Difference between revisions of "1951 AHSME Problems/Problem 44"
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[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
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Latest revision as of 11:26, 5 July 2013
Problem
If 
, where 
are other than zero, then 
equals:
Solution
Note that 
, 
, and 
. Therefore
![\[\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}{2}\]](http://latex.artofproblemsolving.com/f/d/b/fdbbaf08e15d5d2bbb99a9f9c1034faec16e7827.png)
Therefore
![\[\frac{1}{x}=\frac{1}{2a}+\frac{1}{2b}+\frac{1}{2c}-\frac{1}{c}=\frac{1}{2a}+\frac{1}{2b}-\frac{1}{2c}\]](http://latex.artofproblemsolving.com/e/e/4/ee475f57576a66bd88886cf2b4f513be655f284f.png)
A little algebraic manipulation yields that
![\[x=\boxed{\textbf{(E)}\ \frac{2abc}{ac+bc-ab}}\]](http://latex.artofproblemsolving.com/1/7/e/17e768245dacfd70d570557db02efad7fe03c07b.png)
See Also
| 1951 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 43 |
Followed by Problem 45 | |
| 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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
