2004 AMC 10A Problems/Problem 4
Contents
Problem
What is the value of if
?
Solution
is the distance between
and
;
is the distance between
and
.
Therefore, the given equation says is equidistant from
and
, so
.
Alternatively, we can solve by casework (a method which should work for any similar problem involving absolute values of real numbers). If , then
and
, so we must solve
, which has no solutions. Similarly, if
, then
and
, so we must solve
, which also has no solutions. Finally, if
, then
and
, so we must solve
, which has the unique solution
.
Solution 2
We know that either or
. The first equation simplifies to
, which is false, so
. Solving, we get
.
See also
2004 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
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 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.