2001 AIME II Problems/Problem 14
Problem
There are 
complex numbers that satisfy both 
and 
. These numbers have the form 
, where 
and angles are measured in degrees. Find the value of 
.
Contents
Solution
can be written in the form 
. Rearranging, we find that 
Since the real part of 
is one more than the real part of 
and their imaginary parts are equal, it is clear that either 
and 
, or 
and 
- Case 1 :
and 
and 
Setting up and solving equations, 
and 
, we see that the solutions common to both equations have arguments 
and 
. We can figure this out by adding 360 repeatedly to the number 60 to try and see if it will satisfy what we need. We realize that it does not work in the integer values.
- Case 2 :
and 
and 
Again setting up equations (
and 
) we see that the common solutions have arguments of 
and 
Listing all of these values, we find that is equal to 
which is equal to 
degrees. We only want the sum of a certain number of theta, not all of it.
See also
| 2001 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 13 |
Followed by Problem 15 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
