1992 AHSME Problems/Problem 16
Revision as of 20:25, 2 May 2023 by Namelyorange (talk | contribs) (New alternate solution using proportions (+ proof of the property I used in the solution). Delete if you want.)
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
Solution 3
Since we can say that .
Let
Therefore, so --- NamelyOrange
See also
1992 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
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