Difference between revisions of "1984 AIME Problems/Problem 13"
Morgandaciuk (talk | contribs) m |
|||
Line 32: | Line 32: | ||
− | ===Solution 4 | + | ===Solution 4=== |
Recall that <math>\cot^{-1}\theta = \frac{\pi}{2} - \tan^{-1}\theta</math> and that <math>\arg(a + bi) = \tan^{-1}\frac{b}{a}</math>. Then letting <math>w = 1 + 3i, x = 1 + 7i, y = 1 + 13i,</math> and <math>z = 1 + 21i</math>, we are left with | Recall that <math>\cot^{-1}\theta = \frac{\pi}{2} - \tan^{-1}\theta</math> and that <math>\arg(a + bi) = \tan^{-1}\frac{b}{a}</math>. Then letting <math>w = 1 + 3i, x = 1 + 7i, y = 1 + 13i,</math> and <math>z = 1 + 21i</math>, we are left with | ||
Revision as of 22:20, 28 July 2014
Problem
Find the value of
Contents
Solution
Solution 1
We know that so we can repeatedly apply the addition formula, . Let , , , and . We have
,
So
and
,
so
.
Thus our answer is .
Solution 2
Apply the formula repeatedly. Using it twice on the inside, the desired sum becomes . This sum can then be tackled by taking the cotangent of both sides of the inverse cotangent addition formula shown at the beginning.
Solution 3
On the coordinate plane, let , , , , , , , , , and . We see that , , , and . The sum of these four angles forms the angle of triangle , which has a cotangent of , which must mean that . So the answer is .
Solution 4
Recall that and that . Then letting and , we are left with
Expanding , we are left with
See also
1984 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |