Difference between revisions of "1984 AIME Problems/Problem 13"
Morgandaciuk (talk | contribs) m |
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Expanding <math>wxyz</math>, we are left with | Expanding <math>wxyz</math>, we are left with | ||
− | <cmath>( | + | <cmath>(3+i)(7+i)(13+i)(21+i) = (20+10i)(13+i)(21+i)</cmath> |
− | <cmath>= ( | + | <cmath>= (2+i)(13+i)(21+i)</cmath> |
− | <cmath>= | + | <cmath>= (25+15i)(21+i)</cmath> |
− | <cmath>= | + | <cmath>= (5+3i)(21+i)</cmath> |
− | <cmath> | + | <cmath>= (102+68i)</cmath> |
− | <cmath> = 10\frac{ | + | <cmath>= (3+2i)</cmath> |
− | + | <cmath>= 10\cot \tan^{-1}\frac{2}{3}</cmath> | |
+ | <cmath> = 10\frac{3}{2} = \boxed{015}</cmath> | ||
== See also == | == See also == |
Revision as of 12:22, 9 May 2017
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 |