2002 AMC 12A Problems/Problem 24
Contents
Problem
Find the number of ordered pairs of real numbers such that
.
Solution
Let be the magnitude of
. Then the magnitude of
is
, while the magnitude of
is
. We get that
, hence either
or
.
For we get a single solution
.
Let's now assume that . Multiply both sides by
. The left hand side becomes
, the right hand side becomes
. Hence the solutions for this case are precisely all the
rd complex roots of unity, and there are
of those.
The total number of solutions is therefore .
Solution 2
As in the other solution, split the problem into when and when
. When
and
,
, so we must have
and hence
. Since
is restricted to
,
can range from
to
inclusive, which is
values. Thus the total is
.
See Also
2002 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 23 |
Followed by Problem 25 |
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 |