Difference between revisions of "2013 AMC 12A Problems/Problem 25"
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or <math>v^2 + w^2 > (1/2 + v)^2 = 1/4 + v + v^2</math>, <math>w^2 > 1/4 + v</math>, or <math>b^2 > a-1</math>. | or <math>v^2 + w^2 > (1/2 + v)^2 = 1/4 + v + v^2</math>, <math>w^2 > 1/4 + v</math>, or <math>b^2 > a-1</math>. | ||
− | In other words, for all <math>z</math>, <math>f(z)=a+bi</math> satisfies <math>b^2 > a-1</math>, | + | In other words, for all <math>z</math>, <math>f(z)=a+bi</math> satisfies <math>b^2 > a-1</math>, with a unique solution <math>z</math>. Therefore we need to count the number of ordered pairs <math>(a,b)</math> such that integers <math>|a|, |b|\leq 10</math>, and that <math>b^2 \geq a</math>. |
When <math>a\leq 0</math>, there is no restriction on <math>b</math> so there are <math>11\cdot 21 = 231</math> pairs; | When <math>a\leq 0</math>, there is no restriction on <math>b</math> so there are <math>11\cdot 21 = 231</math> pairs; |
Revision as of 21:06, 2 September 2013
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, for all , satisfies , with a unique solution . Therefore we need to count 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.
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 |
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