer.
 
which makes for more direct, less error-prone computations. Substitution gives the desired answer.
  
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=== Solution 3: Complex Numbers ===
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== Solution 3: Complex Numbers ==
 
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  
 
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  
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<cmath>(3+i)(4+i)(5+i)(n+i)</cmath>
 
<cmath>(3+i)(4+i)(5+i)(n+i)</cmath>
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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; then, we can we set them equal to get
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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.
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<cmath>48n - 46 = 48 + 46n.</cmath>
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So we set them equal and expand the product to get
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Therefore, <math>n=\boxed{47}</math>.
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<math>48n - 46 = 48 + 46n.</math>
 
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Therefore, <math>n</math> equals <math>\boxed{47}</math>.
 
  
 
==Solution 4 Sketch ==
 
==Solution 4 Sketch ==

Latest revision as of 21:36, 28 November 2023

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 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; then, we can we set them equal to get \[48n - 46 = 48 + 46n.\] Therefore, $n=\boxed{47}$.

Solution 4 Sketch

You could always just bash out $\sin(a+b+c)$ (where a,b,c,n are the angles of the triangles respectively) using the sum identities again and again until you get a pretty ugly radical and use a triangle to get $\cos(a+b+c)$ and from there you use a sum identity again to get $\sin(a+b+c+n)$ and using what we found earlier you can find $\tan(n)$ by division that gets us $\frac{23}{24}$

~YBSuburbanTea

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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