Difference between revisions of "2008 AIME I Problems/Problem 8"
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Now, <math>\arctan{\dfrac{7}{11}} + \arctan{\dfrac{1}{5}} = \arctan{\dfrac{23}{24}}</math>. | Now, <math>\arctan{\dfrac{7}{11}} + \arctan{\dfrac{1}{5}} = \arctan{\dfrac{23}{24}}</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{ | + | 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>. |
+ | |||
=== Solution 2 (generalization) === | === 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 | ||
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\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}, | \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> | </cmath> | ||
− | 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. |
+ | |||
+ | === 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 | ||
+ | |||
+ | <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. | ||
+ | So we set them equal and expand the product to get | ||
+ | <math>48n - 46 = 48 + 46n.</math> | ||
+ | Therefore, <math>n</math> equals <math>\boxed{47}</math>. | ||
== See also == | == See also == | ||
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[[Category:Intermediate Trigonometry Problems]] | [[Category:Intermediate Trigonometry Problems]] | ||
+ | {{MAA Notice}} |
Revision as of 17:59, 18 March 2020
Problem
Find the positive integer such that
Contents
Solution
Solution 1
Since we are dealing with acute angles, .
Note that , by tangent addition. Thus, .
Applying this to the first two terms, we get .
Now, .
We now have . Thus, ; and simplifying, .
Solution 2 (generalization)
From the expansion of , we can see that and If we divide both of these by , then we have 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, , is the argument of . The sum of these angles is then just the argument of the product
and expansion give us . Since the argument of this complex number is , its real and imaginary parts must be equal. So we set them equal and expand the product to get Therefore, equals .
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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