2013 AMC 12A Problems/Problem 25
Contents
Problem
Let be defined by . How many complex numbers are there such that and both the real and the imaginary parts of are integers with absolute value at most ?
Solution
Suppose . We look for with such that are integers where .
First, use the quadratic formula:
Generally, consider the imaginary part of a radical of a complex number: , where .
.
Now let , then , , .
Note that if and only if . The latter is true only when we take the positive sign, and that ,
or , , or .
In other words, when , the equation has unique solution in the region ; and when there is no solution. Therefore the number of desired solution is the same as the number of ordered pairs such that integers , and that .
When , there is no restriction on so there are pairs;
when , there are pairs.
So there are in total.
Solution 2 (motivated by coordinate geometry)
We consider the function as a mapping from the 2-D complex plane onto itself. We complete the square of .
Now, we must decide the range of based on the domain of , . To do this, we are interested in mapping the boundary line . To make the mapping simpler, let , or .
We intend to map of the line using the function . This transformation is equivalent to the polar equation . Using polar and trig identities, we can restate this equation as the rectangular form of a parabola,
,
where and . So, we conclude that maps the line to the parabola
.
A quick check reveals that the range of is to the left of the parabola, meaning that any point on or to the right of parabola cannot be reached.
Since the problem requires and to both be integers and at most 10, all that remains is counting all points with integer coordinates in the range of . To do this, we employ complementary counting.
The points of interest are and , resulting in a total of points. For lattice points on or to the right of the parabola, there are points for , points for , points for , and point for . Summing it all together, our answer is .
Video Solution by Richard Rusczyk
https://artofproblemsolving.com/videos/amc/2013amc12a/365
See also
2013 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 24 |
Followed by Last Question |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.