Difference between revisions of "2008 AIME I Problems/Problem 8"
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<cmath>(3+i)(4+i)(5+i)(n+i)</cmath> | <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. | + | 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: | ||
+ | 48n - 46 = 48 + 46n. | ||
+ | Therefore, n = 47. | ||
== See also == | == See also == |
Revision as of 16:23, 25 February 2018
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
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: 48n - 46 = 48 + 46n. Therefore, n = 47.
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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