Difference between revisions of "1981 AHSME Problems/Problem 8"
Coolmath34 (talk | contribs) (Created page with "==Problem 8== For all positive numbers <math>x</math>, <math>y</math>, <math>z</math>, the product <cmath> (x+y+z)^{-1}(x^{-1}+y^{-1}+z^{-1})(xy+yz+xz)^{-1}[(xy)^{-1}+(yz)^{-...") |
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| + | {{AHSME box|year=1981|num-b=7|num-a=9}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 13:38, 28 June 2025
Problem 8
For all positive numbers 
, 
, 
, the product
![\[(x+y+z)^{-1}(x^{-1}+y^{-1}+z^{-1})(xy+yz+xz)^{-1}[(xy)^{-1}+(yz)^{-1}+(xz)^{-1}]\]](http://latex.artofproblemsolving.com/b/c/7/bc79c652eb33e7237474c872a5fa9dcb78c0dd79.png)
equals
Solution
Start simplifying:
![\[(\frac{1}{x+y+z})(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})(\frac{1}{xy+yz+xz})(\frac{1}{xy}+\frac{1}{xz}+\frac{1}{yz})\]](http://latex.artofproblemsolving.com/a/2/0/a20e509e7ded217f6df3f7656e5bccea92101567.png)
![\[(\frac{1}{x+y+z})(\frac{xy+yz+xz}{xyz})(\frac{1}{xy+yz+xz})(\frac{x+y+z}{xyz})\]](http://latex.artofproblemsolving.com/b/5/8/b5834de3b51f99167dc7c13fccc0e3a1d37124dc.png)
The 
and 
cancel out:
![\[\frac{1}{(xyz)^2}\]](http://latex.artofproblemsolving.com/8/5/5/855d5fa7b9f9a3829e8758f23ecac12ef70d7038.png)
The answer is 
-edited by coolmath34
See also
| 1981 AHSME (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 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
