1996 AJHSME Problems/Problem 8
Problem
Points and are 10 units apart. Points and are 4 units apart. Points and are 3 units apart. If and are as close as possible, then the number of units between them is
Solution
If and , then by the triangle inequality. In the triangle inequality, the equality is only reached when the "triangle" is really a degenerate triangle, and are collinear.
Simplifying, this means the smallest value can be is .
Applying the triangle inequality on with and , we know that when is minimized. If were larger, then could be larger, but we want the smallest possible, and not the largest. Thus, must be at least , but cannot be smaller than . Therefore, is the answer.
This answer comes when are all on a line, with and .
See also
1996 AJHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
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 AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.