1992 AHSME Problems/Problem 16
Contents
Problem
If for three positive numbers and , all different, then
Solution 1
We have and . Equating the two expressions for gives , so as cannot be for positive and , we must have .
Solution 2
We cross multiply the first and third fractions and the second and third fractions, respectively, for Notice how the first equation can be expanded and rearranged to contain an term. We can divide this by the second equation to get
See also
1992 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
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 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.