https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Connorcha&feedformat=atomAoPS Wiki - User contributions [en]2024-03-29T14:13:12ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2008_AIME_I_Problems/Problem_8&diff=1197402008 AIME I Problems/Problem 82020-03-18T21:59:01Z<p>Connorcha: /* Solution 3: Complex Numbers */</p>
<hr />
<div>== Problem ==<br />
Find the positive integer <math>n</math> such that <br />
<br />
<cmath>\arctan\frac {1}{3} + \arctan\frac {1}{4} + \arctan\frac {1}{5} + \arctan\frac {1}{n} = \frac {\pi}{4}.</cmath><br />
<br />
__TOC__<br />
== Solution ==<br />
=== Solution 1 ===<br />
Since we are dealing with acute angles, <math>\tan(\arctan{a}) = a</math>.<br />
<br />
Note that <math>\tan(\arctan{a} + \arctan{b}) = \dfrac{a + b}{1 - ab}</math>, by tangent addition. Thus, <math>\arctan{a} + \arctan{b} = \arctan{\dfrac{a + b}{1 - ab}}</math>.<br />
<br />
Applying this to the first two terms, we get <math>\arctan{\dfrac{1}{3}} + \arctan{\dfrac{1}{4}} = \arctan{\dfrac{7}{11}}</math>.<br />
<br />
Now, <math>\arctan{\dfrac{7}{11}} + \arctan{\dfrac{1}{5}} = \arctan{\dfrac{23}{24}}</math>.<br />
<br />
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>.<br />
<br />
=== Solution 2 (generalization) === <br />
From the expansion of <math>e^{iA}e^{iB}e^{iC}e^{iD}</math>, we can see that<br />
<cmath><br />
\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,<br />
</cmath><br />
and<br />
<cmath><br />
\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 .<br />
</cmath><br />
If we divide both of these by <math>\cos A \cos B \cos C \cos D</math>, then we have<br />
<cmath><br />
\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},<br />
</cmath><br />
which makes for more direct, less error-prone computations. Substitution gives the desired answer.<br />
<br />
=== Solution 3: Complex Numbers ===<br />
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 <br />
<br />
<cmath>(3+i)(4+i)(5+i)(n+i)</cmath><br />
<br />
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. <br />
So we set them equal and expand the product to get<br />
<math>48n - 46 = 48 + 46n.</math><br />
Therefore, <math>n</math> equals <math>\boxed{47}</math>.<br />
<br />
== See also ==<br />
{{AIME box|year=2008|n=I|num-b=7|num-a=9}}<br />
<br />
[[Category:Intermediate Trigonometry Problems]]<br />
{{MAA Notice}}</div>Connorchahttps://artofproblemsolving.com/wiki/index.php?title=2008_AIME_I_Problems/Problem_8&diff=1197392008 AIME I Problems/Problem 82020-03-18T21:58:40Z<p>Connorcha: /* Solution 1 */</p>
<hr />
<div>== Problem ==<br />
Find the positive integer <math>n</math> such that <br />
<br />
<cmath>\arctan\frac {1}{3} + \arctan\frac {1}{4} + \arctan\frac {1}{5} + \arctan\frac {1}{n} = \frac {\pi}{4}.</cmath><br />
<br />
__TOC__<br />
== Solution ==<br />
=== Solution 1 ===<br />
Since we are dealing with acute angles, <math>\tan(\arctan{a}) = a</math>.<br />
<br />
Note that <math>\tan(\arctan{a} + \arctan{b}) = \dfrac{a + b}{1 - ab}</math>, by tangent addition. Thus, <math>\arctan{a} + \arctan{b} = \arctan{\dfrac{a + b}{1 - ab}}</math>.<br />
<br />
Applying this to the first two terms, we get <math>\arctan{\dfrac{1}{3}} + \arctan{\dfrac{1}{4}} = \arctan{\dfrac{7}{11}}</math>.<br />
<br />
Now, <math>\arctan{\dfrac{7}{11}} + \arctan{\dfrac{1}{5}} = \arctan{\dfrac{23}{24}}</math>.<br />
<br />
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>.<br />
<br />
=== Solution 2 (generalization) === <br />
From the expansion of <math>e^{iA}e^{iB}e^{iC}e^{iD}</math>, we can see that<br />
<cmath><br />
\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,<br />
</cmath><br />
and<br />
<cmath><br />
\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 .<br />
</cmath><br />
If we divide both of these by <math>\cos A \cos B \cos C \cos D</math>, then we have<br />
<cmath><br />
\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},<br />
</cmath><br />
which makes for more direct, less error-prone computations. Substitution gives the desired answer.<br />
<br />
=== Solution 3: Complex Numbers ===<br />
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 <br />
<br />
<cmath>(3+i)(4+i)(5+i)(n+i)</cmath><br />
<br />
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. <br />
So we set them equal and expand the product to get<br />
<math>48n - 46 = 48 + 46n.</math><br />
Therefore, <math>n</math> equals <math>\boxed{047}</math>.<br />
<br />
== See also ==<br />
{{AIME box|year=2008|n=I|num-b=7|num-a=9}}<br />
<br />
[[Category:Intermediate Trigonometry Problems]]<br />
{{MAA Notice}}</div>Connorcha